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Flura [38]
4 years ago
5

A solution contains 10 g of KNO3 per 100 g of water at 30˚C. Is the solution unsaturated, saturated or supersaturated?

Chemistry
1 answer:
anzhelika [568]4 years ago
6 0

Answer:

Explanation:    As you can see, potassium nitrate has s solubility of about  

67 g / 100g H

2

O

at  

40

∘

C

. This means that at  

40

∘

C

, a saturated potassium nitrate solution will contain  

67 g

of dissolved salt for every  

10 0g

of water.

You know that at this temperature, your solution contains  

35 g

of potassium nitrate in  

100 g

of water. This solution will be unsaturated because it contains less potassium nitrate than the maximum amount that can be dissolved.

In order to make a saturated solution, you must get the total mass of potassium nitrate to  

67 g

, which means that you must add

mass of KNO

3

a

to be added

=

67 g

−

35 g

=

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

32 g

a

a

∣

∣

−−−−−−−−

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Sidana [21]

<u>larger is the value of </u>Ksp<u> the higher will the solubility of solid in water.</u>

What is called compound?

  • In chemistry, a compound is a substance made up of two or more different chemical elements combined in a fixed ratio.
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The equation for the dissociation of a solid MX in water is given below

MX(s) ⇄ Mⁿ⁺(aq) + Xⁿ⁻ (aq)

Assume s be the solubility of MX in pure water, then the equilibrium concentrations of ions are

[ Mⁿ⁺] = s

[ Xⁿ⁻ ] = s

The expression for the solubility product constant (Ksp) is as follows

Ksp = [ Mⁿ⁺] [ Xⁿ⁻ ]

       = s²

That is, larger is the value of Ksp the higher will the solubility of solid in water.

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6 0
2 years ago
How many moles are there in 24.00 g of NaCl
Elis [28]

Answer:

The answer to your question is 0.41 moles

Explanation:

Data

moles of NaCl = ?

mass of NaCl = 24 g

Process

To solve this problem just calculate the molar mass of NaCl, and remember that the molar mass of any substance equals to 1 mol.

1.- Calculate the molar mass

NaCl = 23 + 35.5 = 58.5 g

2.- Use proportions and cross multiplication

               58.5 g of NaCl ------------------- 1 mol

               24.0 g               ------------------- x

                     x = (24 x 1) / 58.5

                     x = 0.41 moles

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How many litres (L) of oxygen at STP can be obtained from 110 g of potassium chlorate? [R=0.0821 L.atm/K.mol]
lisabon 2012 [21]
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2 KClO3 (s) ---> 2 KCl (s) + 3 O2 (g)

Moles of KClO3 =  110 / 122.5 = 0.89

Following the balanced chemical equation:
We can say moles of O2 produce =  \frac{3}{2} x moles of KClO3

So, O2 = (3 / 2) x  0.89

= 1.34 moles

So, Volume at STP = nRT / P

T = <span>273.15 K
P = 1 atm

So, V = (1.34 x 0.0821 x 273.15) / 1  =  30.2 L</span>
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Yanka [14]

Answer:

Explanation:

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