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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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goldfiish [28.3K]

Answer:

K = Ka/Kb

Explanation:

P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?

P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka

PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb

K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)

Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)

Kb = [PCl₅]/ ([PCl₃] [Cl₂])

Since [PCl₅] = [PCl₅]

From the Ka equation,

[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)

From the Kb equation

[PCl₅] = Kb ([PCl₃] [Cl₂])

Equating them

Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])

(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)

(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)

Comparing this with the equation for the overall equilibrium constant

K = Ka/Kb

5 0
3 years ago
The Symbol below is used to represent what component in a circuit?
FrozenT [24]
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8 0
3 years ago
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7 0
3 years ago
How many molecules of CF₂Cl₂ are in 45.7 grams of CF₂Cl₂? (Show work)
posledela

Answer:

The answer to your question is: 2.20 x 10 ²³ molecules

Explanation:

Data

mass = 45.7 g

molecules of CF₂Cl₂ = ?

Process

1.- Calculate the mass number of CF₂Cl₂

   C = 12    F =2 x 19    Cl = 2 x 35.5

   total = 12 + 38 + 71

   total = 121 g

2.- Use the Avogradro's number to solve the problem

                       121 g -------------------  6.023 x 10²³ molecules of CF₂Cl₂

                        45.7 g ---------------     x

                        x = (45.7 x 6.023 x 10²³) / 121

                        x = 2.75 x 10²⁵ / 121

                        x = 2.20 x 10 ²³ molecules

4 0
3 years ago
NaHCO3 (s) + HC2H3O2 (aq) = NaC2H3O2 (aq) + H2O (I) + CO2 (g)
Misha Larkins [42]

Moles=volume*concentration   
         =0.1*.83
         =.083 Moles of HC2H3O2
Mole ratio between HC2H3O2 and CO2 is 1:1
This means .083 Moles of CO2

Mass =Moles*Rfm of CO2
         =.083*(12+16+16)
         =3.7grams
8 0
4 years ago
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