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hoa [83]
3 years ago
9

Calculate the number of moles of O2 gas held in a sealed 2.00 L tank at 3.50 atm and 25 ℃.

Chemistry
1 answer:
Leviafan [203]3 years ago
6 0

Answer:

n=0.286mol

Explanation:

Hello,

In this case, we consider oxygen as an ideal gas, for that reason, we use yhe ideal gas equation to compute the moles based on:

PV=nRT\\\\n=\frac{PV}{RT}

Hence, at 3.50 atm and 25 °C for a volume of 2.00 L we compute the moles considering absolute temperature in Kelvins:

n=\frac{3.50atm*2.00L}{0.082\frac{atm*L}{mol*K}(25+273)K} \\\\n=0.286mol

Best regards.

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Calculate the density of argon gas at a pressure of 753 mmHg and a temperature of 35 °C
ycow [4]
PV=nRT
PV= (m/M)RT
PM=(m/v)RT
PM =dRT
d= (PM) ÷(RT)

substitute the the given data to the above equation. d-density, p-pressure, M- molar mass of argon gas, T- temperature. and R is a constant. make sure to substitute its value too.. :)
3 0
3 years ago
A 2.00 L sample of gas at 35C is to be heated at constant pressure until it reaches a volume of 5.25 L. To what Kelvin temperat
love history [14]

Answer:

When the volume increased from 2.00 to 5.25L the new temperature is 808.9 K ( =535.75 °C)

Explanation:

Step 1: Data given

The initial volume of the sample = 2.00 L

The initial temperature = 35 °C =  308 K

The increased volume = 5.25 L

Pressure = constant

Step 2: Calculate the new temperature

V1/T1 = V2/T2

⇒ with V1 = the initial volume = 2.00 L

⇒ with T1 = the initial volume = 308 K

⇒ with V2 = the new volume = 5.25 L

⇒ with T2 = the new temperature

2.00 / 308 = 5.25 / T2

0.00649 = 5.25/T2

T2 = 5.25/ 0.00649

T2 = 808.9 K

When the volume increased from 2.00 to 5.25L the new temperature is 808.9 K ( =535.75 °C)

7 0
2 years ago
A chemist dissolves 716.mg of pure potassium hydroxide in enough water to make up 130.mL of solution. Calculate the pH of the so
Natasha_Volkova [10]

<u>Answer:</u> The pH of the solution is 13.0

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Given mass of KOH = 716. mg = 0.716 g    (Conversion factor:  1 g = 1000 mg)

Molar mass of KOH = 56 g/mol

Volume of solution = 130 mL

Putting values in above equation, we get:

\text{Molarity of solution}=\frac{0.716\times 1000}{56g/mol\times 130}\\\\\text{Molarity of solution}=0.098M

1 mole of KOH produces 1 mole of hydroxide ions and 1 mole of potassium ions

  • To calculate hydroxide ion concentration of the solution, we use the equation:

pOH=-\log[OH^-]

We are given:

[tex[[OH^-]=0.098M[/tex]

Putting values in above equation, we get:

pOH=-\log(0.098)\\\\pOH=1.00

To calculate the pH of the solution, we use the equation:

pOH + pH = 14

So,  pH = 14 - 1.00 = 13.0

Hence, the pH of the solution is 13.0

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3 years ago
Select the true characteristics of respiration.
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Answer:

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can produce alcohol

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