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3 years ago

Quick help I really need help

Chemistry

1 answer:

Sauron [17]3 years ago

8 0

Answer:

CO2 H2o

Explanation:

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9. Given the following reaction:CO (g) + 2 H2(g) CH3OH (g)In an experiment, 0.45 mol of CO and 0.57 mol of H2 were placed in a 1

WITCHER [35]

Answer:

$Keq=11.5$

Explanation:

Hello,

In this case, for the given reaction at equilibrium:

$CO (g) + 2 H_2(g) \rightleftharpoons CH_3OH (g)$

We can write the law of mass action as:

$Keq=\frac{[CH_3OH]}{[CO][H_2]^2}$

That in terms of the change $x$ due to the reaction extent we can write:

$Keq=\frac{x}{([CO]_0-x)([H_2]_0-2x)^2}$

Nevertheless, for the carbon monoxide, we can directly compute $x$ as shown below:

$[CO]_0=\frac{0.45mol}{1.00L}=0.45M\\$

$[H_2]_0=\frac{0.57mol}{1.00L}=0.57M\\$

$[CO]_{eq}=\frac{0.28mol}{1.00L}=0.28M\\$

$x=[CO]_0-[CO]_{eq}=0.45M-0.28M=0.17M$

Finally, we can compute the equilibrium constant:

$Keq=\frac{0.17M}{(0.45M-0.17M)(0.57M-2*0.17M)^2}\\\\Keq=11.5$

Best regards.

3 0

3 years ago

Can someone please please help me with the first two !!!! I really need to turn this in !!!!!

kati45 [8]

Answer:

Be Electron configuration: 1s2 2s2

Be Orbital Diagram: \//\ \//\ (it would be little arrows going up and down to show the spins)

F Electron Configuration: 1s2 2s2 2p5

F Orbital Diagram: \//\ \//\ \//\ \//\ \/

7 0

2 years ago

Please help me I will give you extra points and the brain thing please actually do it.

Shkiper50 [21]

Inertia is the retaliation of an object to change in its velocity

4 0

3 years ago

What is meant by localised orbitals? Explain.

Anni [7]

Localized molecular orbitals are molecular orbitals which are concentrated in a limited spatial region of a molecule, for example a specific bond or a lone lake on a specific atom.

5 0

3 years ago

A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 23.0 mL of

Georgia [21]

NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa

NaOH + CH3COOH → CH3COONa + H2O

Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH

Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH

These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L

Molarity of CH3COOH = 0.0106/0.071 = 0.1493M

CH3COONa = 0.0076 / 0.071 = 0.1070M

pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74.

pH using Henderson - Hasselbalch equation:

pH = pKa + log ([salt]/[acid])

pH = 4.74 + log ( 0.1070/0.1493)

pH = 4.74 + log 0.717

pH = 4.74 + (-0.14)

pH = 4.60.

7 0

3 years ago