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ss7ja [257]
2 years ago
12

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

Chemistry
1 answer:
Brums [2.3K]2 years ago
7 0

<span>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 </span>6,02*10^{23} molecules, if chloroform contains 1,00 * 10^{12} molecules. <span> 
We have:
 
</span>1,00 * 10^{12} * 6,02 * 10^{23} = 6,02 * 10 ^ {35} molecules<span>
 
Now, how many grams are there in the chloroform molecule?

grams -------- molecules</span><span>
119,5 </span>→ 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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The pathway by which carbon is transferred from living organisms to the atmosphere is called:
Phoenix [80]

Answer:

cellular respiration

Explanation:

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Aerobic respiration  (oxidative phosphorylation)

- Use molecular O2.

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3 0
3 years ago
For the following word equations, write it as a chemical equation, then balance it.
Lorico [155]

4K+O2-----------2K2O

7 0
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

HA + H20. ----> H3O+ A-

Initially: C. 0. 0

After : C-Cx. Cx. Cx

Ka= [H3O+][A-]/[HA]

Ka= Cx × Cx/C-Cx

Ka= C²X²/C(1-x)

Ka= Cx²/1-x

Where x is degree of dissociation = 0.1% = 0.001 and c is the concentration =0.2

Ka= 0.2(0.001²)/(1-0.001)

Ka= 2.00×10^-7

Therefore the dissociation constant is

2.00×10^-7

7 0
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
vichka [17]

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

Mn(NO_3)_2\rightarrow Mn^{2+}+2NO_3^{-}

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

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

6 0
3 years ago
A. 1720 kJ<br> B. 125.6 kJ<br> C. 3440 kJ<br> D. 4730 kJ
Feliz [49]

Answer:

Q = 3440Kj

Explanation:

Given data:

Mass of gold = 2kg

Latent heat of vaporization = 1720 Kj/Kg

Energy required to vaporize 2kg gold = ?

Solution:

Equation

Q= mLvap

It is given that heat required to vaporize the one kilogram gold is 1720 Kj thus, for 2 kg

by putting values,

Q= 2kg ×  1720 Kj/Kg

Q = 3440Kj

7 0
3 years ago
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