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Agata [3.3K]
3 years ago
6

What is the number of moles of potassium chloride present in 142 g of kcl?

Chemistry
1 answer:
morpeh [17]3 years ago
5 0

Answer:

98.3!

Explanation:

Hope this helps! :)

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How many moles of atoms are in 3.011 x10^23 atoms mg
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When a solute is added to a solution why does it remain homogeneous?
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<h3>Answer:</h3>

                 When a solute is added to a solution, it remains homogeneous because the solute is soluble in given solvent.

<h3>Explanation:</h3>

                         Homogeneous mixtures, also called true solutions are those mixtures in which the components proportions are same throughout in any given sample. For example, the mixture of table salt (NaCl) and water. When the solution is unsaturated and further NaCl is added to it, it will dissolve the NaCl because the saturation point is still not reached. Remember, as "<em>Like Dissolves Like</em>" NaCl being polar in nature will interact with water molecules and will dissociate into Na⁺ and Cl⁻ ions surrounded by δ- O and δ+ H atoms of water molecules.

<h3>Conclusion:</h3>

                  In order to form a Homogeneous mixture the solution must be unsaturated, solvent must have affinity for incoming solute particles and the size of solute should be equal to 1 Â (Angstrom).

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The air bags in cars are inflated when a collision triggers the explosive, highly exothermic decomposition of sodium azide (NaN3
Oksanka [162]

Answer : The mass of NaN_3 required is, 166.4 grams.

Explanation :

First we have to calculate the moles of nitrogen gas.

Using ideal gas equation:

PV=nRT

where,

P = Pressure of N_2 gas = 1.00 atm

V = Volume of N_2 gas = 113 L

n = number of moles N_2 = ?

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of N_2 gas = 85^oC=273+85=358K

Putting values in above equation, we get:

1.00atm\times 113L=n\times (0.0821L.atm/mol.K)\times 358K

n=3.84mol

Now we have to calculate the moles of sodium azide.

The balanced chemical reaction is,

2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)

From the balanced reaction we conclude that

As, 3 mole of N_2 produced from 2 mole of NaN_3

So, 3.84 moles of N_2 produced from \frac{2}{3}\times 3.84=2.56 moles of NaN_3

Now we have to calculate the mass of NaN_3

\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3

Molar mass of NaN_3 = 65 g/mole

\text{ Mass of }NaN_3=(2.56moles)\times (65g/mole)=166.4g

Therefore, the mass of NaN_3 required is, 166.4 grams.

3 0
3 years ago
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