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Levart [38]
2 years ago
5

Calculate the minimum mass of ammonia needed to produce 396 kg of ammonium sulfate, (NH4)2SO4, and excess sulfuric acid.

Chemistry
1 answer:
goldenfox [79]2 years ago
8 0

Answer:

Mass =  101891.2 g

Explanation:

Given data:

Mass of ammonia needed = ?

Mass of ammonium sulfate produced = 396 Kg = 396 ×1000 = 396000 g

Solution:

Chemical equation:

2NH₃ + H₂SO₄      →      (NH₄)₂SO₄

Number of moles of  (NH₄)₂SO₄:

Number of moles = mass/molar mass

Number of moles = 396000 g / 132.14 g/mol

Number of moles = 2996.8 mol

Now we will compare the moles of ammonia and ammonium sulfate

               (NH₄)₂SO₄          :        NH₃

                      1                   :         2

                  2996.8            :        2/1×2996.8 = 5993.6 mol

Mass of ammonia needed:

Mass = number of moles × molar mass

Mass = 5993.6 mol × 17 g/mol

Mass =  101891.2 g

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Alexeev081 [22]

Answer:

Amount left after 25 days = 12.5 g

Explanation:

Given data:

Mass of sample = 400 g

Half life of sample = 5 days

Mass left after 25 days = ?

Solution:

First of all we will calculate the number of half lives passes in given time period.

Number of half lives = Time elapsed / Half life

Number of half lives = 25 days/ 5 days

Number of half lives = 5

At time zero = 400 g

At 1st half life = 400 g/2 = 200 g

At 2nd half life = 200 g/2 = 100 g

At 3rd half life = 100 g/2 = 50 g

At 4th half life = 50 g/2 = 25 g

At 5th half life = 25 g/2 = 12.5 g

4 0
3 years ago
How many molecules of fructose would you have if you have 0.7 moles of<br> fructose?<br> *
iogann1982 [59]

Answer:

4.214 × 10^23 molecules.

Explanation:

Number of molecules in a substance can be calculated by multiplying the number of moles in that substance by Avagadro's number, which is 6.02 × 10^23.

That is, no. of molecule = n × Avagadro constant

In this case, there are 0.7 moles of fructose. Hence;

number of molecules = 0.7 × 6.02 × 10^23

no. of molecule = 4.214 × 10^23 molecules.

5 0
2 years ago
The coefficients in a chemical equation??
cricket20 [7]
D is the answer I believe.
3 0
3 years ago
What is the pH at the half-equivalence point in the titration of a weak base with a strong acid? The pKb of the weak base is 7.9
ValentinkaMS [17]

Hey there!

Given the reaction:

B + H⁺   => HB⁺


At half-equivalence point :  [B] = [HB⁺]

=> [B] / [HB⁺] = 1

Henderson-Hasselbalch equation :


pH = pKa + log ( [B] ) / ( HB⁺)]

pH = 14 - pKb + log ( 1 )

pH = 14 -  7.95 + 0

pH = 6.05


Answer C


Hope that helps!



5 0
3 years ago
Please help, thank you:))
frez [133]

Explanation:

15 8 O ---> 15 7 N + 0 1 e

Element: Oxygen

Mass Number: 16 u

Decay: Beta

238 92 U ----> 234 90 Th + -4 -2 e

Element: Uranium

Mass Number: 238.051 u

Decay: Alpha

4 0
3 years ago
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