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Dahasolnce [82]
2 years ago
12

What is the mass of a sample of NH3 containing 7.20 × 1024 molecules of NH3? 161 grams 187 grams 203 grams 214 grams

Chemistry
2 answers:
Shkiper50 [21]2 years ago
7 0

Answer:

203 grams

Explanation:

<em>It is known that 1.0 mole of a compound contains Avogadro's number of molecules (6.022 x 10²³). </em>

<em><u>Using cross multiplication:</u></em>

1.0 mol contains → 6.022 x 10²³ molecules.

??? mol contains → 7.2 x 10²⁴ molecules.

∴ The no. of moles of (6.3 x 10²⁴ molecules) of NH₃ = (1.0 mol)(7.2 x 10²⁴ molecules)/(6.022 x 10²³ molecules) = 11.96 mol.

<em>∴ The no. of grams of NH₃ present = no. of moles x molar mass </em>= (11.96 mol)(17.0 g/mol) = <em>203.3 g ≅ 203.0 g.</em>

inn [45]2 years ago
3 0

Answer:

this is correct^

Explanation:

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How many liters of radon gas would be in 3.43 moles at room temperature and pressure (293 K and 100 kPa)?
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Using ideal gas equation,

P\times V=n\times R\times T

Here,

P denotes pressure

V denotes volume

n denotes number of moles of gas

R denotes gas constant

T denotes temperature

The values at STP will be:

P=100 kPa

T=293 K

R=8.314472 L kPa K⁻¹ mol⁻¹

Number of moles of gas=3.43 mole

Putting all the values in the above equation,

V=\frac{3.43\times 8.314\times 293}{100}

V=83.55 L

So the volume will be 83.55 L.

83.55 L of radon gas would be in 3.43 moles at room temperature and pressure (293 K and 100 kPa).

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Answer:

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Explanation:

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14. What is the pH of a 0.24 M solution of sodium propionate, NaC3H502, at 25°C? (For
Murrr4er [49]

Answer:

9.1

Explanation:

Step 1: Calculate the basic dissociation constant of propionate ion (Kb)

Sodium propionate is a strong electrolyte that dissociates according to the following equation.

NaC₃H₅O₂ ⇒ Na⁺ + C₃H₅O₂⁻

Propionate is the conjugate base of propionic acid according to the following equation.

C₃H₅O₂⁻ + H₂O ⇄ HC₃H₅O₂ + OH⁻

We can calculate Kb for propionate using the following expression.

Ka × Kb = Kw

Kb = Kw/Ka = 1.0 × 10⁻¹⁴/1.3 × 10⁻⁵ = 7.7 × 10⁻¹⁰

Step 2: Calculate the concentration of OH⁻

The concentration of the base (Cb) is 0.24 M. We can calculate [OH⁻] using the following expression.

[OH⁻] = √(Kb × Cb) = √(7.7 × 10⁻¹⁰ × 0.24) = 1.4 × 10⁻⁵ M

Step 3: Calculate the concentration of H⁺

We will use the following expression.

Kw = [H⁺] × [OH⁻]

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Step 4: Calculate the pH of the solution

We will use the definition of pH.

pH = -log [H⁺] = -log 7.1 × 10⁻¹⁰ = 9.1

5 0
2 years ago
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