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3 years ago

Which of the following samples contains the greatest number of atoms?

Chemistry

1 answer:

mart [117]3 years ago

7 0

Answer:

C. 11 moles of N2O

Explanation:

A. CO2 exists as a molecular compound. The number of atoms present = 1 atom of Carbon and 2 atoms oxygen = 3 atoms * 9 = 18 atoms

B. Xe exists as an atom. Number of atoms present = 10 * 1 atom = 10 atoms

C. N20 contains 3 atoms; 2 atoms of Nitrogen and 1 atom of oxygen.

Number of atoms present = 3 * 11 = 33 atoms

D. CO contains 2 atoms, 1 each of carbon and oxygen.

Number of atoms present = 12 * 2 = 24 atoms

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Answer:

1.60×10¹⁴ s⁻¹

Explanation:

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3 years ago

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hodyreva [135]

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3 years ago

A student is performing an experiment to study thermodynamics. The student mixes citric acid and sodium bicarbonate in a calorim

seraphim [82]

Explanation :

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3 years ago

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11111nata11111 [884]

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3 years ago

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0

3 years ago