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Rainbow [258]
3 years ago
10

Ammonium perchlorate is a powerful solid rocket fuel, used in the Space Shuttle boosters. It decomposes into nitrogen gas, chlor

ine gas, oxygen gas and water vapor, releasing a great deal of energy. Calculate the moles of water produced by the reaction of of ammonium perchlorate. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.
Chemistry
1 answer:
Sedbober [7]3 years ago
4 0

Question:

Ammonium perchlorate NH₄ClO₄ is a powerful solid rocket fuel, used in the Space Shuttle boosters. It decomposes into nitrogen N₂ gas, chlorine Cl₂ gas, oxygen O₂ gas and water vapor, releasing a great deal of energy. Calculate the moles of water produced by the reaction of 2.5 mol of ammonium perchlorate. Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits.

Answer:

The number of moles of water produced by the reaction of 2.5 mol of ammonium perchlorate is 5 moles (of H₂O).

Explanation:

To solve the question, we are required to know the amount of each reagent and products there are in the balanced chemical reaction as follows

2 NH₄ClO₄ → N₂ + Cl₂ + 2O₂ + 4H₂O

From the chemical reaction stoichiometry, it shows that 2 moles of ammonium perchlorate, NH₄ClO₄ produces 4  moles of H₂O

Therefore 2.5 moles of ammonium perchlorate will produce \frac{4}{2} 2.5 moles of H₂O or 5 moles of H₂O.

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Calculate the percentage difference in the fundamental vibrational wavenumbers of 23Na35Cl and 23Na37Cl on the assumption that t
stealth61 [152]

Answer:

1.089%

Explanation:

From;

ν =1/2πc(k/meff)^1/2

Where;

ν = wave number

meff = reduced mass or effective mass

k = force constant

c= speed of light

Let

ν =1/2πc (k/meff)^1/2  vibrational wave number for 23Na35 Cl

ν' =1/2πc(k'/m'eff)^1/2 vibrational wave number for 23Na37 Cl

The between the two is obtained from;

ν' - ν /ν  = (k'/m'eff)^1/2 - (k/meff)^1/2 / (k/meff)^1/2

Therefore;

ν' - ν /ν = [meff/m'eff]^1/2 - 1

Substituting values, we have;

ν' - ν /ν = [(22.9898 * 34.9688/22.9898 + 34.9688) * (22.9898 + 36.9651/22.9898 * 36.9651)]^1/2  -1

ν' - ν /ν = -0.01089

percentage difference in the fundamental vibrational wavenumbers of 23Na35Cl and 23Na37Cl;

ν' - ν /ν * 100

|(-0.01089)|  × 100 = 1.089%

4 0
3 years ago
What is my theoretical yield (in moles) of Potassium Bromide (KBr) if I start with 40 grams of Iron (II) Bromide [FeBr2]? moles
Verizon [17]
The reaction will be: FeBr2 + K --> KBr + Fe
Balancing gives: FeBr2 + 2K --> 2KBr + Fe
The molar mass of FeBr2 is 55.85 + 2*79.9 = 215.65 g/mol.
We divide 40 g / 215.65 g/mol = 0.185 mol FeBr2
Based on stoichiometry:
(0.185 mol FeBr2)(2 mol KBr/1 mol FeBr2) = 0.370 mol KBr
4 0
4 years ago
HELP a #QUEEN out pls??
Anastaziya [24]

Answer:

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Explanation:

jk jk.........

__5___SiO2 + ___2___CaC2 → __5___Si + ____2___CaO + __4___CO2

4 0
3 years ago
3 A roller coaster car travels up a hill until it reaches the top, slowly goes over, and speeds down the hill and through the nu
STALIN [3.7K]

Answer:

3rd law

Explanation:

8 0
3 years ago
9.0 mol Na2S can from 9.0 mol CuS and 8.0 mol CuSO4 can form 8.0 mol Cus.
ICE Princess25 [194]

Answer:

765.0 grams CuS

Explanation:

The limiting reagent is the reactant which completely reacts before the other reactant(s) is used up. When 9.0 moles Na₂S and 8.0 moles CuSO₄ react, it appears that CuSO₄ is the limiting reagent. You can tell because it results in the production of less product.

You can determine the mass of CuS by multiplying the moles by the molar mass. It is important to arrange the ratio in a way that allows for the cancellation of units.

Molar Mass (CuS): 95.62 g/mol

8.0 moles CuS               95.62 g
-------------------------  x  -----------------------  =  765.0 grams CuS
                                         1 mole

4 0
2 years ago
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