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Anton [14]
3 years ago
8

A sample of baking soda, naHCO3, always contains 27.37% by mass of sodium, 1.20% of hydrogen, 14.30% of carbon, and 57.14% of ox

ygen.
Which law do these data illustrate? State the law.
Chemistry
1 answer:
Dvinal [7]3 years ago
4 0

Answer: Law of definite proportion.

Explanation: Law of definite proportion states that any chemical compound consists of elements in a fixed ratio by their masses. This law is also known as Proust's Law.

We are given a chemical compound NaHCO_3

Assuming that the total mass of 1 mole of NaHCO_3 = 100 grams

where

Mass of Sodium = 27.37 grams

Mass of Hydrogen =  1.20 grams

Mass of Carbon = 14.30 grams

Mass of Oxygen = 57.14 grams

To Calculate the ratio of all the elements, we divide each of the masses by the lowest mass.

\text{Mass ratio of Sodium}=\frac{27.37}{1.20}=22.80\approx 23

\text{Mass ratio of Hydrogen}=\frac{1.20}{1.20}=1

\text{Mass ratio of Carbon}=\frac{14.30}{1.20}=11.91\approx 12

\text{Mass ratio of Oxygen}=\frac{57.14}{1.20}=47.61\approx 48

Mass ratio of elements in NaHCO_3=Sodium:Hydrogen:Carbon:Oxygen

Na : H : C : O = 23 : 1 : 12 : 48

For every NaHCO_3 compound, this mass ratio will always be constant.

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Carbon Monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700K, the equilibrium constant is 5.10. Calculate t
ozzi

Answer: Concentration of CO_2 at equilibrium= 1.386 M

Concentration of H_2 at equilibrium = 1.386 M

Concentration of CO at equilibrium = 0.614 M

Concentration of H_2O at equilibrium= 0.614 M

Explanation:

Moles of CO = 1.00 mole

Moles of H_2O = 1.00 mole

Moles of CO_2 = 1.00 mole

Moles of H_2O = 1.00 mole

Volume of solution = 1.00 L

Initial concentration of CO =\frac{moles}{Volume}=\frac{1.00mol}{1.00L}=1.00M

Initial concentration of H_2O =\frac{moles}{Volume}=\frac{1.00mol}{1.00L}=1.00M

Initial concentration of CO_2 =\frac{moles}{Volume}=\frac{1.00mol}{1.00L}=1.00M

Initial concentration of H_2 =\frac{moles}{Volume}=\frac{1.00mol}{1.00L}=1.00M

The given balanced equilibrium reaction is,

                         CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2

Initial conc.          1.00M          1.00 M          1.00 M     1.00 M

At eqm. conc.     (1.00-x) M   (1.00-x) M   (1.00+x) M   (1.00+x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[CO_2]\times [H_2]}{[CO]\times [H_2O]}

Now put all the given values in this expression, we get :

5.10=\frac{(1.00+x)^2}{(1.00-x)^2}

By solving the term 'x', we get :

x =  0.386

Concentration of CO_2 at equilibrium= (1.00+x) M= (1.00+0.386)= 1.386 M

Concentration of H_2 = (1.00+x) M= (1.00+0.386)= 1.386 M

Concentration of CO = (1.00-x) M = (1.00-0.386) M = 0.614 M

Concentration of H_2O = (1.00-x) M = (1.00-0.386) M = 0.614 M

3 0
3 years ago
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Following are the Balanced Reactions,

Reaction 1:


                            (NH₄)₂CO₃     →    <u>2</u> NH₄⁺  +  CO3⁻²

Reaction 2:

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Reaction 3:

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The answer would be hypotheis since its an theory that isnt proven yet which would involve a scientist to expertiment to make the hypothesis true or valid
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Explanation:

The initial concentrations for a mixture :

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Water at equilibrium = 0.40 M

CH3COOH + C_2H_5OH\rightleftharpoons CH_3CO_2C_2H_5+H_2O

Initially:

0.15 M            0.15 M            0.40 M   0.40 M

At equilibrium

(0.15-x)M       (0.15-x) M     (0.40+x) M   (0.40+x) M

The equilibrium constant is given by expression

K_c=\frac{[CH_3CO_2C_2H_5][H_2O]}{[CH_3COOH][C_2H_5OH]}

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The equilibrium concentrations for a mixture :

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Ethanol at equilibrium = (0.15-x)M = (0.15-0.033) M = 0.117 M

Ethyl acetate at equilibrium = (0.40+x)M = (0.40+0.033) M = 0.433 M

Water at equilibrium = (0.40+x)M = (0.40+0.033) M = 0.433 M

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Answer:

He took a deep breath and splashed some water on his face.

Explanation:

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7 0
2 years ago
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