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3 years ago

What is the molarity of 125 ml of solution containing 5.31 g sodium nitrate

Chemistry

1 answer:

quester [9]3 years ago

8 0

<h3>Answer:</h3>

0.5 moles/L or 0.5 M

<h3>Explanation:</h3>

Molarity is the concentration of a solution in moles per liter.
It is calculated by;

Molarity = Number of moles ÷ Volume

In this case;

Volume = 125 mL

Mass of sodium nitrate = 5.31 g

We are required to calculate the molarity of sodium nitrate

<h3>Step 1: Calculate the number of moles of sodium nitrate </h3>

Moles = Mass ÷ Molar mass

Molar mass of sodium nitrate = 84.9947 g/mol

Therefore,

Moles of NaNO₃ = 5.31 g ÷ 84.9947 g/mol

= 0.0625 moles

<h3>Step 2: Molarity of sodium nitrate </h3>

Remember;

Molarity = Moles ÷ Volume

Volume of NaNO₃ = 0.125 L

Therefore;

Molarity = 0.0625 moles ÷ 0.125 L

= 0.5 Moles/L or 0.5 M

Thus; the concentration of sodium nitrate is 0.5M

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A sample of potassium phosphate octahydrate (K3PO4•8H2O) is heated until 7.93 grams of water are released. How many grams did th

Gekata [30.6K]

The original sample of potassium phosphate octahydrate had a mass of 19.6 grams. When it was heated, it released 7.93 grams of water.

Further Explanation:

For every mole of the compound potassium phosphate octahydrate, there are 8 moles of water of hydration which can be removed from the crystal by heating without altering the chemical composition of the substance.

To determine how much original sample was used, the amount of water released upon heating may be used as well as the mole ratio of the water of hydration with the compound itself following the steps below:

Convert mass of water released to moles.
Use the mole ratio of water of hydration to the compound (8 mol water for every mol of potassium phosphate octahydrate) to get the moles of original sample.
Convert the moles of original sample to grams.

STEP 1: Convert 7.93 g water to moles.

$moles \ of\ H_{2}O \ = 7.93 \ g \ H_{2}O \ (\frac{1 \ mol \ H_{2}O}{18.00 \ g \ H_{2}O})\\\boxed {moles \ of \ H_{2}O \ = 0.4406 \ mol}$

STEP 2: Calculate the moles of original sample using the mole ratio: 1 mol K3PO4 8H2O : 8 mol H2O.

$moles \ of \ K_{3}PO_{4}\ 8H_{2}O \ = 0.4406 \ mol \ H_{2}O \ (\frac{1 \ mol \ K_{3}PO_{4}\ 8H_{2}O \ }{8 \ mol \ H_{2}O})\\\\\boxed {moles \ of \ K_{3}PO_{4}\ 8H_{2}O \ = 0.0551 \ mol}$

STEP 3: Convert the moles of original sample to mass.

$mass \ of \ K_{3}PO_{4}\ 8H_{2}O = 0.0551 \ mol \ K_{3}PO_{4}\ 8H_{2}O \ (\frac{356.3885 \ g}{1 \ mol\ K_{3}PO_{4}\ 8H_{2}O})\\ mass \ of \ K_{3}PO_{4}\ 8H_{2}O \ = 19.637 \ g$

Following the significant figures of the given, the final answer should be:

$\boxed {mass \ of \ K_{3}PO_{4}\ 8H_{2}O = 19.6 \ g}$

Learn More:

Learn more about water of hydration brainly.com/question/6053815
Learn more about mole conversion brainly.com/question/12979299
Learn more about percent hydrate brainly.com/question/12398621

Keywords: water of hydration, hydrate

4 0

3 years ago

Read 2 more answers

A 10.0-milliliter sample of NaOH(aq) is neutralized by 40.0 milliliters of 0.50 M HCl. What is the molarity of the NaOH(aq)?

Ludmilka [50]

$\frac{10 mL}{1000 mL} = 0.01 L$

$\frac{40mL}{1000mL} =0.04L$

$M _{1} V_{1}=M_{2}V_{2}$

$M_{1}(0.01)=(0.50)(0.04)$

$M_{1}(0.01)=0.02$

$M_{1}= \frac{0.02}{0.01}$

$M_{1}=2$

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4 years ago

What is the pressure in a 5.00 L tank with 23.50 grams of oxygen gas<br> at 350 K?

storchak [24]

Answer:

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Explanation:

4 0

4 years ago

The fertilizer ammonium sulfate, (nh4)2so4, is prepared by the reaction between ammonia (nh3) and sulfuric acid: 2nh3(g) + h2so4

iren2701 [21]

The chemical reaction would be:

<span>2nh3(g) + h2so4(aq) = (nh4)2so4(aq)
</span>
To determine the ammonia needed, we use the given amount of the product to be produced and use the chemical equation to relate the substances. We do as follows:

3.60 × 105 kg (nh4)2so4 ( 1 kmol / 132.14 kg ) ( 2 kmol NH3 / 1 kmol (nh4)2so4 ) (17.04 kg / 1 kmol ) = 92846.98 kg NH3 needed

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3 years ago

The first order rate constant for the conversion of cyclobutane to ethylene at 1000.Â°C is 87 1/s. Cyclobutane â 2 ethylene (a)

aliina [53]

Answer:

a) 7.96* 10⁻³ s

b) 0.0112 s

c) 1.645* 10⁻³⁸ g

Explanation:

for the reaction

Cyclobutane (A) → 2 ethylene (B)

the reaction rate (first order )is

-dCa/dt = k*Ca

∫dCa/Ca = - ∫k*dt

ln(Ca/Ca₀) = -k*t → Ca = Ca₀*e^(-k*t)

therefore

a) the half- life represents the time required for the concentration Ca to drop to half of the initial value ( Ca=Ca₀/2) therefore

Ca₀/2 = Ca₀*e^(-k*t) → - ln 2 = -k*t → t = ln 2 / k =ln 2 / ( 87 1/s) = 7.96* 10⁻³ s

b) Ca = Ca₀*e^(-k*t) , for Ca= Wa/(V*M) , where Wa is weight:

Wa = Wa₀*e^(-k*t)

for Wa₀= 4 g and Wa = 4g - 2.5 g = 1.5 g

→ t= (1/k)* ln(Wa₀/Wa) = 1/( 87 1/s) * ln [ 4g/(1.5 g)] = 0.0112 s

c) for Wa₀= 4 g and t=1 s

Wa = Wa₀*e^(-k*t) = 1 g * e^(- 87 (1/s) *1 s )= 1.645* 10⁻³⁸ g ≈ 0

8 0

3 years ago

What is the molarity of 125 ml of solution containing 5.31 g sodium nitrate​

What is the molarity of 125 ml of solution containing 5.31 g sodium nitrate