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Varvara68 [4.7K]

2 years ago

8

A balloon that can hold 85 L of air is inflated with 3.5 moles of gas at a pressure of 1.0 atmosphere. What is the temperature o

f the balloon?

1 answer:

slega [8]2 years ago

7 0

Answer:

295.96 K ( = 22.81 degrees C)

Explanation:

PV = n RT R = .082057 L-Atm / (K-Mol) n = 3.5 P = 1 atm V = 85 liters

(1)(85) = 3.5 (.082057)(T)

solve for T = 295.96 K

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Samantha is using a lever to lift a rock. To life the rock most easily, where should she apply force to the lever

Inessa [10]

Answer:

On the opposite side of the rock.

Explanation:

6 0

3 years ago

A gas at STP has a volume of 1.00 L. If the pressure is doubled and the temperature remains constant, what is the news of the ga

4vir4ik [10]

Answer: PV = nRT

A gas at STP... This means that the temperature is 0°C and pressure is 1 atm.

R is the gas constant which is 0.08206 L*atm/(K*mol)

Rearranging for volume

V = nRT/P

The temperature and number of moles are held constant. This means that this uses Boyle's Law. (The ideal gas law could be manipulated to give us this result when T and n are held constant.)

PV = k

where k is a constant.

This means that

P₁V₁ = k = P₂V₂

P₁V₁ = P₂V₂

(1 atm) * (1 L) = (2 atm) * V₂

V₂ = 0.5 L

The new volume of the gas is 0.5 L.

Explanation:

3 0

2 years ago

What is the composition, in atom percent, of an alloy that consists of a) 5.5 wt% Pb and b) 94.5 wt% of Sn? Assume that the atom

Anastaziya [24]

Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

Explanation :

First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

As, 207.2 g of lead contains $6.022\times 10^{23}$ atoms

So, 5.5 g of lead contains $\frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22}$ atoms

and,

As, 118.71 g of lead contains $6.022\times 10^{23}$ atoms

So, 94.5 g of lead contains $\frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23}$ atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

$\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%$

and,

$\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%$

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

6 0

3 years ago

Jobisdone [24]

Answer:

hello,

first one is 25.15

second is 301.55

Explanation:

honestly if you look up on the internet there is a converter to get you your answers

5 0

3 years ago

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TBESRBTBTRE

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5 0

2 years ago