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4 years ago

I need help on question 5

Chemistry

1 answer:

Naily [24]4 years ago

4 0

Pretty sure the answer is A, but you probably shouldn't use Brainly to answer test questions.

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How has technology helped to reduce the spread of infectious disease?

Serga [27]

Technology is a wide spread human resource that many people use today, technology helps us cure fatal diseases or just the start of symptoms or getting sick. Technology has also helped us learn more about the world and what is
T was like before our time has come.

6 0

4 years ago

1.00 M CaCl2 Density = 1.07 g/mL

Lesechka [4]

Explanation:

Molarity of solution = 1.00 M = 1.00 mol/L

In 1 L of solution 1.00 moles of calcium chloride is present.

Mass of solute or calcium chloride = m

$m = 1 mol\times 111 g/mol = 111 g$

Mass of solution = M

Volume of solution = V = 1L = 1000 mL

Density of solution , d= 1.07 g/mL

$M=d\times V=1.07 g/mL\times 1000 mL=1,070 g$

1) The value of %(m/M):

$\frac{m}{M}\times 100=\frac{111 g}{1,070 g}\times 100=10.37\%$

2) The value of %(m/V):

$\frac{m}{V}\times 100=\frac{111 g}{1000 L}\times 100=11.1\%$

$Molality = \frac{\text{Moles of compound }}{\text{mass of solvent in kg}}$

$Normality=\frac{\text{Moles of compound }}{n\times \text{volume of solution in L}}$

n = Equivalent mass

n = $\frac{\text{molar mass of ion}}{\text{charge on an ion}}$

3) Normality of calcium ions:

Moles of calcium ion = 1 mol (1 $CaCl_2$ mole has 1 mole of calcium ion)

$n=\frac{40 g/mol}{2}=20$

$=\frac{1 mol}{20 g/mol\times 1L}=0.050 N$

4) Normality of chlorine ions:

Moles of chlorine ion = 2 mol (1 $CaCl_2$ mole has 2 mole of chlorine ion)

$n=\frac{35.5 g/mol}{1}=35.5$

$=\frac{2 mol}{35.5 g/mol\times 1L}=0.056 N$

Moles of calcium chloride = 1.00 mol

Mass of solvent = Mass of solution - mass of solute

= 1,070 g - 111 g = 959 g = 0.959 kg ( 1 g =0.001 kg)

5) Molality of the solution :

$\frac{1 mol}{0.959 kg}=1.043 mol/kg$

Moles of calcium chloride = $n_1=1mol$

Mass of solvent = 959 g

Moles of water = $n_2=\frac{959 g}{18 g/mol}=53.28 mol$

Mass of solvent = 959 g

6) Mole fraction of calcium chloride =

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{1mol}{1 mol+53.28 mol}=0.01842$

7) Mole fraction of water =

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{53.28 mol}{1mol+53.28 mol}=0.9816$

8) Mass of solution = m'

Volume of the solution= v = 100 mL

Density of solution = d = 1.07 g/mL

$m'=d\times v=1.07 g/ml\times 100 g= 107 g$

Mass of 100 mL of this solution 107 grams of solution.

9) Volume of solution = V = 100 mL

Mass of solution = M'' = 107 g

Mass of solute = m

The value of %(m/V) of solution = 11.1%

$11.1\%=\frac{m}{100 mL}\times 100$

m = 11.1 g

Mass of solvent = M''- m = 107 g -11.1 g = 95.9 g

95.9 grams of water was present in 100 mL of given solution.

3 0

3 years ago

Determine the number of protons, neutrons, and electrons in the following isotopes that are used in medical diagnoses: (a) atomi

il63 [147K]

Answer:

7-c=760

Explanation:

the answer is 7-c=7698

7 0

4 years ago

Which of the following happens to a molecule of an object when the object is cooled?

Romashka [77]

Loses kinetic energy

8 0

3 years ago

There are 3.45 moles of (NH4)2 SO4, how many grams is this?

blsea [12.9K]

Answer:

455.4 g

Explanation:

Data given:

no. of moles of (NH₄)₂SO₄= 3.45 mol

mass of (NH₄)₂SO₄ = ?

Solution

Formula will be used

no.of moles = mass in grams / molar mass

Rearrange the above equation for mass

mass in grams = no. of moles x molar mass . . . . . . . . (1)

molar mass of (NH₄)₂SO₄

molar mass of (NH₄)₂SO₄ = 2(14 + 4(1)) + 32 + 4(16)

molar mass of (NH₄)₂SO₄ = 2 (14 +4) + 32 + 64

molar mass of (NH₄)₂SO₄ = 2 (18) + 32 + 64

molar mass of (NH₄)₂SO₄ = 36 + 32 + 64 = 132 g/mol

Put values in equation 1

mass in grams = 3.45 mole x 132 g/mol

mass in grams = 455.4 g

So,

mass of (NH₄)₂SO₄ = 455.4 g

8 0

3 years ago