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vodomira [7]
3 years ago
12

For each set of reactions, determine the value of δh2 in terms of δh1. a+b→2c,δh1 1/2a+1/2b→c,δh2=? express your answer in terms

of δh1.
Chemistry
1 answer:
allochka39001 [22]3 years ago
3 0
ΔH2 = - δH1 δH2 = - 2 x δH1 δH2 = 2 x <span>δ</span>H1
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True or False<br> Helium belongs to Noble Metals
Lostsunrise [7]

Answer:

This answer is "True"

6 0
3 years ago
Read 2 more answers
Consider the formation of nitrogen dioxide from nitric
Temka [501]

Answer:

9 L

Explanation:

According to the question , the given reaction is -

2NO(g) + O₂(g)------->2NO₂(g)

Since ,

At STP ,

One mole of a gas occupies the volume of 22.4 L.

Hence , as given in the question -

9 L of NO , i.e .

22.4 L = 1 mol

1 L = 1 / 22.4 mol

9 L = 1 / 22.4  * 9 L = 0.40 mol

From the chemical reaction ,

The Oxygen is in excess , hence NO becomes the limiting reagent , and will determine the moles of product .

Hence ,  

2 moles of NO will produce 2 moles of NO₂.

Therefore ,

0.40 mol of NO will produce 0.40 mol of NO₂.

Hence , the volume of NO₂ can be calculated as -

1 mol = 22.4 L

0.40 mol = 0.40 * 22.4 L = 9 L

6 0
4 years ago
Write the complete balanced equation for the reaction between iron (III) oxide (Fe2O3) and water (H2O). You do not need to make
Alina [70]
Fe2O3+3H2O->2Fe(OH)3
6 0
3 years ago
Why are plastic containers preferred to glass containers for storing chemicals in the laboratory?
Nookie1986 [14]
Plastics are non-corrosive and non-reactive in nature. So they are used for storing chemicals in the laboratory. They are used for strong chemicals because they do not react with chemicals neither do they corrode
7 0
3 years ago
The activation energy of a certain uncatalyzed reaction is 64 kJ/mol. In the presence of a catalyst, the Ea is 55 kJ/mol. How ma
Ksivusya [100]

Answer:

About 5 times faster.

Explanation:

Hello,

In this case, since the Arrhenius equation is considered for both the catalyzed reaction (1) and the uncatalized reaction (2), one determines the relationship between them as follows:

\frac{k_1}{k_2}=\frac{Aexp(-\frac{Ea_1}{RT} )}{Aexp(-\frac{Ea_2}{RT})}  \\\frac{k_1}{k_2}=\frac{exp(-\frac{Ea_1}{RT} )}{exp(-\frac{Ea_2}{RT})}

By replacing the corresponding values we obtain:

\frac{k_1}{k_2}=\frac{exp(-\frac{55000J/mol}{8.314J/molK*673.15K} )}{exp(-\frac{64000J/mol}{8.314J/molK*673.15K} )} =4.8

Such result means that the catalyzed reaction is about five times faster than the uncatalyzed reaction.

Best regards.

4 0
3 years ago
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