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Anastaziya [24]
1 year ago
10

How many moles of solute are present in .75 l of a .89 m (molar) solution?

Chemistry
1 answer:
never [62]1 year ago
8 0

0.6675 moles of solute are present in .75 l of a .89 m (molar) solution.

<h3>Define the molarity of a solution.</h3>

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution.

Given data:

V= 0.75

M=0.89

Molality = \frac{Moles \;solute}{Volume of solution in litre}

0.89 M=  \frac{Moles \;solute}{0.75 L}

Moles= 0.6675

Hence, 0.6675 moles of solute are present in .75 l of a .89 m (molar) solution.

Learn more about the moles here:

brainly.com/question/24081006

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Which of the following molecules would exhibit only London forces?a)CH4, BH3, and CCl4b)H2O, NH3, and CCl4c)PH3, NH3, and CCl4
Alex787 [66]

Answer:

a)CH₄, BH₃, and CCl₄

Explanation:

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The intermolecular force acting in the molecule are induced dipole-dipole forces or London Dispersion forces / van der Waals forces which are the weakest intermolecular force.

Out of the given options, H₂O , NH₃ exhibits hydrogen bonding which is:-

<u>Hydrogen bonding:- </u>

Hydrogen bonding is a special type of the dipole-dipole interaction and it occurs between hydrogen atom that is bonded to highly electronegative atom which is either fluorine, oxygen or nitrogen atom.

Thus option B and C rules out.

<u>Hence, the correct option which represents the molecules which would exhibit only London forces is:- a)CH₄, BH₃, and CCl₄</u>

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How many grams of CH4 will be in a 500ml contain at STP?
ArbitrLikvidat [17]

Answer:

Mass = 0.32 g

Explanation:

Given data:

Mass of CH₄ = ?

Volume of CH₄ = 500 mL (500 mL× 1L/1000 mL= 0.5 L)

Temperature = 273 K

Pressure = 1 atm

Solution:

Volume of CH₄:

500 mL (500 mL× 1L/1000 mL= 0.5 L)

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

By putting values,

1 atm× 0.5 L = n×0.0821 atm.L/ mol.K  × 273 K

0.5 atm.L = n×22.4 atm.L/ mol

n = 0.5 atm.L / 22.4 atm.L/ mol

n = 0.02 mol

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2 years ago
Question 1 (1 point)
Sloan [31]

Answer:

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