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velikii [3]
3 years ago
11

Easy Chem, Will Give brainliest

Chemistry
1 answer:
Crazy boy [7]3 years ago
5 0

Answer:

3.94 L

Explanation:

From the question given above, the following data were obtained:

Mass of O₂ = 5.62 g

Volume of O₂ =?

Next, we shall determine the number of mole present in 5.62 g of O₂. This can be obtained as follow:

Mass of O₂ = 5.62 g

Molar mass of O₂ = 2 × 16 = 32 g/mol

Mole of O₂ =?

Mole = mass / molar mass

Mole of O₂ = 5.62 / 32

Mole of O₂ = 0.176 mole

Finally, we shall determine the volume of 5.62 g (i.e 0.176 mole) of O₂ at STP. This can be obtained as follow:

1 mole of O₂ occupied 22.4 L at STP.

Therefore, 0.176 mole of O₂ will occupy = 0.176 × 22.4 = 3.94 L at STP.

Thus 5.62 g (i.e 0.176 mole) of O₂ occupied 3.94 L at STP

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3 years ago
How many moles are in 3.01 c 10^24 atoms of sodium chloride
Mademuasel [1]

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number of moles = 5 moles

Explanation:

atomic mass of sodium chloride = 23+35.5 = 60.5 g

number of moles = (number of molecules)/(Avagadro number)

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                              =5 moles.

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3 years ago
Which solute would be more effective at lowering the freezing point of water: MgCl2 and KNO3? Explain.
Phantasy [73]

Answer:

AlCl₃.

Explanation:

Adding solute to water causes depression of the boiling point.

The depression in freezing point (ΔTf) can be calculated using the relation:

ΔTf = i.Kf.m,

where, ΔTf is the depression in freezing point.

i is the van 't Hoff factor.

van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

Kf is the molal depression constant of water.

m is the molality of the solution (m = 1.0 m, for all solutions).

(1) NaCl:

i for NaCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

∴ ΔTb for (NaCl) = i.Kb.m = (2)(Kf)(1.0 m) = 2(Kf).

(2) MgCl₂:

i for MgCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

∴ ΔTb for (MgCl₂) = i.Kb.m = (3)(Kf)(1.0 m) = 3(Kf).

(3) NaCl:

i for KBr = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

∴ ΔTb for (KBr) = i.Kb.m = (2)(Kf)(1.0 m) = 2(Kf).

(4) AlCl₃:

i for AlCl₃ = no. of particles produced when the substance is dissolved/no. of original particle = 4/1 = 4.

∴ ΔTb for (CoCl₃) = i.Kb.m = (4)(Kf)(1.0 m) = 4(Kf).

So, the ionic compound will lower the freezing point the most is: AlCl₃

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