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2 years ago

How many moles of oxygen must be placed

Chemistry

2 answers:

aleksklad [387]2 years ago

4 0

Answer:

0.245 on Ed

Explanation:

Annette [7]2 years ago

3 0

0.24 moles of oxygen must be placed in a 3.00 L container to exert a pressure of 2.00 atm at 25.0°C.

The variables given are Pressure, volume and temperature.

Explanation:

Given:

P = 2 atm

V = 3 litres

T = 25 degrees or 298.15 K by using the formula 25 + 273.17 = K

R = 0.082057 L atm/ mole K

n (number of moles) = ?

The equation used is of Ideal Gas law:

PV = nRT

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

Putting the values given for oxygen gas in the Ideal gas equation, we get

n = $\frac{2 x 3}{0.082157 x 298.15}$

= 0.24

Thus, from the calculation using Ideal Gas law it is found that 0.24 moles of oxygen must be placed in a container.

Ideal gas law equation is used as it tells the relation between temperature, pressure and volume of the gas.

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Calculate the pH of a solution that contains 2.7 M HF and 2.7 M HOC6H5. Also, calculate the concentration of OC6H5- in this solu

borishaifa [10]

Answer:

$\large \boxed{\mathbf{1.36; 3.6 \times 10^{\mathbf{-9}}}\textbf{mol/L}}$

Explanation:

The HF is about five million times as strong as phenol, so it will be by far the major contributor of hydronium ions. We can ignore the contribution from the phenol.

1 .Calculate the hydronium ion concentration

We can use an ICE table to organize the calculations.

HF + H₂O ⇌ H₃O⁺ + F⁻

I/mol·L⁻¹: 2.7 0 0

C/mol·L⁻¹: -x +x +x

E/mol·L⁻¹: 2.7 - x x x

$K_{\text{a}} = \dfrac{\text{[H}_{3}\text{O}^{+}] \text{F}^{-}]} {\text{[HF]}} = 7.2 \times 10^{-4}\\\\\dfrac{x^{2}}{2.7 - x} = 7.2 \times 10^{-4}\\\\\text{Check for negligibility of }x\\\\\dfrac{2.7}{7.2 \times 10^{-4}} = 4000 > 400\\\\\therefore x \ll 2.7\\\dfrac{x^{2}}{2.7} = 7.2 \times 10^{-4}\\\\x^{2} = 2.7 \times 7.2 \times 10^{-4} = 1.94 \times 10^{-3}\\x = 0.0441\\\text{[H$_{3}$O$^{+}$]}= \text{x mol$\cdot$L$^{-1}$} = \text{0.0441 mol$\cdot$L$^{-1}$}$

2. Calculate the pH

$\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{0.0441} = \large \boxed{\mathbf{1.36}}$

3. Calculate [C₆H₅O⁻]

C₆H₅OH + H₂O ⇌ C₆H₅O⁻ + H₃O⁺

2.7 x 0.0441

$K_{\text{a}} = \dfrac{0.0441x} {2.7} = 1.6 \times 10^{-10}\\\\0.0441x = 1.6 \times 10^{-10}\\x = \dfrac{1.6 \times 10^{-10}}{0.0441} = \mathbf{3.6 \times 10^{\mathbf{-9}}}\textbf{ mol/L}\\\text{The concentration of phenoxide ion is $\large \boxed{\mathbf{3.6 \times 10^{\mathbf{-9}}}\textbf{ mol/L}}$}$

6 0

3 years ago

Is combining water and powdered drink mix a chemical reaction?

VikaD [51]

Answer:

No, it's a physical reaction.

Explanation:

A chemical change produces new chemical compounds, but combining water and powder is just mixing the powder with the water. It's not a new compound.

I don't know how to really explain, sorry :)

3 0

1 year ago

If 8.500 g CH is burned and the heat produced from the burning is added to 5691 g of water at 21 °C, what is the final

mojhsa [17]

The final temperature = 36 °C

<h3>Further explanation</h3>

The balanced combustion reaction for C₆H₆

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MW C₆H₆ : 78.11 g/mol

mol C₆H₆ :

$\tt \dfrac{8.5}{78.11}=0.109$