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timurjin [86]
3 years ago
5

The reaction to prepare methanol from carbon monoxide and hydrogen ‍ is exothermic. If you wanted to use this reaction to produc

e methanol commercially, would high or low temperatures favor a maximum yield? Explain.
Chemistry
1 answer:
Alex_Xolod [135]3 years ago
8 0

Answer:

The reaction in low temperature will favor maximum yield.

Explanation:

Methanol is mainly produced by hydrogenation of carbon monoxide.It is the simplest alcohol which contain methyl group that is linked with hydroxyl group.Methanol is volatile,colorless,flammable liquid.

Ethanol is less toxic than methanol.Methanol is the precursor of acetic acid,methyl tertiary butyl ether.

During formation of commercial methanol,carbon dioxide and water acts as poisons to alkali metal methoxide catalyst in low temperature in methanol synthesis reaction.

More over,high temperature evaporates methanol.

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Atoms and elements are examples of A. molecules. B. mixtures. C. compounds. D. pure substances.
UkoKoshka [18]

Answer:

atom are very tiny partical

3 0
2 years ago
0.17 M KCHO2 Find the pH
SashulF [63]
I believe the pH is 7. If you would like me to explain comment.
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3 years ago
The compound Xe(CF3)2 decomposes in afirst-order reaction to elemental Xe with a half-life of 30. min.If you place 7.50 mg of Xe
Irina-Kira [14]

Answer:

t=147.24\ min

Explanation:

Given that:

Half life = 30 min

t_{1/2}=\frac{\ln2}{k}

Where, k is rate constant

So,  

k=\frac{\ln2}{t_{1/2}}

k=\frac{\ln2}{30}\ min^{-1}

The rate constant, k = 0.0231 min⁻¹

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given that:

The rate constant, k = 0.0231 min⁻¹

Initial concentration [A_0] = 7.50 mg

Final concentration [A_t] = 0.25 mg

Time = ?

Applying in the above equation, we get that:-

0.25=7.50e^{-0.0231\times t}

750e^{-0.0231t}=25

750e^{-0.0231t}=25

x=\frac{\ln \left(30\right)}{0.0231}

t=147.24\ min

6 0
3 years ago
How many moles of water are produced from 15 moles of oxygen?
Grace [21]

Answer:

30 moles

Explanation:

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I multiply 15 moles O2 by the molar ratio of (hydrogen/oxygen)

15 mol. O2 *  (2 mol. H2/1 mol O2) = 30 moles of water

8 0
3 years ago
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An aldehyde is an organic compound containing a terminal carbonyl group (C = O). This functional group, consisting of a carbon atom bound to a hydrogen atom and an oxygen atom via double bond (the general formula: CHO) is called the aldehyde group. In a reaction of the addition of alcohol to the carbonyl group, it forms hemiacetals. 
On the picture attached it is shown the reaction of alcohol addition to the carbonyl group with the major organic product <span>formed in the reaction.</span>

7 0
3 years ago
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