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sergejj [24]
2 years ago
14

How does the law of conservation of mass apply to this reaction:Mg + 2HCl → H2 + MgCl2?

Chemistry
1 answer:
krok68 [10]2 years ago
5 0
The law of conservation of mass applies to every reaction. In this case, you start with 1 Mg, 2 H, and 2CL and end up with the same five only their bonds have been rearranged, or in other words, they are joined up differently.
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Explanation:

The given data is as follows.

        V_{max} = 6.8 \times 10^{-10} \mu mol/min

          K_{m} = 5.2 \times 10^{-6} M

Now, according to Michaelis-Menten kinetics,

              V_{o} = V_{max} \times [\frac{S}{(S + Km)}]

where, S = substrate concentration = 10.4 \times 10^{-6} M

Now, putting the given values into the above formula as follows.

        V_{o} = V_{max} \times [\frac{S}{(S + Km)}]

        V_{o} = 6.8 \times 10^{-10} \mu mol/min \times [\frac{10.4 \times 10^{-6} M}{(10.4 \times 10^{-6}M + 5.2 \times 10^{-6} M)}]

            V_{o} = 6.8 \times 10^{-10} \mu mol/min \times 0.667

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This means that V_{o} would approache V_{max}.

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What is the chemical formula for ammonium sulfate?
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The percent yield of the calcium hydroxide is 84.5%.

<h3>What is stoichiometry?</h3>

Stoichiometry enables us to obtain the mass of a substance form the equation of the reaction.

The equation of the reaction is;

CaCO3 + 2HCl -----> CaCl2 + CO2 + H2O

Number of moles of X = 40.0 grams/100 g/mol = 0.4 moles

Number of moles of HCl = 850/1000 * 1 M = 0.85 moles

If 1 mole of CaCO3 reacts with 2 moles of HCl

0.4 moles of  CaCO3 reacts with  0.4 moles  * 2 moles/1 mole

= 0.8 moles of HCl

Hence X is the limiting reactant.

The reaction is 1:1 then the amount of CO2  produced is 0.4 moles

Mass of CO2 = 0.4 CO2 * 44 g/mol = 17.6 g

2) The reaction equation is; 2NaOH + CaCO3 --->  Ca(OH)2 + Na2CO3

Number of moles of X = 25.0 grams/100 g/mol =  0.25 moles

Number of moles of NaOH= 40/1000 L * 2 M = 0.08 moles

If 1 mole of X reacts with 2 moles of NaOH

0.25 moles  reacts with   0.25 moles   * 2 moles /1 mole

= 0.5 moles

NaOH is the limiting reactant

2 moles of NaOH produces 1 mole of CO2

0.08 moles of NaOH produces 0.08 moles * 1 mole/2 moles

= 0.04 moles of CO2

Theoretical yield of CO2 =  0.04 moles of CO2 * 74 g/mol = 2.96  g

Percent yield = 2.5 g/ 2.96  g * 100

= 84.5%

Learn ore about percent yield:brainly.com/question/17042787

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