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ale4655 [162]
3 years ago
10

You’re at the zoo and have a big red 1.80 L helium balloon. The barometric pressure today is 785 mmHg. Then you hear the roar of

a lion. Startled, you accidentally release the balloon. It flies away. By the time it reaches the clouds, the atmospheric pressure that high is only 3.00 atmospheres. What would the volume of the balloon up there? (Temperature is constant)
Chemistry
2 answers:
Marianna [84]3 years ago
8 0

Answer:

The volume of the balloon up there will be 0.618 L

Explanation:

Given:

V₁ = 1.8 L

P₁ = 785 mmHg = 1.03 atm

P₂ = 3 atm

Question: What would the volume of the balloon up there, V₂ = ?

According the ideal gas law:

PV = nRT

However, in this case, the number of moles of helium balloon is constant. The temperature will also be assumed to be constant. Therefore, the expression of the ideal gases is as follows:

P₁V₁ = P₂V₂

Solving for V₂

V_{2} =\frac{P_{1}V_{1}  }{P_{2} } =\frac{1.03*1.8}{3} =0.618L

nadezda [96]3 years ago
5 0

Answer:

0.62 L

Explanation:

Step 1:

Data obtained from the question.

Initial Volume (V1) = 1.80 L

Initial pressure (P1) = 785 mmHg

Final pressure (P2) = 3.00 atm

Final volume (V2) =?

Step 2:

Conversion of the pressure in mmHg to atm.

It is important to express the initial and the final pressure in the same unit. Either express both in atm or in mmHg. What ever the case is, we'll still arrive at same answer. Here, we shall be converting from mmHg to atm. This is illustrated below:

760mmHg = 1atm

Therefore, 785 mmHg = 785/760 = 1.03 atm

Step 3:

Determination of the final volume. This is illustrated below.

We shall be applying the Boyle's law equation since the temperature is constant.

P1V1 = P2V2

Initial Volume (V1) = 1.80 L

Initial pressure (P1) = 1.03 atm

Final pressure (P2) = 3.00 atm

Final volume (V2) =?

P1V1 = P2V2

1.03 x 1.8 = 3 x V2

Divide both side by 3

V2 = (1.03 x 1.8) /3

V2 = 0.62 L

Therefore, the new volume of the balloon is 0.62 L

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If volumes are additive and 253 mL of 0.19 M potassium bromide is mixed with 441 mL of a potassium dichromate solution to give a
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Answer:

The concentration of the Potassium Dichromate solution is 0.611 M

Explanation:

First of all, we need to understand that in the final solution we'll have potassium ions coming from KBr and also K2Cr2O7, so we state the dissociation equations of both compounds:

KBr (aq) → K+ (aq) + Br- (aq)

K2Cr2O7 (aq) → 2K+ (aq) + Cr2O7 2- (aq)

According to these balanced equations when 1 mole of KBr dissociates, it generates 1 mole of potassium ions. Following the same thought, when 1 mole of K2Cr2O7 dissociates, we obtain 2 moles of potassium ions instead.

Having said that, we calculate the moles of potassium ions coming from the KBr solution:

0.19 M KBr: this means that we have 0.19 moles of KBr in 1000 mL solution. So:

1000 mL solution ----- 0.19 moles of KBr

253 mL solution ----- x = 0.04807 moles of KBr

As we said before, 1 mole of KBr will contribute with 1 mole of K+, so at the moment we have 0.04807 moles of K+.

Now, we are told that the final concentration of K+ is 0.846 M. This means we have 0.846 moles of K+ in 1000 mL solution. Considering that volumes are additive, we calculate the amount of K+ moles we have in the final volume solution (441 mL + 253 mL = 694 mL):

1000 mL solution ----- 0.846 moles K+

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This is the final quantity of potassium ion moles we have present once we mixed the KBr and K2Cr2O7 solutions. Because we already know the amount of K+ moles that were added with the KBr solution (0.04807 moles), we can calculate the contribution corresponding to K2Cr2O7:

0.587124 final K+ moles - 0.04807 K+ moles from KBr = 0.539054 K+ moles from K2Cr2O7

If we go back and take a look a the chemical reactions, we can see that 1 mole of K2Cr2O7 dissociates into 2 moles of K+ ions, so:

2 K+ moles ----- 1 K2Cr2O7 mole

0.539054 K+ moles ---- x = 0.269527 K2Cr2O7 moles

Now this quantity of potassium dichromate moles came from the respective  solution, that is 441 mL, so we calculate the amount of them that would be present in 1000 mL to determine de molar concentration:

441 mL ----- 0.269527 K2Cr2O7 moles

1000 mL ----- x = 0.6112 K2Cr2O7 moles = 0.6112 M

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Answer:

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