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2 years ago

Calculate the mass of chloroform that contains 1.00X10^12 molecules of chloroform

1 answer:

7 0

Molar mass of chloroform (CHCl3)
C = 1 * 12 = 12 a.m.u
H = 1 * 1 = 1 a.m.u
Cl = 3 * 35.5 = 106.5 a.m.u
Molar Mass (CHCl3) = 12 + 1 + 106.5 = 119.5 g / mol

Knowing that: The value of Avogadro's constant corresponds to approximately $6,02*10^{23}$ molecules, if chloroform contains $1,00 * 10^{12}$ molecules. 
We have:

 $1,00 * 10^{12} * 6,02 * 10^{23} = 6,02 * 10 ^ {35} molecules$ 

Now, how many grams are there in the chloroform molecule?

grams -------- molecules
119,5 → $6,02 * 10^{23}$
x → $6,02 * 10^{35}$

$6,02 * 10 ^ {23} * x = 119,5 * 6,02 * 10 ^{35}$
$6,02 * 10 ^ {23}x = 719,39 * 10^{35}$
$x = \frac{719,39}{6,02}*10^{35-23}$
$\boxed{x = 119,5*10^{12}grams}$

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3 years ago

Calculate the acid dissociation constant Ka of a 0.2 M solution of weak acid that is 0.1% ionized is ________.

mars1129 [50]

Answer: acid dissociation constant Ka= 2.00×10^-7

Explanation:

For the reaction

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2 years ago

Match the following aqueous solutions with the appropriate letter from the column on the right.1. 0.19 m AgNO3 2. 0.17 m CrSO4 3

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Answer:

0.13 m of $Mn(NO_3)_2$ → Highest boiling point

0.19 m of $AgNO_3$ → Second Highest boiling point

0.17 m of $CrSO_4$ → Third highest boiling point

0.31 m Sucrose (nonelectrolyte) → Lowest boiling point

Explanation:

Elevation in boiling is given by :

$\Delta T_b=i\times k_b\times m$

Where :

i = van't Hoff factor

$k_b$ = Molal Elevation constant of solvent

m = molaity of the solution

1) 0.19 m of $AgNO_3$

$AgNO_3\rightarrow Ag^++NO_3^{-}$

i = 2 (electrolyte)

Molality of the solution = 0.19

Elevation is boiling point of solution:

$\Delta T_b=2\times k_b\times 0.19 m$

$\Delta T_b=0.38 m\times k_b$

2) 0.17 m of $CrSO_4$

$CrSO_4\rightarrow Cr^{2+}+SO_4^{2-}$

i = 2 (electrolyte)

Molality of the solution = 0.17

Elevation is boiling point solution :

$\Delta T_b=2\times k_b\times 0.17 m$

$\Delta T_b=0.34 m\times k_b$

3) 0.13 m of $Mn(NO_3)_2$

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i = 3 (electrolyte)

Molality of the solution = 0.13

Elevation is boiling point solution :

$\Delta T_b=3\times k_b\times 0.13 m$

$\Delta T_b=0.39 m\times k_b$

4) 0.31 m Sucrose (nonelectrolyte)

i = 1 ( non electrolyte)

Molality of the solution = 0.31 m

Elevation is boiling point solution :

$\Delta T_b=1\times k_b\times 0.31 m$

$\Delta T_b=0.31 m\times k_b$

Higher the value of elevation in temperature higher will be the boiling point of the solution .

The decreasing order of solution from highest boiling point to lowest boiling point is :

$0.39 m\times k_b>0.38 m\times k_b>0.34 m\times k_b>0.31 m\times k_b$

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3 years ago

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Answer:

Q = 3440Kj

Explanation:

Given data:

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Energy required to vaporize 2kg gold = ?

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Equation

Q= mLvap

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Q= 2kg × 1720 Kj/Kg

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3 years ago