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11111nata11111 [884]
3 years ago
15

A balloon contains 0.140 molmol of gas and has a volume of 2.78 LL . If an additional 0.152 molmol of gas is added to the balloo

n (at the same temperature and pressure), what will its final volume be?
Chemistry
1 answer:
levacccp [35]3 years ago
8 0

Answer:

The final volume will be 5.80 L

Explanation:

Step 1: Data given

Number of moles gas = 0.140 moles

Volume of gas = 2.78 L

Number of moles added = 0.152 moles

Step 2: Calculate the final volume

V1/n1 = V2/n2

⇒ with V1 = the initial volume = 2.78 L

⇒ with n1 = the initial number of moles = 0.140 moles

⇒ with  V2 = The new volume = TO BE DETERMINED

⇒ with n2 = the new number of moles = 0.140 + 0.152 = 0.292 moles

2.78/0.140 = V2 /0.292

V2 = 5.80 L

The final volume will be 5.80 L

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Explanation:

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No. 65, chemistry help<br> Thank you
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Consider the following reaction where K. = 154 at 298 K: 2NO(g) + Brz(9) 2NOBr(g) A reaction mixture was found to contain 2.69x1
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Explanation:

2NO(g) + Br_2(g)\rightleftharpoons 2NOBr(g)

Equilibrium constant of reaction = K=154

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Concentration of NOBr gas = [Br_2]=\frac{9.56\times 10^{-2} mol}{1 L}=9.56\times 10^{-2} M

The reaction quotient is given as:

Q=\frac{[NOBr]^2}{[NO]^2[Br_2]}=\frac{(9.56\times 10^{-2} M)^2}{(2.69\times 10^{-2} M)^2\times 3.85\times 10^{-2} M}

Q=328.06

Q>K

The reaction will go in backward direction in order to achieve an equilibrium state.

1. In order to reach equilibrium NOBr (g) must be produced.  False

2. In order to reach equilibrium K must decrease. False

3. In order to reach equilibrium NO must be produced. True

4. Q. is less than K . False

5. The reaction is at equilibrium. No further reaction will occur. False

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What is the smallest division on this balance?
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0.001 would be the smallest.

Good Luck! :)
4 0
3 years ago
Read 2 more answers
How many grams are in 1 mole of Ar?
Ilia_Sergeevich [38]

Answer: 39.948 grams

Explanation:

The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles Ar, or 39.948 grams

3 0
3 years ago
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