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kykrilka [37]
3 years ago
13

What is the mass of 1.25 mol of zinc ​

Chemistry
1 answer:
Ilya [14]3 years ago
5 0

Answer:

1 mole is equal to 1 moles Zinc, or 65.38 grams.

Explanation:

Could i have branliest, heart and 5 stars thanks!

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At constant temperature and pressure, the coefficients for gaseous species in a valences chemical reaction can be interpreted as
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A solution that is 0.135 M is diluted to make 500.0 mL of a 0.0851 M solution. How many milliliters of the original solution wer
sineoko [7]

Answer:

315mL

Explanation:

Data obtained from the question include the following:

Molarity of stock solution (M1) = 0.135 M

Volume of stock solution needed (V1) =?

Molarity of diluted solution (M2) = 0.0851 M

Volume of diluted solution (V2) = 500mL

The volume of the stock solution needed can be obtain as follow:

M1V1 = M2V2

0.135 x V1 = 0.0851 x 500

Divide both side by 0.135

V1 = (0.0851 x 500) / 0.135

V1 = 315mL

Therefore, the volume of the stock solution needed is 315mL

6 0
3 years ago
Convert 3.01€22 molecules of O²(a) mole of O²) ,volume at S.T.P,no of Oxygen atom. (d) mass​
Marat540 [252]

Answer:

a )0.05 moles of O₂

b) 1.1207 dm³

c) 0.3× 10²³ atoms of oxygen

d) 1.6 g

Explanation:

Given data:

Number of molecules of O₂ = 3.01 × 10²² molecules

Number of moles of O₂ = ?

Volume of oxygen at ATP = ?

Number of oxygen atoms = ?

Mass of oxygen = ?

Solution:

The given problem will solve by using Avogadro number.

"It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance".  The number 6.022 × 10²³ is called Avogadro number.

Moles of oxygen:

1 mole = 6.022 × 10²³  molecules

3.01 × 10²² molecules  × 1 mol  / 6.022 × 10²³  molecules

0.05 moles of O₂

Number of atoms of oxygen:

1 mole = 6.022 × 10²³ atoms

0.05 mol × 6.022 × 10²³ atoms /1 mol

0.3× 10²³ atoms of oxygen

Volume of oxygen:

1 mole of oxygen at STP occupy 22.414 dm³

0.05 mol × 22.414 dm³ / 1mol

1.1207 dm³

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 0.05 mol  × 32 g/mol

Mass = 1.6 g

8 0
3 years ago
Consider the combustion of methanol at some high temperature in a constant-pressure reaction chamber: 2ch3oh (g) + 3o2 (g) \long
Monica [59]

The balanced reaction is

2CH3OH (g) + 3O2 (g) ⟶ 2CO2 (g) + 4H2O (g)

as per equation two moles of methanol gas will react with 3 moles of oxygen

one mole of gas occupies 22.4 L of volume

so the moles and volume goes in same ratio

it means two unit volume of methanol will react with three unit volume of oxygen

therefore 1L of methanol gas will react with 3 /2 L of oxygen

Or 18 L of methanol gas will react with 3 x 18 /2 = 27  L of oxygen

So here oxygen is limiting reagent

As per balanced equation

10L of oxygen will react with = 2 X 10 /3 L of methanol = 6.67 L of methanol gas to give 6.67 L of CO2 gas and 13.33 L of water gas

So overall there will be = 18 - 6.67 L of left out methanol = 11.33 L

And 6.67 L of CO2 + 13.33 L of water = 20 L

Total volume of gas = 11.33+ 20 = 31.33 L

8 0
3 years ago
Read 2 more answers
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