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Fiesta28 [93]
1 year ago
6

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

that would have this mw. write out your logic.
Chemistry
1 answer:
garri49 [273]1 year ago
7 0

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

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The equilibrium constant for the chemical equation is Kp = 5.23 at 191 °C. Calculate the value of the Kc for the reaction at 191
Lapatulllka [165]
To convert from Kp to Kc, you need this formula---> Kp= Kc (RT)^Δn, where Δn= gas moles of product- gas moles of reactants. since you did not give a reaction formula, I can't calculate Δn. but all once you find it out. just plug it. 

Kp= Kc (RT)^Δn------------------> Kc= Kp/[(RT)^Δn]
 Kp= 5.23
R= 0.0821
T= 191 C= 464 K
Δn= ?

Kc= 5.23/ (0.0821 x 464)^Δn= ???

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3 years ago
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PtichkaEL [24]

Answer:

Explanation:

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5 0
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Alinara [238K]

Answer:

See explanation

Explanation:

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As more shells are added and repulsion of inner electrons become more significant, atomic size increases down the group. However, across the period, atomic size decreases due to increase in effective nuclear charge without any increase in the number of shells. This causes increased attraction between the nucleus and the outermost shell thereby decreasing the size of the atom.

Ionization energy decreases down the group because the outermost electron is more shielded by inner electrons making it easier for this outermost electron to be lost. Across the period, ionization energy increases due to increase in effective nuclear charge which makes it more difficult to remove the outermost electron due to increased nuclear attraction.

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Answer:

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Explanation:

This task is a joint variation task involving only direct proportionality:

Direct variation is one in which two variables are in direct proportionality to each other. This means that as one increases, the other variable also increases and vice - versa.

Joint variation is one in which one variable is dependent on two or more variables and varies directly as each of them.

In this exercise:

If a ∝ b and a ∝ c, then a ∝ bc

Taking the above three proportionalities,

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V ∝ a ∝ bc

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Explanation:

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