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2 years ago

What mass (in grams) of silver contains the same number of atoms as 5.59 grams of sulfur?

Chemistry

1 answer:

Amanda [17]2 years ago

7 0

Answer:

18.84 g of silver.

Explanation:

We'll begin by calculating the number atoms present in 5.59 g of sulphur. This can be obtained as follow:

From Avogadro's hypothesis,

1 mole of sulphur contains 6.02×10²³ atoms.

1 mole of sulphur = 32 g

Thus,

32 g of sulphur contains 6.02×10²³ atoms.

Therefore, 5.59 g of sulphur will contain = (5.59 × 6.02×10²³) / 32 = 1.05×10²³ atoms.

From the calculations made above, 5.59 g of sulphur contains 1.05×10²³ atoms.

Finally, we shall determine the mass of silver that contains 1.05×10²³ atoms.

This is illustrated below:

1 mole of silver = 6.02×10²³ atoms.

1 mole of silver = 108 g

108 g of silver contains 6.02×10²³ atoms.

Therefore, Xg of silver will contain 1.05×10²³ atoms i.e

Xg of silver = (108 × 1.05×10²³)/6.02×10²³

Xg of silver = 18.84 g

Thus, 18.84 g of silver contains the same number of atoms (i.e 1.05×10²³ atoms) as 5.59 g of sulfur

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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

For each of the esters provided, identify the alcohol and the carboxylic acid that reacted.

Veronika [31]

Answer:

52. The alcohol USED => methanol, CH3OH

The carboxylic acid USED => propanoic acid, CH3CH2COOH.

53. The alcohol USED => Ethanol, CH2CH3OH

The carboxylic acid USED => Formic acid, HCOOH.

Explanation:

52. To obtain Methyl propanoate, CH3CH2COOCH3, we simply react propanoic, CH3CH2COOH and methanol, CH3OH together as shown below:

CH3CH2COOH + CH3OH —> CH3CH2COOCH3 + H2O

The alcohol used: methanol, CH3OH

The carboxylic acid used: propanoic acid, CH3CH2COOH.

53. To obtain Ethyl methanoate, HCOOCH2CH3, we simply react

Formic acid, HCOOH and ethanol, CH3CH2OH together as show below:

HCOOH + CH3CH2OH —> HCOOCH2CH3 + H2O

The alcohol USED => Ethanol, CH2CH3OH

The carboxylic acid USED => Formic acid, HCOOH.

7 0

3 years ago

The decomposition of SOCl2 is first-order in SOCl2. If the half-life for the reaction is 4.1 hr, how long would it take for the

qaws [65]

4.1 h = 14760 s

t 1/2 = ln 2 / k 

k = rate reaction = 4.97 x 10^-5 

ln 0.045 / 0.36 = - 4.97 x 10^-5 t 

2.08 = 4.97 x 10^-5 t 

t = 41839.9 s = 11 h 37 min 19 s

3 0

3 years ago

Which of the following is evidence of a chemical change?

krek1111 [17]

the formation of a gas

5 0

3 years ago

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Yuliya22 [10]

Answer:

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