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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?

Chemistry
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

#SPJ1

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

#SPJ1

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<u>Answer:</u> 3 neutrons are produced in the above nuclear reaction.

<u>Explanation:</u>

In a nuclear reaction, the total mass and total atomic number remains the same.

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^{244}_{95}\textrm{Am}\rightarrow ^{134}_{53}\textrm{I}+^{107}_{42}\textrm{Mo}+Y^1_0\textrm{n}

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For the following reaction, 4.21 grams of hydrogen gas are allowed to react with 10.6 grams of ethylene (C2H4) . hydrogen(g) + e
sergiy2304 [10]

Answer:a)  11.34 g of ethane (C_2H_6) can be formed

b) C_2H_4 is the limiting reagent

c) 3.44 g of the excess reagent remains after the reaction is complete

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}    

1. \text{Moles of} H_2=\frac{4.21}{2}=2.10moles

2. \text{Moles of} C_2H_4=\frac{10.6}{28}=0.378moles

H_2(g)+C_2H_4(g)\rightarrow C_2H_6(g)

According to stoichiometry :

1 mole of C_2H_4 require 1 mole of H_2

Thus 0.378 moles of C_2H_4 will require=\frac{1}{1}\times 0.378=0.378moles  of H_2

Thus C_2H_4 is the limiting reagent as it limits the formation of product and H_2 is the excess reagent.

moles of H_2 left = (2.10-0.378) = 1.72 moles

mass of H_2 left=moles\times {\text {Molar mass}}=1.72moles\times 2g/mol=3.44g

According to stoichiometry :

As 1 mole of C_2H_4 give = 1 mole of C_2H_6

Thus 0.378 moles of C_2H_4 give =\frac{1}{1}\times 0.378=0.378moles  of C_2H_6

Mass of C_2H_6=moles\times {\text {Molar mass}}=0.378moles\times 30g/mol=11.34g

Thus 11.34 g of ethane is formed.

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