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just olya [345]
3 years ago
13

Two gas samples have the same number of molecules, the same volume, and the same pressure. Which of the following could be true?

A. The first gas sample has a temperature of 273 K, and the second gas sample has a temperature of 0o C. B. The first gas sample has a temperature of 273 K, and the second gas sample has a temperature of 298 K. C. The first gas sample has a temperature of 273 K, and the second gas sample has a temperature of 273o C. D. The first gas sample has a temperature of 273 K, and the second gas sample has a temperature of 100o C.
Chemistry
2 answers:
IRISSAK [1]3 years ago
4 0
The correct answer is (A): The first gas sample has a temperature of 273 K, and the second gas sample has a temperature of 0 C.

bulgar [2K]3 years ago
3 0
Lets name one gas sample as A and other gas sample as B.
we can apply ideal gas law equation for both samples
PV = nRT
P - Pressure of A = Pressure of B
V - volume of A = volume of B
n  - number of molecules of both A and B being equal is equivalent to number of moles of A = number of moles of B
R - universal gas constant
Tᵃ - temperature of A
Tᵇ - temperature of B
for gas A
PV = nRTᵃ  --1)
for gas B
PV = nRTᵇ  ---2)
when we divide both equations 
1 = Tᵃ  / Tᵇ
Tᵃ = Tᵇ
both temperatures are equal 
temperature in Celsius + 273 = temperature in Kelvin
therefore 0 °C = 273 K 
the correct answer is 
A)
<span>The first gas sample has a temperature of 273 K, and the second gas sample has a temperature of 0 </span>°<span>C</span>
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133.74 L

Explanation:

First we <u>convert the given pressures and temperatures into atm and K</u>, respectively:

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  • 40°C ⇒ 40+273.16 = 313.16 K

Then we<u> use the PV=nRT formula to calculate the number of moles of helium in the balloon</u>, using<em> the data of when it was on the ground</em>:

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Then, knowing the value of n, we <u>use PV=nRT once again, this time to calculate V</u> using <em>the data of when the balloon was high up:</em>

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5 0
3 years ago
A True
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g What volume (mL) of the partially neutralized stomach acid was neutralized by NaOH during the titration? (portion of 25.00 mL
Pani-rosa [81]

The question is incomplete, here is the complete question:

What volume (mL) of the partially neutralized stomach acid having concentration 2M was neutralized by 0.01 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)

<u>Answer:</u> The volume of HCl neutralized is 0.125 mL

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl (Stomach acid)

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=2M\\V_1=?mL\\n_2=1\\M_2=0.01M\\V_2=25mL

Putting values in above equation, we get:

1\times 2\times V_1=1\times 0.01\times 25\\\\V_1=\frac{1\times 0.01\times 25}{1\times 2}=0.125mL

Hence, the volume of HCl neutralized is 0.125 mL

3 0
4 years ago
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Answer:

0.43×10²³ atoms

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0.43×10²³ atoms

3 0
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