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vredina [299]
3 years ago
11

Which of the following is true of a gas?

Chemistry
1 answer:
mote1985 [20]3 years ago
6 0

Answer:

d

Explanation:

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12. An electrolysis reaction is
tia_tia [17]

Answer: D.) non-spontaneous.

Explanation:

6 0
2 years ago
Complete combustion of a 0.350 g sample of a compound in a bomb calorimeter releases 14.0 kJ of heat. The bomb calorimeter has a
dexar [7]

Answer:

Its final temperature is 25.8 °C

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

There is a direct proportional relationship between heat and temperature. The constant of proportionality depends on the substance that constitutes the body as on its mass, and is the product of the specific heat by the mass of the body. So, the equation that allows calculating heat exchanges is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (ΔT=Tfinal-Tinitial)

When a body transmits heat there is another that receives it. This is the principle of the calorimeter. Then the heat released by the compound will be equal to the heat obtained by the calorimeter.

In this case, you know:

  • Q= 14 kJ= 14,000 J
  • c= 3.55  \frac{J}{g*C}
  • m=1.20 kg= 1200 g (1 kg=1000 g)
  • Tfinal= ?
  • Tinitial= 22.5 °C

Replacing:

14,000 J= 3.55 \frac{J}{g*C}*1200 g*(Tfinal-22.5C)

Solving:

\frac{14,000J}{3.55\frac{J}{g*C} *1200 g} =T final - 22.5C

3.3=Tfinal - 22.5 C

3.3 + 22.5=Tfinal

Tfinal= 25.8 °C

<u><em>Its final temperature is 25.8 °C</em></u>

3 0
3 years ago
Read 2 more answers
Is bubbles forming in water a chemical change
Stolb23 [73]
It is a physical change

7 0
3 years ago
Read 2 more answers
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
Just vibin yall how about you
Vlada [557]

Answer:

Not much just operating brainly and waiting for my friends :)

4 0
3 years ago
Read 2 more answers
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