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rjkz [21]
2 years ago
14

In one or two sentences, explain why the combined weight of several liquids mixed in an open flask can be less than but never mo

re than the sum of all the weights of the ingredients.
Chemistry
1 answer:
exis [7]2 years ago
6 0

Answer:

j

Explanation:

h

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The following reactions have the indicated equilibrium constants at a particular temperature: N2(g) + O2(g) ⇌ 2NO(g) Kc = 4.3 ×
Anuta_ua [19.1K]

Answer:

Kc=~1.49x10^3^4}

Explanation:

We have the reactions:

A: N_2_(_g_) + O_2_(_g_)  2NO_(_g_)~~~~~~Kc = 4.3x10^-^2^5

B: 2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~~~Kc = 6.4x10^9

Our <u>target reaction</u> is:

4NO_(_g_)  N_2_(_g_) + 2NO_2_(_g_)

We have NO_(_g_) as a reactive in the target reaction and  NO_(_g_) is present in A reaction but in the products side. So we have to<u> flip reaction A</u>.

A: 2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}

Then if we add reactions A and B we can obtain the target reaction, so:

A: 2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}

B: 2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~Kc=6.4x10^9

For the <u>final Kc value</u>, we have to keep in mind that when we have to <u>add chemical reactions</u> the total Kc value would be the <u>multiplication</u> of the Kc values in the previous reactions.

4NO_(_g_)  N_2_(_g_) + 2NO_2_(_g_)~~~Kc=\frac{6.4x10^9}{4.3x10^-^2^5}

Kc=~1.49x10^+^3^4}

3 0
2 years ago
How does the government rely on scientists?
ivann1987 [24]
They rely on scientists for facts and answers
5 0
2 years ago
The equilibrium constant for the chemical equation is Kp = 5.23 at 191 °C. Calculate the value of the Kc for the reaction at 191
Lapatulllka [165]
To convert from Kp to Kc, you need this formula---> Kp= Kc (RT)^Δn, where Δn= gas moles of product- gas moles of reactants. since you did not give a reaction formula, I can't calculate Δn. but all once you find it out. just plug it. 

Kp= Kc (RT)^Δn------------------> Kc= Kp/[(RT)^Δn]
 Kp= 5.23
R= 0.0821
T= 191 C= 464 K
Δn= ?

Kc= 5.23/ (0.0821 x 464)^Δn= ???

7 0
3 years ago
What is the formula for Silver (I) chloride?
Kisachek [45]
The formula for Silver (I) chloride is: AgCl
6 0
3 years ago
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Which method can be used to find the range
lidiya [134]

Answer:

To find the range, first order the data from least to greatest. Then subtract the smallest value from the largest

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2 years ago
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