Final answer:
The balanced reaction between Au3+(aq) and Cu+(aq) that forms Au(s) and Cu2+(aq) is Au3+(aq) + 3Cu+(aq) → Au(s) + 3Cu2+(aq). To determine the standard cell potential, the standard reduction potentials of the associated half-reactions should be used to evaluate spontaneity.
Explanation:
The question at hand involves galvanic cells and the balancing of redox reactions.
For the balanced equation for the reaction between Au3+(aq) and Cu+(aq), the chemically accurate version is:
Au3+(aq) + 3Cu+(aq) → Au(s) + 3Cu2+(aq)
Standard Reduction Potentials and Galvanic Cell Potential
To determine the standard cell potential (E°), you would reference the standard reduction potentials of the half-reactions:
Oxidation: Cu(s) → Cu2+(aq) +2e−
Reduction: Au3+(aq) + 3e− → Au(s)
By adding the reduction potentials of these half-reactions together, one can ascertain whether the reaction is spontaneous under standard state conditions. Specifically, if the standard cell potential (E°) is positive, the reaction is spontaneous.
Sodium hydrogen carbonate reacts with sulfuric acid to produce sodium sulfate, water, and carbon dioxide.
Write the equation out, balance, and tell what kind of reaction it is:
Answer:
2NaHCO3(aq)+H2SO4(aq)→Na2SO4(aq)+2CO2(g)+2H2O(l)
Explanation:
As we know that
acid + carbonate → salt + carbon dioxide + water
So, the general (un-balanced) equation would be-
NaHCO3(aq)+H2SO4(aq)→Na2SO4(aq)+CO2(g)+H2O(l)
Now we will write the net ionic reactions
[tex]HCO_3^-+ H3O^+[/tex] ----> CO2(g)↑+2H2O(l)
[tex]Na ---> Na^+ + e^-[/tex]
[tex]2H^+ + 2e^- ---> H2[/tex]
[tex]SO_4^{2-} ---- SO_4 + 2e^-[/tex]
Adding all the above equation, we get
2NaHCO3(aq)+H2SO4(aq)→Na2SO4(aq)+2CO2(g)+2H2O(l)
The nonvolatile, nonelectrolyte estrogen (estradiol), C18H24O2 (272.4 g/mol), is soluble in benzene C6H6.
How many grams of estrogen are needed to generate an osmotic pressure of 4.45 atm when dissolved in 234 ml of a benzene solution at 298 K.
_________grams estrogen
Answer: 11.6g of estrogen are needed to generate an osmotic pressure of 4.45 atm when dissolved in 234 ml of a benzene solution at 298 K.
Explanation:
To calculate the amount of solute, we use the equation for osmotic pressure, which is:
[tex]\pi=iMRT[/tex]
Or,
[tex]\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT[/tex]
where,
[tex]\pi[/tex] = osmotic pressure of the solution = 4.45 atm
i = Van't hoff factor = 1 (for non-electrolytes)
Let Mass of solute (estrogen) = x g
Volume of solution = 234 mL
R = Gas constant = [tex]0.0821Latmmol^{-1}K^{-1}[/tex]
T = temperature of the solution = [tex]298K[/tex]
Putting values in above equation, we get:
[tex]4.45=1\times \frac{x\times 1000}{272.4\times234}\times 0.0821Latmmol^{-1}K^{-1}\times 298K[/tex]
[tex]x=11.6g[/tex]
Hence, 11.6g of estrogen are needed to generate an osmotic pressure of 4.45 atm when dissolved in 234 ml of a benzene solution at 298 K.
The question is about calculating the quantity of estrogen needed to generate a specific osmotic pressure in a benzene solution using the Van 't Hoff equation. The moles of estrogen needed is calculated first, then converted into grams using the molar mass of estrogen.
Explanation:This question is based on the concept of osmotic pressure and solution chemistry, for a nonvolatile, nonelectrolyte compound in solution. The Van 't Hoff equation (π= nRT/V) can be used to solve the problem, where π refers to the osmotic pressure, n is the amount of solute in moles, R is the gas constant (0.0821 L·atm/K·mol for this problem), T is the temperature in Kelvin, and V is the volume in liters. Given the osmotic pressure (4.45 atm), the temperature (298 K), and the volume (0.234 L), you can find the number of moles of estrogen needed. After calculating the amount in moles, use the molar mass of the estrogen (272.4 g/mol) to find the mass in grams. Hence, the quantity of estrogen needed can be calculated.
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Draw the curved arrow mechanism for the formation of an acetal from acidic methanol and 4-methylpentan-2-one in the fewest steps. When given the choice, draw the arrows that lead to the resonance structures with full octets around each atom other than hydrogen. Do not show any inorganic byproducts or counterions. Reagents needed for each step are provided in the boxes.
Answer:
See explanation below
Explanation:
This reaction is known as Ketone hydrolisis in acid medium. This involves the formation of an hemi cetal, and then, the acetal. This is often used to convert ketones or aldehydes in ethers.
The first step involves the reaction with the acid. The carbonile reacts with the acid and forms an alcohol there. The next step is the reaction of the alcohol, in this case, the methanol to form the hemi cetal. Then in the third step, we repeat the first step, using acid to turn the OH group into a great leaving group such water. Then the water leaves the molecule, leaving the space wide open in the next step for methanol, and the acetal is formed.
See picture for the curved arrow mechanism
Protonation of 4-methylpentan-2-one creates a positive carbon center. First, methanol attacks this center, then a proton transfer occurs. After water is lost forming an oxonium ion, a second methanol attacks the intermediate and deprotonation results in an acetal.
Explanation:In order to draw the curved arrow mechanism for the formation of an acetal from acidic methanol and 4-methylpentan-2-one, we proceed as follows:
The first step in this reaction is protonation of the carbonyl oxygen in the 4-methylpentan-2-one by the acid, creating a more positive carbon center susceptible to nucleophilic attack. The oxygen of the methanol then attacks the carbonyl carbon, followed by proton transfer to the methanol.The resulting complex then loses water, forming an oxonium ion.Finally, another methanol molecule attacks the positively charged intermediate, leading to deprotonation and the formation of the acetal.Learn more about Curved Arrow Mechanism here:https://brainly.com/question/31973720
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1. 3A + 2B + C + 2D + 20 kJ
a) pressure is decreased_____
b) temperature is raised____
c) D is removed from the system_____
Final answer:
The reaction in question demonstrates the application of Le Chatelier's Principle, with shifts in equilibrium occurring in response to decreased pressure, increased temperature, and the removal of a reactant.
Explanation:
The reaction presented is dealing with changes in reaction conditions in a chemical equilibrium situation. This is directly related to Le Chatelier's Principle, which states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. In this scenario:
Decreased pressure (arrow "b") will cause the equilibrium to shift towards the side with more gas molecules to increase the pressure.Increase in temperature (arrow "a") will cause the equilibrium to shift towards the endothermic direction, in this case to the left, absorbing the added heat.Removal of D from the system will cause the equilibrium to shift towards the right to produce more D, according to Le Chatelier's Principle.The diagram represents an energy pyramid. At each successive tophic level from
1 to 3, the amount of energy available to the next higher level
A)decrease
B)remains the same
C)increases
HELP PLZZZZ
this is my final test to to determine if I pass my grade
pls answer fast
Answer:
decrease
Explanation:
How many moles of carbon atoms are in 18 g?
Answer:
= 1.5 moles of carbon
Explanation:
Molar mass of Carbon = 12g
12g of carbon are weighed by 1 mole of carbon
18g of carbon will be weighed by [(18÷12)×1]
= 1.5 moles
The number of moles of carbon atoms in 18 grams is approximately 1.498 moles.
Explanation:
To calculate the number of moles of carbon atoms in 18 grams, we need to use the molar mass of carbon. The molar mass of carbon is 12.01 g/mol. We can use the formula:
Number of moles = Mass / Molar mass
So, for 18 grams of carbon:
Number of moles = 18 g / 12.01 g/mol = 1.498 moles
Therefore, there are approximately 1.498 moles of carbon atoms in 18 grams.
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a 0.258 g piece of potassium solid is placed inot water inside a coffee cup calorimeter resulting in a vigorous reaction. assume a total volume of 100 ml for the resulting solution. the temperature of the solution changes from 22 to 25.1 due to the reaction. how much heat in kj is generated per gram of potassium for this reaction? assume the density of the solution after the reaction is the same as the density of water
Answer:
5 kJ/g
Explanation:
There are two energy flows in this reaction.
q₁ = heat from reaction
q₂ = heat to warm the solution
q₁ + q₂ = 0
m₁ΔH + m₂CΔT = 0
Data:
m₁ = 0.258 g
V₂ = 100 mL
C = 4.184 J°C⁻¹g⁻¹
T_i = 22 °C
T_f = 25.1 °C
Calculations
(a) Mass of solution
[tex]\text{Mass} = \text{100 mL} \times \dfrac{\text{1.00 g}}{\text{1 mL}} = \text{100 g}[/tex]
(b) ΔT
ΔT = T_f - T_i = 25.1 °C - 22 °C = 3.1°C
(c) ΔH
[tex]\begin{array}{ccccl}m_{1}\Delta H & +& m_{2}C \Delta T& = &0\\\text{0.258 g}\times \Delta H& + & \text{100 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 3.1 \, ^{\circ}\text{C} & = & 0\\0.258 \Delta H \text{ g} & + & \text{1300 J} & = & 0\\&&0.258 \Delta H \text{ g} & = & \text{-1300 J} & & \\& &\Delta H & = & \dfrac{\text{-1300 J}}{\text{0.258 g}}\\\\& & & = & \text{-5000 J/g}\\& & & = & \textbf{-5 kJ/g}\\\end{array}[/tex]
[tex]\text{The reaction produces $\large \boxed{\textbf{5 kJ}}$ per gram of potassium.}[/tex]
Note: The answer can have only one significant figure because you measured the initial temperature of the water only to the nearest degree.
From the calculation, the heat generated from the solution is -194.4 kJ/mol
What is a calorimeter?A calorimeter is an instrument that is used to measure heat.
Now we know that number of moles of the potassium = 0.258 g /39 g/mol = 0.0066 moles
Total mass present = 0.258 g + 100 g = 100.258 g
Temperature change = 25.1°C - 22°C = 3.1°C
Now;
H = -(100.258 * 4.128 * 3.1)/ 0.0066
= -194.4 kJ/mol
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Sometimes in lab we collect the gas formed by a chemical reaction over water (see sketch at right). This makes it easy to isolate and measure the amount of gas produced. Wala Suppose the H, gas evolved by a certain chemical reaction taking place at 40.0 °C is collected over water, using an apparatus something like that in the sketch, and the final volume of gas in the collection tube is measured to be 80.0 mL. Sketch of a gas-collection apparatus.
Calculate the mass of H, that is in the collection tube. Round your answer to 2 significant digits. You can make any normal and reasonable assumption about the reaction conditions and the nature of the gases. XS ?
Answer:
The correct answer is 0.00582 grams.
Explanation:
In order to solve the question, let us consider the vapor pressure of H2O, as hydrogen gas is collected over water, therefore, we have to consider the vapor pressure of water in the given case. Let us assume that the pressure is 760 torr or 1 atm.
It is known that the vapor pressure of water at 40 degree C is 53.365 torr (Based on the data).
Therefore, the pressure of H2 will be,
P = 760-55.365 = 704.635 torr or 704.635/760 = 0.9272 atm
The volume of the hydrogen gas collected in the tube is 80 ml or 0.08 L
Temperature in Kelvin will be 40+273 = 313 K
To calculate the moles of hydrogen (H2) gas, there is a need to use the ideal gas equation, that is, PV= nRT, in this R is the gas constant, whose value is 0.0821 L atm/molK, and n is the moles of the gas.
By inserting the values in the equation we get:
PV = nRT
n = PV/RT = 0.9272 *0.08 / 0.0821 * 313
n = 0.00289 moles
The mass of H2 will be moles * molar mass = 0.00289 * 2.016
= 0.00582 grams.
The heart is an example of which level of organization within an organism?
O
cell
organ
tissue
organ system
O
O
Predict where the largest jump between successive ionization energies occurs for K . between the third and fourth ionization energies between the fourth and fifth ionization energies between the first and second ionization energies between the second and third ionization energies Predict where the largest jump between successive ionization energies occurs for Ca . between the first and second ionization energies between the third and fourth ionization energies between the fourth and fifth ionization energies between the second and third ionization energies
Answer:
K-between the first and second ionization energies
Ca-between the second and third ionization energies
Explanation:
Ionization energy is the energy required to remove an electron from a gaseous atom or ion. The first ionization energy of an atom is the energy required to remove one mole of electrons from one mole of isolated gaseous atoms.
The second ionization energy is usually greater in magnitude than the first ionization energy. However, when the electrons on the valence shell of the atom are exhausted, further ionization will involve the core electrons of the atom. This requires the removal of electrons from a completely filled Shell, this process requires a very large amount of energy. The energy required is far larger than the energy required to remove valence electrons from the outermost shell of the atom. This sudden tremendous increase in the magnitude of ionization energy as electrons are removed from a completely filled shell is known as a jump.
For calcium, a jump occurs after the second electron has been removed, the third electron must be removed from a completely filled inner shell. For potassium, a jump occurs after the first electron has been removed. The second electron must be removed from a completely filled inner shell.
For potassium (K), the largest jump between successive ionization energies is between the first and second. For calcium (Ca), the largest jump between successive ionization energies occurs between the second and third.
Explanation:For potassium (K), the largest jump between successive ionization energies occurs between the first and second ionization energies. The first ionization energy involves removing an electron from the higher energy level (4s), while the second involves removing an electron from a lower energy level (3p), so a much stronger attraction exists between the nucleus and the electron to be removed in the second ionization.
For calcium (Ca), the largest jump between successive ionization energies occurs between the second and third ionization energies. The third ionization requires removal of an electron from a lower energy level, which is closer to the nucleus and thus retains a stronger attraction. This removal requires significantly more energy than the first two ionizations, which involve removing the outermost 4s electrons.
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How many joules of heat are lost by 1000g of granite as it cools from 41.2 Celsius to -12.9 Celsius?
Answer: The amount of heat released is 42739 Joules
Explanation:
The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.
[tex]Q=m\times c\times \Delta T[/tex]
Q = Heat released =?
c = heat capacity of granite = [tex]0.790J/g^0C[/tex]
Initial temperature = [tex]T_i[/tex] = [tex]41.2^0C[/tex]
Final temperature of the calorimeter = [tex]T_f[/tex] = [tex]-12.9^0C[/tex]
Change in temperature ,[tex]\Delta T=T_f-T_i=(-12.9-41.2)^0C=-54.1^0C[/tex]
Putting in the values, we get:
[tex]Q=1000g\times 0.790J/g^0C\times -54.1^0C=-42739J[/tex]
As heat comes out to be negative, that means the heat has been released and the amount of heat released is 42739 Joules
For the reaction: N2(g) + 2 O2(g) ⇌ 2 NO2(g), Kc = 8.3 × 10-10 at 25°C. What is the concentration of N2 gas at equilibrium when the concentration of NO2 is twice the concentration of O2 gas? For the reaction: N2(g) + 2 O2(g) ⇌ 2 NO2(g), Kc = 8.3 × 10-10 at 25°C. What is the concentration of N2 gas at equilibrium when the concentration of NO2 is twice the concentration of O2 gas? 4.2 × 10-10 M 2.1 × 10-10 M 2.4 × 109 M 4.8 × 109 M 1.7 x 10 -9 M
Answer: Concentration of N₂ is 4.8.[tex]10^{9}[/tex] M.
Explanation: [tex]K_{c}[/tex] is a constant of equilibrium and it is dependent of the concentrations of the reactants and the products of a balanced reaction. For
N2(g) + 2 O2(g) ⇄ 2 NO2(g)
[tex]K_{c}[/tex] = [tex]\frac{[NO2]^{2} }{[N2][O2]^{2} }[/tex]
From the question concentration of NO2 is twice of O2:
[NO2] = 2[O2]
Substituting this into [tex]K_{c}[/tex]:
[tex]K_{c}[/tex] = [tex]\frac{[2O2]^{2} }{[N2][O2]^{2} }[/tex]
8.3.[tex]10^{-10}[/tex] = [tex]\frac{4O2^{2} }{[N2].O2^{2} }[/tex]
[N2] = [tex]\frac{4O2^{2} }{8.3.10^{-10}.O2^{2} }[/tex]
[N2] = [tex]\frac{4}{8.3.10^{-10} }[/tex]
[N2] = 4.8.[tex]10^{9}[/tex]
The concentration of N2 in the equilibrium is [N2] = 4.8.[tex]10^{9}[/tex]M.
The concentration of [tex]N_2[/tex] is [tex]4.8.10^9M[/tex]
Calculation of concentration:Since k_e represent the equilibrium constant and based on the concentrations of the reactants and the products of a balanced reaction.
Also, the reaction: N2(g) + 2 O2(g) ⇌ 2 NO2(g), And, the concentration of NO2 is twice the concentration of O2 gas
Kc = 8.3 × 10-10 at 25°C.
NO2 is twice of O2.
Now
[tex]8.310^{-10} = \frac{4O_2^2}{N_2O_2^2} \\\\N_2 = \frac{4O_2^2}{8.3.10^{-10}O_2^2}\\\\ = 4\div 8.3.10^{-10}\\\\= 4.8.10^9[/tex]
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6) How many valence electrons does an atom of Cu possess?
ОА) 2
Ов) 9
Ос) 11
OD) 3
OE) 1
Choose the aqueous solution that has the highest boiling point. These are all solutions of nonvolatile solutes and you should assume ideal van't Hoff factors where applicable. Choose the aqueous solution that has the highest boiling point. These are all solutions of nonvolatile solutes and you should assume ideal van't Hoff factors where applicable. 0.100 m C6H12O6 0.100 m NaCl 0.100 m AlCl3 0.100 m MgCl2
Answer:
0.100 m AlCl3 will have the highest boiling point
Explanation:
Step 1: Data given
The molal boiling point elevation constant for water is 0.51°C/m
Since those are all aqueous solutions, the have the same molal boiling point elevation constant
Step 2:
0.100 m C6H12O6
ΔT = i*Kb*m
⇒with ΔT is the boiling point elevation = TO BE DETERMINED
⇒with i = the van't Hoff factr = 1
⇒with Kb = The molal boiling point elevation constant for water is 0.51°C/m
⇒with m = the molality = 0.100m
ΔT = 1 * 0.51 * 0.100
ΔT = 0.051 °C
0.100 m NaCl
ΔT = i*Kb*m
⇒with ΔT is the boiling point elevation = TO BE DETERMINED
⇒with i = the van't Hoff factr = Na+ + Cl- = 2
⇒with Kb = The molal boiling point elevation constant for water is 0.51°C/m
⇒with m = the molality = 0.100m
ΔT =2 * 0.51 * 0.100
ΔT = 0.102 °C
0.100 m AlCl3
ΔT = i*Kb*m
⇒with ΔT is the boiling point elevation = TO BE DETERMINED
⇒with i = the van't Hoff factr = Al^3+ + 3Cl- = 4
⇒with Kb = The molal boiling point elevation constant for water is 0.51°C/m
⇒with m = the molality = 0.100m
ΔT =4 * 0.51 * 0.100
ΔT = 0.204 °C
0.100 m MgCl2
ΔT = i*Kb*m
⇒with ΔT is the boiling point elevation = TO BE DETERMINED
⇒with i = the van't Hoff factr = Mg^2+ +2Cl- = 3
⇒with Kb = The molal boiling point elevation constant for water is 0.51°C/m
⇒with m = the molality = 0.100m
ΔT =3 * 0.51 * 0.100
ΔT = 0.153 °C
0.100 m AlCl3 will have the highest boiling point
Calculate the concentration in M) of hydroxide ions in a solution with a pOH of 2.468
A) 2.94 x 102
B) 3.40 x 1011
OC) 3.40 x 10-3
OD) 2.94 x 10-12
Answer:
C) 3.40 x 10-3
Explanation:
Hello,
In this case, for the given pOH, we are able to compute the concentration of hydrixide ions by applying the following formula:
[tex]pOH=-log([OH^-])[/tex]
[tex][OH^-]=10^{-pOH}=10^{-2.468}\\[/tex]
[tex][OH^-]=0.0034M=3.4x10^{-3}M[/tex]
Therefore, answer is C).
Best regards.
Answer:
3.40 × 10⁻³ M
Explanation:
The pOH scale is used to express the acidity or basicity of a solution.
If pOH < 7, the solution is basic.If pOH = 7, the solution is neutral.If pOH > 7, the solution is acid.The pOH of this solution is 2.468, so it is basic. We can calculate the concentration of hydroxide ions in the solution using the following expression.
[tex]pOH = -log [OH^{-} ]\\\[[OH^{-}] = antilog-pOH = antilog-2.468 = 3.40 \times 10^{-3} M[/tex]
In which solution will thymol blue indicator appear blue?
Answer:
0.1 M KOH
Explanation:
The thymol blue indicator will appear blue in a basic solution, particularly in a solution with a pH greater than 8.0. A diluted sodium hydroxide solution is an example of such a basic environment where thymol blue would turn blue.
The thymol blue indicator will appear blue in a basic solution. Specifically, thymol blue changes color from yellow to blue over a pH range of approximately 8.0 to 9.6. This means that for thymol blue to exhibit a blue color, it needs to be in a solution with a pH greater than 8.0, indicating a basic environment.
An example of a solution in which thymol blue would appear blue is a diluted sodium hydroxide (NaOH) solution, as NaOH is a strong base that would increase the pH of the solution.
2 H2(g) + O2(g) → 2 H2O(g)
How many liters of water can be made from 55 grams of oxygen gas and an excess of hydrogen at STP?
a
56 L
b
77 L
c
35 L
d
67 L
Answer:
77 L of water can be made.
Explanation:
Molar mass of [tex]O_{2}[/tex] = 32 g/mol
So, 55 g of [tex]O_{2}[/tex] = [tex]\frac{55}{32}[/tex] mol of [tex]O_{2}[/tex] = 1.72 mol of [tex]O_{2}[/tex]
As hydrogen is present in excess amount therefore [tex]O_{2}[/tex] is the limiting reagent.
According to balanced equation, 1 mol of [tex]O_{2}[/tex] produces 2 mol of [tex]H_{2}O[/tex].
So, 1.72 mol of [tex]O_{2}[/tex] produce [tex](2\times 1.72)[/tex] mol of [tex]H_{2}O[/tex] or 3.44 mol of [tex]H_{2}O[/tex].
Let's assume [tex]H_{2}O[/tex] gas behaves ideally at STP.
Then, [tex]P_{H_{2}O}.V_{H_{2}O}=n_{H_{2}O}.R.T[/tex] , where P, V, n, R and T represents pressure, volume, no. of moles, gas constant and temperature in kelvin scale respectively.
At STP, pressure is 1 atm and T is 273 K.
Here, [tex]n_{H_{2}O}[/tex] = 3.44 mol and R = 0.0821 L.atm/(mol.K)
So, [tex](1atm)\times V_{H_{2}O}=(3.44mol)\times (0.0821L.atm.mol^{-1}.K^{-1})\times (273K)[/tex]
[tex]\Rightarrow[/tex] [tex]V_{H_{2}O}=77L[/tex]
Option (b) is correct.
Final answer:
Using stoichiometry, 55 grams of oxygen gas can produce 77 liters of water at STP, considering the molar volume of a gas at STP is 22.4 liters per mole.
Explanation:
To determine how many liters of water can be made from 55 grams of oxygen gas and an excess of hydrogen at STP, we must first convert the mass of oxygen to moles and then use stoichiometry to find the volume of water produced. The balanced chemical equation for the reaction is:
2 H2(g) + O2(g) → 2 H2O(g)
At STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 liters. Therefore, the moles of oxygen (O2) can be calculated using its molar mass (32 g/mol):
Moles of O2 = 55 g / 32 g/mol = 1.71875 moles
The balanced equation shows us that 1 mole of O2 produces 2 moles of H2O.
Total moles of H2O produced = 1.71875 moles of O2 × 2 = 3.4375 moles
Volume of H2O (at STP) = 3.4375 moles × 22.4 L/mol = 76.92 L
Therefore, when rounded to the nearest whole number, 77 liters of water can be made from 55 grams of oxygen gas and an excess of hydrogen at STP, which corresponds to option b.
Determine the electron geometry (eg), molecular geometry (mg), and polarity of SO2. Determine the electron geometry (eg), molecular geometry (mg), and polarity of SO2. eg=trigonal planar, mg=bent, polar eg=tetrahedral, mg=bent, polar eg=linear, mg=linear, nonpolar eg=trigonal pyramidal, mg=trigonal pyramidal, polar eg=tetrahedral, mg=tetrahedral, nonpolar
Answer:
See explaination
Explanation:
The electrons geometry shows the special distribution of the electrons around of the central atom of the molecule.
The molecular geometry shows the special distribution of the atoms that form the molecule.
Please kindly check attachment for further solution.
What is the result when 12 grams of H2 and 28 grams of N2 react to completion at STP.
3H2(g) + N2 --> 2 NH3
Answer:
The correct answer is 89.6 L
Explanation:
We have the following chemical equation and the molar masses for the reaction:
3H₂(g) + N₂ --> 2 NH₃
6 g 28 g 34 g
That means that 3 moles of H₂ (6 g) reacts with 1 mol of N₂ (28 g) and gives 2 moles of NH₃ (34 g). In order to calculate how many liters of NH₃ result from the reaction of 12 grams of H₂ and 28 grams of N₂, we have to first figure out which reactant is the limiting reactant. According to the equation, if 6 grams of H₂ reacts with 28 g of N₂, and we have 12 grams:
6 g H₂------- 28 g N₂
12 g H₂-------- X = 12 g H₂ x 28 g N₂/6 g H₂ = 56 g N₂
We need 56 g of N₂ but we have 28 g of N₂, so N₂ is the limiting reactant. With the limiting reactant we can calculate the moles of product (NH₃) we will obtain:
We have 28 g N₂ -----> 28 g/14 g/mol = 2 moles N₂
1 mol N₂ ----------- 2 moles NH₃
2 mol N₂ --------- X = 2 mol N₂ x 2 moles NH₃/1 mol N₂ = 4 mol NH₃
Finally, we convert the moles of NH₃ to liters:
1 mol gas at STP = 22.4 L
Liters NH₃ obtained = 4 moles NH₃ x 22.4 L/1 mol = 89.6 L
Final answer:
When 12 grams of H2 and 28 grams of N2 react completely according to the equation 3H2(g) + N2(g) → 2NH3(g), 34.08 grams of NH3 are produced, with some H2 remaining unreacted due to it being in excess.
Explanation:
The question asks about the result when 12 grams of H2 and 28 grams of N2 react to completion according to the balanced chemical equation 3H2(g) + N2(g) -> 2NH3(g). This equation signifies that one mole of nitrogen gas (N2) reacts with three moles of hydrogen gas (H2) to produce two moles of ammonia (NH3). Given the molar masses (N2 = 28.02 g/mol, H2 = 2.02 g/mol), we can determine that the initial amounts provided are excess H2. Precisely, 28 grams of N2 is one mole, and 12 grams of H2 is six moles, which is enough to fully react according to the stoichiometry of the equation. The reaction of 28.02 g of N2 and 6.06 g of H2 produces 34.08 g of NH3, according to mass conservation principles. Since more H2 is provided than required, only 6.06 g will be consumed, leaving excess H2 unreacted. Therefore, the reaction yields 34.08 grams of NH3.
Convert the Haworth projection of the carbohydrate below into its corresponding Fischer projection (in the standard format with the most oxidized carbon at the top) and chair conformation (correctly drawing equatorial and axial bonds) of the opposite anomer. Then answer the questions in the problem. Upload a photo of your answer as a jpeg file. If we can't read it, we can't grade it!
Answer:
See explaination
Explanation:
H O / I MOH the OH da Anomer My nequatorial bovel SOH niematerial and CH2OH Yr axial boud - H o H o . H (Auro) 2 flip . note: 1)Fischer projection : L- Sugar ( OH group is in the left side ) at C-5 carbon (bottom most chiral center )
In the Haworth projection given above in the qsn it is Beta anomer( -CH2OH & C-OH are in same side i.e. Cis to each other ). It is a L -sugar .
I omitted C2 -H bond (axial ) in the chair conformation of alpha anomer.
See attachment for further solution
Transitioning from a Haworth to a Fischer projection, every carbon atom becomes a line end or intersection point. Correct position and orientation of the hydroxyl group determines the chair configuration. Carbohydrates, which contain carbon, hydrogen, and oxygen atoms in a 1:2:1 ratio, have monosaccharide, disaccharide, and polysaccharide subtypes.
Explanation:The question is asking you to convert a Haworth projection of a carbohydrate into a Fischer projection and a chair conformation of its opposite anomer. This involves understanding several concepts in organic chemistry. The Haworth projection is a common way of representing cyclic sugars, which are typically found in ring forms in aqueous solutions. The Fischer projection is used to represent the stereochemistry of the molecule. In converting the Haworth projection to a Fischer projection, each carbon atom in the ring becomes the end of a line or a point where lines intersect in the Fischer projection. The most oxidized carbon (typically the one attached to an oxygen atom) is placed at the top.
Next, to draw the chair conformation, you must understand axial and equatorial bonds, which describe the orientation of chemical groups around the carbon atoms in the ring. Axial bonds are vertical and shunted to the side, while equatorial bonds are more horizontal and radiate outwards. For the opposite anomer, if the original has the hydroxyl group below the plane in an alpha position, the opposite anomer would have it above the plane in a beta position.
Carbohydrates bear the stoichiometric formula (CH₂O)n, indicating a carbon to hydrogen to oxygen ratio of 1:2:1. Classified into monosaccharides, disaccharides, and polysaccharides, carbohydrates are a crucial chemical group in biology.
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A chemist titrates 220.0 mL of a 0.1917M propionic acid (HC2H,CO2) solution with 0.1787 M KOIH solution at 25 °C. Calculate the pH at equivalence. The pK of propionic acid is 4.89.
Answer:
pH = 8.93
Explanation:
In this case, this titration is the case of a weak acid and a strong base. Now, at the equivalence point, it's supposed that we have the same moles of each reactant in solution, and we will expect that the pH would have to be 7. However, as the acid is pretty weak, there's a little difference in the solution because of the grade of dissociation of the acid, and the pH will be higher than 7. To know this, we first need to calculate the volume of added base:
M₁V₁ = M₂V₂
With this expression, let's calculate the volume of the base:
V₂ = M₁V₁ / M₂
V₂ = 0.1917 * 220 / 0.1787 = 236 mL
So, at the equivalence point, 236 mL are needed to neutralize this reaction. As the moles are the same for each reactant, we just need to calculate the concentration of the acid in this part. This will be the sum between the initial volume of acid and the calculated volume of base:
V of solution = 236 + 220 = 456 mL or 0.456 L
Then, the new concentration of the acid is:
[C₂H₅COOH] = 0.1917 * 0.220 / 0.456 = 0.0924 M
Now, the reaction with the base is the following:
C₂H₅COOH + KOH --------> C₂H₅COOK + H₂O
This means that in the equivalence point we have the propionic potassium and water, so, if take this and dissociates into it's ions we can calculate the pH of the solution:
C₂H₅COO⁻ + H₂O <-------> C₂H₅COOH + OH⁻
With this reaction in solution in the equivalence point, we just need the Kb of propionate ion, and this can be calculated with the value of the pKa of the acid:
Ka = 10^(-pKa)
Ka = 1.29x10⁻⁵
Now the value of Kb can calculated using the following expression:;
Kb = Kw / Ka ---> replacing we have
Kb = 1x10⁻¹⁴ / 1.29x10⁻⁵
Kb = 7.75x10⁻¹⁰
Now, with this value and the above reaction we can write an ICE chart to calculate the [OH⁻] and then, the pH of solution:
C₂H₅COO⁻ + H₂O --------> C₂H₅COOH + OH⁻ Kb = 7.75x10⁻¹⁰
i) 0.0924 0 0
e) 0.0924-x x x
The Kb expression:
Kb = [C₂H₅COOH] [OH⁻] / [C₂H₅COO⁻]
7.75x10⁻¹⁰ = x² / 0.0924-x ---> Kb is very small, so this substraction can be neglected to just 0.0924 assuming x will be very small too.
7.75x10⁻¹⁰ = x² / 0.0924
7.75x10⁻¹⁰ * 0.0924 = x²
x = [OH⁻] = 8.46x10⁻⁶ M
With this value, we can calculate pOH and then the pH:
pOH = -log(8.46x10⁻¹⁰) = 5.07
Finally the pH:
pH = 14 - pOH
pH = 14 - 5.07
pH = 8.93In a laboratory experiment, students synthesized a new compound and found that when 11.09 grams of the compound were dissolved to make 180.9 mL of a diethyl ether solution, the osmotic pressure generated was 3.88 atm at 298 K. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight they determined for this compound
Answer:
The molecular weight of this compound is 386.4 g/mol
Explanation:
Step 1: Data given
MAss of the compound = 11.09 grams
Volume of diethyl ether = 180.9 mL
Osmotic pressure = 3.88 atm
Temperature = 298 K
The compound = nonvolatile and non-electrolyte
Step 2: Calculate molar concentration
π = i*M*R*T
⇒with π = the osmotic pressure = 3.88 atm
⇒with i = the van't Hoff factor = 1
⇒with C = the molar concentration = = TO BE DETERMINED
⇒ with R = the gas constant = 0.08206 L*atm/mol*K
⇒with T = 298 K
C = 3.88 / (0.08206*298)
C = 0.1587 M
Step 3: Calculate moles compound
C = moles / volume
moles = 0.1587 M * 0.1809 L
Moles compound = 0.0287 moles
Step 4: Calculate molecular weight of the compound
Molar mass = mass / moles
Molar mass compound = 11.09 grams / 0.0287 moles
Molar mass compound = 386.4 g/mol
The molecular weight of this compound is 386.4 g/mol
To find pH of hydrochloric acid.
Lab data: on the attached file
Acid used to titrate 25.0mL of 0.150M potassium hydroxide.
Answer:
1.22
Explanation:
Step 1:
We'll begin by writing the balanced equation for the reaction. This is given below:
HCl + KOH —> KCl + H2O
From the balanced equation above,
The mole ratio of acid (nA) = 1
The mole ratio of base (nB) = 1
Step 2:
Data obtained from the question. This includes the following:
Volume of acid (Va) =
(62.35 + 62.40)/2 = 62.38 mL
Molarity of acid (Ma) =.?
Volume of base (Vb) = 25 mL
Molarity of base (Mb) = 0.150M
Step 3:
Determination of the molarity of the acid.
This is illustrated below:
MaVa/MbVb = nA/nB
Ma x 62.38/ 0.15 x 25 = 1
Cross multiply to express in linear form
Ma x 62.38 = 0.15 x 25
Divide both side by 62.38
Ma = (0.15 x 25) /62.38
Ma = 0.06M
The molarity of the acid is 0.06M
Step 4:
Determination of the concentration of Hydrogen ion, [H+] in the acid. This is illustrated below:
Hydrochloric acid (HCl) will dissociate to produce hydrogen ion as follow:
HCl —> H+ + Cl-
From the above equation,
1 mole of HCl produced 1mole of H+.
Therefore, 0.06M HCl will also produce 0.06M H+.
The concentration of Hydrogen ion, [H+] is 0.06M
Step 5:
Determination of the pH of HCl. This is illustrated below:
pH = – Log [H+]
[H+] = 0.06M
pH = – Log 0.06
pH = 1.22
Therefore, the pH of HCl is 1.22
A sample of argon has a volume of 1.2 L at STP. If the temperature is increased to 21 c and the pressure is lowered to 0.80 atm, what will the new
Answer:
The new volume is 1.62 L
Explanation:
Boyle's law says:
"The volume occupied by a given gas mass at constant temperature is inversely proportional to the pressure." It is expressed mathematically as:
Pressure * Volume = constant
o P * V = k
Charles's law is a law that says that when the amount of gas and pressure are kept constant, the ratio between volume and temperature will always have the same value:
[tex]\frac{V}{T}=k[/tex]
Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the gas pressure increases. And when the temperature is decreased, the gas pressure decreases. So this law indicates that the quotient between pressure and temperature is constant.
Gay-Lussac's law can be expressed mathematically as follows:
[tex]\frac{P}{T}=k[/tex]
Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law.
[tex]\frac{P*V}{T}=k[/tex]
Having an initial state 1 and a final state 2 it is possible to say that:
[tex]\frac{P1*V1}{T1} =\frac{P2*V2}{T2}[/tex]
Standard temperature and pressure (STP) indicate pressure conditions P = 1 atm and temperature T = 0 ° C = 273 ° K. Then:
P1= 1 atmV1= 1.2 LT1= 273 °KP2= 0.80 atmV2= ?T2= 21°C= 294 °KReplacing:
[tex]\frac{1 atm* 1.2 L}{273K} =\frac{0.8 atm*V2}{294K}[/tex]
Solving:
[tex]V2=\frac{1 atm*1.2 L}{273 K} *\frac{294 K}{0.8 atm}[/tex]
V2= 1.62 L
The new volume is 1.62 L
g The rate of a reaction typically increases as the temperature increases because: The rate of a reaction typically increases as the temperature increases because: the A term in the Arrhenius equation increases. the activation energy decreases. the activation energy increases. the fraction of molecules with kinetic energy greater than Ea increases. the molecules make more collisions with the wall of the reaction vessel.
Answer:
The fraction of molecules with kinetic energy greater than Ea increases.
Explanation:
In the scenario above, it is primarily by the fraction of molecules with kinetic energy higher or bigger that Ea increases.
Naturally chemical reactions are everywhere around you and inside you. In addition to the biochemical reactions that allow you to transform the molecules you eat and breathe into usable energy, there are industrial laboratories in cities around the world producing chemicals as well as products that rely on chemicals for their manufacture.
In addition to the product or products produced and having a proper supply of reactants, is how quickly a reaction can be expected to proceed. This can have an impact on safety, product quality and other outcomes.
Final answer:
The reaction rate increases with temperature because more molecules possess the necessary kinetic energy to overcome the activation energy, leading to a higher number of effective collisions and thus an increased rate of reaction.
Explanation:
The rate of a reaction typically increases as the temperature increases because the fraction of molecules with kinetic energy greater than the activation energy (Ea) increases. As the temperature rises, molecules move more rapidly, leading to a higher frequency of collisions. More importantly, these collisions are more energetic which means more of them have enough energy to surmount the activation energy barrier. The Arrhenius equation helps in understanding this phenomenon, predicting that the rate constant (k) increases with temperature since a larger fraction of reactant molecules has enough kinetic energy to overcome Ea, resulting in an increased reaction rate.
What mass of NaOH must be used to prepare 2.5 L of a 0.010 M solution
Answer:
[tex]m = 1\,g[/tex]
Explanation:
The molarity is the ratio of the amount of moles solvent to volume of the solute.
[tex]0.010\,M = \frac{0.010\,moles\,NaOH}{1\,L}[/tex]
The quantity of solute is determined by simple rule of three:
[tex]n = \left(\frac{2.5\,L}{1\,L} \right)\cdot (0.010\,mole)[/tex]
[tex]n = 0.025\,moles[/tex]
The molecular weight of NaOH is [tex]39.997\,\frac{g}{mole}[/tex], the mass of solute is:
[tex]m = 1\,g[/tex]
The rate constant for this second‑order reaction is 0.190 M − 1 ⋅ s − 1 0.190 M−1⋅s−1 at 300 ∘ C. 300 ∘C. A ⟶ products A⟶products How long, in seconds, would it take for the concentration of A A to decrease from 0.820 M 0.820 M to 0.340 M?
Answer:
9.1 seconds
Explanation:
Given that for a second order reaction
1/[A]t = kt + 1/[A]o
Where [A]t= concentration at time = t= 0.340M
[A]o= initial concentration = 0.820M
k= rate constant for the reaction=0.190m-1s-1
t= time taken for the reaction (the unknown)
Hence;
(0.340)^-1 = 0.190×t + (0.820)^-1
t= (0.340)^-1 - (0.820)^-1/0.190
t= 9.1 seconds
Hence the time taken for the concentration to decrease from 0.840M to 0.340M is 9.1 seconds.
Final answer:
To calculate the time it takes for the concentration of compound A to decrease in a second-order reaction, the integrated rate law 1/[A] - 1/[A]0 = kt is used with the provided rate constant and initial and final concentrations.
Explanation:
Calculating Time for a Second-Order Reaction:
The question pertains to the time it takes for the concentration of compound A to decrease from 0.820 M to 0.340 M in a second-order reaction with a rate constant of 0.190 M-1·s-1 at 300 °C. The integrated rate law for a second-order reaction is given by:
1/[A] - 1/[A]0 = kt
Where [A]0 is the initial concentration, [A] is the concentration at time t, k is the rate constant, and t is the time. Using the provided concentrations, we can solve for t:
1/0.340 M - 1/0.820 M = (0.190 M-1·s-1)t
After calculating the left side, we can isolate t:
t = (1/0.340 M - 1/0.820 M) / (0.190 M-1·s-1)
Hence:
(0.340)^-1 = 0.190×t + (0.820)^-1
t= (0.340)^-1 - (0.820)^-1/0.190
t= 9.1 seconds
Hence the time taken for the concentration to decrease from 0.840M to 0.340M is 9.1 seconds.
An isothermal chromatogram at 90 °C shows an elution order of 1-pentanol followed by ethylene diamine followed by diethylene glycol. But in order to reduce the overall time of the experiment, a temperature programmed chromatogram is tried, beginning at 60 °C and increasing to 190 °C over 5 minutes then holding at 190 °C for 4 minutes. Predict the elution order of the compounds using the temperature programmed conditions.
Answer:
The new elution order expected will be the following:
ethylene diamine1-pentanoldiethylene glycolPlease see below for details and explanation.
Explanation:
Which compound will elute first depends on a number of factors. The compound with the lowest boiling point will elute before another compound with a higher boiling point and so on. By extension, the volatility of the compound will also be considered when predicting elution order. Thirdly, how the solutes interact with each other during the stationary phase. I've listed the boiling points below:
1-pentanol 138 °C
ethylene diamine 116 °C
diethylene glycol 245 °C
The advantage of using temperature programmed chromatogram is that it changes retention times (time needed for the solute to pass through the column). And it will be according to the respective boiling points.
Hope that answers the question, have a great day!
When of benzamide are dissolved in of a certain mystery liquid , the freezing point of the solution is less than the freezing point of pure . Calculate the mass of ammonium chloride that must be dissolved in the same mass of to produce the same depression in freezing point. The van't Hoff factor for ammonium chloride in . Be sure your answer has a unit symbol, if necessary, and round your answer to significant digits.
Given question is incomplete. The complete question is as follows.
When 72.8 g of benzamide ([tex]C_{7}H_{7}NO[/tex]) are dissolved in 600 g of a certain mystery liquid X, the freezing point of the solution is [tex]6.90^{o}C[/tex] less than the freezing point of pure X. Calculate the mass of ammonium chloride [tex](NH_{4}Cl)[/tex] that must be dissolved in the same mass of X to produce the same depression in freezing point. The van't Hoff factor i = 70 for ammonium chloride in X. Be sure your answer has a unit symbol, if necessary, and round your answer to significant digits.
Explanation:
The given data is as follows.
Mass of solute (benzamide), [tex]w_{B}[/tex] = 72.8 g
Mass of solvent (X), [tex]w_{A}[/tex] = 600 g
[tex]\Delta T_{f} 6.90^{o}C[/tex]
Molar mass of benzamide, [tex]M_{w_{B}}[/tex] = 121.14 g/mol
We know that,
[tex]\Delta T_{f} = k_{f} \times X \times m[/tex] (for non-dissociating)
[tex]6.90 = k_{f} \times \frac{72.8 \times 1000}{121.14 \times 600}[/tex] ...... (1)
For other experiment, when [tex]NH_{4}Cl[/tex] is taken :
Mass of [tex]NH_{4}Cl[/tex], ([tex]w_{NH_{4}Cl}[/tex]) = ?
Molar mass of [tex]NH_{4}Cl[/tex] = 53.491 g/mol
Mass of solvent (X) = 600 g
[tex]\Delta T_{f} = 6.90^{o}C[/tex]
i = Van't Hoff factor = 1.70
As, [tex]\Delta T_{f} = i \times k_{f} \times m[/tex]
[tex]6.90 = 1.70 \times k_{f} \times \frac{w_{NH_{4}Cl} \times 1000}{53.491 \times 600}[/tex] ........... (2)
Now, we will divide equation (1) by equation (2) as follows.
[tex]w_{NH_{4}Cl} \times 1 = \frac{72.8 \times 53.491}{1.70 \times 121.14}[/tex]
= 18.90 g
Therefore, we can conclude that the mass of ammonium chloride [tex](NH_{4}Cl)[/tex] that must be dissolved in the same mass of X to produce the same depression in freezing point is 18.90 g.
Final answer:
The mass of ammonium chloride required to produce the same freezing point depression as benzamide can't be calculated without additional data. The van't Hoff factor is crucial for understanding the effects of ionic compounds in freezing point depression calculations.
Explanation:
To calculate the mass of ammonium chloride that must be dissolved to produce the same depression in freezing point as benzamide, we will need to use the freezing point depression concept, which states that the change in freezing point (ΔTf) is equal to the molal freezing point depression constant of the solvent (Kf) multiplied by the molality (m) of the solution. The given data is as follows.
Mass of solute (benzamide), w_(B) = 72.8 g
Mass of solvent (X), w_(A) = 600 g
\Delta T_(f) 6.90^(o)C
Molar mass of benzamide, M_{w_(B)} = 121.14 g/mol
We know that,
\Delta T_(f) = k_(f) * X * m (for non-dissociating)
6.90 = k_(f) * (72.8 * 1000)/(121.14 * 600) ...... (1)
For other experiment, when NH_(4)Cl is taken :
Mass of NH_(4)Cl, (w_{NH_(4)Cl}) = ?
Molar mass of NH_(4)Cl = 53.491 g/mol
Mass of solvent (X) = 600 g
\Delta T_(f) = 6.90^(o)C
i = Van't Hoff factor = 1.70
As, \Delta T_(f) = i * k_(f) * m
6.90 = 1.70 * k_(f) * \frac{w_{NH_(4)Cl} * 1000}{53.491 * 600} ........... (2)
Now, we will divide equation (1) by equation (2) as follows.
w_{NH_(4)Cl} * 1 = (72.8 * 53.491)/(1.70 * 121.14)
= 18.90 g
Therefore, we can conclude that the mass of ammonium chloride (NH_(4)Cl) that must be dissolved in the same mass of X to produce the same depression in freezing point is 18.90 g.
To prepare a 2 M solution of potassium nitrate (KNO3), which quantities must be measured? The mass of the and the volume of the must be measured.
Answer:
The mass of the solute and the volume of the solution.
Explanation:
Hello,
In this case, given the formula of molarity:
[tex]M=\frac{n_{solute}}{V_{solution}}[/tex]
In such a way, since the moles could not be directly measured, we must measure the mass of the solute and by using its molar mass, one could compute its moles. Moreover, since the solution is composed by the solvent (typically water) and the solute, we consequently must measure the volume of the solution needed for the preparation of such concentration-known solution. In such a way, we can actually prepare the required solution.
Best regards.
For the formation of 2 M solution of potassium nitrate, mass and volume of the solution has been measured.
Molarity can be defined as the mass of solute present in a liter of solution. The molarity has been used for the determination of the concentration of the compounds.
It can be expressed as mol/L. The molarity (M) has expression:
[tex]M=\rm \dfrac{solute\;mass}{solute\;molar\;mass}\;\times\;Solution\;Volume[/tex]
For the formation of 2 M potassium nitrate solution, the mass of the solute and the volume of the solution has to be measured.
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