Determine the expression for the equilibrium constant, KcKcK_c, for the reaction by identifying which terms will be in the numer
1 answer:
Answer: Hello your question lacks some details below is the complete question
answer :
Numerator = CH₃OH
Denominator = (CO ) ( H₂)²
Explanation:
CO + 2H₂ ⇄ CH₃OH ( formation of methanol )
hence for the equilibrium constant Kc
Numerator = CH₃OH
Denominator = (CO ) ( H₂)²
<u><em>Placing into the Bin </em></u>
Numerator : CH₃OH
Denominator : (CO ) ( H₂)²
Not Used : ( CO )² , H₂ , ( CH₃OH ) ²
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Answer:
The bismuth sample.
Explanation:
The specific heat of a substance (might not be a metal) is the amount of heat required for heating a unit mass of this substance by unit temperature (e.g.,
.) The formula for specific heat is:
,
where
is the amount of heat supplied.
is the mass of the sample.
is the increase in temperature.
In this question, the value of (amount of heat supplied to the metal) and
(mass of the metal sample) are the same for all four metals. To find
(change in temperature,) rearrange the equation:
,
.
In other words, the change in temperature of the sample, can be expressed as a fraction. Additionally, the specific heat of sample,
, is in the denominator of that fraction. Hence, the value of the fraction would be the largest for sample with the smallest specific heat.
Make sure that all the specific heat values are in the same unit. Find the one with the smallest specific heat: bismuth (.) That sample would have the greatest increase in temperature. Since all six samples started at the same temperature, the bismuth sample would also have the highest final temperature.
Answer: The volume of the sample after the reaction takes place is 29.25 L.
Explanation:
The given reaction equation is as follows.
So, moles of product formed are calculated as follows.
Hence, the given data is as follows.
= 0.17 mol,
= 0.255 mol
= 19.5 L,
As the temperature and pressure are constant. Hence, formula used to calculate the volume of sample after the reaction is as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that the volume of the sample after the reaction takes place is 29.25 L.