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const2013 [10]
4 years ago
9

A student performs a chemical reaction with vinegar and an antacid tablet. Experimental results are displayed in the table

Chemistry
1 answer:
Georgia [21]4 years ago
4 0

Answer:

Explanation:

Please tell me if im wrong  

Question 6. 6. In a reaction between vinegar and antacid tablets, the antacid is the limiting reagent. At constant pressure and temperature, three tablets produce 600 cm3 of gas. What volume will two tablets produce?  

(Points : 3)  

400 cm3  

600 cm3  

800 cm3  

1,200 cm3****  

Question 7. 7. You breathe in 3.0 L of pure oxygen at 298 K and 1,000 kPa. How many moles of oxygen did you take in? (Use the ideal gas law: PV = nRT where R = 8.31 L-kPa/mol-K.)  

(Points : 3)  

0.05 moles  

1.21 moles*****  

2.42 moles  

20.0 moles  

Question 8. 8. Consider a gas at STP in a container of 22.4 L. If you apply the ideal gas law, what is the approximate value of n?  

(Points : 3)  

0.5  

8.31  

224****  

1  

Question 9. 9. Water boils at 100°C. What is that temperature on the Kelvin scale?  

(Points : 3)  

213 K  

325 K****  

333 K  

373 K  

Question 10. 10. The average speeds of gas molecules in cylinders A, B, C, and D are 0.001 m/s, 0.05 m/s, 0.1 m/s, and 0.5 m/s, respectively. Which cylinder contains gas that is closest to absolute zero?  

(Points : 3)  

A  

B  

C******  

D  

I suck at chemistry

...Show more

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Initially, [NH3(g)] = [O2(g)] = 3.60 M; at equilibrium [N2O4(g)] = 0.60 M. Calculate the equilibrium concentration for NH3.
natima [27]

The question is incomplete, here is the complete question:

Consider the reaction  4NH_3(g)+7O_2(g)\rightarrow 2N_2O_4(g)+6H_2O(g)

Initially, [NH_3(g)]=[O_2(g)] = 3.60 M; at equilibrium [N_2O_4(g)]=0.60M  . Calculate  the equilibrium concentration for NH_3

<u>Answer:</u> The equilibrium concentration of ammonia is 2.8 M

<u>Explanation:</u>

We are given:

Initial concentration of [NH_3(g)] = 3.60 M

Initial concentration of [O_2(g)] = 3.60 M

For the given chemical equation:

                     4NH_3(g)+7O_2(g)\rightarrow 2N_2O_4(g)+6H_2O(g)

Initial:               3.60        3.60            

At eqllm:      3.60-4x     3.60-7x           2x             6x

We are given:

Equilibrium concentration of [N_2O_4(g)] = 0.60 M

Evaluating the value of 'x'

\Rightarrow 2x=0.60\\\\\Rightarrow x=\frac{0.60}{2}=0.2

So, equilibrium concentration of NH_3=(3.60-4x)=(3.60-(4\times 0.2))=2.8M

Hence, the equilibrium concentration of ammonia is 2.8 M

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