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ankoles [38]
2 years ago
5

What is the percent composition of (NH4)2S?

Chemistry
1 answer:
enyata [817]2 years ago
3 0

(NH₄)₂S

%N=  41.18%

%H =11.76%

%S=47.06%

Mg(NO₃)₂

%Mg=32.43%

%N=18.92%

%O=(64.86%

<h3>Further explanation</h3>

Given

(NH₄)₂S

Mg(NO₃)₂

Required

The percent composition

Solution

(NH₄)₂S MW=68 g/mol

%N=2. Ar N  / MW (NH₄)₂S x 100%

%N= (2.14/68) x 100% = 41.18%

%H =(8.1/68) x 100%=11.76%

%S=(32/68)x 100%=47.06%

Mg(NO₃)₂ MW=148 g/mol

%Mg=(2.24/148)x100%=32.43%

%N=(2.14/148)x100%=18.92%

%O=(6.16/148)x100%=64.86%

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The normal boiling point of bromine is 58.8°C, and its enthalpy of vaporization is 30.91 kJ/mol. What is the approximate vapor p
saul85 [17]

Answer : The vapor pressure of bromine at 10.0^oC is 0.1448 atm.

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of bromine at 10.0^oC = ?

P_2 = vapor pressure of propane at normal boiling point = 1 atm

T_1 = temperature of propane = 10.0^oC=273+10.0=283.0K

T_2 = normal boiling point of bromine = 58.8^oC=273+58.8=331.8K

\Delta H_{vap} = heat of vaporization = 30.91 kJ/mole = 30910 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{1atm}{P_1})=\frac{30910J/mole}{8.314J/K.mole}\times (\frac{1}{283.0K}-\frac{1}{331.8K})

P_1=0.1448atm

Hence, the vapor pressure of bromine at 10.0^oC is 0.1448 atm.

4 0
3 years ago
The Sun heats land and water, how is this related to wind and ocean currents?
Yuki888 [10]

Answer:

"The sun warms up parts of the oceans. Warm waters rise just like warm air rises. So, as the warmer ocean waters begin to rise in a particular area, the cooler ocean waters from a different area will move in to replace the warmer ocean waters, and this creates our ocean currents."

Explanation:

Hope this is helpful :)

8 0
2 years ago
A large container holds 2 liters of water. 45.0 grams of NaCl is added to the container. What is the molarity of sodium chloride
djverab [1.8K]

Answer:

THE MOLARITY OF SODIUM CHLORIDE IN THE CONTAINER IS 0.3846 M.

Explanation:

Molarity of a solution is the number of moles of solute per dm3  of solution.

Mass concentration = Molar concentration * Molar mass

1. calculate the mass concentration;

Mass conc. = 45 g in 2 L

= 45 g in 2 dm3

In 1 dm3, the mass will be 45 / 2

= 22.5 g/dm3 of NaCl.

2. Calculate the molar mass;

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Molar mass = ( 23 + 35.5 ) g/mol

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3. calculate the molarity

Molarity = mss concentration / molar mass

Molarity = 22.5 g/dm3 / 58.5 g/mol

Molarity = 0.3846 mol/dm3 of NaCl.

The molarity of sodium chloride in the container is 0.3846 mol/dm3

4 0
3 years ago
Calculate the molarity of each of the following solutions. Use the periodic table if necessary. A salt solution with a volume of
Andru [333]

The molarity of the NaCl in the 250 ml slat solution is 2.8M.

<h3>What is molarity?</h3>

The molarity of a solution is defined as the concentration of a solute per unit volume of the solution.

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M = \dfrac{n}{V}\\\\\\M = \dfrac{0.70}{0.25} = 2.8 M

Thus, the molarity is 2.8 M.

Learn more about molarity

brainly.com/question/14693949

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Stolb23 [73]
The percent by volume of ethanol in the solution 30% ((150/500)*100%) based on the Law of Conservation of Mass. The object's mass before a reaction and after a reaction is equal according to the Law of Conservation of Mass. The dissolved ethanol become the part of the solution but it does not decrease its volume.
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