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spayn [35]
4 years ago
14

How many moless is 4.91x10^22 molecules of H3PO4?

Chemistry
1 answer:
Lynna [10]4 years ago
6 0
1 mol = 6.023x10^23 number of molecules (Avogadro's number)

1 : 6.023x10^23
X : 4.91x10^22

(6.023x10^23)X = 4.91x10^22

X = 4.91x10^22/6.023x10^23

X = 0.082 Moles
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Defined the following convection
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a 4.4 g sample of gas occupies 2.24 l of volume at stp. without thinking too hard, what is the mw of the gas, and name two gases
garri49 [273]

As a result, gas's molecular weight is 44g/mol. It is the gas's (Carbon dioxide) molecular weight.

What is Molecular Weight?

The total atomic weights of the atoms in a molecule are measured by its molecular weight. To calculate stoichiometry in chemical equations and reactions, chemists employ molecular weight. M.W. or MW are two frequent abbreviations for molecular weight. Atomic mass units (amu), Daltons, or a unitless expression can be used to indicate molecular weight (Da).

The mass of the isotope carbon-12, which is given a value of 12 amu, serves as the reference point for defining both atomic weight and molecular weight. Because there are many carbon isotopes, the atomic weight of carbon is not exactly 12.

A mole of any gas at STP takes up a volume of 22.4l

at STP, a gas fills a volume of 2.24l

Calculating the quantity of moles of gas in step two

Consequently, the amount of gas in moles

= 0.1 moles

This is equivalent to carbon dioxide's molecular weight.

=44g/mol

As a result, gas's molecular weight is 44g/mol. It is the gas's (Carbon dioxide) molecular weight.

Learn more about Molecular Weight from given link

brainly.com/question/837939

#SPJ4

7 0
1 year ago
A mixture of 0.307 M Cl 2 , 0.465 M F 2 , and 0.706 M ClF is enclosed in a vessel and heated to 2500 K . Cl 2 ( g ) + F 2 ( g )
monitta

<u>Answer:</u> The equilibrium concentration of chlorine gas, fluorine gas and ClF gas is 0.159 M, 0.317 M and 1.002 M respectively.

<u>Explanation:</u>

We are given:

Initial concentration of chlorine gas = 0.307 M

Initial concentration of fluorine gas = 0.465 M

Initial concentration of ClF gas = 0.706 M

The given chemical equation follows:

                            Cl_2(g)+F_2(g)\rightleftharpoons 2ClF(g)

<u>Initial:</u>                  0.307       0.465       0.706

<u>At eqllm:</u>           0.307-x    0.465-x       0.706+2x

The expression of K_c for above equation follows:

K_c=\frac{[ClF]^2}{[Cl_2][F_2]}

We are given:

K_c=20.0

Putting values in above equation, we get:

20.0=\frac{(0.706+2x)^2}{(0.307-x)(0.465-x)}\\\\x=0.148,0.993

Neglecting the value of x = 0.993 because the equilibrium concentrations of chlorine and fluorine gases will become negative, which is not possible

So, equilibrium concentration of chlorine gas = (0.307 - x) = [0.307 - 0.148] = 0.159 M

Equilibrium concentration of fluorine gas = (0.465 - x) = [0.465 - 0.148] = 0.317 M

Equilibrium concentration of ClF gas = (0.706 + 2x) = [0.706 + 2(0.148)] = 1.002 M

Hence, the equilibrium concentration of chlorine gas, fluorine gas and ClF gas is 0.159 M, 0.317 M and 1.002 M respectively.

5 0
3 years ago
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