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3 years ago

How much water would I need to add to 500 mL of a 2.4 M KCl solution to make a 1.0 M solution?

Chemistry

1 answer:

GalinKa [24]3 years ago

3 0

Answer:

700 mL of water

Explanation:

This is the perfect example of dilation calculations. Along with this concept we have a formula c = n ( solute ) / V ( solution ). Let us first solve for n by changing this equation to isolate the solute,

n ( KCL ) = 2.4 mol / L * 500 * 10 ^ - 3 L,

n ( KCL ) = 1.2 moles ( KCL )

Knowing the amount of moles of potassium chloride, we have to now identify how much is present in the target solution,

V = 1 .2 moles / ( 1.0 moles / L )

V = 1.2 L = 1200 mL

_______________________________________________________

Vadded = 1200 - 500 = 700 mL

Hope that helps!

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Calculate the solubility of copper(II) hydroxide, Cu(OH)2, in g/L

Oduvanchick [21]

Answer:

Ksp = [ Cu+² ] [ OH-] ²

molar mass Cu(oH )2 ==> M= 63.546 (1) + 16 (2) + 1 (2) = 97.546 g/mol

Ksp = [ Cu+² ] [ OH-] ²

Ksp [ cu (OH)2 ] = 2.2 × 10-²⁰

|__________|___Cu+² __|_2OH-____|

|Initial concentration(M)|___0__|_0______|

|Change in concentration(M)|_+S |__+2S__|

|Equilibrium concentration(M)|_S _|2S___|

Ksp = [ Cu+² ] [ OH-] ²

2.2 ×10-²⁰ = (S)(2S)²= 4S³

$s = \sqrt[3]{ \frac{2.2 \times {10}^{ - 20} }{4} } = 1.8 \times {10}^{ - 7}$

S = 1.8 × 10-⁷ M

The molar solubility of Cu(OH)2 is 1.8 × 10-⁷ M

Solubility of Cu (OH)2 =

$Cu (OH)2 = \frac{1.8 \times {10}^{ - 7} mol \:Cu (OH)2 }{1L} \times \frac{97.546 \: g \: Cu (OH)2}{1 \: mol \: Cu (OH)2} \\ = 1.75428 \times 10 ^{ - 5}$

<h3>Solubility of Cu (OH)2 = 1.75428 × 10 -⁵ g/ L</h3>

I hope I helped you^_^

8 0

3 years ago

Particle quantity/number of Cu(NO3)2

Murrr4er [49]

Answer:

Cupid nitrate is what I'm going for

8 0

3 years ago

Which equation represents a neutralization reaction?

Harrizon [31]

Answer: HCI + KOH → KCI + H20

Explanation:

HCI(aq) + KOH(aq) → KCI(aq) + H20(l)

Acid + base → Salt + Water.

The above is a neutralization reaction in which an acid, aqeous HCl reacts completely with an appropriate amount of a base, aqueous KOH to produce salt, aqueous KCl and water, liquid H2O only.

This is a neutralization reaction since, the hydrogen ion, H+, from the HCl is neutralized by the hydroxide ion, OH-, from the KOH to form the water molecule, H2O and salt, KCl only.

3 0

3 years ago

Water has a boiling point of 100.0°C and a Kb of 0.512°C/m. What is the boiling

Reil [10]

Answer:

The boiling point of a 8.5 m solution of Mg3(PO4)2 in water is 394.91 K.

Explanation:

The formula for molal boiling Point elevation is :

$\Delta T_{b} = iK_{b}m$

$\Delta T_{b}$ = elevation in boiling Point

$K_{b}$ = Boiling point constant( ebullioscopic constant)

m = molality of the solution

i = Van't Hoff Factor

Van't Hoff Factor = It takes into accounts,The abnormal values of Temperature change due to association and dissociation .

In solution Mg3(PO4)2 dissociates as follow :

$Mg_{3}(PO_{4})_{2}\rightarrow 3Mg^{2+} + 2 PO_{4}^{3-}$

Total ions after dissociation in solution :

= 3 ions of Mg + 2 ions of phosphate

Total ions = 5

i = Van't Hoff Factor = 5

m = 8.5 m

$K_{b}$ = 0.512 °C/m

Insert the values and calculate temperature change:

$\Delta T_{b} = iK_{b}m$

$\Delta T_{b} = 5\times 0.512\times 8.5$

$\Delta T_{b} = 21.76 K$

Boiling point of pure water = 100°C = 273.15 +100 = 373.15 K

$\Delta T_{b} = T_{b} - T_{b}_{pure}$

$T_{b}_{pure}$ = 373.15 K[/tex]

21.76 = T - 373.15

T = 373.15 + 21.76

T =394.91 K

8 0

3 years ago

What class of enzyme catalyzes addition of oxygen?

bulgar [2K]

Answer:

The answer is Lyase

Explanation:

Any enzyme that catalyzes the addition or removal of the elements water (hydrogen, oxygen), ammonia (nitrogen, hydrogen), or carbon dioxide (carbon, oxygen) at double bonds, as defined in physiology. Decarboxylases, for example, remove carbon dioxide from amino acids, while dehydrases eliminate water.

8 0

3 years ago