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AveGali [126]
4 years ago
5

A sealed isothermal container initially contained 2 moles of CO gas and 3 moles of H2 gas. The following reversible reaction occ

ured: CO(g) + 2H2(g) ---> CH3OH(g). At equilibrium, there was 1 mole of CH3OH in the container. What was the total number of moles of gas present in the container at equilibrium.
A.) 1
B.) 2
C.) 3
D.) 4
Chemistry
1 answer:
zhenek [66]4 years ago
3 0

Answer:

(c) Total Number of moles = 3

Explanation:

To solve this question we have to follow the stoichiometry of the reaction and the fact that at equilibrium we have 1 mol of CH₃OH using the mnemonic helper RICE ( Reaction, Initial, Change, Equilibrium) as follows:

R             CO(g) + 2 H₂ (g)   ⇄  CH₃OH (g)

I (mol)        2             3                     0

C (mol)       -x         x -2x               + x

E (mol)      2-x        3-2x                   x

But we are told that at equilibrium x = mol CH₃OH = 1 , then at equilibrium we will have

E (mol)       2-(1)      3-2(1)             1

E (mol)          1            1                 1

Therefore, the total number of moles of gas present in the container at equilibrium was 3

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Contact [7]

Answer:

The pH of the solution is 11.48.

Explanation:

The reaction between NaOH and HCl is:

NaOH  +  HCl  →  H₂O  +  NaCl

From the reaction of 3.60x10⁻³ moles of NaOH and 5.95x10⁻⁴ moles of HCl we have that all the HCl will react and some of NaOH will be leftover:

n_{NaOH}} = n_{i_{NaOH}} - n_{HCl} = 3.60 \cdot 10^{-3} moles - 5.95 \cdot 10^{-4} moles = 3.01 \cdot 10^{-3} moles

Now, we need to find the concentration of the OH⁻ ions.

[OH^{-}] = \frac{n_{NaOH}}{V}

Where V is the volume of the solution = 1.00 L                

[OH^{-}] = \frac{n_{NaOH}}{V} = \frac{3.01 \cdot 10^{-3} moles}{1.00 L} = 3.01 \cdot 10^{-3} mol/L

Finally, we can calculate the pH of the solution as follows:

pOH = -log([OH^{-}]) = -log(3.01 \cdot 10^{-3}) = 2.52

pH + pOH = 14

pH = 14 - pOH = 14 - 2.52 = 11.48

Therefore, the pH of the solution is 11.48.

I hope it helps you!

3 0
3 years ago
The number of energy levels to which an electron can jump depends on the
bija089 [108]
It usually depends on the amount of energy that an electron has or gains
8 0
4 years ago
The student's lab manual says to mix some of his Na2CO3 solution with an aqueous solution of copper(II) sulfate (CuSO4)
lord [1]

Explanation:

When the student mixed the solution sodium carbonate with solution of copper(II) sulfate ; Copper Hydroxocarbonate , sodium sulfate and carbon dioxide gas was obtained as a products.

The balanced chemical reaction

2Na_2CO_3+2CuSO_4\rightarrow Cu_2(OH)_2CO_3+2Na_2SO_4+CO_2

Where:

Cu_2(OH)_2CO_3 = Copper(II) Hydroxocarbonate

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3 0
4 years ago
1. Calculate the quantity of glucose needed to prepare (a) 250 mL of a 100 mM of
weeeeeb [17]

Answer:

a. mass of glucose required = 4.50 g

b. mass of glucose required = 0.05 g

c. mass of glucose required = 25g

Explanation:

1a. molar mass of glucose = 180.156 g/mol

volume of solution = 250 mL or 250 mL * 1 L/1000 mL = 0.25 L

concentration of solution = 100 mM or 100mM * 1 M/1000 mM = 0.10 M

amount in moles = mass/ molar mass = concentration * volume

mass in grams = concentration * volume * molar mass

mass = 0.10 * 0.25 * 180.156

mass of glucose required = 4.50 g

b. mass concentration of 0.05 % solution = 0.05 g/100 ml = 0.5 g/L

volume of solution = 100 mL = 0.1 L

mass in grams = mass concentration * volume

mass in grams =  0.5 g/L * 0.1 L

mass of glucose required = 0.05 g

c. mass concentration of solution = 50 mg/ mL = 0.05 g/ 0.001 L = 50 g/L

volume of solution = 500 mL = 0.5 L

mass of solute required = concentration * volume

mass required = 50 g/L * 0.5 L

mass of glucose required = 25g

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