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2 years ago

Calculate the mole fraction of cai2 in an aqueous solution prepared by dissolving 0.400 moles of cai2 in 850.0 g of water.

1 answer:

7 0

1) Formulas:

a) mole fraction of component 1, X1

X1 = number of moles of compoent 1 / total number of moles

b) Molar mass = number grams / number of moles => number of moles = number of grams / molar mass

2) Application

Number of moles of CaI2 = 0.400

Molar mass of water = 18.0 g/mol

Number of moles of water: 850.0 g / 18.0 g/mol = 47.22 mol

Total number of moles = 0.400 + 47.22 =47.62

Molar fraction of CaI2 = 0.400 / 47.62 = 0.00840

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Amino acids are the building blocks of proteins. The simplest amino acid is glycine (H2NCH2COOH). Draw a Lewis structure for gly

VashaNatasha [74]

Answer:

Explanation:

The lewis structure (indicating all the atoms and patterns provided as hint in the question) of glycine can be seen in the attachment below. While the chemical structure of glycine can be seen below

H₂N - C - C =O

| \

H OH

The structure (of glycine) above provides a "fair idea" of how the lewis structure will be.

3 0

2 years ago

During a titration the following data were collected. A 10. mL portion of an unknown monoprotic acid solution was titrated with

Liula [17]

Answer:

8.0 moles

Explanation:

Since the acid is monoprotic, 1 mole of the acid will be required to stochiometrically react with 1 mole of NaOH.

Using the formula: $\frac{concentration of acid X volume of acid}{concentration of base X volume of base} = \frac{mole of acid}{mole of base}$

Concentration of acid = ?

Volume of acid = 10 mL

Concentration of base = 1.0 M

Volume of base = 40 mL

mole of acid = 1

mole of base = 1

Substitute into the equation:

$\frac{concentration of acid X 10}{1.0 X 40} = \frac{1}{1}$

Concentration of acid = 40/10 = 4.0 M

To determine the number of moles of acid present in 2.0 liters of the unknown solution:

Number of moles = Molarity x volume

molarity = 4.0 M

Volume = 2.0 Liters

Hence,

Number of moles = 4.0 x 2.0 = 8 moles

8 0

2 years ago

What is the concentration of a solution made with 0.150 moles of KOH in 400.0 mL of solution?

otez555 [7]

Answer:

$concentration = \frac{0.15}{0.4}=0.375 mol/L$

Explanation:

Concentration: i is defined as the mole per litre.

$concentration = \frac{mole}{volume in L}$

mole=0.15

volume=400 ml=0.4 litre

$concentration = \frac{0.15}{0.4}=0.375 mol/L$

6 0

2 years ago

Website

vredina [299]

Answer:

volume in L = 0.25 L

Explanation:

Given data:

Mass of Cu(NO₃)₂ = 2.43 g

Volume of KI = ?

Solution:

Balanced chemical equation:

2Cu(NO₃)₂ + 4KI → 2CuI + I₂ + 4KNO₃

Moles of Cu(NO₃)₂:

Number of moles = mass/ molar mass

Number of moles = 2.43 g/ 187.56 g/mol

Number of moles = 0.013 mol

Now we will compare the moles of Cu(NO₃)₂ with KI.

Cu(NO₃)₂ : KI

2 : 4

0.013 : 4 × 0.013=0.052 mol

Volume of KI:

<em>Molarity = moles of solute / volume in L</em>

volume in L = moles of solute /Molarity

volume in L = 0.052 mol / 0.209 mol/L

volume in L = 0.25 L

6 0

2 years ago

What is the molality of sodium chloride in solution that is 13.0% by mass sodium chloride and that has a density of 1.10 g/ml?

8090 [49]

Answer is: molality od sodium chloride is 2,55 mol/kg.
V(solution) = 100 ml.
m(solution) = d(solution) · V(solution).
m(solution) = 1,10 g/ml · 100 ml.
m(solution) = 110 g.
ω(NaCl) = 13,0% = 0,13.
m(NaCl) = ω(NaCl) · m(solution).
m(NaCl) = 0,13 · 110 g.
m(NaCl) = 14,3 g.
n(NaCl) = m(NaCl) ÷ M(NaCl).
n(NaCl) = 14,3 g ÷ 58,5 g/mol.
n(NaCl) = 0,244 mol.
m(H₂O) = 110 g - 14,3 g.
m(H₂O) = 95,7 g = 0,0957 kg.
b(NaCl) = n(NaCl) ÷ m(H₂O).
b(NaCl) = 0,244 mol ÷ 0,0957 kg.
b(NaCl) = 2,55 mol/kg.

3 0

2 years ago