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myrzilka [38]
3 years ago
10

How many molecules are there in 2.30 grams of NH3

Chemistry
1 answer:
mixer [17]3 years ago
7 0
<span>Avogadro's number represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. This number can be used to convert the number of atoms or molecules into number of moles. We calculate as follows:

2.30 g NH3 (1 mol / 17.03 g ) (</span>6.022 x 10^23 molecules / 1 mol ) = 8.13x10^22 molecules
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How many milliliters of nitrogen, N2, would have to be collected at 99.19 kPa and 28oC to have a sample containing 0.015 moles o
Semmy [17]

Answer:

378mL

Explanation:

The following data were obtained from the question:

Pressure (P) = 99.19 kPa

Temperature (T) = 28°C

Number of mole (n) = 0.015 mole

Volume (V) =...?

Next, we shall convert the pressure and temperature to appropriate units. This is illustrated below:

For Pressure:

101.325 KPa = 1 atm

Therefore, 99.19 kPa = 99.19/101.325 = 0.98 atm

For Temperature:

T(K) = T(°C) + 273

T(°C) = 28°C

T(K) = 28°C + 273 = 301K.

Next we shall determine the volume of N2. The volume of N2 can be obtained by using the ideal gas equation as shown below:

PV = nRT

Pressure (P) = 0.98 atm

Temperature (T) = 301K

Number of mole (n) = 0.015 mole

Gas constant (R) = 0.0821atm.L/Kmol.

Volume (V) =...?

0.98 x V = 0.015 x 0.0821 x 301

Divide both side by 0.98

V = (0.015 x 0.0821 x 301) /0.98

V = 0.378 L

Finally, we shall convert 0.378 L to millilitres (mL). This is illustrated below:

1L = 1000mL

Therefore, 0.378L = 0.378 x 1000 = 378mL

Therefore, the volume of N2 collected is 378mL

4 0
3 years ago
Show All Work For Brainliest
sattari [20]

Answer:

1) <u>16.8 L CO2</u>

<u>2) 36.96 L NH3</u>

<u>3) </u> <u>9.88 L CO2 </u>

<u>4) 56.99 L H2O</u>

Explanation:

How many liters of carbon dioxide gas will be produced when 75.0 g of calcium carbonate decomposes to form calcium oxide when at STP?

CaCO3 → CaO + CO2

Moles calcium carbonate = 75.0 grams / 100.09 g/mol

Moles calcium carbonate = 0.750 moles

For 1 mol CaCO3 we'll have 1 mol CaO and 1 mol CO2

For 0.750 moles CaCO3 we'll have 0.750 moles CO2

1 mol = 22.4 L

0.750 moles CO2 = 0.750 *22.4 L =<u> 16.8 L CO2</u>

2. Hydrogen gas reacts with 23.1 g of nitrogen gas to produce ammonia (NH3). What volume of ammonia will be produced at STP?

3H2 + N2 → 2NH3

Moles N2 = 23.1 grams / 28.0 g/mol

Moles N2 = 0.825 moles

For 3 moles H2 we need 1 mol N2 to produce 2 moles NH3

For 0.825 moles N2 we'll have 2*0.825 = 1.65 moles NH3

1 mol = 22.4 L

1.65 mol = 1.65 * 22.4 L = <u>36.96 L NH3</u>

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3. Iron (III) oxide reacts with carbon monoxide to form iron and carbon dioxide. How many liters of carbon dioxide will be produced from 23.5 g of iron (III) oxide when at STP?

Fe2O3 + 3CO → 2Fe + 3CO2

Moles Fe2O3 = 23.5 grams / 159.69 g/mol

Moles Fe2O3 = 0.147 moles

For 1 mol Fe2O3 we need 3 moles CO to produce 2 moles Fe and 3 moles CO2

For 0.147 moles Fe2O3 we'll have 3*0.147 = 0.441 moles CO2

1 mol = 22.4 L

0.441 moles = 22.4 * 0.441 = <u>9.88 L CO2 </u>

<u />

<u />

<u />

4.How many liters of water vapor would be produced in the combustion of 12.5L of ethane, C2H6 at STP?

2C2H6 + 7O2 →4CO2 + 6H2O

22.4 L = 1 mol

12.5 L = 0.848 moles C2H6

For 2 moles C2H6 we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

For 0.848 moles C2H6 we'll have 3*0.848 =  2.544 moles H2O

1 mol = 22.4 L

2.544 moles = 22.4 L * 2.544 = <u>56.99 L H2O</u>

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D - ingestion

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