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3 years ago

How many moles of gas does it take to occupy 120 L at a pressure of 233 kPa and a temperature of 340 K?

Chemistry

2 answers:

damaskus [11]3 years ago

6 0

Answer:

The correct answer is option D.

Explanation:

Using ideal gas equation:

PV = nRT

where,

P = Pressure of gas = 233 kPa

1 atm = 101.325 kPa

$233kPa=\frac{233}{101.325} atm=2.30 atm$

V = Volume of gas = 120 L

n = number of moles of gas = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 340 K

Putting values in above equation, we get:

$2.30 atm\times 120L=n\times (0.0821L.atm/mol.K)\times 340K\\\\n=9.88 mole\approx 9.9 mole$

natka813 [3]3 years ago

5 0

A would be correct have a nice day.

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K = °C + 273 A 4.1 L sample of gas is held at 25 °C. If the gas expands to 6.8 L, what is the final temperature?

gregori [183]

Answer:

221 °C

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 4.1 L

Initial temperature (T₁) = 25 °C

= 25 °C + 273

= 298 K

Final volume (V₂) = 6.8 L

Final temperature (T₂) =?

The final temperature of the gas can be obtained as follow:

V₁ / T₁ = V₂ / T₂

4.1 / 298 = 6.8 / T₂

Cross multiply

4.1 × T₂ = 298 × 6.8

4.1 × T₂ = 2026.4

Divide both side by 4.1

T₂ = 2026.4 / 4.1

T₂ ≈ 494 K

Finally, we shall convert 494 K to celcius temperature. This can be obtained as follow:

°C = K – 273

K = 494

°C = 494 – 273

°C = 221 °C

Thus the final temperature of the gas is 221 °C

6 0

3 years ago

The air we breathe is not considered by chemists to be an element. why not

ohaa [14]

Because Air itself is not an element. it is made up by different elements such as Oxygen and nitrogen. since its made up of a mix of different elements, its rather a homogeneous mixture :)

7 0

3 years ago

Read 2 more answers

Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . Suppose 6.9 g

satela [25.4K]

Answer:

$5.6gNa_2SO_4$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2NaOH(aq)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)$

Therefore, since the masses of both of the reactants are given, one computes the available moles of sulfuric acid and those moles of it consumed by the sodium hydroxide as shown below:

$n_{H_2SO_4}^{available}=6.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.0704molH_2SO_4\\n_{H_2SO_4}^{consumed\ by\ NaOH}=3.14gNaOH*\frac{1molNaOH}{40gNaOH}*\frac{1molH_2SO_4}{2molNaOH}=0.04molH_2SO_4$

In such a way, since there is more available sulfuric acid than it that is consumed, the sodium hydroxide is the limiting reagent, consequently, the maximum mass of sodium sulfate turns out:

$m_{Na_2SO_4}=0.04molH_2SO_4*\frac{1molNa_2SO_4}{1molH_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4}=5.6gNa_2SO_4$

Best regards.

5 0

3 years ago

If you have 120. mL of a 0.100 M TES buffer at pH 7.55 and you add 3.00 mL of 1.00 M HCl, what will be the new pH? (The pKa of T

Simora [160]

Answer:

The new pH after adding HCl is 7.07

Explanation:

The formula for calculating pH of a buffer is

pH = pKa + log([Conjugate base]/[Acid])

<u>Before adding HCl,</u>

7.55 = 7.55 + log([Conjugate base]/[Acid])

⇔ log([Conjugate base]/[Acid]) = 0

⇔ [Conjugate base] = [Acid] = 1/2 x 0.100 = 0.05 M

⇒ Mole of Conjugate base = Mole of Acid = 0.05 M x 0.12 mL = 0.006 mol

<u>After adding HCl (3.00 mL, 1.00 M)</u>

⇒ Mole of HCl = 0.003 x 1 = 0.003 mol)

New volume solution is 120 m L+ 3 mL = 123 mL

HCl is a strong acid, it will convert the conjugate base to acid form, or we can express

Mole of new Conjugate base = 0.006 - 0.003 = 0.003 mol

⇒ Concentration = 0.003/0.123 M

Mole of new Acid form = 0.006 + 0.003 = 0.009 mol

⇒ Concentration = 0.009/0.123 M

Use the formula

pH = pKa + log([Conjugate base]/[Acid])

= 7.55 + log(0.003 / 0.009) = 7.07

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3 years ago

How many moles are in 155 grams of aluminum

Aliun [14]

Answer:

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Explanation:

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