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tresset_1 [31]
2 years ago
9

The pressure of a sample of helium is 1.556 atm in a 268.5 mL container. If the container is compressed to 112.4 mL without chan

ging the temperature, what is the new pressure?
a. 3.72
b. 0.651
c. 1.94e4
d. 277
Chemistry
1 answer:
Rainbow [258]2 years ago
3 0

Answer:

a.  3.72 [atm]

Explanation:

For a gas at constant temperature, (with no change in number of molecules of the gas), we can apply Boyle's Law:  P_1V_1=P_2V_2

(1.556[atm])(268.5[mL])=P_2(112.4[mL])

\dfrac{(1.556[atm])(268.5[mL\!\!\!\!\!\!\!\!{--}])}{112.4[mL \!\!\!\!\!\!\!\!{--}]}=\dfrac{P_2(112.4[mL]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----})}{112.4[mL]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----}}

3.716957[atm]=P_2

It seems like the answer should have 4 significant figures since all of the other quantities have 4 significant figures, but the closest answer choice of those provided is a.  3.72

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Answer:

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Explanation:

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6 0
3 years ago
A compound has the formula X2O3 where x is an unknown element. If the compound is 24.26% oxygen, what element does x represent?
Vinil7 [7]
<h3>Answer:</h3>

Element X is Arsenic (Ar)

<h3>Explanation:</h3>

We are given the compound X₂O₃

The compound is 24.26% Oxygen

We are required to identify element X

<h3>Step 1: Determine the mass of Oxygen in the compound </h3>

Atomic mass of Oxygen is 16.0 g

There are three oxygen atoms in the compound

Therefore;

Mass of oxygen in the compound = 3 × 16

                                                        = 48 g

<h3>Step 2: Total mass of the compound </h3>

Total percentage = 100%

But, Oxygen in the compound  is 24.26 % and has a mass of 48 g

Therefore;

Mass of the compound = (48 g ÷ 24.26 ) × 100

                                      = 197.857 g

<h3>Step 3 mass of X in the compound </h3>

Since we know the mass of oxygen in the compound and the mass of the compound we can determine the mass of X

Mass of X in the compound = Mass of the compound - Mass of Oxygen in the compound

                                               = 197.857 g - 48 g

                                               = 149.857 g

But there are two atoms of X in the compound.

Therefore, molar mass of X = 149.857 g ÷ 2

                                             = 74.929 g

<h3>Step 4: Identity of element X</h3>

We then we need to identify the element with a molar mass of 74.929 g

Therefore, the element represented above is Arsenic which has an atomic mass of 74.92.

Thus, our compound is Ar₂O₃

5 0
3 years ago
HELP
Semenov [28]

C, 0.746 mol Ag.

1 mol Ag = 6.022 x 1023 atoms of Ag -> 4.49 x 1023 atoms of Ag x 1mol Ag/6.022 x 1023 atoms -> 0.746 mol Ag

7 0
3 years ago
A sample of xenon occupies a volume of 736 mL at 2.02 atm and 1 °C. If the volume is changed to 416 mL and the temperature is ch
kozerog [31]

Answer:

\large \boxed{\text{4.63 atm}}

Explanation:

To solve this problem, we can use the Combined Gas Laws:

\dfrac{p_{1}V_{1} }{n_{1}T_{1}} = \dfrac{p_{2}V_{2} }{n_{2}T_{2}}

Data:

p₁ = 2.02 atm; V₁ = 736 mL; n₁ = n₁; T₁ =    1 °C

p₂ = ?;             V₂ = 416 mL; n₂ = n₁; T₂ =  82 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = (   1 + 273.15) K = 274.15 K

T₂ = (82 + 273.15) K = 355.15 K

(b) Calculate the new pressure

\begin{array}{rcl}\dfrac{p_{1}V_{1}}{n_{1} T_{1}} & = & \dfrac{p_{2}V_{2}}{n_{2} T_{2}}\\\\\dfrac{\text{2.02 atm}\times \text{736 mL}}{n _{1}\times \text{274.15 K}} & = &\dfrac{p_{2}\times \text{416 mL}}{n _{1}\times \text{355.15 K}}\\\\\text{5.423 atm} & = &1.171{p_{2}}\\p_{2} & = & \dfrac{\text{5.423 atm}}{1.171}\\\\ & = & \textbf{4.63 atm} \\\end{array}\\\text{The  new pressure will be $\large \boxed{\textbf{4.63 atm}}$}}

6 0
3 years ago
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