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frozen [14]
3 years ago
12

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 chlorine produced by the reaction of of ammonium perchlorate. Be sure your answer has a unit symbol, if necessary, and round it to significant digits.
Chemistry
1 answer:
Lorico [155]3 years ago
7 0

The question is incomplete , complete question is :

Ammonium perchlorate NH_4ClO_4 is a powerful solid rocket fuel, used in the Space Shuttle boosters. It decomposes into nitrogen gas, chlorine gas, oxygen gas and water vapor, releasing a great deal of energy. Calculate the moles of chlorine 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:

1.2 moles of chlorine produced by the reaction 2.5 moles of ammonium perchlorate.

Explanation:

2NH_4ClO_4(s)\rightarrow N_2(g)+Cl_2(g)+2O_2(g)+4H_2O(g)

Moles of ammonium perchlorate = 2.5 moles

According to reaction, 2 moles of ammonium perchlorate on decomposition gives 1 mole of chlorine gas, then 2.5 moles of ammonium perchlorate wil give:

\frac{1}{2}\times 2.5 mol=1.25 mol\approx 1.2 of chlorine gas

1.2 moles of chlorine produced by the reaction 2.5 moles of ammonium perchlorate.

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2. Balance the equation below, and answer the following question: What volume of chlorine gas, measured at STP, is needed to com
Juliette [100K]
Balanced equation: 2Na(s) + Cl₂(g) ---> 2NaCl(s)

when we have STP conditions, we can use this conversion: 1 mol = 22.4 L

first, we have to convert grams to molecules using the molar mass, and then use mole to mole ratio from the balanced equation. 

molar mass of Na= 23.0 g/mol
 ratio: 2 mol Na= 1 mol Cl₂ (based on coefficients of balanced equation)

calculations:

6.25 g Na ( \frac{1 mol Na}{23.0 g} ) ( \frac{1 mol Cl_2}{2 mol Na} ) ( \frac{22.4 L}{1 mol} )= 3.04 L
8 0
3 years ago
What is the mass of 1.2 x 1023 atoms of arsenic?
Gre4nikov [31]

Answer:

14.93 g

Explanation:

First we <u>convert 1.2 x 10²³ atoms of arsenic (As) into moles</u>, using <em>Avogadro's number</em>:

  • 1.2 x 10²³ atoms ÷ 6.023x10²³ atoms/mol = 0.199 mol As

Then we can<u> calculate the mass of 0.199 moles of arsenic</u>, using its<em> molar mass</em>:

  • 0.199 mol * 74.92 g/mol = 14.93 g

Thus, 1.2x10²³ atoms of arsenic weigh 14.93 grams.

6 0
3 years ago
How many moles are 21.67 L of NH4CI?
Gnoma [55]

Answer:

0.967mole

Explanation:

Given parameters:

Volume of NH₄Cl  = 21.67L

Unknown:

Number of moles  = ?

Solution:

If we assume that the volume was taken at standard temperature and pressure,

 Then;

  Number of moles  = \frac{volume }{22.4L}  

 Number of moles  = \frac{21.67}{22.4}   = 0.967mole

8 0
3 years ago
Explain what you think a compound is
Kryger [21]

Answer:

composed of two or more parts, elements, or ingredients:

Explanation:

6 0
3 years ago
The cell potential of a redox reaction occurring in an electrochemical cell under any set of temperature and concentration condi
avanturin [10]

Answer : The actual cell potential of the cell is 0.47 V

Explanation:

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given redox reaction is :

Ni^{2+}(aq)+Zn(s)\rightarrow Ni(s)+Zn^{2+}(aq)

The balanced two-half reactions will be,

Oxidation half reaction : Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction : Ni^{2+}+2e^-\rightarrow Ni

The expression for reaction quotient will be :

Q=\frac{[Zn^{2+}]}{[Ni^{2+}]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(0.0141)}{(0.00104)}=13.6

The value of the reaction quotient, Q, for the cell is, 13.6

Now we have to calculate the actual cell potential of the cell.

Using Nernst equation :

E_{cell}=E^o_{cell}-\frac{RT}{nF}\ln Q

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 316 K

n = number of electrons in oxidation-reduction reaction = 2 mole

E^o_{cell} = standard electrode potential of the cell = 0.51 V

E_{cell} = actual cell potential of the cell = ?

Q = reaction quotient = 13.6

Now put all the given values in the above equation, we get:

E_{cell}=0.51-\frac{(8.314)\times (316)}{2\times 96500}\ln (13.6)

E_{cell}=0.47V

Therefore, the actual cell potential of the cell is 0.47 V

4 0
3 years ago
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