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abruzzese [7]
3 years ago
9

Please help with all 3 parts!

Chemistry
2 answers:
Murljashka [212]3 years ago
8 0

Answer:

1:Part A.

\bold{42.2 g C_{12}H_{22}O_{11} \:in \:528 g H₂O}

Mass Percent=\bold{\frac{Mass\: of \:Solute}{Mass\: of \:Solution}×100\%}

=\frac{42.2}{528}*100\%=\bold{\underline{7.99\: or \:8\%}}

Part B.

\bold{198\:m g\: C_{6}H_{12}O_{6} \:in\:4.71 g\: H₂O}

mass of solute: 198mg

mass of solvent :4.71g=4710g

Mass Percent=\bold{\frac{Mass\: of \:Solute}{Mass\: of \:Solution}×100\%}

=\frac{198}{4710}*100\%=\bold{\underline{4.20\%}}

Part C.

\bold{8.85 g NaCl \:in \:190 g\: H₂O}

Mass Percent=\bold{\frac{Mass\: of \:Solute}{Mass\: of \:Solution}×100\%}

=\frac{8.85}{190}*100\%=\bold{\underline{4.66\%}}

Alecsey [184]3 years ago
6 0

Answer:

It will help you !!!!!!!!!!

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3 years ago
Measure out 2.87 moles of sodium chloride (Nacl) into a clean dry cup. ​
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6 0
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4 0
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A 31.1 g wafer of pure gold, initially at 69.3 _c, is submerged into 64.2 g of water at 27.8 _c in an insulated container. what
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Given:
Ma = 31.1 g, the mass of gold
Ta = 69.3 °C, the initial temperature of gold
Mw = 64.2 g, the mass of water
Tw = 27.8 °C, the initial temperature of water 

Because the container is insulated, no heat is lost to the surroundings.
Let T °C be the final temperature.

From tables, obtain
Ca = 0.129 J/(g-°C), the specific heat of gold
Cw = 4.18 J/(g-°C), the specific heat of water

At equilibrium, heat lost by the gold - heat gained by the water.
Heat lost by the gold is
Qa = Ma*Ca*(T - Ta)
      = (31.1 g)*(0.129 J/(g-°C)(*(69.3 - T °C)- 
      = 4.0119(69.3 - T) j
Heat gained by the water is
Qw = Mw*Cw*(T-Tw)
       = (64.2 g)*(4.18 J/(g-°C))*(T - 27.8 °C)
       = 268.356(T - 27.8)

Equate Qa and Qw.
268.356(T - 27.8) = 4.0119(69.3 - T)
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Answer: 28.4 °C

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