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3 years ago

The solvent is usually referred to as the component of a solution which is present as

2 answers:

8 0

The solvent is usually referred to as the component of a solution which is present as the one with the larger quantity and in most cases as the liquid which dissolves a solid. In a solution, there are two components namely the solvent and the solute. The solute is the one in smaller amount.

Serjik [45]3 years ago

5 0

Answer: Larger proportion

Explanation:

A binary solution is made up of two components , one of which is solute and another is solvent.

A solute is defined as the component which is present in smaller proportion. It can be in solid, liquid or gaseous phase.

A solvent is defined as the component which is present in larger proportion in a solution. It can be in solid, liquid or gaseous phase.

Hence, solvent is present in larger proportion.

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LenKa [72]

Answer:

Explanation: 15

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3 years ago

In calcium fluoride, what is the closets noble gas that each ion becomes?

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2 years ago

Which of the changes gas to solid gas to liquid, liquid to solid liquid to gas,solid to liquid solid to gas will involve an outp

Verdich [7]

Answer:

Liquid to solid liquid to gas

Explanation:

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2 years ago

How do you calculate the mass of a product when the amounts of more than one reactant are given?

chubhunter [2.5K]

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3 0

2 years ago

The density of potassium, which has the BCC structure, is 0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate

Nimfa-mama [501]

Answer:

lattice parameter = 5.3355x10^-8 cm

atomic radius = 2.3103x10^-8 cm

Explanation:

known data:

p=0.855 g/cm^3

atomic mass = 39.09 g/mol

atoms/cell = 2 atoms

Avogadro number = 6.02x10^23 atom/mol

a) the lattice parameter:

Since potassium has a cubic structure, its volume is equal to:

v = [(atoms/cell)x(atomic mass)/(p)x(Avogadro number)]

substituting values:

v =[(2)x(39.09)/(0.855x6.02x10^23)]=1.5189x10^-22 cm^3

but as the cell volume is

a^3 =v

$a=\sqrt[3]{v}=\sqrt[3]{1.5189x10^{-22} } = 5.3355x10^-8$ cm

for a BCC structure, the atomic radius is equal to

$r=\frac{ax\sqrt{3} }{4}=\frac{5.3355x10^{-8}x\sqrt{3} }{4}=2.3103x10^{-8}cm$

7 0

2 years ago