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2 years ago

According to the law of conservation of mass, how much zinc was present in the zinc carbonate?

Chemistry

1 answer:

oksano4ka [1.4K]2 years ago

6 0

Answer : The correct option is, 104 grams.

Explanation :

Law of conservation of mass : It states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

The balanced chemical reaction will be,

$Ca+ZnCO_3\rightarrow CaCO_3+Zn$

According to the law of conservation of mass,

Total mass of reactant side = Total mass of product side

Total mass of $Ca+ZnCO_3$ = Total mass of $CaCO_3+Zn$

Mass of $Ca$ + Mass of $ZnCO_3$ = Mass of $CaCO_3$ + Mass of $Zn$

As we are given :

Mass of $Ca$ = 64 g

Mass of $ZnCO_3$ = 192 g

Mass of $CaCO_3$ = 152 g

So,

Mass of $Ca$ + Mass of $ZnCO_3$ = Mass of $CaCO_3$ + Mass of $Zn$

64 g + 192 g = 152 g + Mass of $Zn$

Mass of $Zn$ = (64 + 192) - 152

Mass of $Zn$ = 104 g

Therefore, the mass of zinc present in the zinc carbonate was 104 grams.

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A pair of students found the temperature of 100. g of water to be 26.5°C. They then dissolved 8.89 g of AgNO3 in the water. When

bekas [8.4K]

Explanation:

Initial temperature of the water = $T_i=26.5^oC$

Final temperature of the water = $T_f=23.7^oC$

a) Change in temperature of the water,ΔT = $T_f-T_i$

$\Delta T=23.7^oC-26.5^oC=-2.8^oC$

Change in temperature of the water,ΔT is -2.8°°C.

b) Endothermic reaction : Reaction in which heat absorbed and the temperature if the surrounding is decreased.

Exothermic reaction : Reaction in which heat released and the temperature if the surrounding is increased.

On dissolving silver nitrate in water the temperature of the water decreased 2.8 degree Celsius which means that energy water was absorbed by solid silver nitrate to get dissolve in water. Hence, endothermic reaction.

(f) Mass of silver nitrate = 8.89 g

Moles of silver nitrate = $\frac{8.89 g}{170 g/mol}=0.05229 mol$

$AgNO_3(aq)\rightarrow Ag^+(aq)+NO_3^-(aq)$

1 mole of silver nitrate gives 1 mole of silver ion i.e. cation.

Then 0.05229 moles of silver nitrate will give:

$\frac{1}{1}\times 0.05229 mol=0.05229 mol$ silver ions.

0.05229 moles of cations are produced.

1 mole of silver nitrate gives 1 mole of nitrate ion i.e. anion .

Then 0.05229 moles of silver nitrate will give:

$\frac{1}{1}\times 0.05229 mol=0.05229 mol$ nitrate ions.

0.05229 moles of anion are produced.

5 0

2 years ago

Calculate the mass (g) of each product formed when 43.82 g of diborane (B₂H₆)

guajiro [1.7K]

Stoichiometry:

The area of study in chemistry concerned with the quantities of chemical species produced or required for a given chemical reaction is stoichiometry. The stoichiometric coefficients of the balanced chemical equation are often used to relate the amounts of substances to one another.

Evaluation :

The balanced chemical equation for the reaction is given as follows:

$B_{2} H_{6} + 6H_{2} O$ → $2H_{3} BO_{3} + 3H_{2}$

By using the molar mass of the given chemical species, along with the stoichiometric coefficients of the balanced chemical equation, we determine the mass of each product as:

43.82 g $B_{2} H_{6}$ × $\frac{1 Mol B_{2} H_{6} }{27.66 g}$ × $\frac{3molH_{2} }{1 molB_{2}H_{6} }$ × $\frac{2g}{1 mol H_{2} } = 9.51 g H_{2}$

43.82 g $B_{2} H_{6}$ × $\frac{1 mol B_{2} H_{6} }{27.66 g}$ × $\frac{1 mol H_{3}BO_{3} }{1 mol B_{2} H_{6} }$ × $\frac{61.8g}{1 mol H_{3} BO_{3} } = 196 gH_{3} BO_{3}$