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kondaur [170]
3 years ago
5

in the reaction Mg (s) + 2HCl (aq) -> H2 (g) + MGCl (aq), how many moles of hydrogen gas will be produced from 75.0 millilite

rs of 1.0 M HCL in an access of Mg?
Chemistry
1 answer:
cricket20 [7]3 years ago
7 0
Step One
Find the moles of HCl
C = 1.0 mol/L
V = 75 mL = 75 / 1000 = 0.75 L 
Note the concentration is in Liters.

<em>Formula</em>
n = C*V 

<em>Solve</em>
n = 1.0 * 0.075 
n = 0.075 moles

Step two 
Adjust the number of moles of HCl
How every mol of H2 produced, you will require 2 mol of HCl

\frac{2 mol HCl}{1 mol H_2} =  \frac{0.075\text{mol of HCl given}}{x}

2/1 = 0.075/x  Cross multiply
2x = 0.075      Divide by 2
x = 0.075 / 2
x = 0.0375      moles of H2  <<<<<<<<< Answer
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find the volume of an item using a scale and waterA crown made of alloy of gold and silver has a volume of 60cm3and mass of 1050
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Answer:

The mass of gold in crown is 960.61 g.And  the volume of gold is 49.77 cm^3.

Explanation:

Density is defined as mass of the substance present in unit volume of the substance.

Mass of gold in an alloy of crown= m

Mass of silver in an alloy of crown = M

Volume  gold in an alloy of crown= v

Volume of silver in an alloy of crown = V

Density of gold = 19.3 g/cm^3

Density of silver = 10.5 g/cm^3

Volume of the crown = 60 cm^3

60 cm^3=v+V

60 cm^3=\frac{m}{19.3 g/cm^3}+\frac{M}{10.5 g/cm^3}..(1)

Mass of eh crown = 1050 g

1050g =m+M..(2)

Solving equation (1) and (2):

m = 960.61 g

M = 89.39 g

Volume of the gold = v = \frac{960.61 g}{19.3 g/cm^3}=49.77 cm^3

Volume of silver = V = 60 cm^3-v=60 cm63-49.77 cm63=10.23 cm^3

The mass of gold in crown is 960.61 g.And  the volume of gold is 49.77 cm^3.

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