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3 years ago

What is the molality (m) of a solution that contains 76.5 g of KCl dissolved in 85.0 g of water?

Chemistry

1 answer:

ladessa [460]3 years ago

5 0

Answer:

76.5g KCl/74.55 grams per mole Kcl = x

molality= x/.085 kg H2O

Explanation:

well remember molality is moles of solute/kilograms of solvent. So it's the moles of KCl over 85 g of h20 converted into kg. if this makes sense.

You might be interested in

Which of the following samples contains the greatest number of atoms?

mart [117]

Answer:

C. 11 moles of N2O

Explanation:

A. CO2 exists as a molecular compound. The number of atoms present = 1 atom of Carbon and 2 atoms oxygen = 3 atoms * 9 = 18 atoms

B. Xe exists as an atom. Number of atoms present = 10 * 1 atom = 10 atoms

C. N20 contains 3 atoms; 2 atoms of Nitrogen and 1 atom of oxygen.

Number of atoms present = 3 * 11 = 33 atoms

D. CO contains 2 atoms, 1 each of carbon and oxygen.

Number of atoms present = 12 * 2 = 24 atoms

7 0

3 years ago

Grace wanted to find out the best conditions for growing lettuce plants.

stiks02 [169]

Explanation:

the investigation lasts for 7 days.

hope this helps you.

6 0

3 years ago

Calculemos la formula empirica y molecular de carbonato de hierro Fe2(co3)3.Fe.38.3%C.12.3%.0.4 92% masa 292g

DENIUS [597]

Respuesta:

292 g / mol;

Por favor, consulte la explicación.

Explicación:

El número de átomos de cada elemento en el compuesto: Fe2 (co3) 3.

Fe = 2; C = 3; 0 = 3 * 3 = 9

Fe2 = 112 g / mol

C = 12 g / mol

O = 16 g /

Masa molecular = ((112 + (12 * 3) + (16 * 9)

= (112 + 36 + 144)

= 292 g / mol.

Total = 2 + 3 + 9 = 14

Fe2 = 112/292 * 100% = 38,356%

C = 36/292 * 100% = 12,328 * '

O = (16 * 9) / 292 * 100% = 49,32%

7 0

3 years ago

I need help pls, and fast.

Vadim26 [7]

Answer:

1) Liquid forms drops that are dome-shaped

2) low surface tension

3) low viscosity

4) Liquid is thick and pours very slowly

Explanation:

It makes sense just use the stuff that's already in the table. It usually works.

7 0

4 years ago

When 50 ml of 1.000x10^-1m pb(no3)2 solution was added to 50 ml of 1.000x10^-1m nai solution?

podryga [215]

Balanced chemical reaction: Pb(NO₃)₂ (aq) + 2NaI(aq) → 2PbI₂(s) + 2NaNO₃(aq).

V(Pb(NO₃)₂) = 50 mL ÷ 1000 mL = 0.05 L, volume of solution.

c(Pb(NO₃)₂) = 0.1 mol/L; concentration of solution.

n(Pb(NO₃)₂) = c(Pb(NO₃)₂) · V(Pb(NO₃)₂).

n(Pb(NO₃)₂) = 0.1 mol/L · 0.05 L.

n(Pb(NO₃)₂) = 0.005 mol.

n(NaI) = c(NaI) · V(NaI).

n(NaI) = 0.1 mol/L · 0.05 L.

n(NaI) = 0.005 mol; amount of substance.

From chemical reaction: n(Pb(NO₃)₂) : n(NaI) = 1 : 2.

n(Pb(NO₃)₂) = 0.005 mol ÷ 2.

n(Pb(NO₃)₂) = 0.0025 mol; number of moles Pb(NO₃)₂ used.

n(NaI) = 0.005 mol; number of moles NaI used.

The limiting reagent is Pb(NO₃)₂.

n(PbI₂) = 0.005 mol.

m(PbI₂) = n(PbI₂) · M(PbI₂).

m(PbI₂) = 0.005 mol · 461 g/mol.

m(PbI₂) = 2.305 g; the theoretical yield of PbI₂.

3 0

3 years ago