All categories

2 years ago

How many atoms of zirconium are in 0.3521 mol of zirconium?

Chemistry

1 answer:

lora16 [44]2 years ago

3 0

Answer:

2.12×10²³ atoms.

Explanation:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ atoms. This simply means that 1 mole of zirconium also 6.02×10²³ atoms.

Thus, we can obtain the number of atoms present in 0.3521 mole of zirconium as follow:

1 mole of zirconium also 6.02×10²³ atoms.

Therefore, 0.3521 mole of zirconium will contain = 0.3521 × 6.02×10²³ = 2.12×10²³ atoms.

Therefore, 0.3521 mole of zirconium contains 2.12×10²³ atoms.

You might be interested in

What is the change in enthalpy associated with the combustion of 530 g of methane (CH4)?Report your answer in scientific notatio

Veseljchak [2.6K]

Answer:

$-3.0\times 10^4 kJ$ is the change in enthalpy associated with the combustion of 530 g of methane.

Explanation:

$CH_4+2O_2\rightarrow CO_2+2H_2O,\Delta H_{comb}=-890.8 kJ/mol$

Mass of methane burnt = 530 g

Moles of methane burnt = $\frac{530 g}{16 g/mol}=33.125 mol$

Energy released on combustion of 1 mole of methane = -890.8 kJ/mol

Energy released on combustion of 33.125 moles of methane :

$-890.8 kJ/mol\times 33.125 mol=-29,507.75 kJ=-2.950775\times 10^4 kJ$

$-2.950775\times 10^4 kJ\approx -3.0\times 10^4 kJ$

$-3.0\times 10^4 kJ$ is the change in enthalpy associated with the combustion of 530 g of methane.

7 0

2 years ago

An article about half-lives describes a daughter isotope. What is a daughter isotope?

Kipish [7]

A Daughter Isotope is...
===
The remaining nuclide left over from radioactive decay.
===
Hope this helps! :)

7 0

2 years ago

The graph shown could represent which of these? A) A book released from a height with the dotted line representing potential ene

jekas [21]

Hey ur amswer is C

I hope this helps

3 0

2 years ago

Read 2 more answers

What is the empirical formula of a compound which contains 84.4% c and 15.6% h by mass?

podryga [215]

Answer: The empirical formula for the given compound is $CH_2$

Explanation : Given,

Percentage of C = 84.4 %

Percentage of H = 15.6 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 84.4 g

Mass of H = 15.6 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{84.4g}{12g/mole}=7.03moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{15.6g}{1g/mole}=15.6moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 7.03 moles.

For Carbon = $\frac{7.03}{7.03}=1$

For Hydrogen = $\frac{15.6}{7.03}=2.22\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 2

Hence, the empirical formula for the given compound is $C_1H_2=CH_2$

7 0

2 years ago

How many mm are equal to 4.75 x 10-2 m?

kherson [118]

Answer:

47.5 mm

Explanation:

5 0

2 years ago