Answer:
![\Delta T=1.286\°C](https://tex.z-dn.net/?f=%5CDelta%20T%3D1.286%5C%C2%B0C)
![m=2.51mol/kg](https://tex.z-dn.net/?f=m%3D2.51mol%2Fkg)
![n=0.0503mol](https://tex.z-dn.net/?f=n%3D0.0503mol)
![M=59.8g/mol](https://tex.z-dn.net/?f=M%3D59.8g%2Fmol)
Explanation:
Hello,
In case, for the boiling point raise we can write:
![T_2-T_1=imK_b](https://tex.z-dn.net/?f=T_2-T_1%3DimK_b)
Whereas T2 accounts for the boiling point of the solution which is 101.286 °C, T1 the volume of pure water which is 100.000 °C, i the van't Hoff factor that for this problem is 1 due to the solute's non-volatility, m the molality of the solute and Kb the boiling point constant that is 0.512 °C/m. In such a way, the change in the temperature is:
![\Delta T=T_2-T_1=101.286\°C-100.000\°C=1.286\°C](https://tex.z-dn.net/?f=%5CDelta%20T%3DT_2-T_1%3D101.286%5C%C2%B0C-100.000%5C%C2%B0C%3D1.286%5C%C2%B0C)
The molality is computed from the boiling point raise:
![m=\frac{T_2-T_1}{Kb}=\frac{1.286\°C}{0.512\°C/m}\\ \\m=2.51mol/kg](https://tex.z-dn.net/?f=m%3D%5Cfrac%7BT_2-T_1%7D%7BKb%7D%3D%5Cfrac%7B1.286%5C%C2%B0C%7D%7B0.512%5C%C2%B0C%2Fm%7D%5C%5C%20%20%5C%5Cm%3D2.51mol%2Fkg)
The moles are computed by multiplying the molality by the kilograms of water as the solvent (0.02002g):
![n=2.51mol/kg*0.02002kg\\\\n=0.0503mol](https://tex.z-dn.net/?f=n%3D2.51mol%2Fkg%2A0.02002kg%5C%5C%5C%5Cn%3D0.0503mol)
And the molecular mass by dividing the mass of the solute by its moles:
![M=\frac{3.005g}{0.0503mol}\\ \\M=59.8g/mol](https://tex.z-dn.net/?f=M%3D%5Cfrac%7B3.005g%7D%7B0.0503mol%7D%5C%5C%20%5C%5CM%3D59.8g%2Fmol)
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The planets speed as they orbit the sun
B) Morrison and Franscioni's research done to create the Frisbee.
Answer:
The mole is a key unit in chemistry. The molar mass of a substance, in grams, is numerically equal to one atom’s or molecule’s mass in atomic mass units.
Explanation:
I, m right?
Explanation:
Copper sulphate reacts with potassium iodide to form cuprous iodide and iodine.
2CuSO
4
+4KI→Cu
2
I
2
↓+I
2
+2K
2
SO
4
Thus, CuI
2
is not formed in this reaction.
Hence, the option B is incorrect and option A is correct.
The liberated iodine is titrated with sodium thiosulphate to form sodium tetrathionate.
2Na
2
S
2
O
3
+I
2
→Na
2
S
4
O
6
+2NaI
Iodine is reduced and sodium thiosulphate is oxidized.