CHEMISTRY HIGH SCHOOL

A 25.5 liter balloon holding 3.5 moles of carbon dioxide leaks. How many grams of carbon dioxide escaped before the container could be sealed if the new volume of the balloon is 15.4L?

Answers

Answer 1
Answer: Assumptions made: 
1. Temperature and pressure is constant. 
2. Ideal gas is involved. 

The variables involved in this problem is the moles of the gas and the volume. We find their relationship from the ideal gas law. PV=nRT. With all other concept besides n and V is constant, Volume is directly proportional to the amount of gas inside the balloon. Solving this is as simple as ratio and proportion. 

25.5L/3.5 moles = =15.4L/X moles 
X= 2.11 moles. 
Answer 2
Answer:

Answer:

61 grams

Explanation:

If the number of moles of CO₂ in the balloon is given to be 3.5 moles, the mass of CO₂ in the balloon will be

number of moles = mass ÷ molar mass

Hence, mass = number of moles × molar mass

Molar mass of CO₂ is 44g. This is 44 because the atomic mass of carbon is 12 while that of oxygen is 16. Thus, 12 + (16 × 2) = 44 g

mass=  3.5 moles × 44g

mass = 154g (which is the initial mass)

When the mass of CO₂ in the 25.5 liter balloon is 154g, the mass of CO₂ in the balloon when the volume of CO₂ in the balloon is 15.4 liter will be X

To get X,

25.5 L ⇒ 154g

15.4 L ⇒ X

cross-multiply, and

X = (15.4 × 154) ÷ 25.5

X = 93.00 grams (which is the final mass)

93.00 grams of CO₂ was left in the balloon, hence the mass of CO₂ that escaped will be: initial mass minus final mass

= 154g - 93g

= 61g

The mass of CO₂ that escaped is 61 grams

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Answers

Answer: Option (d) is the correct answer.

Explanation:

When two or more different substances are mixed together then it results in the formation of a mixture.

Mixture are of two types, that is, homogeneous mixture and heterogeneous mixture.

In homogeneous mixture, the constituent particles are distributed evenly throughout the mixture.

Whereas in heterogeneous mixture, the constituent particles are non-uniformly distributed.

Thus, we can conclude that mixtures are classified based on the distribution of particles in them.
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Answers

There are 7 hydrogens in that formula
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Answers

I believe it is a physical change. Crumpling a sheet of paper does not change any properties of the paper and it does not form a new substance it must be a physical change. It is only the shape or form of the paper that is being changed. Hope this answers the question.
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Answers

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If a sample of a gas at 25.2oC has a volume of 536 mL at 0.838 atm, what will its volume be if the pressure is increased to 0.937 atm? A) 477.78 mL B) 563.21 mL C) 503.88 mL D) 613.52 mL

Answers

Final answer:

The volume of the gas sample would be 477.78 mL.

Explanation:

The student is asking about the change in volume of a gas sample when the pressure is increased. This can be solved using Boyle's law, which states that the pressure and volume of a gas are inversely proportional at constant temperature. The equation for Boyle's law is: P1 * V1 = P2 * V2

We are given the initial conditions: T1 = 25.2°C, V1 = 536 mL, and P1 = 0.838 atm. We need to find V2 when P2 = 0.937 atm. Using the Boyle's law equation, the volume can be calculated as: V2 = (P1 * V1) / P2

Plugging in the values, we get: V2 = (0.838 atm * 536 mL) / 0.937 atm = 477.78 mL

Therefore, the volume of the gas sample would be 477.78 mL (Option A) when the pressure is increased to 0.937 atm.

Learn more about Boyle's law here:

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1. For each of the following numbers, by how many places must the decimal point bemoved to express the number in scientific notation?
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0.0000089911

Answers

The number that you have shown the exponent would be negative and I believe it would be 6 places so 8.9911 x 10^-6
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