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Harman [31]
4 years ago
9

Methylhydrazine, CH6N2, is commonly used as a liquid rocket fuel. The heat of combustion of methylhydrazine is −1.30 × 103 kJ/mo

l. How much heat is released when 202.7 g of methylhydrazine is burned
Chemistry
1 answer:
Step2247 [10]4 years ago
7 0

Answer:

There is 5720 J of heat released.

Explanation:

Step 1: Data given

Molar mass of CH6N2 = 46.07 g/mol

The heat of combustion of methylhydrazine is −1.30 × 103 kJ/mol

Mass of methylhydrazine = 202.7 grams

Step 2: Calculate number of moles

Moles CH6N2 = mass of CH6N2 / Molar mass CH6N2

Moles CH6N2 = 202.7 grams / 46.07 g/mol

Moles CH6N2 = 4.4 moles

Step 3: Calculate heat released when 202.7 grams of CH6N2 is burned

q = 1300 J/mol * 4.4 moles

q = 5720 J

(Since there is heat released, ΔH = negative)

There is 5720 J released.

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Explanation:

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3 years ago
Explain how the number of chromosomes per cell is cut in half during meiosis in which the diploid parent cell produces haploid d
Igoryamba
<span>The number of chromosomes per cell is cut in half during meiosis </span>by separation of homologous chromosomes<span> in a </span>diploid cell<span> which become </span>haploid cells.

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3 0
3 years ago
How many moles of h2 can be formed if a 3.25g sample of Mg reacts with excess HCl
kogti [31]

Answer:

0.134 moles of H₂ can be formed if a 3.25g sample of Mg reacts with excess HCl

Explanation:

The balanced reaction is:

Mg + 2 HCl → MgCl₂ + H₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles react:

  • Mg: 1 mole
  • HCl: 2 moles
  • MgCl₂: 1 mole
  • H₂: 1 mole

Being:

  • Mg: 24. 31 g/mole
  • H: 1 g/mole
  • Cl: 35.45 g/mole

the molar mass of the compounds participating in the reaction is:

  • Mg: 24.31 g/mole
  • HCl: 1 g/mole + 35.45 g/mole= 36.45 g/mole
  • MgCl₂: 24.31 g/mole + 2*35.45 g/mole= 95.21 g/mole
  • H₂: 2*1 g/mole= 2 g/mole

Then, by stoichiometry of the reaction, the following quantities of mass participate in the reaction:

  • Mg: 1 mole* 24.31 g/mole= 24.31 g
  • HCl: 2 moles* 36.45 g/mole= 72.9 g
  • MgCl₂: 1 mole* 95.21 g/mole= 95.21 g
  • H₂: 1 mole* 2 g/mole= 2 g

Then you can apply the following rule of three: if by stoichiometry 24.31 grams of Mg form 1 mole of H₂, 3.25 grams of Mg how many moles of H₂ will they form?

moles of H_{2} =\frac{3.25 grams of Mg*1 mole of H_{2} }{24.31 grams of Mg}

moles of H₂= 0.134

<u><em>0.134 moles of H₂ can be formed if a 3.25g sample of Mg reacts with excess HCl</em></u>

7 0
4 years ago
What is the theoretical yield of SO3 produced by 8.96 g of S?
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Answer: Theoretical yield of SO_3 produced by 8.96 g of S is 33.6 g

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}

\text{Moles of} S=\frac{8.96g}{32g/mol}=0.28moles

The balanced chemical equation is:  

2S+3O_2\rightarrow 2SO_3  

According to stoichiometry :

2 moles of S produce =  3 moles of SO_3

Thus 0.28 moles of S will produce=\frac{3}{2}\times 0.28=0.42moles  of SO_3  

Mass of SO_3=moles\times {\text {Molar mass}}=0.42moles\times 80g/mol=33.6g

Thus theoretical yield of SO_3 produced by 8.96 g of S is 33.6 g

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