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kolezko [41]
3 years ago
6

Enter your answer in the provided box. Liquid methanol (CH3OH) can be used as an alternative fuel in pickup and SUV engines. An

industrial method for preparing it uses the catalytic hydrogenation of carbon monoxide: CO(g) + 2H2(g) CH3OH(l) How much heat (in kJ) is released when 15.0 L of CO at 85°C and 112 kPa reacts with 17.0 L of H2 at 75°C and 744 torr?
Chemistry
1 answer:
andreyandreev [35.5K]3 years ago
3 0

Answer: E=∆H*n= -40.6kj

Explanation:

V(CO) =15L=0.015M³

P=11200Pa

T=85C=358.15K

PV=nRT

n=(112000×0.015)/(8.314×358.15)

n(Co)= 0.564mol

V(Co)= 18.5L = 0.0185m³

P=744torr=98191.84Pa

T= 75C = 388.15k

PV=nRT

n= (99191.84×0.0185)/(8.314×348.15)

n(H2) = 0.634mol

n(CH30H) =1/2n(H2)=1/2×0.634mol

=0.317mol

∆H =∆Hf{CH3OH}-∆Hf(Co)

∆H=-238.6-(-110.5)

∆H = 128.1kj

E=∆H×n=-40.6kj.

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3 years ago
Will GIVE BRAINLIEST --A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water. Answe
leva [86]

Answer:

0.5188 M or 0.5188 mol/L

Explanation:

Concentration is calculated as <u>molarity</u>, which is the number of moles per litre.

***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".

The formula for molarity is:

c = \frac{n}{V}

n = moles (unit mol)

V = volume (unit L)

<u>Find the molar mass (M) of potassium hydroxide.</u>

M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}

M_{KOH} = 56.106 \frac{g}{mol}

<u>Calculate the moles of potassium hydroxide.</u>

n_{KOH} = \frac{14.555 g}{1}*\frac{1mol}{56.106g}

n_{KOH} = 0.25941(9)mol

Carry one insignificant figure (shown in brackets).

<u>Convert the volume of water to litres.</u>

V = \frac{500.0mL}{1}*\frac{1L}{1000mL}

V = 0.5000L

Here, carrying an insignificant figure doesn't change the value.

<u>Calculate the concentration.</u>

c = \frac{n}{V}

c = \frac{0.25941(9)mol}{0.5000 L}              

c = 0.5188(3) \frac{mol}{L}         <= Keep an insignificant figure for rounding

c = 0.5188 \frac{mol}{L}              <= Rounded up

c = 0.5188M               <= You use the unit "M" instead of "mol/L"

The concentration of this standard solution is 0.5188 M.

7 0
3 years ago
In a titration of 47.41 mL of 0.3764 M ammonia with 0.3838 M aqueous nitric acid, what is the pH of the solution when 47.41 mL +
Volgvan

<u>Answer:</u> The pH of the solution is 1.136

<u>Explanation:</u>

To calculate the moles from molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

  • <u>For ammonia:</u>

Molarity of ammonia = 0.3764 M

Volume of ammonia = 47.41 mL = 0.04741 L   (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

0.3764mol/L=\frac{\text{Moles of ammonia}}{0.04741L}\\\\\text{Moles of ammonia}=0.01784mol

  • <u>For nitric acid:</u>

Molarity of nitric acid = 0.3838 M

Volume of ammonia = (47.41 + 10.00) mL = 57.41 mL= 0.05741 L

Putting values in above equation, we get:

0.3838mol/L=\frac{\text{Moles of nitric acid}}{0.05741L}\\\\\text{Moles of nitric acid}=0.02203mol

After the completion of reaction, amount of nitric acid remained = 0.022 - 0.0178 = 0.0042 mol

For the reaction of ammonia with nitric acid, the equation follows:

                       NH_3+HNO_3\rightarrow NH_4NO_3

At t=0             0.0178   0.022

Completion        0     0.0042        0.0178

As, the solution of the reaction is made from strong acid which is nitric acid and the conjugate acid of weak base which is ammonia. So, the pH of the reaction will be based totally on the concentration of nitric acid.

To calculate the pH of the reaction, we use the equation:

pH=-\log[H^+]

where,

[H^+]=\frac{0.0042mol}{0.05741L}=0.0731M

Putting values in above equation, we get:

pH=-\log(0.0731)\\\\pH=1.136

Hence, the pH of the solution is 1.136

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