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Natasha2012 [34]
3 years ago
6

Evaluate 0.411 + 9.9 + 0.8 and round the answer appropriately,

Chemistry
1 answer:
Lemur [1.5K]3 years ago
8 0

Answer:

the answer is 11.111 and when you round it to the nearest one it’s 11, but to the nearest tenth it’s 10

Explanation:

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When conducting a medication house cleaning, and after sorting all of the client's drugs into two groups (drugs currently being
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After sorting the drugs into two groups, it is very important to check with the client, who is the one using the drugs, in order to ensure that the drugs have been properly sorted and no mistake was made. After ensuring that the drug has been accurately sorted then the ones that the client is no longer using can be put away in an appropriate safe location in order to avoid mix up.
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What does high enthalpies of fusion and vaporizing means​
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Solution X is a strong base with a pH of 12. Solution X is mixed with solution Y,and the pH of the resulting mixture is 8.Based
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Which of the following properties of water allows an insect to walk on water?
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At a certain temperature, 0.3411 0.3411 mol of N 2 N2 and 1.661 1.661 mol of H 2 H2 are placed in a 2.50 2.50 L container. N 2 (
Sveta_85 [38]

<u>Answer:</u> The equilibrium constant for the above reaction is 1.31

<u>Explanation:</u>

We are given:

Initial moles of nitrogen gas = 0.3411 moles

Initial moles of hydrogen gas = 1.661 moles

Equilibrium moles of nitrogen gas = 0.2001 moles

For the given chemical reaction:

                   N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)

<u>Initial:</u>         0.3411      1.661

<u>At eqllm:</u>     0.3411-x  1.661-3x      2x

Evaluating for 'x', we get:

\Rightarrow (0.3411-x)=0.2001\\\\\Rightarrow x=0.3411-0.2001=0.141

Volume of the container = 2.50 L

The expression of K_c for the above equation follows:

K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}

We are given:

[NH_3]=\frac{2\times 0.141}{2.50}=0.1128M

[N_2]=\frac{0.2001}{2.5}=0.08004M

[H_2]=\frac{1.661-(3\times 0.141)}{2.5}=0.4952M

Putting values in above expression, we get:

K_c=\frac{(0.1128)^2}{0.08004\times (0.4952)^3}\\\\K_c=1.31

Hence, the equilibrium constant for the above reaction is 1.31

5 0
3 years ago
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