Answer: The correct answer is Option A.
Explanation:
We are given:
4.5 wt % of Pb means that 4.5 grams of lead is present in 100 g of alloy.
95.5 wt % of Sn means that 95.5 grams of tin is present in 100 g of alloy.
To calculate the atom percent of any compound in a mixture, we use the equation:
[tex]\text{atom }\%=\frac{\text{Moles of compound}\times N_A}{\text{Total number of moles of mixture}\times N_A}\times 100[/tex]
where,
[tex]N_A[/tex] = Avogadro's number
Moles of a compound is given by the formula:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
For Lead:Given mass of lead = 4.5 g
Molar mass of lead = 207.19 g/mol
[tex]\text{Atom percent of lead}=\left(\frac{\frac{4.5g}{207.17g/mol}\times N_A}{(\frac{4.5g}{207.17g/mol}+\frac{95.5g}{118.71g/mol})\times N_A}\right)\times 100\\\\\text{Atom percent of lead}=2.6\%[/tex]
For Tin:Given mass of tin = 95.5 g
Molar mass of tin = 118.71 g/mol
[tex]\text{Atom percent of Tin}=\left(\frac{\frac{95.5g}{118.71g/mol}\times N_A}{(\frac{4.5g}{207.17g/mol}+\frac{95.5g}{118.71g/mol})\times N_A}\right)\times 100\\\\\text{Atom percent of Tin}=97.4\%[/tex]
Hence, the correct answer is Option A.
The composition, in atom percent, of an alloy consisting of 4.5 wt% Pb and 95.5 wt% Sn is 2.6 at% Pb and 97.4 at% Sn. The correct answer is (A).
The atomic weights for Pb and Sn are 207.19 g/mol and 118.71 g/mol, respectively.
To determine the atom percent, we need to follow these steps:
Calculate the moles of Pb and Sn in 100 g of alloy.Moles of Pb: 4.5 g / 207.19 g/mol = 0.0217 mol
Moles of Sn: 95.5 g / 118.71 g/mol = 0.8047 mol
Calculate the total moles of atoms in the alloy.Total moles = 0.0217 mol (Pb) + 0.8047 mol (Sn) = 0.8264 mol
Calculate the atom percent for each element.Atom percent of Pb: (0.0217 mol / 0.8264 mol) * 100 ≈ 2.6 at% Pb
Atom percent of Sn: (0.8047 mol / 0.8264 mol) * 100 ≈ 97.4 at% Sn
Thus, the correct answer is (A) 2.6 at% Pb and 97.4 at% Sn.
When 40.0 mL of 1.00 M H2SO4 is added to 80.0 mL of 1.00 M NaOH at 20.00°C in a coffee cup calorimeter, the temperature of the aqueous solution increases to 29.20°C. If the mass of the solution is 120.0 g and the specific heat of the calorimeter and solution is 4.184 J/g • °C, how much heat is given off in the reaction? (Ignore the mass of the calorimeter in the calculation.) Use q=mCp(tiangle)t
4.62 kJ
10.0 kJ
14.7 kJ
38.5 kJ
Answer:
[tex]\boxed{\text{4.62 kJ}}[/tex]
Explanation:
There are two heat transfers to consider:
[tex]\begin{array}{ccccc}\text{Heat released by reaction} & + &\text{heat absorbed by water} & =& 0\\q_{1}& + & q_{2} & = & 0\\q_{1}& + & mC_{p}\Delta T & = & 0\\\end{array}[/tex]
Calculate q₂
m = 120.0 g
C = 4.184 J·°C⁻¹g⁻¹
T₂ = 29.20 °C
T₁ = 20.00 °C
ΔT = T₂ - T₁ =(29.20 – 20.00) °C =9.20 °C
q₂ = 120.0 g × 4.184 J·°C⁻¹g⁻¹ × 9.20 °C = 4620 J = 4.62 kJ
Calculate q₁
q₁ + 4.62 kJ = 0
q₁ = -4.62 kJ
The negative sign shows that heat is given off.
[tex]\text{The reaction gives off }\boxed{\textbf{4.62 kJ}}[/tex]
Answer:
A. 4.62kJ
Explanation:
Identify the oxidized substance, the reduced substance, the oxidizing agent, and the reducing agent in the redox reaction. Mg(s)+Cl2(g)⟶Mg2+(aq)+2Cl−(aq) Mg(s)+Cl2(g)⟶Mg2+(aq)+2Cl−(aq) Which substance gets oxidized? MgMg Cl−Cl− Mg2+Mg2+ Cl2Cl2 Which substance gets reduced? Cl2Cl2 Mg2+Mg2+ MgMg Cl−Cl− What is the oxidizing agent? Mg2+Mg2+ Cl−Cl− MgMg Cl2Cl2 What is the reducing agent? Cl2Cl2 MgMg Cl−Cl− Mg2+
In the reaction Mg + Cl₂-> MgCl₂, Mg is oxidized, and Cl₂ is reduced. Mg serves as the reducing agent, and Cl₂ is the oxidizing agent.
In the redox reaction Mg(s) + Cl₂(g)
ightarrow MgCl₂(s), we can identify the substances that are oxidized and reduced by looking at the changes in oxidation states. Magnesium (Mg) starts with an oxidation number of 0 in elemental form and increases to +2 when it forms Mg⁺², indicating that Mg is oxidized. Chlorine (Cl₂) begins with an oxidation number of 0 and is reduced to -1 in Cl-, showing that Cl₂ is reduced.
The substance that gets oxidized works as the reducing agent, which in this case is Mg. The substance that gets reduced acts as the oxidizing agent, which is Cl₂ in this reaction. Therefore, Mg is the reducing agent because it provides electrons, and Cl₂ is the oxidizing agent because it accepts electrons, facilitating the oxidation of Mg.
The compound known as diethyl ether, commonly referred to as ether, contains carbon, hydrogen, and oxygen. A 1.501 g1.501 g sample of ether was combusted in an oxygen rich environment to produce 3.565 g3.565 g of CO2(g)CO2(g) and 1.824 g1.824 g of H2O(g)H2O(g) . Insert subscripts to complete the empirical formula of ether. empirical formula: CHO
Answer: The empirical formula of the ether will be [tex]C_4H_{10}O[/tex]
Explanation:
The chemical equation for the combustion of ether follows:
[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]
where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.
We are given:
Mass of [tex]CO_2=3.565g[/tex]
Mass of [tex]H_2O=1.824g[/tex]
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:In 44g of carbon dioxide, 12 g of carbon is contained.
So, in 3.565 g of carbon dioxide, [tex]\frac{12}{44}\times 3.565=0.972g[/tex] of carbon will be contained.
For calculating the mass of hydrogen:In 18g of water, 2 g of hydrogen is contained.
So, in 1.824 g of water, [tex]\frac{2}{18}\times 1.824=0.202g[/tex] of hydrogen will be contained.
Mass of oxygen in the compound = (1.501) - (0.972 + 0.202) = 0.327 gTo formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.Moles of Carbon =[tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.972g}{12g/mole}=0.081moles[/tex]
Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.202g}{1g/mole}=0.202moles[/tex]
Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.327g}{16g/mole}=0.0204moles[/tex]
Step 2: Calculating the mole ratio of the given elements.For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0204 moles.
For Carbon = [tex]\frac{0.081}{0.0204}=3.97\approx 4[/tex]
For Hydrogen = [tex]\frac{0.202}{0.0204}=9.9\approx 10[/tex]
For Oxygen = [tex]\frac{0.0204}{0.0204}=1[/tex]
Step 3: Taking the mole ratio as their subscripts.The ratio of C : H : O = 4 : 10 : 1
Hence, the empirical formula for the given compound is [tex]C_4H_{10}O_1=C_4H_{10}O[/tex]
Magnesium and nitrogen react in a combination reaction to produce magnesium nitride: 3 Mg + N2→ Mg3N2 In a particular experiment, a 5.65-g sample of N2 reacts completely. The mass of Mg consumed is ________ g.
Answer: The mass of magnesium consumed will be 14.731 g.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
Given mass of nitrogen gas = 5.65 g
Molar mass of nitrogen gas = 28 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of }N_2=\frac{5.65g}{28g/mol}=0.202mol[/tex]
For the given chemical equation:
[tex]3Mg+N_2\rightarrow Mg_3N_2[/tex]
By Stoichiometry of the reaction:
1 mole of nitrogen gas reacts with 3 moles of magnesium.
So, 0.202 moles of nitrogen gas will react with = [tex]\frac{3}{1}\times 0.202=0.606mol[/tex] of magnesium.
Now, calculating the mass of magnesium by using equation 1, we get:
Moles of magnesium = 0.606 moles
Molar mass of magnesium = 24.31 g/mol
Putting values in equation 1, we get:
[tex]0.606mol=\frac{\text{Mass of magneisum}}{24.31g/mol}\\\\\text{Mass of magnesium}=14.731g[/tex]
Hence, the mass of magnesium consumed will be 14.731 g.
According to the following balanced reaction, how many moles of HNO3 are formed from 8.44 moles of NO2 if there is plenty of water present? 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g)
To determine the number of moles of HNO3 formed from 8.44 moles of NO2, we need to use the mole ratio from the balanced chemical equation.
Explanation:To determine the number of moles of HNO3 formed from 8.44 moles of NO2, we need to use the mole ratio from the balanced chemical equation. The ratio of HNO3 to NO2 is 2:3, which means for every 3 moles of NO2, 2 moles of HNO3 are formed. Therefore, we can calculate:
(8.44 mol NO2) / (3 mol NO2) * (2 mol HNO3) = 5.63 mol HNO3
So, 5.63 moles of HNO3 are formed from 8.44 moles of NO2.
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To find the moles of HNO3 formed, we can use the stoichiometric coefficients and set up a proportion. The answer is 5.63 moles HNO3.
Explanation:To determine the number of moles of HNO3 formed from 8.44 moles of NO2, we can use the stoichiometric coefficients in the balanced equation. From the equation, we see that 3 moles of NO2 produce 2 moles of HNO3. Therefore, if we have 8.44 moles of NO2, we can set up the following proportion:
(3 moles NO2) / (2 moles HNO3) = (8.44 moles NO2) / (x moles HNO3)
Solving for x, we find that x = (2 moles HNO3) * (8.44 moles NO2) / (3 moles NO2) = 5.63 moles HNO3.
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alculate the concentration of H3O⁺in a solution that contains 5.5 × 10-5M OH⁻at 25°C. Identify the solution as acidic, basic, or neutral.A) 1.8 × 10-10M, basicB) 1.8 × 10-10M, acidicC) 5.5 × 10-10M, neutralD) 9.2 × 10-1M, acidicE) 9.2 × 10-1M, basic
Answer:
[H₃O⁺] = 1.82 x 10⁻¹⁰M
Explanation:
[H₃O⁺][OH⁻] = Kw = 1.0 x 10⁻¹⁴ = [H₃O⁺](5.5 x 10⁻⁵) => [H₃O⁺] = (1.0 x 10⁻¹⁴/5.5 x 10⁻⁵)M = 1.82 x 10⁻¹⁰M
When two solutions that differ in solute concentration are placed on either side of a selectively permeable membrane, and osmosis is allowed to take place, the water will ______. exhibit a net movement to the side with lower water concentration exhibit a net movement to the side with higher water concentration exhibit a net movement to the side with lower solute concentration exhibit an equal movement in both directions across the membrane not cross the membrane
Answer:
In osmosis, the water has a net movement to the side with lower water concentration.
Explanation:
It moves in the direction of a solution with higher solute ( therefore lower water) concentration.
Answer:
Exhibit a net movement to the side with lower water concentration.
Explanation:
Hello,
Osmosis is a process by which the equilibrium between the concentration of a solute placed at two sides separated by a permeable membrane is reached by moving the solvent's molecules.
For the considered example, it is seen that the water will exhibit a net movement to the side with lower water concentration as long as the place with more water will have a higher solute's concentration which implies that the water move from such side to other one in order to equal both the solutions concentrations.
Best regards.
For the reaction KClO2⟶KCl+O2 KClO2⟶KCl+O2 assign oxidation numbers to each element on each side of the equation. K in KClO2:K in KClO2: K in KCl:K in KCl: Cl in KClO2:Cl in KClO2: Cl in KCl:Cl in KCl: O in KClO2:O in KClO2: O in O2:O in O2: Which element is oxidized? KK OO ClCl Which element is reduced? OO KK Cl
Answer:
K in KClO2 = +1
Cl in KClO2 = +3
O in KClO2 = -2
K in KCl = +1
Cl in KCl = -1
O in O2 = 0
Chlorine is going from +3 to -1 so it is being reduced
Oxygen is going from -2 to 0 so it is being oxidized
Explanation:
Potassium is a constant +1
Chlorine could be -1, +1, +3, +5, or +7
Oxygen can be either -2 or +2
Every reactant has to equal zero if there is not a given charge.
Each element can only have certain charges, for example in the previous answer K = +5. Potassium can only be the charge of +1
Chlorine was reduced and oxygen was oxidized in the reaction as shown.
The oxidation number of an atom in a compound is the charge that the atom appears to have as determined by a certain set of rules.
On the left hand side, the oxidation numbers of the elements are;
K = +1
Cl = +3
O = -2
On the right hand side;
K = +1
Cl = -1
O = zero
The element that was oxidized is oxygen. Its oxidation number increased from -2 to zero. The element that was reduced is chlorine. Its oxidation number decreased from +3 to -1.
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Of the following gases, ________ will have the greatest rate of effusion at a given temperature. Of the following gases, ________ will have the greatest rate of effusion at a given temperature. NH3 HCl CH4 Ar HBr
Answer: From the given gases, the greatest rate of effusion is of [tex]CH_4[/tex]
Explanation:
Rate of effusion of a gas is determined by a law known as Graham's Law.
This law states that the rate of effusion or diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:
[tex]\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}[/tex]
It is visible that molar mass is inversely related to rate of effusion. So, the gas having lowest molar mass will have the highest rate of effusion.
For the given gases:
Molar mass of [tex]NH_3=17g/mol[/tex]
Molar mass of [tex]HCl=36.5g/mol[/tex]
Molar mass of [tex]CH_4=16g/mol[/tex]
Molar mass of [tex]Ar=40g/mol[/tex]
Molar mass of [tex]HBr=81g/mol[/tex]
The molar mass of methane gas is the lowest. Thus, it will have the greatest rate of effusion.
Hence, the greatest rate of effusion is of [tex]CH_4[/tex]
Answer:
CH4
CH4
Explanation:
How do you calculate vapor pressure of water above a solution prepared by adding 24 g of lactose (C12H22O11) to 200 g of water at 338 K? (Vapor-pressure of water at 338 K 187.5 torr.)
Answer:
VP(solution) = 186.3 Torr
Explanation:
Given 24g Lactose (C₁₂H₂₂O₁₁) IN 200g H₂O @338K (65°C) & 187.5Torr
VP(soln) = VP(solvent) - [mole fraction of solute(X)·VP(solvent)] => Raoult's Law
VP(H₂O)@65°C&187.5Torr = 187.5Torr
moles Lactose = (24g/342.3g/mol) = 0.0701mole Lactose
moles Water = (200g/18g/mol) = 11.11mole Water
Total moles = (11.11 + 0.0701)mole = 11.181mole
mole fraction Lac = n(lac)/[n(lac) + n(H₂O)] = (0.0701/11.181) = 6.27 x 10⁻³
VP(solution) = 187.5Torr - (6.27 x 10⁻³)187.5Torr = 186.3Torr
Final answer:
To find the vapor pressure of water above a lactose solution, calculate the mole fraction of water in the solution using moles of lactose and water, then apply Raoult’s Law by multiplying the mole fraction by the vapor pressure of pure water at 338 K, resulting in a vapor pressure of 186.3 torr.
Explanation:
To calculate the vapor pressure of water above a solution prepared by adding 24 g of lactose (C12H22O11) to 200 g of water at 338 K, we use Raoult’s Law. Raoult's Law states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent in the solution. First, calculate the mole fraction of water in the solution:
Moles of lactose = mass (g) / molar mass (g/mol). For lactose, molar mass = 342.3 g/mol, so moles of lactose = 24 g / 342.3 g/mol = 0.0701 mol.
Moles of water = mass (g) / molar mass (g/mol). For water, molar mass = 18.015 g/mol, so moles of water = 200 g / 18.015 g/mol = 11.10 mol.
Total moles = moles of lactose + moles of water = 0.0701 mol + 11.10 mol = 11.1701 mol.
Mole fraction of water = moles of water / total moles = 11.10 / 11.1701 = 0.9937.
Finally, calculate the vapor pressure of water: Vapor pressure = mole fraction of water × vapor pressure of pure water = 0.9937 × 187.5 torr = 186.3 torr.
Give the set of reactants (including an alkyl halide and a nucleophile) that could be used to synthesize the following ether: Draw the molecules on the canvas by choosing buttons from the Tools (for bonds and charges), Atoms, and Templates toolbars, including charges where needed.
CH3CH2OCH2CH2CHCH3
|
CH3
To synthesize the given ether, you would need ethyl bromide as the alkyl halide and ethoxide as the nucleophile.
Explanation:To synthesize the ether CH3CH2OCH2CH2CHCH3CH3, you would need an alkyl halide and a nucleophile. One possible set of reactants that could be used is ethyl bromide (CH3CH2Br) as the alkyl halide and ethoxide (CH3CH2O-) as the nucleophile. The reaction can be represented as:
CH3CH2Br + CH3CH2O- → CH3CH2OCH2CH2CHCH3CH3 + Br-
Which of the following statements is completely correct? a. NH3 is a weak base, and H2CO3 is a strong acid. b. H2CO3 is a strong acid, and NaOH is a strong base. c. NH3 is a weak base, and HCl is a strong acid. d. H2CO3 is a weak acid, and NaOH is a weak base. e. NH3 is a strong base, and HCl is a weak acid.
Answer: The correct answer is Option c.
Explanation:
Weak acid is defined as the acid which does not get completely dissociated into its ions when dissolved in water. For Example: [tex]CH_3COOH,H_2CO_3[/tex] etc..
Strong acid is defined as the acid which gets completely dissociated into its ions when dissolved in water. For Example: [tex]HCl,HNO_3[/tex] etc..
Weak base is defined as the base which does not get completely dissociated into its ions when dissolved in water. For Example: [tex]NH_3,NH_4OH[/tex] etc..
Strong base is defined as the base which gets completely dissociated into its ions when dissolved in water. For Example: [tex]NaOH,KOH[/tex] etc..
From the above information, it is clearly visible that the correct answer is Option c.
The correct statement is that C. NH3 is a weak base, and HCl is a strong acid.
Explanation:When a weak base, such as ammonia (NH3), reacts with a strong acid, like hydrochloric acid (HCl), a chemical reaction occurs. The acid donates protons (H+) to the base, forming water and the conjugate acid of the weak base. For example, in the reaction between NH3 and HCl, [tex]NH3 + HCl -- NH4+ + Cl-,[/tex] ammonium chloride is formed. The resulting solution is acidic due to the presence of excess H+ ions.
Thus, the correct statement among the options is c. NH3 is a weak base, and HCl is a strong acid. Ammonia (NH3) is a weak base because it donates only a partial amount of its lone pair of electrons. Hydrochloric acid (HCl) is a strong acid because it completely ionizes in water to produce a high concentration of hydrogen ions.
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Calculate the change in pH when 5.00 mL of 0.100 M HCl is added to 75.0 mL of a buffer solution that is 0.100 M in NH3 and 0.100 M in NHaCI.
The first part of the answer is in the photo attached. I hope i can attach a second photo...
To find the change in pH, use the Henderson-Hasselbalch equation with the initial concentrations of NH3 and NH4+ to find the initial pH. Adjust concentrations after addition of HCl to find the final pH. Subtract initial pH from final to find change.
Explanation:The question asks us to calculate the change in pH when 5.00 ml of 0.100 M HCl is added to 75.0 ml of a buffer solution that is 0.100 M in NH3 and 0.100 M in NH4Cl. This is a problem related to acid-base chemistry. We have a strong acid (HCl) being added to a buffer solution composed of NH3 (a weak base) and NH4Cl (the conjugate acid of the base).
First, it's important to use the Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]), where pKa is the -log of the Ka, [A-] is the concentration of base (NH3 in this case) and [HA] is the concentration of conjugate acid (NH4+ in this case). Plug the initial concentrations into the equation to find initial pH. Then, calculate the moles of HCl being added and adjust the concentrations of NH4+ and NH3 accordingly. Using the new concentrations in Henderson-Hasselbalch equation will give the final pH. The difference between the initial and final pH is the change in pH.
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The explosive nitroglycerin (C3H5N3O9) decomposes rapidly upon ignition or sudden impact according to the following balanced equation: 4 C3H5N3O9 (l) → 12 CO2 (g) + 10 H2O (g) + 6 N2 (g) + O2 (g) ΔrxnHo = −5678 kJ Calculate the standard enthalpy of formation (ΔfHo) for nitroglycerin. The enthalpy of formation of CO2 (g) is -393.5 kJ/mol. The enthalpy of formation of H2O (g) is -241.8 kJ/mol.
Answer: The [tex]\Delta H_f[/tex] for [tex]C_3H_5N_3O_9[/tex] in the reaction is -365.5 kJ/mol.
Explanation:
Enthalpy change of a reaction is defined as the difference in enthalpy of all the products and the reactants each multiplied with their respective number of moles. The equation that is used to calculate the enthalpy change of a reaction is:
[tex]\Delta H_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}][/tex]
For the given chemical reaction:
[tex]4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+10H_2O(g)+6N_2(g)+O_2(g)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H_{rxn}=[(12\times \Delta H_f_{(CO_2)})+(10\times \Delta H_f_{(H_2O)})+(6\times \Delta H_f_{(N_2)})+(1\times \Delta H_f_{(O_2)})]-[(4\times \Delta H_f_{(C_3H_5N_3O_9)})][/tex]
We are given:
[tex]\Delta H_f_{(H_2O)}=-241.8kJ/mol\\\Delta H_f_{(N_2)}=-0kJ/mol\\\Delta H_f_{(CO_2)}=-393.5kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol\\\Delta H_{rxn}=-5678kJ[/tex]
Putting values in above equation, we get:
[tex]-5678=[(12\times (-393.5))+(10\times (-241.8))+(6\times (0))+(1\times (0))]-[(4\times \Delta H_f_{(C_3H_5N_3O_9)})]\\\\\Delta H_f_{(C_3H_5N_3O_9)}=-365.5kJ/mol[/tex]
Hence, the [tex]\Delta H_f[/tex] for [tex]C_3H_5N_3O_9[/tex] in the reaction is -365.5 kJ/mol.
The standard enthalpy of formation (ΔfHo) for nitroglycerin which decomposes by the given equation can be calculated using the formula for enthalpy change and the provided enthalpies of formation for CO2(g) and H2O(g). Nitrogen and Oxygen are in their standard states so their enthalpies of formation are zero. The calculated ΔfHo for Nitroglycerin is about -364.6 kJ/mol.
Explanation:To calculate the standard enthalpy of formation (ΔfHo) for nitroglycerin, we first need to know that the formation reactions are those that produce 1 mol of a substance from its elements in their standard states. Here, the combustion reaction for nitroglycerin given is: 4 C3H5N3O9 (l) → 12 CO2 (g) + 10 H2O (g) + 6 N2 (g) + O2 (g). And the provided ΔrxnHo = −5678 kJ.
Also, the standard enthalpy of formations for CO2(g) and H2O(g) are given as -393.5 kJ/mol and -241.8 kJ/mol, respectively. But, the enthalpy of formation for elements in their standard state (here N2 and O2) is zero.
Using the equation: ΔrxnHo = Σ ΔfHo(products) - Σ ΔfHo(reactants), and the knowledge of stoichiometry we get: -5678 kJ = [12(-393.5 kJ/mol) + 10(-241.8 kJ/mol) + 6 * 0 + 1 * 0] - [4 * ΔfHo (C3H5N3O9)]
By calculating we get the ΔfHo for Nitroglycerin = -364.6 kJ/mol (approx)
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1. According to the following balanced chemical equation, how many moles of iron will react with 0.455 moles of chlorine? 2Fe(s) + 3Cl2(g) → 2FeCl3(s)
Answer:
[tex]\boxed{ \text{0.303 mol}}[/tex]
Explanation:
(a) Balanced equation
2Fe + 3Cl₂ ⟶2FeCl₃
(b) Calculation
You want to convert moles of Cl₂ to moles of Fe.
The molar ratio is 2 mol Fe:3 mol Cl₂
[tex]\text{Moles of Fe} =\text{0.455 mol Cl$_{2}$} \times \dfrac{\text{2 mol Fe}}{\text{3 mol Cl$_{2}$}} = \textbf{0.303 mol Fe}\\\\\boxed{ \textbf{0.303 mol of Fe}}\text{ will react.}[/tex]
Automobile antifreeze is a mixture that consists mostly of water and ethylene glycol (C2H4(OH)2). What is the volume percent the ethylene glycol in the antifreeze mixture if it is created by mixing 3.00 gal of ethylene glycol with 2.00 gal water?
Answer:
60.00% is the volume percent the ethylene glycol in the antifreeze mixture.
Explanation:
Volume of the ethylene glycol in the mixture = 3.00 gal
Volume of the water in the mixture = 2.00 gal
Total volume of the mixture =3.00 gal + 2.00 gal = 5.00 gal
Volume percent the ethylene glycol :
[tex]\frac{\text{Volume of solute}}{\text{Volume of solution or mixture}}\times 100[/tex]
[tex]\%=\frac{3.00 gal}{5.00 gal}\times 100=60.00\%[/tex]
60.00% is the volume percent the ethylene glycol in the antifreeze mixture.
If 100.0 mL of 0.453 M Na2SO4 are added to 100.0 mL of 0.907 M Pb(NO3)2, how many grams of PbSO4 can be produced?
Na2SO4(aq)+Pb(NO3)2(aq)⟶2NaNO3(aq)+PbSO4(s)
The mass of PbSO₄ that could be produced is 13.74 g
From the question,
We are to determine the mass of PbSO₄ that could be produced from the reaction
The given balanced chemical equation for the reaction is
Na₂SO₄(aq) + Pb(NO₃)₂(aq) ⟶ 2NaNO₃(aq) + PbSO₄(s)
This means
1 mole of Na₂SO₄ reacts with 1 mole of Pb(NO₃)₂ to produce 2 moles of NaNO₃ and 1 mole of PbSO₄
Now, we will determine the number of moles of each reactant present
For Na₂SO₄Volume = 100.0 mL = 0.1 L
Concentration = 0.453 M
Using the formula
Number of moles = Concentration × Volume
∴ Number of moles of Na₂SO₄ = 0.453 × 0.1
Number of moles of Na₂SO₄ = 0.0453 mol
For Pb(NO₃)₂Volume = 100.0 mL = 0.1 L
Concentration = 0.907 M
∴ Number of moles of Pb(NO₃)₂ = 0.907 × 0.1
Number of moles of Pb(NO₃)₂ = 0.0907 mol
Since, 1 mole of Na₂SO₄ reacts with 1 mole of Pb(NO₃)₂
Then,
0.0453 mole of Na₂SO₄ will react with 0.0453 mole of Pb(NO₃)₂
Now, from the balanced chemical equation
1 mole of Na₂SO₄ reacts with 1 mole of Pb(NO₃)₂ to produce 1 mole of PbSO₄
Then,
0.0453 mole of Na₂SO₄ will react with 0.0453 mole of Pb(NO₃)₂ to produce 0.0453 mole of PbSO₄
∴ The number of moles of PbSO₄ produced is 0.0453 mole
Now, for the mass of PbSO₄ produced
From the formula
Mass = Number of moles × Molar mass
Molar mass of PbSO₄ = 303.26 g/mol
∴ Mass of PbSO₄ produced = 0.0453 × 303.26
Mass of PbSO₄ produced = 13.737678 g
Mass of PbSO₄ produced ≅ 13.74 g
Hence, the mass of PbSO₄ that could be produced is 13.74 g
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The question involves the stoichiometric calculation during a chemical reaction. In the reaction, Pb(NO₃)₂ and Na₂SO₄ react to produce PbSO₄. Given the volumes and molarities, we determined Na₂SO₄ to be the limiting reactant. Therefore, approximately 13.72 g of PbSO₄ can be produced.
Explanation:This question is trying to understand a stoichiometric calculation involved in a chemical reaction. We start by determining the limiting reactant. To do this, divide the number of moles of each reactant by the stoichiometric coefficient in the balanced chemical equation. The reactant that returns the smallest number is the limiting reactant.
In this case, the reaction uses one mole of Pb(NO₃)₂ and one mole of Na₂SO₄ to produce PbSO₄, therefore the molar ratio is 1:1. Pb(NO₃)₂ has a higher molarity, hence it will not be the limiting factor. Therefore, Na₂SO₄ is the limiting reactant. Considering this, we calculate that (0.453 moles/L × 0.1 L = 0.0453 moles of Na₂SO₄) are present.
To find the mass of PbSO₄ that can be produced, we use the molar mass of PbSO₄ (303.26 g/mol)—based on the atomic masses of lead, sulfur, and oxygen. From this we can calculate the mass as follows: 0.0453 moles × 303.26 g/mol = 13.72398 g of PbSO₄.
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A reaction in which A, B, and C react to form products is zero order in A, one-half order in B, and second order in C. By what factor does the reaction rate change if [A] is doubled (and the other reactant concentrations are held constant)?
Answer: There will be no change in rate.
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
[tex]A+B+C\rightarrow Products[/tex]
[tex]Rate=k[A]^x[B]^y[C]^z[/tex]
k= rate constant
x = order with respect to A = 0
y = order with respect to B = [tex]\frac{1}{2}[/tex]
z= order with respect to C = 2
Thus [tex]Rate=k[A]^0[B]^{\frac{1}{2}}[C]^2[/tex]
Given : when [A] is doubled and the other reactant concentrations are held constant.
Thus the new rate law is [tex]Rate'=k[2A]^0[B]^{\frac{1}{2}}[C]^2[/tex]
[tex]Rate'=k[2]^0[A]^0[B]^{\frac{1}{2}}[C]^2[/tex]
[tex]Rate'=k[A]^0[B]^{\frac{1}{2}}[C]^2[/tex] [tex](2^0=1)[/tex]
[tex]Rate'=Rate[/tex]
Thus the reaction rate would not change.
Doubling the concentration of A has no effect on the reaction rate because the reaction is zero order in A. The rate law for the reaction is rate = k[B]^1/2[C]², indicating independence from [A].
The question pertains to chemical kinetics, specifically how the reaction rate changes with respect to changes in reactant concentrations. Since the reaction is zero order with respect to A, doubling the concentration of A ([A]) will have no effect on the rate of the reaction. The rate law for this reaction could be represented as rate =k[B]^1/2[C]², which clearly shows that the reaction rate is independent of the concentration of A. Therefore, the reaction rate remains unchanged if the concentration of A is doubled while keeping B and C constant.
An aqueous solution of hydroiodic acid is standardized by titration with a 0.186 M solution of calcium hydroxide. If 26.5 mL of base are required to neutralize 20.3 mL of the acid, what is the molarity of the hydroiodic acid solution? M hydroiodic acid
Answer: The molarity of hydroiodic acid in the titration is 0.485 M.
Explanation:
To calculate the molarity of acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HI[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is [tex]Ca(OH)_2[/tex]
We are given:
[tex]n_1=1\\M_1=?M\\V_1=20.3mL\\n_2=2\\M_2=0.186M\\V_2=26.5mL[/tex]
Putting values in above equation, we get:
[tex]1\times M_1\times 20.3=2\times 0.186\times 26.5\\\\M_1=0.485M[/tex]
Hence, the molarity of hydroiodic acid is 0.485M.
Final answer:
The molarity of the hydroiodic acid solution is calculated using the stoichiometry of its reaction with calcium hydroxide and the volumes and molarity of the titrant, resulting in a molarity of 0.485 M for the hydroiodic acid.
Explanation:
To find the molarity of the hydroiodic acid solution, we need to first understand the reaction occurring between hydroiodic acid (HI) and calcium hydroxide (Ca(OH)₂). The balanced chemical equation for this reaction is: 2HI(aq) + Ca(OH)₂(aq) → CaI₂(aq) + 2H₂O(l). This equation indicates that two moles of HI react with one mole of Ca(OH)₂.
Using the titration information provided: 26.5 mL of 0.186 M Ca(OH)₂ were required to neutralize 20.3 mL of HI. From this, we can calculate the moles of Ca(OH)₂ used (Moles = Molarity × Volume in L), which is (0.186 M) × (0.0265 L) = 0.004929 moles of Ca(OH)₂. Given the stoichiometry of the reaction, there are twice as many moles of HI, so 0.009858 moles of HI were neutralized.
To find the molarity of the hydroiodic acid solution, we use the formula Molarity = Moles/Volume in L. Here, it is 0.009858 moles / 0.0203 L = 0.485 M HI.
The value of ΔH° for the reaction below is -126 kJ. The amount of heat that is released by the reaction of 10.0 g of Na2O2 with water is ________ kJ. 2Na2O2 (s) + 2H2O (l) → 4NaOH (s) + O2 (g)
Approximately -8 kJ of heat is released when 10 grams of Na2O2 react with water, based on the provided ΔH° value for the reaction.
Explanation:The value of ΔH° (-126 kJ) is the enthalpy change of the reaction when 2 moles of Na2O2 react with water to form 4 moles of NaOH and O2. The molar mass of Na2O2 is about 78.0 g/mol. Therefore, 10.0 g of Na2O2 is roughly equivalent to 0.128 moles. Given that the ΔH° value provided is for 2 moles, the heat released by the reaction of 0.128 moles would be (0.128/2) x -126 kJ which is approximately -8 kJ. So, the amount of heat that is released by the reaction of 10.0 g of Na2O2 with water is roughly -8 kJ.
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Each of the following reactions is allowed to come to equilibrium and then the volume is changed as indicated. Predict the effect (shift right, shift left, or no effect) of the indicated volume change. Drag the appropriate items to their respective bins.CO(g) + H2O(g) <=> CO2(g) + H2(g)
(volume is decreased)
PCl3(g) + Cl2(g) <=> PCl5(g)
(volume is increased)
CaCO3(s)<=> CaO(s) + CO2(g)
(volume is increased)
No net shift in the reactants and products is an equilibrium constant. In the first reaction, there is no effect, in second shifts to the left and third shifts to right.
What is the equilibrium shifts?Equilibrium shift can be explained by the rules of Le and the change in the volume affects the system as the number of the particles gets varied.
For the first reaction [tex]\rm CO(g) + H_{2}O(g) \Leftrightarrow CO_{2}(g) + H_{2}(g)[/tex] when the volume is decreased then the pressure will increase then the equilibrium will shift towards the fewer and then more number of the gas moles and hence, will have no net effect.
In the second reaction [tex]\rm PCl_{3}(g) + Cl_{2}(g) \Leftrightarrow PCl_{5}(g)[/tex] when the volume is increased then the pressure will decrease and the reaction will shift towards more moles of the gas present that is the left side.
In the third reaction, [tex]\rm CaCO_{3}(s) \Leftrightarrow CaO(s) + CO_{2}(g)[/tex] when the volume is increased then the pressure will decrease and the equilibrium will shift towards the side where more moles are present. Hence the increase in the volume will shift the equilibrium towards the right.
Therefore, a decrease in volume will have no effect, while the increase in volume will shift the reaction towards the left and right.
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Determine the limiting reactant (LR) and the mass (in g) of nitrogen that can be formed from 50.0 g N204 and 45.0 g N2H4. Some possibly useful molar masses are as follows: N2O4 92.02 g/mol, N2H4 32.05 g/mol N204) 2 N2H4(1)3 N2(g) + 4 H2O(g)
Answer: The mass of nitrogen gas produced will be 45.64 grams.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
For [tex]N_2O_4[/tex]Given mass of [tex]N_2O_4=50.0g[/tex]
Molar mass of [tex]N_2O_4=92.02g/mol[/tex]
Putting values in equation 1, we get:
[tex]\text{Moles of }N_2O_4=\frac{50g}{92.02g/mol}=0.543mol[/tex]
For [tex]N_2H_4[/tex]Given mass of [tex]N_2H_4=45.0g[/tex]
Molar mass of [tex]N_2O_4=32.05g/mol[/tex]
Putting values in equation 1, we get:
[tex]\text{Moles of }N_2H_4=\frac{45g}{32.05g/mol}=1.40mol[/tex]
For the given chemical reaction:
[tex]N_2O_4(l)+2N_2H_4(l)\rightarrow 3N_2(g)+4H_2O(g)[/tex]
By stoichiometry of the reaction:
1 mole of [tex]N_2O_4[/tex] reacts with 2 moles of [tex]N_2H_4[/tex]
So, 0.543 moles of [tex]N_2O_4[/tex] will react with = [tex]\frac{2}{1}\times 0.543=1.086moles[/tex] of [tex]N_2H_4[/tex]
As, the given amount of [tex]N_2H_4[/tex] is more than the required amount. Thus, it is considered as an excess reagent.
Hence, [tex]N_2O_4[/tex] is the limiting reagent.
By Stoichiometry of the reaction:
1 mole of [tex]N_2O_4[/tex] produces 3 moles of nitrogen gas
So, 0.543 moles of [tex]N_2O_4[/tex] will produce = [tex]\frac{3}{1}\times 0.543=1.629moles[/tex] of nitrogen gas.
Now, calculating the mass of nitrogen gas from equation 1, we get:
Molar mass of nitrogen gas = 28.02 g/mol
Moles of nitrogen gas = 1.629 moles
Putting values in equation 1, we get:
[tex]1.629mol=\frac{\text{Mass of nitrogen gas}}{28.02g/mol}\\\\\text{Mass of nitrogen gas}=45.64g[/tex]
Hence, the mass of nitrogen gas produced will be 45.64 grams.
Calculate the enthalpy change associated with the conversion of 25.0 grams of ice at -4.00 °C to water vapor at 110.0 °C. The specific heats of ice, water, and steam are 2.09 J/g-K, 4.18 J/g-K, and 1.84 J/g-K, respectively. For , ΔHfus = 6.01 kJ/mol and ΔHvap = 40.67 kJ/mol.
The total heat energy required to convert 25 grams of ice at -4.00°C to steam at 110.0 °C can be calculated in several steps involving warming of ice to 0°C, melting of ice, heating water to 100°C, vaporization of water, and heating of steam to the final temperature. Each step requires the use of specific heat values, molar heats of fusion and vaporization, and the mass of the starting ice chunk. The total energy calculated would be the sum of these steps, yields 75579 Joules.
Explanation:The question is about calculating the total energy, or enthalpy change, needed to convert ice at -4.00°C to steam at 110.0°C. We need to take into account the warming of ice, the melting of ice, the heating of water, the vaporization of water, and the heating of steam. This process involves several steps and requires using the principle of conservation of energy and heat transfer equations.
First, the ice is warmed to 0°C : ΔH1 = mcΔT = (25.0 grams) (2.09 J/g-K)(4 K) = 209 J
Then, the ice is melted: ΔH2 = n(ΔHfus) = (25.0 grams/18.015 g/mol)(6.01 kJ/mol)= 8.33 kJ = 8330 J
The water is heated to 100°C: ΔH3 = mcΔT = (25.0 grams) (4.18 J/g-K)(100 K) = 10450 J
The water is vaporized: ΔH4 = n(ΔHvap) = (25.0 grams/18.015 g/mol)(40.67 kJ/mol) = 56.13 kJ = 56130 J
Finally, the steam is heated to 110°C: ΔH5 = mcΔT = (25.0 grams) (1.84 J/g-K)(10 K) = 460 J
The total energy change is the sum of all these changes: ΔHtotal = ΔH1 + ΔH2 + ΔH3 + ΔH4 + ΔH5 = 209J + 8330J + 10450J + 56130J + 460J = 75579 J
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Given the following equation: |4 NH3 (g) + 5 О2 (g) —> 4 NO (g) + 6 H20 () How many grams of H20 is produced if 21.1 grams of NH3 reacts with 73.9 grams of O2?
Answer: The mass of water that can be formed are 33.48 g.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ....(1)
For ammonia gas:Given mass of ammonia gas = 21.1 g
Molar mass of ammonia gas = 17 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of ammonia}=\frac{21.1g}{17g/mol}=1.24mol[/tex]
For oxygen gas:Given mass of oxygen gas = 73.9 g
Molar mass of oxygen gas = 32 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of oxygen gas}=\frac{73.9g}{32g/mol}=2.30mol[/tex]
For the given chemical equation:
[tex]4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(l)[/tex]
By Stoichiometry of the reaction:
4 moles of ammonia gas reacts with 5 moles of oxygen gas.
So, 1.24 moles of ammonia gas will react with = [tex]\frac{5}{4}\times 1.24=1.55moles[/tex] of oxygen gas.
As, given amount of oxygen gas is more than the required amount. Thus, it is considered as an excess reagent.
So, ammonia gas is considered as a limiting reagent because it limits the formation of products.
By Stoichiometry of the above reaction:
4 moles of ammonia gas is producing 6 moles of water.
So, 1.24 moles of ammonia gas will produce = [tex]\frac{6}{4}\times 1.24=1.86moles[/tex] of water.
Now, calculating the mass of water by using equation 1, we get:
Moles of water = 1.86 moles
Molar mass of water = 18 g/mol
Putting all the values in equation 1, we get:
[tex]1.86mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=33.48g[/tex]
Hence, the mass of water that can be formed are 33.48 g
The tendency of water molecules to stick together is referred to as ______. A) adhesion B) polarity C) cohesion D) transpiration E) evaporation
Answer:
Cohesion
Explanation:
Think of it like this. The water molecules STICK TOGETHER, so they COoperate.
COhesion COoperate
Cohesion is the tendency of water molecules to stick together, due to hydrogen bonding. It is a unique property of water that plays a vital role in living organisms and the environment.
Explanation:
The tendency of water molecules to stick together is referred to as cohesion. This process occurs due to hydrogen bonding between the water molecules, where the slightly positive hydrogen of one water molecule is attracted to the slightly negative oxygen of a neighboring water molecule. Thus, creating a force that holds these molecules together. It is one of the unique properties of water that contributes to its important role in living organisms and the environment. Some other terms related to water are adhesion, which is the tendency of water to stick to other substances; polarity, which is the property of having oppositely charged ends; transpiration, which is the process by which water evaporates from the leaves of plants; and evaporation, which is the change of state from liquid to gas.
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If 5.00 grams of aluminum react with an excess of copper (II) sulfate and the percentage yield is 63.4%, what is the mass of the copper produced? The other product is aluminum sulfate.
Answer : The mass of copper produced will be, 11.796 grams
Explanation : Given,
Mass of [tex]Al[/tex] = 5 g
Molar mass of [tex]Al[/tex] = 26.98 g/mole
Molar mass of [tex]Cu[/tex] = 63.66 g/mole
First we have to calculate the moles of [tex]Al[/tex].
[tex]\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}=\frac{5g}{26.98g/mole}=0.185moles[/tex]
Now we have to calculate the moles of [tex]Cu[/tex].
The balanced chemical reaction is,
[tex]2Al+3CuSO_4\rightarrow Al_2(SO_4)_3+3Cu[/tex]
From the balanced reaction we conclude that
As, 2 moles of [tex]Al[/tex] react to give 3 moles of [tex]Cu[/tex]
So, 0.185 moles of [tex]Al[/tex] react to give [tex]\frac{3}{2}\times 0.185=0.2775[/tex] moles of [tex]Cu[/tex]
Now we have to calculate the mass of [tex]Cu[/tex].
[tex]\text{Mass of }Cu=\text{Moles of }Cu\times \text{Molar mass of }Cu[/tex]
[tex]\text{Mass of }Cu=(0.2775mole)\times (63.66g/mole)=17.66g[/tex]
The theoretical yield of Cu = 17.66 grams
Now we have to calculate the actual yield of Cu.
[tex]\%\text{ yield of }Cu=\frac{\text{Actual yield of }Cu}{\text{Theoretical yield of }Cu}\times 100[/tex]
Now put all the given values in this formula, we get the actual yield of Cu.
[tex]63.4=\frac{\text{Actual yield of }Cu}{17.66}\times 100[/tex]
[tex]\text{Actual yield of }Cu=11.796g[/tex]
Therefore, the mass of copper produced will be, 11.796 grams
3. The rate law for the reaction NH4+(aq) + NO2–(aq) → N2(g) + 2H2O(l) is given by rate = k[NH4+][NO2–]. At 25ºC, the rate constant is 3.0 × 10–4/ M · s. Calculate the rate of the reaction at this temperature if [NH4+] = 0.26 M and [NO2–] = 0.080 M. (5 points)
Answer:
The rate of the reaction is [tex]6.24\times 10^{-6} M/s[/tex].
Explanation:
[tex]NH4^+(aq) + NO_2^-(aq)\rightarrow N_2(g) + 2H_2O(l)[/tex]
Concentration of [tex][NH_4^{+}]=0.26 M[/tex]
Concentration of [tex][NO_2^{-}]=0.080 M[/tex]
Rate constant of the reaction = k= [tex]3.0\times 10^{-4} M^{-1} s^{-1}[/tex]
[tex]R= k[NH_{4}^+][NO_{2}^-][/tex]
[tex]R=3.0\times 10^{-4} M^{-1} s^{-1}\times 0.26 M\times 0.080 M[/tex]
[tex]R=6.24\times 10^{-6} M/s[/tex]
The rate of the reaction is [tex]6.24\times 10^{-6} M/s[/tex].
Answer:
Rate of Reaction = 6.24 x 10–6 M/s
Explanation:
Rate of reaction = k[NH4+][NO2–]
Concentration of [NH4+] = 0.26 M
Concentration of [NO2–] = 0.080 M
k= 3.0 × 10–4/ M · s
Rate of Reaction = (3.0 × 10–4/ M · s)( 0.26 M)(0.080 M)
A monoprotic acid is an acid that donates a single proton to the solution. Suppose you have 0.140 g of a monoprotic acid dissolved in 35.0 mL of water. This solution is then neutralized with 14.5 mL of 0.110 M NaOH. What is the molar mass of the acid?
Answer:
molar mass HA = 87.8 g/mole
Explanation:
Given 0.140g HA + 14.5ml(0.110M NaOH) => NaA + H₂O
Rxn is a 1:1 rxn ration => moles HA neutralized = moles NaOH used
=> 0.140g/molar mass of HA = (0.110M)(0.0145L)
=> molar mass of HA = (0.140g)/[(0.110moles/L)(0.0145L)] = 87.8 g/mole
Answer: The molar mass of monoprotic acid is 87.72 g/mol
Explanation:
To calculate the concentration of acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of monoprotic acid
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.
We are given:
[tex]n_1=1\\M_1=?M\\V_1=35.0mL\\n_2=1\\M_2=0.110M\\V_2=14.5mL[/tex]
Putting values in above equation, we get:
[tex]1\times M_1\times 35.0=1\times 0.110\times 14.5\\\\x=\frac{1\times 0.110\times 14.5}{1\times 35.0}=0.0456M[/tex]
To calculate the molecular mass of acid, we use the equation used to calculate the molarity of solution:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
We are given:
Molarity of solution = 0.0456 M
Given mass of acid = 0.140 g
Volume of solution = 35.0 mL
Putting values in above equation, we get:
[tex]0.0456M=\frac{0.140\times 1000}{\text{Molar mass of acid}\times 35}\\\\\text{Molar mass of acid}=\frac{0.140\times 1000}{0.0456\times 35}=87.72g/mol[/tex]
Hence, the molar mass of monoprotic acid is 87.72 g/mol
The initial concentration of fluoride ions in an aqueous solution is 2.00 M and the initial concentration of Al3+ ions is 0.15 M. After the solution has reached equilibrium what is the concentration ofAl3+, F- , and AlF6 3- ? Kf for [AlF6] 3- = 4.0 x 1019 .
Answer: [tex][Al^{3+}][/tex] = 1.834 M
[tex][F^-][/tex] = 0.004 M
[tex][AlF_6^{3-}][/tex] = 0.166 M
Explanation:
[tex]Al^{3+}+6F^-\rightleftharpoons AlF_6^{3-}[/tex]
Initial concentration of [tex]Al^{3+}[/tex] = 0.15 M
Initial concentration of [tex]F^-[/tex] = 2.0 M
The given balanced equilibrium reaction is,
[tex]Al^{3+}+6F^-\rightleftharpoons AlF_6^{3-}[/tex]
Initial conc. 2 M 0.15 M 0
At eqm. conc. (2-x) M (1-6x) M (x) M
The expression for equilibrium constant for this reaction will be,
[tex]K_f=\frac{[AlF_6^{3-}]}{[Al^{3+}][F^-]^6}[/tex]
Now put all the given values in this expression, we get :
[tex]4.0\times 10^{19}=\frac{(x)}{(2-x)\times (1-6x)^6}[/tex]
By solving the term 'x', we get :
[tex]x=0.166[/tex]
[tex][Al^{3+}][/tex] = (2-x) = 2-0.166 = 1.834 M
[tex][F^-][/tex] = (1-6x) = 1-6(0.1660)= 0.004 M
[tex][AlF_6^{3-}][/tex] = x = 0.166 M
A gas mixture contains 1.20 g N2 and 0.77 g O2 in a 1.65-L container at 15 ∘C. Part A Calculate the mole fraction of N2. Express your answer using two significant figures. X1 X 1 = nothing Request Answer Part B Calculate the mole fraction of O2. Express your answer using two significant figures. X2 X 2 = nothing Request Answer Part C Calculate the partial pressure of N2. Express your answer using two significant figures. P1 P 1 = nothing atm Request Answer Part D Calculate the partial pressure of O2. Express your answer using two significant figures. P2 P 2 = nothing atm Request Answer Provide Feedback
Explanation:
Moles of nitrogen gas = [tex]n_1=\frac{1.20 g}{28 g/mol}=0.0428 mol[/tex]
Moles of oxygen gas = [tex]n_2=\frac{0.77 g}{32 g/mol}=0.0240 mol[/tex]
Mole fraction of nitrogen gas=[tex]\chi_1=\frac{n_1}{n_1+n_2}[/tex]
[tex]\chi_1=\frac{0.0428 mol}{0.0428 mol+0.0240 mol}=0.6407\approx 0.64[/tex]
Mole fraction of oxygen gas=[tex]\chi_2=1-\chi_1=1-0.6407=0.3593\approx 0.36[/tex]
Total umber of moles in container :
n =[tex]n_1+n_2[/tex]= 0.0428 mol + 0.0240 mol = 0.0668 mol
Volume of the container = V = 1.65 L
Temperature of the container = T = 15°C = 288.15 K
Total pressure in the container = P
Using an ideal gas equation:
[tex]PV=nRT[/tex]
[tex]P=\frac{0.0668 mol\times 0.0821 atm L/mol k\times 288.15 K}{1.65 L}[/tex]
P = 0.9577 atm
Partial pressure of nitrogen gas = [tex]p^{o}_1[/tex]
Partial pressure of nitrogen gas = [tex]P^{o}_2[/tex]
Partial pressure of nitrogen gas and oxygen gas can be calculated by using Dalton's law of partial pressure:
[tex]p^{o}_i=p_{total}\times \chi_i[/tex]
[tex]p^{o}_1=P\times \chi_1=0.6135 atm\approx 0.61 atm[/tex]
[tex]p^{o}_2=P\times \chi_2=0.3441 atm\approx 0.34 atm[/tex]