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BigorU [14]
3 years ago
10

Suppose of sodium chloride is dissolved in of a aqueous solution of ammonium sulfate. Calculate the final molarity of sodium cat

ion in the solution. You can assume the volume of the solution doesn't change when the sodium chloride is dissolved in it. Round your answer to significant digits.
Chemistry
1 answer:
mariarad [96]3 years ago
5 0

Answer:

0.103 M.

Explanation:

The balanced equation of reaction is given below;

2NaCl + (NH4)2SO4 -----------------> Na2SO4 + 2NH4Cl

The parameters given from the question are; mass of Sodium chloride, NaCl = 0.197 g, the volume of ammonium sulphate is 100 mL = 0.1 litres, concentration of ammonium sulfate = 54.0mM = 0.054 mol/L.

Step one: Calculate the number of moles of NaCl by using the formula below;

Number of moles,n= mass/ mass number.

Number of moles,n = 0.917/ 58.5.

Number of moles, n= 0.0157 g/mol.

Step two: find the number of moles of ammonium sulfate.

Hence, the number of moles of ammonium sulfate, n = concentration (mol/L) of ammonium sulfate × volume.

Therefore, the number of moles of ammonium sulfate, n = 0.054 × 0.1.

the number of moles of ammonium sulfate, n = 0.0054 mol.

Step three: Calculate the excess moles of NaCl.

That is; 0.0157 - 0.0054 = 0.0103 mol.

Step four: Calculate the final molarity of sodium cation in the solution.

[Na^+] = number of moles/ total volume.

[Na^+] = 0.0103 mol/ 0.1.

0.0103 mol= 0.103 M.

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A 825 g iron block is heated to 352 degrees C and is placed in an insulated container (of negligible heat capacity) containing 4
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Answer : The final equilibrium temperature of the water and iron is, 537.12 K

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

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T_f = final temperature of water and iron = ?

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

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