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lutik1710 [3]
2 years ago
14

Why can molar mass be used as a conversion factor

Chemistry
2 answers:
Gre4nikov [31]2 years ago
5 0

From the relative atomic mass of each element, it is possible to determine each element's molar mass by multiplying the molar mass constant (1 g/mol) by the atomic weight of that particular element. The molar mass value can be used as a conversion factor to facilitate mass-to-mole and mole-to-mass conversions.

Pachacha [2.7K]2 years ago
3 0

Answer:

A substances molar mass is determined by increasing its relative nuclear mass by the molar mass constanttt (1 g/mol). So the molar mass consistent can be utilized to change mass over to moles. So by duplicating a given mass by the molar mass, the measure of moles of the substance can be determined.

Explanation:

From the general nuclear mass of every component, it is feasible to decide every component's molar mass by duplicating the molar mass steady (1 g/mol) by the nuclear load of that specific element.The molar mass worth can be utilized as a change factor to work with mass-to-mole and mole-to-mass transformations.

Brainliest?

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For the reaction of hydrogen with iodine
fenix001 [56]

Answer:

r_{H_2} = \frac{-1}{2} r_{HI}

Explanation:

Hello!

In this case, considering the given chemical reaction:

H_2(g) + I_2(g) \rightarrow 2HI(g)

Thus, by applying the law of rate proportions, we can write:

\frac{1}{-1} r_{H_2} = \frac{1}{-1}r_{i_2} = \frac{1}{2} r_{HI}

Whereas the stoichiometric coefficients of reactants are negative due their disappearance and that of the product is positive due to its appearance. In such a way, when we relate the rate of disappearance of hydrogen gas to the rate of formation of hydrogen iodide, we obtain:

r_{H_2} = \frac{-1}{2} r_{HI}

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8 0
2 years ago
Determine the poh of a 0.348 m ba(oh)2 solution at 25°c.
vladimir1956 [14]

<u>Given:</u>

Concentration of Ba(OH)2 = 0.348 M

<u>To determine:</u>

pOH of the above solution

<u>Explanation:</u>

Based on the stoichiometry-

1 mole of Ba(OH)2 is composed of 1 mole of Ba2+ ion and 2 moles of OH- ion

Therefore, concentration of OH- ion = 2*0.348 = 0.696 M

pOH = -log[OH-] = - log[0.696] = 0.157

Ans: pOH of 0.348M Ba(OH)2 is 0.157

6 0
3 years ago
Read 2 more answers
Which is heavier? 30kg or 60lb?
Maurinko [17]
30kg think of it like this imagine you are moving to your college dorm would you have you carry the box if weights or will you carry your books instead
4 0
3 years ago
3 NO2(g) + H2O(ℓ) −→
gulaghasi [49]
The balanced equation for the above reaction is as follows;
3NO₂ + H₂O --> 2HNO₃ + NO
stoichiometry of NO₂ to NO is 3:1
molar volume is where 1 mol of any gas occupies a volume of 22.4 L
volume of gas is directly proportional to number of moles of gas.
therefore stoichiometry can be applied for volume as well.
volume ratio of NO₂ to NO is 3:1
volume of NO₂ reacted - 854 L
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7 0
3 years ago
When 229.0 J of energy is supplied as heat to 3.00 mol of Ar(g) at constant pressure the temperature of the sample increases by
bazaltina [42]

Answer:

The molar heat capacity at constant volume is 21.62 JK⁻¹mol⁻¹

The molar heat capacity at constant pressure is 29.93 JK⁻¹mol⁻¹

Explanation:

We can calculate the molar heat capacity at constant pressure from

C_{p,m} = \frac{C_{p} }{n}

Where C_{p,m} is the molar heat capacity at constant pressure

{C_{p} } is the heat capacity at constant pressure

and n is the number of moles

Also {C_{p} } is given by

{C_{p} } = \frac{\Delta H}{\Delta T}

Hence,

C_{p,m} = \frac{C_{p} }{n} becomes

C_{p,m} = \frac{\Delta H }{n \Delta T}

From the question,

\Delta H = 229.0 J

n = 3.00 mol

\Delta T = 2.55 K

Hence,

C_{p,m} = \frac{\Delta H }{n \Delta T} becomes

C_{p,m} = \frac{229.0}{(3.00) (2.55)}

C_{p,m} = 29.93 JK⁻¹mol⁻¹

This is the molar heat capacity at constant pressure

For, the molar heat capacity at constant volume,

From the formula

C_{p,m} = C_{v,m} + R

Where C_{v,m} is the molar heat capacity at constant volume

and R is the gas constant (R = 8.314 JK⁻¹mol⁻¹)

Then,

C_{v,m} = C_{p,m}  - R

C_{v,m} = 29.93 - 8.314

C_{v,m} = 21.62 JK⁻¹mol⁻¹

This is the molar heat capacity at constant volume

5 0
3 years ago
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