All categories

1 year ago

How many moles of Chromium is in 4.41 ×10^24 atoms

Chemistry

1 answer:

zalisa [80]1 year ago

3 0

7.32 moles of Chromium is present in 4.41 × 10²⁴ atoms.

<h3>How to find the number of moles ?</h3>

Number of moles = $\frac{\text{Given number of atoms}}{\text{Avogadro's Number}}$

<h3>What is Avogadro's Number ?</h3>

Avogadro's number is the number of particles in one mole of substance. 6.022 × 10²³ is known as Avogadro's Constant / Avogadro's Number.

Avogadro's Number = 6.022 × 10²³

Now put the values in above formula we get

Number of moles = $\frac{\text{Given number of atoms}}{\text{Avogadro's Number}}$

= $\frac{4.41 \times 10^{24}}{6.022 \times 10^{23}}$

= 7.32 moles

Thus from the above conclusion we can say that 7.32 moles of Chromium is present in 4.41 × 10²⁴ atoms.

Learn more about the Avogadro's Number here: brainly.com/question/1581342

#SPJ1

You might be interested in

What is decomposision reaction

Virty [35]

Decomposition reactions are a type of chemical reaction that involves breaking down a compound into smaller compounds or individual elements. These reactions often involve an input of energy in the form of heat, light, or electricity to break down the compounds.

3 0

3 years ago

Which of the following bonds is the most reactive? A. C—C B. N—H C. O—O D. C—H

USPshnik [31]

I think the answer is b

6 0

2 years ago

If an individual is eating a diet in which 25% of their ATP production is a result of utilizing carbohydrates, 50% of their ATP

NISA [10]

Answer:

Oxidative phosphorylation proceeds with the formation of energy laden molecules i.e; carbondioxide and water.

Therefore, Total CO₂ production is directly related to VCO₂ = R x VO₂

where, R is the respiratory quotient varing among 0.7 to 1.0 according to the energy intake (ATP) ie 0.25 of the total diet consumed .

VO₂ is, as mentioned above arterial venous oxygen difference = 6.2ml/dl

therefore, VCO₂ = 0.25 x 6.2

= 1.55 ml/dl

ie; VO₂ : VCO₂ = 6.2 : 1.55.

Explanation:

5 0

2 years ago

The equilibrium constant for the reaction

FinnZ [79.3K]

Answer: The concentrations of $Cl_2$ at equilibrium is 0.023 M

Explanation:

Moles of $Cl_2$ = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{10g}{71g/mol}=0.14mol$

Volume of solution = 1 L

Initial concentration of $Cl_2$ = $\frac{0.14mol}{1L}=0.14M$

The given balanced equilibrium reaction is,

$COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)$

Initial conc. 0.14 M 0 M 0M

At eqm. conc. (0.14-x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[CO]\times [Cl_2]}{[COCl_2]}$

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

$4.63\times 10^{-3}=\frac{x)^2}{(0.14-x)}$

By solving the term 'x', we get :

x = 0.023 M

Thus, the concentrations of $Cl_2$ at equilibrium is 0.023 M

7 0

3 years ago

Consider the formula for glucose: C6H12O6. What does this indicate about the relationship of the reactants to glucose?

stellarik [79]

The correct option is D.

The reactants that combine together to form glucose are carbon dioxide, water and energy from the sun. Six molecules of carbon dioxide combine with six molecules of water in the presence of sunlight to form glucose. The chemical equation for the reaction is given below"

6CO2 + 6H2O + Sunlight = C6H12O6.

6 0

2 years ago

Read 2 more answers