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user100 [1]
3 years ago
10

The following reaction was carried out in a 3.25 L reaction vessel at 1100 K:

Chemistry
1 answer:
kipiarov [429]3 years ago
3 0

Answer : The value of reaction quotient Q is, 0.498

Explanation:

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given balanced chemical reaction is,

C(s)+H_2O(g)\rightarrow CO(g)+H_2(g)

The expression for reaction quotient will be :

Q=\frac{[CO][H_2]}{[H_2O]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now we have to calculate the concentration of H_2O,CO\text{ and }H_2

\text{Concentration of }H_2O=\frac{\text{Moles of }H_2O}{\text{Volume of solution}}=\frac{13.3mol}{3.25L}=4.09M

and,

\text{Concentration of }CO=\frac{\text{Moles of }CO}{\text{Volume of solution}}=\frac{3.40mol}{3.25L}=1.05M

and,

\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{6.30mol}{3.25L}=1.94M

Now put all the given values in the above expression, we get:

Q=\frac{(1.05)\times (1.94)}{(4.09)}=0.498

Thus, the value of reaction quotient Q is, 0.498

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25.0 mL of a hydrofluoric acid solution of unknown concentration is titrated with 0.200 M NaOH. After 20.0 mL of the base soluti
lesantik [10]

Answer:

[HF]₀ = 0.125M

Explanation:

NaOH + HF => NaF + H₂O

Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3.   This is 0.089M NaF and 0.001M HF remaining.

=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.

                HF  ⇄    H⁺    +      F⁻

C(eq)       [HF]     10⁻³M      0.089M (<= soln after adding 20ml 0.200M NaOH)

Ka = [H⁺][F⁻]/[HF]₀ => [HF]₀ = [H⁺][F⁻]/Ka

[HF]₀ = (0.001)(0.089)/(7.1 x 10⁻⁴) M = 0.125M

6 0
3 years ago
Fill in the Blanks: A molecular compound is made up of two or more _______ atoms. The type
Marizza181 [45]

Answer:

1.molecules of atoms

2.chemical bond

3.electrons

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8 0
3 years ago
9.05 mol of oxygen gas to g of oxygen gas
NARA [144]
Molar mass of oxygen gas:

O₂ = 16 * 2 = 32.0 g/mol

1 mole O₂ -------------- 32.0
9.05 mole O₂ ---------- ?

Mass = 9.05 * 32.0

Mass = 289.6 g of O₂

hope this helps!
7 0
3 years ago
What would happen to the volume of the container if the pressure is increased by a factor of 2​
telo118 [61]

Answer:

volume of the container will decreases if pressure increases.

Explanation:

According to Boyle's law:

Pressure is inversely proportional to volume which means if pressure of a gas increases the volume of the gas will decreases as gas molecules will collide and come closer forcefully so volume will decreases. And its formula for determining volume and pressure is:

<em>PV=nRT</em>

where "R" is a ideal gas constant

"T" is temperature and

"n" is number of particles given in moles while "V" is volume and "P" is pressure.

8 0
3 years ago
4.60 mL of 0.1852 M HNO3 is titrated to the phenolphthalein indicator endpoint with 27.35 mL of a KOH solution. What is the mola
Kamila [148]

Answer:

M_{base}=0.0311M

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve for the molarity of the KOH solution by knowing this base react in a 1:1 mole ratio with nitric acid, HNO3; thus, we can write the following equation, as their moles are the same at the endpoint:

n_{acid}=n_{base}

Which in terms of molarities and volumes is:

M_{acid}V_{acid}=M_{base}V_{base}

Thus, we solve for the molarity of the base (KOH) to obtain:

M_{base}=\frac{M_{acid}V_{acid}}{V_{base}} \\\\M_{base}=\frac{4.60mL*0.1852M}{27.35mL}\\\\M_{base}=0.0311M

Regards!

4 0
2 years ago
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