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Nezavi [6.7K]
3 years ago
12

A 0.3146-g sample of a mixture of nacl(s) and kbr(s) was dissolved in water. the resulting solution required 55.00 ml of 0.08765

m agno3(aq) to precipitate the cl–(aq) and br–(aq) as agcl(s) and agbr(s). calculate the mass percentage of nacl(s) in the mixture.
Chemistry
1 answer:
dlinn [17]3 years ago
4 0

The balanced chemical equation representing the reaction of NaCl with AgNO3 is,

NaCl (aq) + AgNO_{3}(aq) --> AgCl (s) + NaNO_{3}(aq)

The balanced chemical equation representing the reaction of KBr with AgNO3 is,

KBr (aq) + AgNO_{3}(aq) --> AgBr(s) + KNO_{3}(aq)

Moles of AgNO_{3} = 0.08765 \frac{mol}{L} * 55.0 mL * \frac{1 L}{1000 mL} =   0.00482075 mol AgNO_{3}

0.00482075 mol AgNO_{3} completely removes Chlorides and bromides in the sample.

Mole ratio of Ag^{+}to Cl^{-} is 1:1

Mole ratio of Ag^{+}to Br^{-} is 1:1

So, the total moles of chloride and bromide in the sample = 0.00482075 mol

Let mass of NaCl be x g

Mass of KBr be y g

Total mass of sample = 0.3146 g

=> x + y = 0.3146 g

x = 0.3146 -y

Total number of moles of NaCl + KBr = 0.00482075 mol

\frac{x}{58.44g/mol} +\frac{y}{119.0g/mol}= 0.00482075 mol

\frac{0.3146-y}{58.44g/mol} +\frac{y}{119.0 g/mol} = 0.00482075 mol

\frac{119(0.3146-y) + 58.44y}{6954.36}=0.00482075

y = 0.0646

Therefore mass of KBr = 0.0646 g

Mass of NaCl = 0.3146 - 0.0646 g = 0.250 g

Mass % of NaCl in the sample = \frac{0.250 g NaCl}{0.3146 g sample} * 100 = 79.5 %


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Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

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In this case, you know:

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  • Specific heat of metal= ?

For water:

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Replacing in the expression to calculate heat exchanges:

For metal: Qmetal= Specific heat of metal× 50 g× (11.08 C - 45 C)

For water: Qwater=  1.035 \frac{cal}{gC} × 250 g× (11.08 C - 10 C)

If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.

Then, the heat that the gold gives up will be equal to the heat that the water receives. Therefore:

- Qmetal = + Qwater

- Specific heat of metal× 50 g× (11.08 C - 45 C)= 1.035 \frac{cal}{gC} × 250 g× (11.08 C - 10 C)

Solving:

- Specific heat of metal× 50 g× (-33.92 C)= 1.035 \frac{cal}{gC} × 250 g× 1.08 C

Specific heat of metal× 1696 g×C= 279.45 cal

Specific heat of metal= \frac{279.45 cal}{1696 gC}

<u><em>Specific heat of metal= 0.165 </em></u>\frac{cal}{gC}

Finally, the specific heat of metal is 0.165 \frac{cal}{gC}.

Learn more about calorimetry:

brainly.com/question/11586486

brainly.com/question/24724338

brainly.com/question/14057615

brainly.com/question/24988785

#SPJ1

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