Answer : The percent yield of the reaction is, 91.32 %
Explanation : Given,
Mass of [tex]C_6H_5CH_3[/tex] = 1 Kg = 1000 g
Molar mass of [tex]C_6H_5CH_3[/tex] = 92.14 g/mole
Molar mass of [tex]C_6H_5COOH[/tex] = 122.12 g/mole
First we have to calculate the moles of [tex]C_6H_5CH_3[/tex].
[tex]\text{Moles of }C_6H_5CH_3=\frac{\text{Mass of }C_6H_5CH_3}{\text{Molar mass of }C_6H_5CH_3}=\frac{1000g}{92.14g/mole}=10.85mole[/tex]
Now we have to calculate the moles of [tex]C_6H_5COOH[/tex].
The balanced chemical reaction will be,
[tex]2C_6H_5CH_3+3O_2\rightarrow 2C_6H_5COOH+2H_2O[/tex]
From the balanced reaction, we conclude that
As, 2 moles of [tex]C_6H_5CH_3[/tex] react to give 2 moles of [tex]C_6H_5COOH[/tex]
So, 10.85 moles of [tex]C_6H_5CH_3[/tex] react to give 10.85 moles of [tex]C_6H_5COOH[/tex]
Now we have to calculate the mass of [tex]C_6H_5COOH[/tex]
[tex]\text{Mass of }C_6H_5COOH=\text{Moles of }C_6H_5COOH\times \text{Molar mass of }C_6H_5COOH[/tex]
[tex]\text{Mass of }C_6H_5COOH=(10.85mole)\times (122.12g/mole)=1325.002g[/tex]
The theoretical yield of [tex]C_6H_5COOH[/tex] = 1325.002 g
The actual yield of [tex]C_6H_5COOH[/tex] = 1.21 Kg = 1210 g
Now we have to calculate the percent yield of [tex]C_6H_5COOH[/tex]
[tex]\%\text{ yield of }C_6H_5COOH=\frac{\text{Actual yield of }C_6H_5COOH}{\text{Theoretical yield of }C_6H_5COOH}\times 100=\frac{1210g}{1325.002g}\times 100=91.32\%[/tex]
Therefore, the percent yield of the reaction is, 91.32 %
What are the resulting coefficients when you balance the chemical equation for the combustion of ethane, C2H6? In this reaction, ethane is burned in the presence of oxygen (O2) to form carbon dioxide (CO2) and water (H2O). ____C2H6(g)+____O2(g)→____CO2(g)+____H2O(g)
Answer: The coefficients for balancing the given chemical equation are 2, 7, 4 and 6
Explanation:
Every balanced chemical equation follows law of conservation of mass.
This law states that mass can neither be created nor be destroyed, but it can only be transformed from one form to another form. This also means that total number of individual atoms on the reactant side must be equal to the total number of individual atoms on the product side.
The given balanced chemical equation follows:
[tex]2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O[/tex]
On reactant side:
Number of carbon atoms = 4
Number of hydrogen atoms = 12
Number of oxygen atoms = 14
On product side:
Number of carbon atoms = 4
Number of hydrogen atoms = 12
Number of oxygen atoms = 14
Hence, the coefficients for balancing the given chemical equation are 2, 7, 4 and 6
An open flask sitting in a lab fridge looks empty, but it is actually filled with a mixture of gases called air. If the flask volume is 2.50 L, and the air is at standard temperature and pressure, how many gaseous molecules does the flask contain?
Answer: [tex]0.67\times 10^{23}[/tex] molecules are contained in the flask.
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
Standard condition of temperature (STP) is 273 K and atmospheric pressure is 1 atm respectively.
According to the ideal gas equation:'
[tex]PV=nRT[/tex]
P= Pressure of the gas = 1 atm
V= Volume of the gas = 2.50 L
T= Temperature of the gas = 273 K
R= Value of gas constant = 0.0821 Latm/K mol
[tex]n=\frac{PV}{RT}=\frac{1\times 2.50L}{0.0821 \times 273}=0.11moles[/tex]
1 mole of gas contains=[tex]6.022\times 10^{23}[/tex] molecules
0.11 moles of gas contains=[tex]\frac{6.022\times 10^{23}}{1}\times 0.11=0.67\times 10^{23}[/tex] molecules
[tex]0.67\times 10^{23}[/tex] molecules are contained in the flask.
In which main energy level does the 'd' sublevel first appear? K (first main energy level) L (second main energy level) M (third main energy level) N (fourth main energy level)
Answer:
M (third main energy level)
Explanation:
The third main energy level bears the first appearance of the 'd' sublevel. The principal quantum number(n) depicts the main energy levels in which an orbital is located. It takes values of n=1,2,3,4,5..... and it can be represented by the shells k,l,m,n.......
The subshells in these main orbitals are represented by s,p,d and f. For the K shell, the principal quantum number is m and its sublevel notations are s,p and d. This is where the d-sublevel first appears.
The 'd' sublevel first appears in the third main energy level (M). This is because the electron configuration in atoms is organized into main energy levels and sublevels, which define an electron's distance from the nucleus and energy.
Explanation:The 'd' sublevel first appears in the third main energy level, also labeled as M. The electron configuration in atoms is organized into main energy levels and sublevels, which help define an electron's relative distance from the nucleus and its energy. The main energy levels are generally labeled K, L, M, N, etc., starting from the nucleus. Each energy level has one or more sublevels: s, p, d, f. The 's' sublevel appears in all main energy levels, the 'p' sublevel starts from the second (L), and the 'd' sublevel commences from the third (M) main energy level. Therefore, the 'd' sublevel does not exist in the first (K) or second (L) main energy levels.
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The equilibrium constant, Kc, for the following reaction is 1.29×10-2 at 600 K. COCl2(g) CO(g) + Cl2(g) Calculate the equilibrium concentrations of reactant and products when 0.323 moles of COCl2(g) are introduced into a 1.00 L vessel at 600 K.
Answer : The equilibrium concentrations of [tex]COCl_2,CO\text{ and }Cl_2[/tex] are, 0.2646, 0.0584 and 0.0584.
Explanation : Given,
Moles of [tex]COCl_2[/tex] = 0.323 mole
Volume of solution = 1 L
Initial concentration of [tex]COCl_2[/tex] = 0.323 M
Let the moles of [tex]CO\text{ and }Cl_2[/tex] be, 'x'. So,
Concentration of [tex]CO[/tex] = x M
Concentration of [tex]Cl_2[/tex] = x M
The given balanced equilibrium reaction is,
[tex]COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)[/tex]
Initial conc. 0.323 M 0 0
At eqm. conc. (0.323-x) M (x M) (x M)
The expression for equilibrium constant for this reaction will be,
[tex]K_c=\frac{[CO][Cl_2]}{[COCl_2]}[/tex]
Now put all the given values in this expression, we get :
[tex]1.29\times 10^{-2}=\frac{(x)\times (x)}{(0.323-x)}[/tex]
By solving the term 'x', we get :
x = 0.0584
Thus, the concentrations of [tex]COCl_2,CO\text{ and }Cl_2[/tex] at equilibrium are :
Concentration of [tex]COCl_2[/tex] = (0.323-x) M = (0.323-0.0584) M = 0.2646 M
Concentration of [tex]CO[/tex] = x M = 0.0584 M
Concentration of [tex]Cl_2[/tex] = x M = 0.0584 M
Therefore, the equilibrium concentrations of [tex]COCl_2,CO\text{ and }Cl_2[/tex] are, 0.2646, 0.0584 and 0.0584.
What are the major products produced in the combustion of C10H22 under the following conditions? Write balanced chemical equations for each. a. An excess of oxygen b. A slightly limited oxygen supply c. A very limited supply of oxygen d. The compound is burned in air
Answer: The chemical reactions are given below.
Explanation:
Combustion reaction is defined as the chemical reaction in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide gas and water molecule.
[tex]\text{hydrocarbon}+O_2\rightarrow CO_2+H_2O[/tex]
If supply of oxygen gas is limited, it is known as incomplete combustion and carbon monoxide gas is also produced as a product.
For a: An excess of oxygenHere, complete combustion reaction takes place. The chemical equation follows:
[tex]2C_{10}H_{22}+31O_2\rightarrow 20CO_2+22H_2O[/tex]
For b: A slightly limited oxygen supplyHere, incomplete combustion takes place and carbon monoxide is also formed.
[tex]C_{10}H_{22}+13O_2\rightarrow 5CO+5CO_2+11H_2O[/tex]
For c: A very limited supply of oxygenHere, incomplete combustion takes place and only carbon monoxide with water are formed as the products.
[tex]2C_{10}H_{22}+21O_2\rightarrow 20CO+22H_2O[/tex]
For d: The compound is burned in airWhen a compound is burned in air, it means that unlimited supply of oxygen is there. So, complete combustion reaction takes place and carbon dioxide gas is formed as a product.
[tex]2C_{10}H_{22}+31O_2\rightarrow 20CO+22H_2O[/tex]
Hence, the chemical reactions are given below.
Answer:
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Explanation:
The balanced combustion reaction for C6H6 is 2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)+6542 kJ If 8.600 g C6H6 is burned and the heat produced from the burning is added to 5691 g of water at 21 ∘ C, what is the final temperature of the water?
In this chemistry problem, we first calculate the heat produced by the combustion of C6H6, and then we use the heat equation to find the final temperature of the water which comes out to be 53.14 °C.
Explanation:The subject of this question pertains to the field of chemistry, specifically physical chemistry dealing with energy changes in chemical reactions, and this seems to be a high school-level problem based on the complexities involved. The principle used here is q=mc∆T, which allows us to calculate how much heat is transferred when the water temperature changes.
First, the amount of heat produced by the combustion of 8.6 g of C6H6 should be calculated based on its enthalpy change (heat produced per mole). Given the heat of combustion, -6542 kJ for 2 moles of C6H6, the heat of combustion for 8.6g of C6H6 can be calculated by (8.6 g / (78.11 g/mol) mol ) * (-6542 kJ / 2 mol) = -227.11 kJ.
The heat absorbed by the water can then be calculated using the heat equation: q=mc∆T. Here, m = mass of water = 5691 g, c = specific heat capacity of water = 4.18 J/g°C. The heat is converted to kilojoules: q = -227.11 kJ * 1000 = -227110 J. Hence, the equation becomes -227110 = 5691*4.18*∆T. After rearranging and solving, the final temperature (T) will be 53.14 °C.
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Enough of a monoprotic weak acid is dissolved in water to produce a 0.0172 M solution. If the pH of the resulting solution is 2.39 at 20 °C, determine the pKa for the acid.
Answer:
9.7 x 10⁻⁴
Explanation:
HA ⇄ H⁺ + A⁻
C(eq) 0.0174 10⁻²·³⁹ 10⁻²·³⁹
=0.0041M =0.0041M
Ka = [H⁺][A⁻]/[HA] = (0.0014)²/(0.0174) = 9.7 x 10⁻⁴
Phosgene (carbonyl chloride), COCl2, is an extremely toxic gas that is used in manufacturing certain dyes and plastics. Phosgene can be produced by reacting carbon monoxide and chlorine gas at high temperatures: CO(g)+Cl2(g)⇌COCl2(g) Carbon monoxide and chlorine gas are allowed to react in a sealed vessel at 480 ∘C . At equilibrium, the concentrations were measured and the following results obtained: Gas Partial Pressure (atm) CO 0.820 Cl2 1.27 COCl2 0.230 What is the equilibrium constant, Kp, of this reaction?
Answer:
0.2208 is the equilibrium constant,[tex]K_p[/tex].
Explanation:
Equilibrium constant is defined as ratio of concentration of products to the concentration of reactants raised to the power equal to their stoichiometric coefficients in balanced chemical equation. It is expressed as [tex]K_c[/tex]
If the equilibrium is in gaseous phase then instead of concentration take partial pressure of each compound.The It is expressed as [tex]K_p[/tex].
[tex]CO(g)+Cl_2(g)\rightarrow COCl_2(g)[/tex]
Partial pressure of the[tex] p_{[CO]} = 0.820 atm[/tex]
Partial pressure of the[tex] p_{[Cl_2]} = 1.27 atm[/tex]
Partial pressure of the [tex]p_{[COCl_2]} = 0.230 atm[/tex]
The expression of an equilibrium constant is given as:
[tex]K_p=\frac{p_{[COCl_2]}}{p_{[CO]}p_{[Cl_2]}}[/tex]
[tex]K_p=\frac{0.230 atm}{0.820 atm \times 1.27 atm}=0.2208 [/tex]
Aluminum chloride can be formed from its elements:
(i) 2Al(s) + 3Cl2 (g) ⟶ 2AlCl3 (s) ΔH° = ? Use the reactions here to determine the ΔH° for reaction (i):
(ii) HCl(g) ⟶ HCl(aq) ΔH(ii) ° = −74.8 kJ
(iii) H2 (g) + Cl2 (g) ⟶ 2HCl(g) ΔH(iii) ° = −185 kJ
(iv) AlCl3 (aq) ⟶ AlCl3 (s) ΔH(iv) ° = +323 kJ/mol
(v) 2Al(s) + 6HCl(aq) ⟶ 2AlCl3 (aq) + 3H2 (g) ΔH(v) ° = −1049 kJ
Answer: The [tex]\Deltas H^o_{formation}[/tex] for the reaction is -1406.8 kJ.
Explanation:
Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.
The chemical reaction for the formation reaction of [tex]AlCl_3[/tex] is:
[tex]2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3 (s)[/tex] [tex]\Delta H^o_{formation}=?[/tex]
The intermediate balanced chemical reaction are:
(1) [tex]HCl(g)\rightarrow HCl(aq.)[/tex] [tex]\Delta H_1=-74.8kJ[/tex] ( × 6)
(2) [tex]H_2(g)+Cl_2(g)\rightarrow 2HCl(g)[/tex] [tex]\Delta H_2=-185kJ[/tex] ( × 3)
(3) [tex]AlCl_3(aq.)\rightarrow AlCl_3(s)[/tex] [tex]\Delta H_3=+323kJ[/tex] ( × 2)
(4) [tex]2Al(s)+6HCl(aq.)\rightarrow 2AlCl_3(aq.)+3H_2(g)[/tex] [tex]\Delta H_4=-1049kJ[/tex]
The expression for enthalpy of formation of [tex]AlCl_3[/tex] is,
[tex]\Delta H^o_{formation}=[6\times \Delta H_1]+[3\times \Delta H_2]+[2\times \Delta H_3]+[1\times \Delta H_4][/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{formation}=[(-74.8\times 6)+(-185\times 3)+(323\times 2)+(-1049\times 1)]=-1406.8kJ[/tex]
Hence, the [tex]\Deltas H^o_{formation}[/tex] for the reaction is -1406.8 kJ.
The change in enthalpy for the reaction 2Al(s) + 3Cl2(g) ⟶ 2AlCl3(s) is -832.2 kJ, calculated by manipulating and summing the enthalpy changes of the given reactions using Hess's Law.
Explanation:The change in enthalpy (∆H°) of the reaction 2Al(s) + 3Cl2(g) ⟶ 2AlCl3(s) can be calculated using Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for each step in the reaction. In this case, we want to find the enthalpy change of reaction (i) using reactions (ii) - (v).
To achieve this, let's manipulate the reactions as follows:
1. Reverse reaction (ii) and multiply by 6. This will give it the same number of HCl (moles) as in reaction (v): 6HCl(aq) ⟶ 6HCl(g), ∆H′ = +74.8 kJ * 6 = +448.8 kJ.
2. Multiply reaction (iii) by 3 to match the number of H2 (moles) and HCl in reaction (v): 3H2 (g) + 3Cl2 (g) ⟶ 6HCl(g), ∆H′′ = -185 kJ * 3 = -555 kJ.
Reaction (iv) remains unchanged: AlCl3 (aq) ⟶ AlCl3 (s), ∆H′′′ = +323 kJ.
3. Using reaction (v) in the forward direction: 2Al(s) + 6HCl(aq) ⟶ 2AlCl3 (aq) + 3H2 (g), ∆H′′′′ = -1049 kJ.
Summing these manipulated reactions will give you reaction (i): ∆H° = ∆H′ + ∆H′′ + ∆H′′′ + ∆H′′′′ = 448.8 kJ - 555 kJ + 323 kJ - 1049 kJ = -832.2 kJ
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A single electron can have the following set of quantum numbers: n=3, l=2, ml=0, ms= -1/2.
True or False
Answer:
True => (3, 2, 0, -1/2) => 3d₀² electron
Explanation:
Studies indicate that the methyl radical is trigonal planar. Based on this, which of the following best describes the methyl radical? The carbon is sp2 hybridized and the unpaired electron occupies an sp2 orbital. The carbon is sp2 hybridized and the unpaired electron occupies a 2p orbital. The carbon is sp3 hybridized and the unpaired electron occupies an sp3 orbital. The carbon is sp3 hybridized and the unpaired electron occupies a 2p orbital.
Answer:
The carbon is sp² hybridized and the unpaired electron occupies a 2p orbital.
Explanation:
The central C atom has a trigonal planar geometry, so it is sp² hybridized.
The sp² bonds are formed by the hybridization of the 2s orbital with two 2p orbitals.
The unpaired electron goes into the unhybridized 2p orbital.
Iodine is prepared both in the laboratory and commercially by adding Cl2(g)Cl2(g) to an aqueous solution containing sodium iodide. 2NaI(aq)+Cl2(g)⟶I2(s)+2NaCl(aq) 2NaI(aq)+Cl2(g)⟶I2(s)+2NaCl(aq) How many grams of sodium iodide, NaI,NaI, must be used to produce 89.1 g89.1 g of iodine, I2?I2? mass: g NaI
Answer : The mass of sodium iodide used to produced must be, 105.22 grams.
Explanation : Given,
Mass of [tex]I_2[/tex] = 89.1 g
Molar mass of [tex]I_2[/tex] = 253.8 g/mole
Molar mass of [tex]NaI[/tex] = 149.89 g/mole
First we have to calculate the moles of [tex]I_2[/tex].
[tex]\text{Moles of }I_2=\frac{\text{Mass of }I_2}{\text{Molar mass of }I_2}=\frac{89.1g}{253.8g/mole}=0.351moles[/tex]
Now we have to calculate the moles of [tex]NaI[/tex].
The balanced chemical reaction is,
[tex]2NaI(aq)+Cl_2(g)\rightarrow I_2(s)+2NaCl(aq)[/tex]
From the balanced chemical reaction, we conclude that
As, 1 mole of [tex]I_2[/tex] obtained from 2 moles of [tex]NaI[/tex]
So, 0.351 moles of [tex]I_2[/tex] obtained from [tex]2\times 0.351=0.702[/tex] moles of [tex]NaI[/tex]
Now we have to calculate the mass of [tex]NaI[/tex].
[tex]\text{Mass of }NaI=\text{Moles of }NaI\times \text{Molar mass of }NaI[/tex]
[tex]\text{Mass of }NaI=(0.702mole)\times (149.89g/mole)=105.22g[/tex]
Therefore, the mass of sodium iodide used to produced must be, 105.22 grams.
To produce 89.1 grams of iodine (I2), 178.2 grams of sodium iodide (NaI) must be used.
Explanation:To find the number of grams of sodium iodide (NaI) needed to produce 89.1 g of iodine (I2), we need to use the stoichiometric coefficients from the balanced chemical equation:
2NaI(aq) + Cl2(g) ⟶ I2(s) + 2NaCl(aq)
From the equation, we can see that 2 moles of NaI react to produce 1 mole of I2. The molar mass of NaI is 149.89 g/mol. We can set up a proportion:
(2 mol NaI / 1 mol I2) = (x g NaI / 89.1 g I2)
Solving for x:
x = (2 mol NaI / 1 mol I2) * (89.1 g I2 / 1 mol I2) = 178.2 g NaI
Therefore, 178.2 grams of sodium iodide (NaI) must be used to produce 89.1 grams of iodine (I2).
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You want to remove as much CO2 gas as possible from a water solution. Which of the following treatments would?
a. cool the solution
b. filter the solution
c. boil the solution
d. aerate the solution
What is the net ionic equation of the reaction of MgSO4 with Pb(NO3)2? Express you answer as a chemical equation including phases. View Available Hint(s)
Answer : The net ionic equation will be,
[tex]Pb^{2+}(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)[/tex]
Explanation :
In the net ionic equations, we are not include the spectator ions in the equations.
Spectator ions : The same number of ions present on reactant and product side which do not participate in a reactions.
The given balanced ionic equation will be,
[tex]MgSO_4(aq)+Pb(NO_3)_2(aq)\rightarrow Mg(NO_3)_2(aq)+PbSO_4(s)[/tex]
The ionic equation in separated aqueous solution will be,
[tex]Mg^{2+}(aq)+SO_4^{2-}(aq)+Pb^{2+}(aq)+2NO^{3-}(aq)\rightarrow PbSO_4(s)+Mg^{2+}(aq)+2NO^{3-}(aq)[/tex]
In this equation, [tex]Mg^{2+}\text{ and }NO_3^-[/tex] are the spectator ions.
By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.
The net ionic equation will be,
[tex]Pb^{2+}(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)[/tex]
A 47.1 g sample of a metal is heated to 99.0°C and then placed in a calorimeter containing 120.0 g of water (c = 4.18 J/g°C) at 21.4°C. The final temperature of the water is 24.5°C. Which metal was used?
Answer : The metal used was iron (the specific heat capacity is [tex]0.44J/g^oC[/tex]).
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
[tex]q_1=-q_2[/tex]
[tex]m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)[/tex]
where,
[tex]C_1[/tex] = specific heat of metal = ?
[tex]C_1[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]
[tex]m_1[/tex] = mass of metal = 47.1 g
[tex]m_2[/tex] = mass of water = 120 g
[tex]T_f[/tex] = final temperature of water = [tex]24.5^oC[/tex]
[tex]T_1[/tex] = initial temperature of metal = [tex]99^oC[/tex]
[tex]T_2[/tex] = initial temperature of water = [tex]21.4^oC[/tex]
Now put all the given values in the above formula, we get
[tex]47.1g\times c_1\times (24.5-99)^oC=-120g\times 4.18J/g^oC\times (24.5-21.4)^oC[/tex]
[tex]c_1=0.44J/g^oC[/tex]
Form the value of specific heat of metal, we conclude that the metal used in this was iron.
Therefore, the metal used was iron (the specific heat capacity is [tex]0.44J/g^oC[/tex]).
What is the correct net ionic equation, including all coefficients, charges, and phases, for the following set of reactants? Assume that the contribution of protons from H2SO4 is near 100 %.Ba(OH)2(aq)+H2SO4(aq)→
Final answer:
The net ionic equation for the reaction between barium hydroxide and sulfuric acid is Ba2+(aq) + SO42-(aq) → BaSO4(s), which shows the formation of an insoluble precipitate of barium sulfate.
Explanation:
The correct net ionic equation for the reaction between barium hydroxide and sulfuric acid is as follows:
Ba2+(aq) + SO42−(aq) → BaSO4(s)
In this equation, Ba2+(aq) represents the barium ion in aqueous solution, SO42−(aq) denotes the sulfate ion in aqueous solution, and BaSO4(s) is the precipitate of barium sulfate. It is important to remember that net ionic equations must be balanced by both mass and charge, and in this case, the equation follows these rules with a product that is insoluble in water.
given a condensed structural formula, write a line-angle formula for (CH3)2CHCH(CH3)2
Answer : The line-angle formula for [tex](CH_3)_2CHCH(CH_3)_2[/tex] is shown below.
Explanation :
Condensed structural formula : It is a formula in which the lines are used between the bonded atoms and the atoms are also shown in the structural formula.
Line-angle formula : In the line-angle formula, the carbon atoms are represented at the corner and ends of the lines with some angle and the hydrogen atoms are hidden.
The line-angle formula for [tex](CH_3)_2CHCH(CH_3)_2[/tex] is shown below.
The line-angle formula for (CH3)2CHCH(CH3)2 is CH3 CH3 CH2 CH3 CH3.
Explanation:The line-angle formula for (CH3)2CHCH(CH3)2 can be written as:
CH3 CH3
|
CH2
|
CH3 CH3
This formula represents the structural arrangement of atoms in the molecule, with each line representing a bond between carbon atoms. The ends of the lines represent carbon atoms, and any additional atoms or groups attached to those carbon atoms are shown as branches off the main line.
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Calculate the enthalpy of the following reaction: 4 B (s) + 3 O2 (g) → 2 B2O3 (s) given the following pertinent information: (A) B2O3 (s) + 3 H2O (g) → 3 O2 (g) + B2H6 (g), ΔHoA = +2035 kJ (B) 2 B (s) + 3 H2 (g) → B2H6 (g), ΔHoB = +36 kJ (C) H2 (g) + LaTeX: \frac{1}{2} 1 2 O2 (g) → H2O (l), ΔHoC = −285 kJ (D) H2O (l) → H2O (g), ΔHoD = +44 kJ
Answer:
4B + 3O₂ => 2B₂O₃; ΔH° = -3673Kj
Explanation:
Work these type problems in pairs of rxns… That is, add Rxn-1 & Rxn-2 => Rxn-1,2; then add Rxn-3 to Rxn-1,2 => Rxn- 1,2,3. Rxn-4 is not needed to obtain target rxn.
Target Rxn => 4B + 3O₂ => 2B₂O₃
Given …
(1) B₂O₃ + 2H₂O => 3O₂ + B₂H₆
=> reverse and double
=> 2B₂H₆ + 6O₂ => 2B₂O₃ + 6H₂O
(2) 2B + 3H₂ => B₂H₆ => double and add to Rxn-1 => 4B + 6H₂ => 2B₂H₆
2B₂H₆ + 6O₂ => 2B₂O₃ + 6H₂O
4B + 6H₂ => 2B₂H₆
________________________
∑(1,2) 4B + 6O₂ + 6H₂ => 2B₂O₃ + 6H₂O; ΔH°₁₂ = -2035Kj + (+72Kj) = -1963Kj
(3) => H₂ + ½O₂ => H₂O
=> reverse and multiply by 6, then add to (1,2) => 6H₂O => 6H₂ + 3O₂
6H₂O => 6H₂ + 3O₂
4B + 6O₂ + 6H₂ => 2B₂O₃ + 6H₂O
________________________
∑[(1,2,3) 4B + 3O₂ => 2B₂O₃; ΔH°₁₂₃ = -1963Kj + 6(-285Kj) = -3673Kj
The pH of a solution is measured eight times by one operator using the same instrument. She obtains the following data: 7.15, 7.20, 7.18, 7.19, 7.21, 7.20, 7.16, and 7.20. (a) Calculate the sample mean. Round your answer to 3 decimal places.
Answer:
7.186
Explanation:
The mean is the average of some given data from a sample point.
To calculate the mean, we use the formula below:
Mean = ∑fx/∑f
Where f = frequency
x = sample data
From the given pH of the solutions, we can form a table:
x f fx
7.15 1 7.15
7.16 1 7.16
7.18 1 7.18
7.19 1 7.19
7.20 3 21.6
7.21 1 7.21
Now ∑fx = 7.15 +7.16 +7.18 + 7.19 + 7.20 + 7.21 = 57.49
∑f = 1 + 1 + 1 + 1 + 3 + 1 = 8
The mean = [tex]\frac{57.49}{8}[/tex] = 7.18625 = 7.186
Given the following information: Mass of proton = 1.00728 amu Mass of neutron = 1.00866 amu Mass of electron = 5.486 × 10^-4 amu Speed of light = 2.9979 × 10^8 m/s Calculate the nuclear binding energy (absolute value) of 3Li^6. which has an atomic mass of 6.015126 amu. J/mol.
Answer: The nuclear binding energy of the given element is [tex]2.938\times 10^{12}J/mol[/tex]
Explanation:
For the given element [tex]_3^6\textrm{Li}[/tex]
Number of protons = 3
Number of neutrons = (6 - 3) = 3
We are given:
[tex]m_p=1.00728amu\\m_n=1.00866amu\\A=6.015126amu[/tex]
M = mass of nucleus = [tex](n_p\times m_p)+(n_n\times m_n)[/tex]
[tex]M=[(3\times 1.00728)+(3\times 1.00866)]=6.04782amu[/tex]
Calculating mass defect of the nucleus:
[tex]\Delta m=M-A\\\Delta m=[6.04782-6.015126)]=0.032694amu=0.032694g/mol[/tex]
Converting this quantity into kg/mol, we use the conversion factor:
1 kg = 1000 g
So, [tex]0.032694g/mol=0.032694\times 10^{-3}kg/mol[/tex]
To calculate the nuclear binding energy, we use Einstein equation, which is:
[tex]E=\Delta mc^2[/tex]
where,
E = Nuclear binding energy = ? J/mol
[tex]\Delta m[/tex] = Mass defect = [tex]0.032694\times 10^{-3}kg/mol[/tex]
c = Speed of light = [tex]2.9979\times 10^8m/s[/tex]
Putting values in above equation, we get:
[tex]E=0.032694\times 10^{-3}kg/mol\times (2.9979\times 10^8m/s)^2\\\\E=2.938\times 10^{12}J/mol[/tex]
Hence, the nuclear binding energy of the given element is [tex]2.938\times 10^{12}J/mol[/tex]
The binding energy of the lithium nucleus is 2.94 * 10^14 J/mol.
What is binding energy?The term binding energy refers to the energy that hold the nucleons in the atom together.
We know that the atomic mass of the Li is 6.015126 amu. Note that there are three protons and three neutrons. Hence;
Mass of protons= 3(1.00728 amu) = 3.02184
Mass of neutrons = 3(1.00866 amu) = 3.02598
Mass defect = (3.02184 + 3.02598) - 6.015126 amu = 0.032694 amu = 0.032694 g/mol
Now;
E = mc^2 = (0.032694 g/mol * (3 * 10^8)^2) = 2.94 * 10^14 J/mol
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Are hydrocarbons approved for retrofit applications
Hydroxyapatite, Ca5(PO4)3(OH), is the main mineral component of dental enamel, dentin, and bone, and thus has many medical uses. Coating it on metallic implants (such as titanium alloys and stainless steels) helps the body accept the implant. In the form of powder and beads, it is used to fill bone voids, which encourages natural bone to grow into the void. Hydroxyapatite is prepared by adding aqueous phosphoric acid to a dilute slurry of calcium hydroxide. (a) Write a balanced equation for this preparation. (b) What mass of hydroxyapatite could form from 100. g of 85% phosphoric acid and 100. g of calcium hydroxide?
Answer:
(a) 5Ca(OH)₂ + 3H₃PO₄ = Ca₅(PO₄)₃(OH) + 9H₂O.
(b) 135.62 g.
Explanation:
(a) Write a balanced equation for this preparation.
The balanced equation for5Ca(OH)₂ + 3H₃PO₄ = Ca₅(PO₄)₃(OH) + 9H₂O,
that 5 mol of Ca(OH)₂ react with 3 mol of H₃PO₄ to produce 1 mol of Ca₅(PO₄)₃(OH) "hydroxyapatite" and 9 mol of H₂O.
(b) What mass of hydroxyapatite could form from 100.0 g of 85% phosphoric acid and 100.0 g of calcium hydroxide?
Firstly, we need to calculate the no. of moles of 100.0 g of 85% phosphoric acid and 100.0 g of calcium hydroxide:The no. of moles of 100.0 g of 85% phosphoric acid:
∵ The percent of phosphoric acid = 85%.
∴ The actual mass of phosphoric acid in solution = 85.0 g.
∴ no. of moles of phosphoric acid = mass/molar mass = (85.0 g)/(97.994 g/mol) = 0.87 mol.
The no. of moles of 100.0 g of calcium hydroxide:
∴ no. of moles of calcium hydroxide = mass/molar mass = (100.0 g)/(74.093 g/mol) = 1.35 mol.
From the stichiometry, Ca(OH)₂ react with H₃PO₄ with (5: 3) molar ratio.
∴ 1.35 mol of calcium hydroxide "limiting reactant" react completely with 0.813 mol of phosphoric acid "excess reactant" with (5: 3) molar ratio.
Now, we can find the no. of moles of produced Ca₅(PO₄)₃(OH) "hydroxyapatite":
Using cross multiplication:
5 mol of Ca(OH)₂ produce → 1 mol Ca₅(PO₄)₃(OH), from stichiometry.
1.35 mol of Ca(OH)₂ produce → ??? mol Ca₅(PO₄)₃(OH).
∴ The no. of moles of produced Ca₅(PO₄)₃(OH) "hydroxyapatite" = (1.35 mol)(1 mol)/(5 mol) = 0.27 mol.
Finally, we can get the mass of produced Ca₅(PO₄)₃(OH) "hydroxyapatite" = (no. of moles)(molar mass) = (0.27 mol)(502.3 g/mol) = 135.62 g.
The equation of the reaction is:
5Ca(OH)₂ + 3H₃PO₄ ---> Ca₅(PO₄)₃(OH) + 9H₂O.Based on the data provided, the mass of hydroxyapatite formed is 135.62 g.
What are minerals?Minerals are substances which contain metallic elements combined with other elements found in the earth's crust.
Example of a mineral is Hydroxyapatite, Ca5(PO4)3(OH).
The balanced equation for the preparation of hydroxyapatite is given below:
5Ca(OH)₂ + 3H₃PO₄ ---> Ca₅(PO₄)₃(OH) + 9H₂O,that 5 mol of Ca(OH)₂ react with 3 mol of H₃PO₄ to produce 1 mol of Ca₅(PO₄)₃(OH) and 9 mol of H₂O.
The mass of hydroxyapatite prepared from 100.0 g of 85% phosphoric acid and 100.0 g of calcium hydroxide? is calculated as follows:
100.0 g of 85% phosphoric acid contains 100 * 85/100 g of phosphoric acid = 85.0 g.
moles of phosphoric acid in 85 g = mass/molar mass
molar mass of phosphoric acid = 97.994 g/mol
moles of phosphoric acid in 85 g = 85.0 g/97.994 g/mol
moles of phosphoric acid in 85 g = 0.87 moles
moles of calcium hydroxide in 100.0 g = mass/molar mass
molar mass of calcium hydroxide = 74.093 g/mol
moles of calcium hydroxide in 100.0 g = 100.0 g/74.093 g/mol
moles of calcium hydroxide in 100.0 g = 1.35 moles
From the equation of the reaction, Ca(OH)₂ react with H₃PO₄ in a 5: 3 molar ratio.
5 mol of Ca(OH)₂ produce 1 mole Ca₅(PO₄)₃(OH)
1.35 mol of Ca(OH)₂ will produce 1.35 * 1/5 moles of Ca₅(PO₄)₃(OH) = 0.27 moles
mass of 0.27 moles of hydroxyapatite = no. of moles * molar mass
molar mass of hydroxyapatite = 502.3 g/mol
mass of hydroxyapatite = 0.27 mol * 502.3 g/mol
mass of hydroxyapatite = 135.62 g
Therefore, the mass of hydroxyapatite produced is 135.62 g
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Directions: Using the definition of molarity, the given balanced equations, and stoichiometry, solve the following problems.
Sodium chloride solution and water react to produce sodium hydroxide and chlorine gas according to the following balanced equation: 2NaCl(aq) + 2H20(l) <-> 2Na0H(aq) + Cl2(g)
a. How many liters of 0.4 M sodium chloride do you need in order to have 3.0 moles of chlorine gas?
b. Find the number of moles of water needed to produce 3.0 L of chlorine gas at STP.
Answer:
15L of 0.40M NaCl(aq) solution
Explanation:
2NaCl(aq) + 2H₂O(l) → 2NaOH(aq) + Cl₂(g)
2Na⁺(aq) + 2Clˉ(aq) + 2H₂O(l) → 2Na⁺(aq) + 2OHˉ(aq) + Cl₂(g)
Na⁺(aq) is a spectator ion in the given reaction and does not enter into the reaction process…
Net Ionic Equation is then 2Clˉ(aq) + 2H₂O(l) → 2OHˉ(aq) + Cl₂(g)
From Rxn, 2 moles Clˉ(aq) is needed to produce 1 mole of Cl₂(g)
Therefore, 6 moles Clˉ(aq) is needed to produce 3 moles of Cl₂(g)
That is, 6 moles NaCl(aq) → 6 moles Clˉ(aq) = 0.40M x V(NaCl)liters
V(NaCl) liters = 6 moles Clˉ(aq)/(0.40mole/liter) = 15 liters of 0.40M NaCl(aq)
7.5 liters of 0.4M sodium chloride solution are needed to produce 3.0 moles of chlorine gas, and 0.134 moles of water are needed to produce 3.0 liters of chlorine gas at STP.
Explanation:The question is related to molarity and stoichiometry. Let's solve part (a) and (b).
Part A:
Molarity is represented by moles of solute / liters of solution. The balanced equation gives a 1:1 ratio between NaCl and Cl2. The question gives us 3.0 moles of Cl2 gas, implying that we need the same moles of NaCl.
So, Moles = Molarity x Volume, Volume = Moles / Molarity, Volume = 3.0 moles / 0.4 M = 7.5 liters of NaCl solution are needed.
Part B:
For part (b), we see from the stoichiometry of the balanced reaction that the number of moles of water is equal to the number of moles of chlorine. According to the ideal gas law (PV=nRT), at standard temperature and pressure (STP), 1 mole of any gas occupies 22.4L. So, 3.0L of Cl2 gas corresponds to 3.0 L / 22.4 L/mole = 0.134 moles. Therefore, you'll need 0.134 moles of water.
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Dumbledore decides to gives a surprise demonstration. He starts with a hydrate of Na2CO3 which has a mass of 4.31 g before heating. After he heats it he finds the mass of the anhydrous compound is found to be 3.22 g. He asks everyone in class to determine the integer x in the hydrate: Na2CO3·xH2O; you should do this also. Round your answer to the nearest integ
Answer:
Na₂CO₃.2H₂O
Explanation:
For the hydrated compound, let us denote is by Na₂CO₃.xH₂O
The unknown is the value of x which is the amount of water of crystallisation.
Given values:
Starting mass of hydrate i.e Na₂CO₃.xH₂O = 4.31g
Mass after heating (Na₂CO₃) = 3.22g
Mass of the water of crystallisation = (4.31-3.22)g = 1.09g
To determine the integer x, we find the number of moles of the anhydrous Na₂CO₃ and that of the water of crystallisation:
Number of moles = [tex]\frac{mass }{molar mass }[/tex]
Molar mass of Na₂CO₃ =[(23x2) + 12 + (16x3)] = 106gmol⁻¹
Molar mass of H₂O = [(1x2) + (16)] = 18gmol⁻¹
Number of moles of Na₂CO₃ = [tex]\frac{3.22}{106}[/tex] = 0.03mole
Number of moles of H₂O = [tex]\frac{1.09}{18}[/tex] = 0.06mole
From the obtained number of moles:
Na₂CO₃ H₂O
0.03 0.06
Simplest
Ratio 0.03/0.03 0.03/0.06
1 2
Therefore, x = 2
After heating, the mass loss corresponds to 1.09 g of water, which, when compared to the moles of anhydrous [tex]\( \text{Na}_2\text{CO}_3 \)[/tex], gives a mole ratio of approximately 2:1, making [tex]\( x = 2 \)[/tex].
To determine the integer x in the hydrate [tex]\( \text{Na}_2\text{CO}_3 \cdot x \text{H}_2\text{O} \)[/tex], we need to follow these steps:
1. Calculate the mass of the water lost during heating:
[tex]\[\text{Mass of water} = \text{Mass of hydrate} - \text{Mass of anhydrous compound}\][/tex]
Given:
[tex]\[\text{Mass of hydrate} = 4.31 \, \text{g}\][/tex]
[tex]\[\text{Mass of anhydrous compound (Na}_2\text{CO}_3\text{)} = 3.22 \, \text{g}\][/tex]
Thus:
[tex]\[\text{Mass of water} = 4.31 \, \text{g} - 3.22 \, \text{g} = 1.09 \, \text{g}\][/tex]
2. Calculate the moles of anhydrous [tex]\( \text{Na}_2\text{CO}_3 \)[/tex]:
The molar mass of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex] is calculated as follows:
[tex]\[\text{Molar mass of Na}_2\text{CO}_3 = 2 \times 23.0 + 12.0 + 3 \times 16.0 = 46.0 + 12.0 + 48.0 = 106.0 \, \text{g/mol}\][/tex]
[tex]\[\text{Moles of Na}_2\text{CO}_3 = \frac{\text{Mass of Na}_2\text{CO}_3}{\text{Molar mass of Na}_2\text{CO}_3} = \frac{3.22 \, \text{g}}{106.0 \, \text{g/mol}} = 0.0304 \, \text{mol}\][/tex]
3. Calculate the moles of water lost:
The molar mass of water [tex](\( \text{H}_2\text{O} \))[/tex] is:
[tex]\[\text{Molar mass of H}_2\text{O} = 2 \times 1.0 + 16.0 = 18.0 \, \text{g/mol}\][/tex]
[tex]\[\text{Moles of water} = \frac{\text{Mass of water}}{\text{Molar mass of H}_2\text{O}} = \frac{1.09 \, \text{g}}{18.0 \, \text{g/mol}} = 0.0606 \, \text{mol}\][/tex]
4. Determine the mole ratio of water to [tex]\( \text{Na}_2\text{CO}_3 \)[/tex]:
[tex]\[x = \frac{\text{Moles of water}}{\text{Moles of Na}_2\text{CO}_3} = \frac{0.0606 \, \text{mol}}{0.0304 \, \text{mol}} = 1.993 \approx 2\][/tex]
Thus, the integer x in the hydrate[tex]\( \text{Na}_2\text{CO}_3 \cdot x \text{H}_2\text{O} \)[/tex]is 2.
When 10.g of CH3COOH is combusted in a sealed calorimeter, it releases enough energy to heat 2000. g of water from 23.5 °C to 34.3 °C. a. Calculate the energy released per 10 g of CH3COOH. b. Calculate the energy released per mole of CH3COOH.
Answer:
a. 90.288 kJ.
b. - 54.06 kJ/mol.
Explanation:
a. Calculate the energy released per 10 g of CH₃COOH.
We can calculate the amount of heat (Q) released to water using the relation:Q = m.c.ΔT,
where, Q is the amount of heat released to water (Q = ??? J).
m is the mass of water (m = 2000.0 g).
c is the specific heat capacity of solution (c = 4.18 J/g.°C).
ΔT is the difference in T (ΔT = final temperature - initial temperature = 34.3°C - 23.5°C = 10.8°C).
∴ Q = m.c.ΔT = (2000.0 g)(4.18 J/g.°C)(10.8°C) = 90288 J = 90.288 kJ.
b. Calculate the energy released per mole of CH₃COOH.
To find ΔH:∵ ΔH = Q/n
no. of moles of CH₃COOH (n) = mass/atomic mass = (10.0 g)/((60.052 g/mol) = 0.167 mol.
∴ ΔH = - Q/n = - (90.288 kJ)/(0.167 mol) = - 54.06 kJ/mol.
The negative sign is not from calculation, but it is an indication that the reaction is exothermic.
The biochemistry that takes place inside cells depends on various elements, such as sodium, potassium, and calcium, that are dissolved in water as ions. These ions enter cells through narrow pores in the cell membrane known as ion channels. Each ion channel, which is formed from a specialized protein molecule, is selective for one type of ion. Measurements with microelectrodes have shown that a 0.30-nm-diameter potassium ion (K+) channel carries a current of 1.8 pA. How many potassium ions pass through if the ion channel opens for 1.0 ms? What is the current density in the ion channel?
Answer:
Explanation:
The biochemistry that takes place inside cells depends on various elements, such as sodium, potassium, and
calcium, that are dissolved in water as ions. These ions enter cells through narrow pores in the cell membrane
known as ion channels. Each ion channel, which is formed from a specialized protein molecule, is selective
for one type of ion. Measurements with microelectrodes have shown that a 0.30-nm-diameter potassium ion (K+) channel carries a current of 1.8 pA.
Part A. How many potassium ions pass through if the ion channel opens for 1.0 ms?
Part B. What is the current density in the ion channel?
Solution: In 1.0 ms, the charge that passes through is
Q = I ∆t =
( 1.8 × 10−12 A ) (1.0 × 10−3 s )
= 1.8 × 10−15 C
Since each ion has a +1 charge (measured in electron charges), this represents
NK+ = Q e = (1.8 × 10−15 C ) (1.60 × 10−19 C) = 11250
The current density is calculated from the current and the size of the channel.
J = I A = ( 1.8 × 10−12 A ) ( π (0.30 × 10−9 m/2)2 ) = 2.55 × 107 A/m2
A) The number of potassium ions that will pass through the ion channel; 11250
B) The current density in the ion channel = 2.55 × 10⁷ A/m²
Given data;
Measurement with microelectrodes = 0.30-nm- diameter
K⁺ = 1.8 pA
a) calculate the number of potassium ions that passes through the ion channel
given that the channel opens for 1.0 ms
first step ; determine the value of charge ( Q )
Q = I*Δt = ( 1.8 × 10⁻¹² A )*(1.0 × 10⁻³ s ) = 1.8 × 10⁻¹⁵ C
where ; I = 1.8 × 10⁻¹² A
Δt = 1.0 × 10⁻³ s
next step : determine number of K⁺ ions passing through
NK⁺ = Q*e = ( 1.8 × 10⁻¹⁵ C )*( 1.60 × 10⁻¹⁹ C) = 11250
∴ number of K⁺ ions passing through = 11250 ions.
B) determine the current density in the ion channel
current density = current * size of channel
= ( 1.8 * 10⁻¹² ) * ( π * ( 0.30 * 10⁻⁹ )²
= 2.55 × 10⁻⁷ A/m²
Hence we can conclude that the number of potassium ions that will pass through the ion channel; 11250 and The current density in the ion channel = 2.55 × 10⁷ A/m²
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When titrating a strongmonoprotic acid and KOH at 25°C, theA) pH will be less than 7 at the equivalence point.B) pH will be greater than 7 at the equivalence point.C) titration will require more moles of base than acid to reach the equivalence point.D) pH will be equal to 7 at the equivalence point.E) titration will require more moles of acid than base to reach the equivalence
Answer:
D => pH will be equal to 7 at equivalence point
Explanation:
For Strong Acid + Strong Base titrations, pH = 7 as neither ion of the salt produced will undergo hydrolysis as would weak electrolyte titrations.
BOTH CASES ARE PRESENTED FOR CONTRAST ...
Stong monoprotic acid being titrated with NaOH ...
=> HX + NaOH => NaCl + H₂O
=> NaCl => Na⁺ + Cl⁻
=> Na⁺ + H₂O => No Rxn ( formation of NaOH will not occur as a strong electrolyte prefers to remain 100% ionized)
=> X⁻ + H₂O => No Rxn (formation of HX will not occur as a strong electrolyte prefers to remain 100% ionized)
This leaves only the Auto Ionization of Water as the reaction affecting the pH of the solution at the equivalence point of a strong acid + strong base titration. That is ...
HOH ⇄ H⁺ + OH⁻ & [H⁺] = [OH⁻] = 1 x 10⁻⁷M
pH = -log[H⁺] = -log(1 x 10⁻⁷) = -(-7) = 7
----------------------------------
Weak Acid + Strong Base titration => pH > 7 at equivalence point
Assume => HA = weak acid
=> HA + NaOH => NaA + H₂O
=> NaA => Na⁺ + A⁻ & A⁻ is the conjugate base of a weak acid HA
=> Na⁺ + H₂O => No Rxn ( formation of NaOH will not occur as a strong electrolyte prefers to remain 100% ionized)
=> A⁻ + H₂O => HA + OH⁻ => Excess OH⁻ at equivalence pt => pH > 7
-------------------------------------
Weak Base + Strong Acid titration => pH < 7 at equivalence point
Weak Bases => ammonia (NH₃) or ammonia derivatives (RNH₂)* in water.
(ammonia in water) => :NH₃ + H₂O => NH₄OH ⇄ NH₄⁺ + OH⁻
(ammonia derivative in water) => RN:-H₂ + H₂O => RNH₃OH ⇄ RNH₃⁺ + OH⁻
Titration of weak base with strong acid ...
=> NH₄OH + HX => NH₄X + H₂O
=> NH₄X => NH₄⁺ + X⁻
=> X⁻ + H₂O => No Rxn (formation of HX will not occur as a strong electrolyte prefers to remain 100% ionized)
=> NH₄⁺ + HOH ⇄ NH₄OH + H⁺ => Weak base is in molecular form with excess hydronium ions (H₃O⁺ = H⁺) at equivalence point => pH < 7.
----------------------
*RNH₂ is a primary amine used in the illustration, but the above process will also occur for secondary (R₂N:-H) and tertiary amines (R₃N:) in water also.
When 1.50g of Ba is added to 100g of water in a container open tothe atmosphere, the reaction shown below occurs and the temperatureof the resulting solution rises from 22 degrees to 33.10 degrees.If the specific heat of the solution is 4.18J/(g*C), calculatedelta H for the reaction, as written.Ba (s)+2H2O(l) yields Ba(OH)2(aq)+H2
Answer:
- 431.15 kJ/mol.
Explanation:
Firstly, we can calculate the amount of heat (Q) released by the solution using the relation:Q = m.c.ΔT,
where, Q is the amount of heat released from the solution (Q = ??? J).
m is the mass of solution (m = 1.5 g + 100 g = 101.5 g).
c is the specific heat capacity of solution (c = 4.18 J/g.°C).
ΔT is the difference in T (ΔT = final temperature - initial temperature = 33.1°C - 22°C = 11.1°C).
∴ Q = m.c.ΔT = (101.5 g)(4.18 J/g.°C)(11.1°C) = 4709.4 J.
To find ΔH:∵ ΔH = Q/n
no. of moles of Ba (n) = mass/atomic mass = (1.50 g)/(137.3270 g/mol) = 0.011 mol.
∴ ΔH = - Q/n = (4709.4 J)/(0.011 mol) = - 431.15 kJ/mol.
The negative sign is not from calculation, but it is an indication that the reaction is exothermic.
Sodium sulfate dissolves as follows: Na2SO4(s) → 2Na+(aq) + SO42- (aq). How many moles of Na2SO4 are required to make 1.0 L of solution in which the Na concentration is 0.10 M?
Answer: The number of moles of [tex]Na_2SO_4[/tex] is 0.05 moles.
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
We are given:
Molarity of solution = 0.10 mol/L
Volume of solution = 1 L
Putting values in above equation, we get:
[tex]0.10mol/L=\frac{\text{Moles of sodium}}{1.0L}\\\\\text{Moles of sodium}=0.10mol[/tex]
The chemical reaction for the ionization of sodium sulfate follows the equation:
[tex]Na_2SO_4(s)\rightarrow 2Na^+(aq.)+SO_4^{2-}(aq.)[/tex]
By Stoichiometry of the reaction:
2 moles of sodium ions are produced by 1 mole of sodium sulfate
So, 0.10 moles of sodium ions will be produced by = [tex]\frac{1}{2}\times 0.1=0.05moles[/tex] of sodium sulfate.
Hence, the number of moles of [tex]Na_2SO_4[/tex] is 0.05 moles.
Given that the density of the saturated solution is found to be 1.16 g/mL. The molar mass of copper sulfate pentahydrate is 249.68 g/mol, calculate grams of copper sulfate pentahydrate that will dissolve in 100 g of water at 0oC (show calculations for full credit)
Explanation:
Since, it is given that density is 1.16 grams per milliliter and molar mass of copper sulfate pentahydrate is 249.68 g/mol.
Now, as we known that density is the amount of mass present in a unit volume.
Mathematically, Density = [tex]\frac{\text{molar mass}}{volume}[/tex]
Hence, calculate the volume as follws.
Density = [tex]\frac{\text{molar mass}}{volume}[/tex]
1.16 g/mL = [tex]\frac{249.68 g/mol}{volume}[/tex]
volume = 215.24 mL
As, 215.24 mL can dissolve in 249.68 g/mol of complex. So, in 100 mL volume amount dissolved will be calculated as follows.
[tex]\frac{249.68 g/mol}{215.24 mL} \times 100[/tex]
= 116 grams
Thus, we can conclude that 116 grams of copper sulfate pentahydrate that will dissolve in 100 g of water at 0[tex]^{o}C[/tex].
To calculate the grams of copper sulfate pentahydrate that dissolve in 100 g of water at 0°C, solubility data at that temperature is needed, which is not provided in the question.
Explanation:The student is asking to calculate the grams of copper sulfate pentahydrate that will dissolve in 100 grams of water at 0°C, given that the density of the saturated solution is 1.16 g/mL and the molar mass of copper sulfate pentahydrate is 249.68 g/mol. Since the volume of solution can be derived from the mass of the water and the density of the solution, we can calculate the amount of copper sulfate pentahydrate dissolved in this volume by using dimensional analysis. However, to perform this calculation, we need the solubility data for copper sulfate pentahydrate at 0°C, which is not provided in the question. Without this information, we cannot proceed with the calculation.