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-BARSIC- [3]
3 years ago
11

Ammonium phosphate ((NH4)3 PO4) is an important ingredient to many fertilizers. It can be made by reacting phosphoric acid (H3 P

O4) with ammonia (NH3)
What mass of ammonia phosphate is produced by the reaction of 5.24 g of ammonia?
Chemistry
1 answer:
lara31 [8.8K]3 years ago
4 0

Answer:

15.35 g of (NH₄)₃PO₄

Explanation:

First we need to look at the chemical reaction:

3 NH₃ + H₃PO₄ → (NH₄)₃PO₄

Now we calculate the number of moles of ammonia (NH₃):

number of moles = mass / molecular wight

number of moles = 5.24 / 17 = 0.308 moles of NH₃

Now from the chemical reaction we devise the following reasoning:

if         3 moles of NH₃ are produce 1 mole of (NH₄)₃PO₄

then   0.308 moles of NH₃ are produce X moles of (NH₄)₃PO₄

X = (0.308 × 1) / 3 = 0.103 moles of (NH₄)₃PO₄

mass = number of moles × molecular wight

mass = 0.103 × 149 = 15.35 g of (NH₄)₃PO₄

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Mercury is the only metal that is a liquid at room temperature. When mercury vapor is inhaled, it is readily absorbed by the lun
Lapatulllka [165]

Answer:

P = 0.0166 mm Hg

Explanation:

To solve this question, we need to use the Clausius Clapeyron equation, which is a commonly used expression to calculate vapour pressure at a given temperature. We have the enthalpy of vaporization of the mercury, so, let's write the equation:

Clausius Clapeyron equation:

Ln (P₂ / P₁) = (-ΔHv / R)(1/T₂ - 1/T₁)    (1)

Where:

R: universal constant of gases (8.314 J / K.mol)

P₂: Vapour pressure at 43°C (or 316 K)

P₁: Pressure of mercury at the boiling point (1 atm)

T₂: temperature at 43 °C

T₁: Boiling point of mercury (357 °C or 630 K)

As we are given the boiling point of the mercury, we can safely assume that the pressure at this point is 1 atm, becuase remember that when a sustance boils, is because it's internal pressure has reached the atmospherical pressure of 1 atm. With this clear, all we just need to do is solve for P₂. We are going to do this very slowly so you can understand the process. First let's replace the given data:

Ln (P₂ / 1) = (-59100 J/mol / 8.314 J / K.mol) (1/316 - 1/630)

Ln P₂ = -7108.49 * (3.16x10⁻³ - 1.59x10⁻³)

Ln P₂ = -7108.49 * (1.51x10⁻³)

Ln P₂ = -10.7338

P₂ = 10⁽⁻¹⁰°⁷³³⁸⁾

P₂ = 2.18x10⁻⁵ atm

We can express this value in mm Hg and it will be:

P₂ = 2.18x10⁻⁵ * 760

<h2>P₂ = 0.0166 mm Hg</h2>

Hope this helps

8 0
3 years ago
According to Table I, which equation represents a change resulting in the greatest quantity of energy released?
love history [14]
The answer is 3. The releasing of energy means exothermic reaction. So the ΔH should be negative. And the greatest quantity of energy released means that the greatest number. So according to the table I, the answer is 3.
8 0
3 years ago
Read 2 more answers
In this experiment, we will be performing a titration with a buret. place the steps in order. 1. record the ph when 0.0 ml of na
77julia77 [94]

I am guessing that your solutions of HCl and of NaOH have approximately the same concentrations. Then the equivalence point will occur at pH 7 near 25 mL NaOH.

The steps are already in the correct order.

1. Record the pH when you have added 0 mL of NaOH to your beaker containing 25 mL of HCl and 25 mL of deionized water.

2. Record the pH of your partially neutralized HCl solution when you have added 5.00 mL of NaOH from the buret.

3. Record the pH of your partially neutralized HCl solution when you have added 10.00 mL, 15.00 mL and 20.00 mL of NaOH.

4. Record the NaOH of your partially neutralized HCl solution when you have added 21.00 mL, 22.00 mL, 23.00 mL and 24.00 mL of NaOH.

5. Add NaOH one drop at a time until you reach a pH of 7.00, then record the volume of NaOH added from the buret ( at about 25 mL).

6. Record the pH of your basic HCl-NaOH solution when you have added 26.00 mL, 27.00 mL, 28.00 mL, 29.00 mL and 30.00 mL of NaOH.

7. Record the pH of your basic HCl-NaOH solution when you have added 35.00 mL, 40.00 mL, 45.00 mL and 50.00 mL of NaOH from your 50mL buret.

4 0
3 years ago
A generic weak acid with formula HA has a Ka = 2.76 x 10-8. Calculate the Kb for the conjugate base of the acid.
Reika [66]

Answer:

3.62x10⁻⁷ = Kb

Explanation:

The acid equilibrium of a weak acid, HX, is:

HX + H₂O ⇄ X⁻ + H₃O⁺

Where Ka = [X⁻] [H₃O⁺] / [HX]

And basic equilibrium of the conjugate base, is:

X⁻ + H₂O ⇄ OH⁻ + HX

Where Kb = [OH⁻] [HX] / [X⁻]

To convert Ka to Kb we must use water equilibrium:

2H₂O ⇄ H₃O⁺ + OH⁻

Where Kw = 1x10⁻¹⁴ = [OH⁻] [H₃O⁺]

Thus, we can obtain:

Kw = Ka*Kb

Solving for Kb:

Kw / Ka = Kb

1x10⁻¹⁴ /  2.76x10⁻⁸ =

3.62x10⁻⁷ = Kb

4 0
3 years ago
How many electrons would be found in an isotope of silicon 29
kenny6666 [7]
14 electrons15 neutrons
5 0
3 years ago
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