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Whitepunk [10]
3 years ago
14

Delta H equals 98.8 KJ, and Delta S equals 141.5 J/K. Is this reaction spontaneous or nonspontaneous at high and low pressure

Chemistry
1 answer:
Softa [21]3 years ago
8 0
A is the answer

Like that high and low
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as 100 milliliters of 0.10 molar KOH is added to 100 milliliters of 0.10 molar HCL at 298 K, the pH of the res?A. decrease to 3B
Mila [183]

Answer:

C. increase to 7.

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

KOH+HCl\rightarrow KCl+H_2O

Thus, the molar relationship is 1 to 1, therefore, the moles are:

n_{HCl}=0.1mol/L*0.1L=0.01molHCl\\n_{KOH}=0.1mol/L*0.1L=0.01molKOH\\

Thus, since the entire hydrogen ions are neutralized, the pH C. increase to 7.

Best regards.

8 0
3 years ago
If a cell wants to tell another cell what to do, it will send the message through
horsena [70]

Answer:

it will send the message through the axon

6 0
3 years ago
How many moles of hydrogen gas will be produced if 5.00 moles of zinc reactions with an excess amount of sulfuric acid
harkovskaia [24]

Answer:

Moles of Hydrogen produced is 5 moles

Explanation:

The balanced Chemical equation for reaction between zinc and sulfuric acid is :

Zn(s) + H_{2}SO_{4}(aq) \rightarrow ZnSO_{4}(aq) + H_{2}(g)

This equation tells that ; when  1 mole of Zn react with 1 mole of sulfuric acid, it produces 1 mole of zinc sulfate and 1 mole of hydrogen.

Since sulfuric acid is in excess so Zinc is the limiting reagent

(Limiting reagent : Substance which get consumed when the reaction completes, limiting reagent helps in predicting the amount of products formed)

Limiting reagent (Zn) will decide the amount of Hydrogen produced

Zn(s) + H_{2}SO_{4}(aq) \rightarrow ZnSO_{4}(aq) + H_{2}(g)

1\ mole\ zinc\rightarrow 1\ mole\ H_{2}

So,

5\ mole\ zinc\rightarrow 5\ mole\ H_{2}

Hence moles of Hydrogen produced is 5 moles

3 0
3 years ago
When methyloxirane is treated with HBr, the bromide ion attacks the less substituted position. However, when phenyloxirane is tr
konstantin123 [22]

Answer:

See explanation and picture below

Explanation:

First, in the case of methyloxirane (Also known as propilene oxide) the mechanism that is taking place there is something similar to a Sn2 mechanism. Although a Sn2 mechanism is a bimolecular substitution taking place in only step, the mechanism followed here is pretty similar after the first step.

In both cases, the H atom of the HBr goes to the oxygen in the molecule. You'll have a OH⁺ in both. However, in the case of methyloxirane the next step is a Sn2 mechanism step, the bromide ion will go to the less substitued carbon, because the methyl group is exerting a steric hindrance. Not a big one but it has a little effect there, that's why the bromide will rather go to the carbon with more hydrogens. and the final product is formed.

In the case of phenyloxirane, once the OH⁺ is formed, the next step is a Sn1 mechanism. In this case, the bond C - OH⁺ is opened on the side of the phenyl to stabilize the OH. This is because that carbon is more stable than the carbon with no phenyl. (A 3° carbon is more stable than a 2° carbon). Therefore, when this bond opens, the bromide will go there in the next step, and the final product is formed. See picture below for mechanism and products.

4 0
3 years ago
How to do my homework? i’ll give 20 points to anyone who will do it!
tensa zangetsu [6.8K]

what is it ??? ill edit this

3 0
3 years ago
Read 2 more answers
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