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Sav [38]
3 years ago
5

Ammonia reacts with sulfuric acid to produce the important fertilizer, ammonium hydrogen sulfate.

Chemistry
1 answer:
Liula [17]3 years ago
5 0

Answer:

404.8g of (NH4)HSO4 is produced.

Explanation:

Step 1:

Data obtained from the question. This include the following:

Temperature (T) = 10°C = 10°C + 273 = 283K

Pressure (P) = 110KPa = 110/101.325 = 1.09atm

Volume (V) = 75L

Step 2:

Determination of the number of mole of ammonia, NH3.

The number of mole (n) of ammonia, NH3 can be obtained by using the ideal gas equation. This is illustrated below:

Note:

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

Number of mole (n) =?

PV = nRT

1.09 x 75 = n x 0.0821 x 283

Divide both side by 0.0821 x 283

n = (1.09 x 75) /(0.0821 x 283)

n = 3.52 moles

Step 3:

Determination of the number of mole ammonium hydrogen sulfate produced from the reaction.

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

NH3 + H2SO4 —> (NH4)HSO4

From the balanced equation above,

1 mole of NH3 produced 1 mole of (NH4)HSO4

Therefore, 3.52 moles of NH3 will also produce 3.52 moles of (NH4)HSO4.

Therefore, 3.52 moles of ammonium hydrogen sulfate, (NH4)HSO4 is produced.

Step 4:

Conversion of 3.52 moles of ammonium hydrogen sulfate, (NH4)HSO4 to grams. This is illustrated below:

Molar Mass of (NH4)HSO4 = 14 + (4x1) + 1 + 32 + (16x4) = 115g/mol

Number of mole of (NH4)HSO4 = 3.52 moles

Mass of (NH4)HSO4 =..?

Mass = mole x molar Mass

Mass of (NH4)HSO4 = 3.52 x 115 = 404.8g

Therefore, 404.8g of (NH4)HSO4 is produced.

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Calculate the percentage of the void space out of the total volume occupied by 1 mole of water molecules. The density of water i
gtnhenbr [62]

Answer:

The percentage of the void space out of the total volume occupied is 93.11%

Explanation:

A mole of water contains 2 atoms of hydrogen and 1 atom of oxygen.

To calculate the volume of a mole of water, we calculate 2 times the volume of the hydrogen atom and 1 times the volume of the oxygen atom

Let's calculate this one after the other.

For the hydrogen, formula for the volume will be

V_{hydrogen = 2 × 4/3 × π × r_{H}^{3}

where r_{H}^{3} = 37 pm which is read as 37 picometer (1 picometer = 10^-12 m) = 37 × 10^{-12} meters

Volume of the hydrogen = 8/3 × (37 × 10^-12)^3 = 4.05 * 10^-31

we multiply this by the avogadro's number = 6.02 * 10^23

= 4.05 * 10^-31 * 6.02 * 10^23 = 2.6 * 10^-8 m^3

We do same for thr oxygen, but this time we do not multiply the volume of the oxygen by 2 as we have only one atom of oxygen

Volume of oxygen = 4/3 * π * (73 * 10^-12) ^3 * avogadro's number = 9.81 * 10^-7 m^3

adding both volumes together, we have 1.24 * 10^-6 m^3 or simply 1.24 ml ( 0.01 m = 1 ml)

Dividing the molar mass of one mole of water by its density, we can get the volume of 1 mole of water

= (18g/mol)/(1 g/ml) = 18 ml/mol

Now we proceed to calculate the volume of void = Total volume - volume of molecule = 18 - 1.24 = 16.76 ml

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7 0
3 years ago
Solve these.. . . . .​
KonstantinChe [14]

Answer:

Please find the ten balanced equations below

Explanation:

According to the question, a balanced equation must contain equal number of atoms on both sides of the equation. To balance an equation, coefficients are used. The following are the balanced equation:

1. C + O2 → CO2 (balanced)

2. 2H2 + O2 → 2H2O (balanced)

3. 2Na + 2H2O → 2NaOH + H2 (balanced)

4. 4Na + O2 → 2Na2O (balanced)

5. 2Na + Cl2 → 2NaCl (balanced)

6. NaOH + HCl → NaCl + H2O (balanced)

7. NaOH + HNO3 → NaNO3 + H2O (balanced)

8. 2NaOH + H2SO4 → Na2SO4 + 2H2O (balanced)

9. Na2CO3 + 2HCl → 2NaCl + H2O + CO2 (balanced)

10. 2NaOH + CO2 → Na2CO3 + H2O (balanced)

7 0
3 years ago
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Brrunno [24]

Answer:

H=mc∆T

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H=1254J

6 0
3 years ago
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