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3 years ago

In an aqueous solution of potassium chloride the solute is

Chemistry

1 answer:

Goryan [66]3 years ago

7 0

Answer:

Potassium chloride

Explanation:

A solution is formed by a solvent and one or more solutes.

The solvent is the species that is in major proportion and usually defines the state of aggregation of the solution, while the solute/s is/are in minor proportion.

Also, water is known as the universal solvent, so in any solution containing water, it is considered as the solvent.

Then, in an aqueous solution of potassium chloride the solute is potassium chloride.

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4 years ago

Which is a physical property of a gas?

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4 years ago

Read 2 more answers

How many moles of na2co3 are necessary to reach stoichiometric quantities with cacl2

lbvjy [14]

0.0102 moles Na₂CO₃ = 1.08g of Na₂CO₃ is necessary to reach stoichiometric quantities with cacl2.

<h3>Explanation:</h3>

Based on the reaction

CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃

1 mole of CaCl₂ reacts per mole of Na₂CO₃

we have to calculate how many moles of CaCl2•2H2O are present in 1.50 g

We must calculate the moles of CaCl2•2H2O using its molar mass (147.0146g/mol) in order to answer this issue.
These moles, which are equal to moles of CaCl2 and moles of Na2CO3, are required to obtain stoichiometric amounts.
Then, we must use the molar mass of Na2CO3 (105.99g/mol) to determine the mass:

<h3>Moles CaCl₂.2H₂O:</h3>

1.50g * (1mol / 147.0146g) = 0.0102 moles CaCl₂.2H₂O = 0.0102moles CaCl₂

Moles Na₂CO₃:

0.0102 moles Na₂CO₃

Mass Na₂CO₃:

0.0102 moles * (105.99g / mol) = 1.08g of Na₂CO₃ are present

Therefore, we can conclude that 0.0102 moles Na₂CO₃ is necessary.to reach stoichiometric quantities with cacl2.

To learn more about stoichiometric quantities visit:

<h3>brainly.com/question/28174111</h3>

#SPJ4

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2 years ago

Complete this neutralization equation:<br> H2SO4 + Al(OH)3

elena-s [515]

Answer:

H2SO4 + Al(OH)3 = Al2(SO4)3 + H2O

Explanation:

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3 years ago

Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit

ser-zykov [4K]

<u>Answer:</u>

<u>For a:</u> The edge length of the unit cell is 314 pm

<u>For b:</u> The radius of the molybdenum atom is 135.9 pm

<u>Explanation:</u>

<u>For a:</u>

To calculate the edge length for given density of metal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density = $10.28g/cm^3$

Z = number of atom in unit cell = 2 (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell =?

Putting values in above equation, we get:

$10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm$