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just olya [345]

1 year ago

5

When solutions of silver nitrate and sodium chloride are mixed, silver chloride precipitates out of the solution according to th

e equation:
AgNO3(aq)+NaCl(aq)→AgCl(s)+NaNO3(aq)

mass of AgCl =46.6 g

The reaction described in Part A required 3.95L of sodium chloride. What is the concentration of this sodium chloride solution?

2 answers:

Luba_88 [7]1 year ago

7 0

The concentration of sodium chloride would be 0.082 M.

<h3>Define the molarity of a solution.</h3>

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.

Stoichiometric calculations

From the equation of the reaction,

$AgNO_3(aq)+NaCl(aq)$ → $AgCl(s)+NaNO_3(aq)$

The ratio of AgCl produced to NaCl required is 1:1.

Mole of 46.6 g AgCl produced = $\frac{46.6}{43.32}$ = 0.325 moles

Equivalent mole of NaCl = 0.325 moles.

Molarity of 0.325 moles, 3.95 L NaCl

Molarity = $\frac{Moles \;solute}{Volume}$

Molarity = $\frac{0.325}{3.95}$

Molarity = 0.082 M

More on stoichiometric calculations can be found here:

brainly.com/question/27287858

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kipiarov [429]1 year ago

3 0

The concentration of the sodium chloride would be 0.082 M

<h3>Stoichiometric calculations</h3>

From the equation of the reaction, the ratio of AgCl produced to NaCl required is 1:1.

Mole of 46.6 g AgCl produced = 46.6/143.32 = 0.325 moles

Equivalent mole of NaCl = 0.325 moles.

Molarity of 0.325 moles, 3.95 L NaCl = mole/volume = 0.325/3.95 = 0.082 M

More on stoichiometric calculations can be found here: brainly.com/question/27287858

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Korolek [52]

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<h3>How to determine the volume occupied by the gas in a balloon </h3>

Let suppose that <em>flammable</em> hydrogen behaves ideally. GIven the molar mass ( $M$ ), in kilograms per kilomole, and mass of the gas ( $m$ ), in kilograms. The volume occupied by the gas ( $V$ ), in cubic centimeters, is found by the equation of state for <em>ideal</em> gases:

$V = \frac{m\cdot R_{u}\cdot T}{P\cdot M}$ (1)

Where:

$R_{u}$ - Ideal gas constant, in kilopascal-cubic meters per kilomole-Kelvin.
$T$ - Temperature, in Kelvin
$P$ - Pressure, in kilopascals

If we know that $m = 0.239\,kg$ , $R_{u} = 8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K}$ , $T = 273.15\,K$ , $P = 101.325\,kPa$ and $M = 2.02\,\frac{kg}{kmol}$ , then the volume of the balloon is:

$V = \frac{(0.239\,kg)\cdot \left(8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K} \right)\cdot (273.15\,K)}{\left(101.325\,kPa\right)\cdot \left(2.02\,\frac{kg}{kmol} \right)}$

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The volume of the balloon is approximately 2652 liters. $\blacksquare$

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4 0

1 year ago

The combustion of ethene in the presence of excess oxygen yields carbon dioxide and water: c2h4 (g) + 3o2 (g) → 2co2 (g) + 2h2o

meriva

Answer:

$\boxed{-267.5}$

Explanation:

You can calculate the entropy change of a reaction by using the standard molar entropies of reactants and products.

The formula is

$\Delta_{r} S^{\circ} = \sum_n {nS_{\text{products}}^{\circ} - \sum_{m} {mS_{\text{reactants}}^{\circ}}}$

The equation for the reaction is

C₂H₄(g) + 3O₂(g) ⟶ 2CO₂(g) + 2H₂O(ℓ)

ΔS°/J·K⁻¹mol⁻¹ 219.5 205.0 213.6 69.9

$\Delta_{r} S^{\circ} = (2\times213.6 + 2\times69.9) - (1\times219.5 + 3\times205.0)\\\\= 567.0 - 834.5 = \boxed{-267.5 \text{ J}\cdot\text{K}^{-1} \text{mol}^{-1}}$

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3 years ago

What physical effect can change the boiling point of a substance?

Artemon [7]

Pressure can affect the boiling pressure of a substance

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2 years ago

Read 2 more answers

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Ksju [112]

Answer:

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The enthalpy of fusion of solid n-butane is 4.66 kJ/mol. Calculate the energy required to melt 58.3 g of solid n-butane.

adelina 88 [10]

Answer : The energy required to melt 58.3 g of solid n-butane is, 4.66 kJ

Explanation :

First we have to calculate the moles of n-butane.

$\text{Moles of n-butane}=\frac{\text{Mass of n-butane}}{\text{Molar mass of n-butane}}$

Given:

Molar mass of n-butane = 58.12 g/mole

Mass of n-butane = 58.3 g

Now put all the given values in the above expression, we get:

$\text{Moles of n-butane}=\frac{58.3g}{58.12g/mol}=1.00mol$

Now we have to calculate the energy required.

$Q=\frac{\Delta H}{n}$

where,

Q = energy required

$\Delta H$ = enthalpy of fusion of solid n-butane = 4.66 kJ/mol

n = moles = 1.00 mol

Now put all the given values in the above expression, we get:

$Q=\frac{4.66kJ/mol}{1.00mol}=4.66kJ$

Thus, the energy required to melt 58.3 g of solid n-butane is, 4.66 kJ

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