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Nataliya [291]
3 years ago
13

How does atomic mass differ from atomic number

Chemistry
2 answers:
valentinak56 [21]3 years ago
5 0
One is weight and one is amount
Lina20 [59]3 years ago
3 0
The atomic mass is the average mass of all the isotopes. The atomic number is the number of protons in the nucleus of an atom. In an uncharged atom the atomic number is also equal to the number of electrons. 

Hope this helps!
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Choose the number of significant figures indicated. 0.06
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The space occupied by any sample of matter is called its______.
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if you react 100g of ammonium chloride with excess calcium oxide, what Is the theoretical yeild (in grams) of ammonia? when you
Maurinko [17]

Answer:

31.78 grams

25.55%

Explanation:

The balanced reaction for ammonium chloride with calcium oxide will be:

2NH4Cl + Ca(OH)2 ---> CaCl2 + 2NH3 + 2H2O

The molecular weight for ammonium chloride(NH4Cl ) is 53.49g/mol, while the molecular weight for ammonium(NH3) is 17g/mol. The number of theoretical yield of ammonia from 100g of ammonium chloride will be:

100g / (53.49g/mol) * 2/2  * 17g/mol= 31.78 grams

If the actual yield is 8.12g, the percent yield will be: 8.12g/31.78g * 100% =25.55%

4 0
3 years ago
Consider the reaction. 2 HBr(g) ¡ H2(g) + Br2(g) a. Express the rate of the reaction in terms of the change in concentration of
Studentka2010 [4]

Answer :

(A) The rate expression will be:

Rate=-\frac{1}{2}\frac{d[HBr]}{dt}=+\frac{d[H_2]}{dt}=+\frac{d[Br_2]}{dt}

(B) The average rate of the reaction during this time interval is, 0.00176 M/s

(C) The amount of Br₂ (in moles) formed is, 0.0396 mol

Explanation :

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The given rate of reaction is,

2HBr(g)\rightarrow H_2(g)+Br_2(g)

The expression for rate of reaction :

\text{Rate of disappearance of }HBr=-\frac{1}{2}\frac{d[HBr]}{dt}

\text{Rate of disappearance of }H_2=+\frac{d[H_2]}{dt}

\text{Rate of formation of }Br_2=+\frac{d[Br_2]}{dt}

<u>Part A:</u>

The rate expression will be:

Rate=-\frac{1}{2}\frac{d[HBr]}{dt}=+\frac{d[H_2]}{dt}=+\frac{d[Br_2]}{dt}

<u>Part B:</u>

\text{Average rate}=-\frac{1}{2}\frac{d[HBr]}{dt}

\text{Average rate}=-\frac{1}{2}\frac{(0.512-0.600)M}{(25.0-0.0)s}

\text{Average rate}=0.00176M/s

The average rate of the reaction during this time interval is, 0.00176 M/s

<u>Part C:</u>

As we are given that the volume of the reaction vessel is 1.50 L.

\frac{d[Br_2]}{dt}=0.00176M/s

\frac{d[Br_2]}{15.0s}=0.00176M/s

[Br_2]=0.00176M/s\times 15.0s

[Br_2]=0.0264M

Now we have to determine the amount of Br₂ (in moles).

\text{Moles of }Br_2=\text{Concentration of }Br_2\times \text{Volume of solution}

\text{Moles of }Br_2=0.0264M\times 1.50L

\text{Moles of }Br_2=0.0396mol

The amount of Br₂ (in moles) formed is, 0.0396 mol

8 0
3 years ago
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