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Alexandra [31]
3 years ago
14

maye Iesult TUI di2... Warm-Up Exercises for Chapter 2 - Acid-Base Reactions roblem 2.68 Part C Which of the following reactions

has the most favorable equilibrium constant? © CH3OH + NH3 = CH2O +NH 0 CH3CH2OH + NH3 =CH3CH2O +NH CH3CH2OH + CH3NH2 = CH3CH,0- + CH3NH; All of them have the same equillibrium constant. Submit Request Answer
Chemistry
1 answer:
denis-greek [22]3 years ago
3 0
Use apex learning for homework! Gives you all the answers and step by step help
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A 56.8g sample of aluminum is heated from 79.5°C to 143.7°C. The specific heat capacity of aluminum is 0.900 J/(g*K). Calculate
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Answer:

I think the answer is A.

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4 years ago
In terms of entropy and energy, systems in nature tend to undergo changes toward
andrew-mc [135]
Number 2 lower entropy and higher entropy
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Is a strap of leather a pure substance
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I'd say no because the only pure substances are elements
4 0
3 years ago
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Use the balanced equation from number four above to determine the limiting reactant if we had a mixture of 5.0 moles of Fe and 4
Tom [10]

Answer:

2Fe + 3O₂  →  2Fe₂O₃

Limiting reactant: O₂

2Fe + O₂ → 2FeO

Limiting reactant: Fe

Explanation:

2Fe + 3O₂  →  2Fe₂O₃

2Fe + O₂ → 2FeO

These are the possible reactions:

In first case, 2 moles of Fe need 3 mol of oyxgen to react

If I have 5 moles of Fe, I will need (5 .3)/2 = 7.5 moles of O₂

Then, the oxygen is my limiting ( I only have 4 moles)

3 moles of O₂ need 2 moles of Fe

If I have 4 moles of O₂, I will need (4 .2)/3 = 2.66 moles (I have 5)

Fe, is the reactant in excess.

For second case, 2 moles of Fe need 1 mol of O₂, to react.

If I have 5 moles of Fe, I will need ( 5 .1) / 2 =2.5 moles of O₂

I have 4 moles of oxygen, so now it is my excess.

1 mol of O₂ need 2 moles of Fe, to react

If I have 4 moles of O₂, I will need the double of amount, 8 moles of Fe.

I have 5 moles, then the Fe is my limtiing reactant.

8 0
3 years ago
Suppose the reaction between nitrogen and hydrogen was run according to the amounts presented in Part A, and the temperature and
andrew11 [14]

Explanation:

Assuming that moles of nitrogen present are 0.227 and moles of hydrogen are 0.681. And, initially there are 0.908 moles of gas particles.

This means that, for N_{2}(g) + 3H_{2}(g) \rightarrow 2NH_{3}

 moles of N_{2} + moles of H_{2} = 0.908 mol

Since, 2 moles of N_{2} = 2 \times 0.227 = 0.454 mol

As it is known that the ideal gas equation  is PV = nRT

And, as the temperature and volume were kept constant, so we can write

        \frac{P(_in)}{n_(in)} = \frac{P_(final)}{n_(final)}

          \frac{10.4}{0.908} = \frac{P_(final)}{0.454
}

       P_(final) = 10.4 \times \frac{0.454}{0.908}

                            = 5.2 atm

Therefore, we can conclude that the expected pressure after the reaction was completed is 5.2 atm.

7 0
3 years ago
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