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3 years ago

Calculate the mass of one mole of these substances.

Chemistry

1 answer:

pogonyaev3 years ago

6 0

Answer:

a. 53.5 g/mol

b. 80.06 g/mol

c. 133.33 g/mol

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

Reading a Periodic Table
Molar Mass - 1 mol per <em>x</em> grams substance

Explanation:

<u>Step 1: Define</u>

a. NH₄Cl

b. NH₄NO₃

c. AlCl₃

<u>Step 2: Find masses</u>

Molar Mass of N - 14.01 g/mol

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Al - 26.98 g/mol

Molar Mass of Cl - 35.45 g/mol

<u>Step 3: Calculate compound masses</u>

Molar Mass of NH₄Cl - 14.01 g/mol + 4(1.01 g/mol) + 35.45 g/mol = 53.5 g/mol

Molar Mass of NH₄NO₃ - 2(14.01 g/mol) + 4(1.01 g/mol) + 3(16.00 g/mol) = 80.06 g/mol

Molar Mass of AlCl₃ - 26.98 g/mol + 3(35.45 g/mol) = 133.33 g/mol

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Answer:

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3 years ago

An ideal gas sealed in a rigid 4.86-L cylinder, initially at pressure Pi=10.90 atm, is cooled until the pressure in the cylinder

seraphim [82]

Answer:

$\Delta H=-11897J$

Explanation:

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In this case, it is widely known that for isochoric processes, the change in the enthalpy is computed by:

$\Delta H=\Delta U+V\Delta P$

Whereas the change in the internal energy is computed by:

$\Delta U=nCv\Delta T$

So we compute the initial and final temperatures for one mole of the ideal gas:

$T_1= \frac{P_1V}{nR}=\frac{10.90atm*4.86L}{0.082*n}=\frac{646.02K }{n} \\\\T_2= \frac{P_2V}{nR}=\frac{1.24atm*4.86L}{0.082*n}=\frac{73.49K }{n}$

Next, the change in the internal energy, since the volume-constant specific heat could be assumed as ³/₂R:

$\Delta U=1mol*\frac{3}{2} (8.314\frac{J}{mol*K} )*(73.49K-646.02K )=-7140J$

Then, the volume-pressure product in Joules:

$V\Delta P=4.86L*\frac{1m^3}{1000L} *(1.24atm-10.90atm)*\frac{101325Pa}{1atm} \\\\V\Delta P=-4756.96J$

Finally, the change in the enthalpy for the process:

$\Delta H=-7140J-4757J\\\\\Delta H=-11897J$

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3 years ago

When the fuel mixture contained in a 1.49 l tank, stored at 750 mmhg and 298 k, undergoes complete combustion, how much heat is

Ket [755]

Get to know first how many moles in the gas:n = pV/RT= (1.013*10^5*750/760) Pa *1.49*10^-3 m^3/(8.314 J/(molK)*298) n = 0.0601 moles.
The combustion energies are 889 kJ/mol (methane) and 2 220 kJ (propane) x = moles methane, y = moles propane
x*889 + y*2220 = 778 x + y = 0.0601----------- x = 0.267784 moles = 0.267784*100/0.0601 = 44.6 % y = 0.243216 moles = 0.243216*100/0.0601 = 55.4 %

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3 years ago

PLSS HELP ASAP it's due

BartSMP [9]

<h3>Answer:</h3>

87.02%

<h3>Explanation:</h3>

Percent yield of a product in a chemical equation is the ratio of actual or experimental yield to theoretical yield expressed as percentage.

In this case we are given;

Mass of Zinc as 7.23 g

Actual volume of Hydrogen gas produced as 2.16 L

We are required to calculate the percentage yield of Hydrogen gas;

<h3>Step 1: Write a balanced equation for the reaction</h3>

The balanced equation for the reaction between Zinc metal and Hydrochloric acid is given by;

Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

<h3 /><h3>Step 2: Moles of Zinc metal that reacted </h3>

Moles are given by dividing mass with molar mass

Moles = Mass ÷ Molar mass

Molar mass of Zinc = 65.38 g/mol

Therefore;

Number of moles = 7.23 g ÷ 65.38 g/mol

= 0.1106 moles

<h3>Step 3: Calculate the number of moles of Hydrogen gas produced</h3>

From the equation 1 mole of Zinc results in the formation of 1 mole of Hydrogen gas.

Therefore, Moles of hydrogen gas = Moles of Zinc × 1

= 0.1106 moles × 1

= 0.1106 moles

<h3>Step 4: Calculate the theoretical volume of Hydrogen gas produced.</h3>

At STP, 1 mole of a gas occupies 22.4 Liters

Therefore;

Volume of Hydrogen = Number of moles × 22.4 L

= 0.1106 mole × 22.4 L

= 2.477 L

<h3>Step 5: Calculate the percent yield </h3>

Percent yield = (Actual yield ÷ Theoretical Yield) × 100%

= (2.16 L ÷ 2.477 L)× 100%

= 87.20%

Thus, the percent yield of hydrogen gas produced is 87.02%

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Answer:

Heat, solutes and high temperature.

Explanation:

A supersaturated solution can be formed by dissolving solute more solute in solvent by increasing temperature of the solution. A supersaturated solution contains more quantity of solutes than can be dissolved in the solvent at room temperature. A solution may remain supersaturated until the solution has high temperature and when the temperature started lower, the extra dissolve solutes begin undissolved and remain suspended in the solution.

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