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Svetach [21]
3 years ago
15

The combustion of a sample butane, C4H10 (lighter fluid) produced 2.46 grams of water. how much oxygen was used up in grams

Chemistry
1 answer:
Afina-wow [57]3 years ago
8 0

The balanced combustion reaction for butane is;

2C4H10 + 13O2 → 8CO2 + 10H2O

Molar mass of water = 18.02 g/mol

Molar mass of oxygen = 32 g/mol

 

2.46g H2O(1 mol H2O/18.02g)(13 mol O2/10 mol H2O)(32g O2/1 mol O2) = 5.68g O2

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1. Which statement is accurate?
valkas [14]

1. Answer:

B) An allele is one form of a gene.

Explanation:

  • <em><u>An allele refers to an alternative form of a gene. In other words it is one of the form of a gene.</u></em>
  • <em><u>A gene may have two alleles, where one of the allele is a dominant allele and the other is a recessive allele. </u></em>
  • A homozygous organisms may be dominant or recessive, a homozygous recessive contains two recessive alleles while a homozygous dominant contains two dominant alleles.
  • A heterozygous organisms contain one dominant allele and one recessive allele.

2. Answer;

B) Alleles

Explanation;

  • <em><u>According to Mendel genes are inherited in pairs of alleles that behave in a dominant and recessive pattern. </u></em>
  • <em><u>Alleles are alternative forms of a gene. </u></em>
  • A gene may have two alleles, where one of the allele is a dominant allele and the other is a recessive allele.
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8 0
4 years ago
WHAT HAPPENS WHEN ELECTRIC CURRENT PASSES THROUGH ACIDIFIED WATER.
mario62 [17]

Answer:

When an electric current is passed through acidified water, it decomposes to give hydrogen and oxygen gas. The hydrogen gas is obtained at the cathode and the oxygen gas is obtained at the anode

8 0
3 years ago
Read 2 more answers
Ag2S + Al(s) = Al2S3 + Ag(s) (unbalanced)
Dovator [93]

Answer:

1. 0.97 V

2. Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/Ag_(_s_)

Explanation:

In this case, we can start with the <u>half-reactions</u>:

Ag^+~_(_a_q_)->~Ag_(_s_)

Al_(_s_)~->~Al^+^3~_(_a_q_)

With this in mind we can <u>add the electrons</u>:

Ag^+~_(_a_q_)+~e^-~->~Ag_(_s_)  <u>Reduction</u>

Al_(_s_)~->~Al^+^3~_(_a_q_)+~3e^-~ <u>Oxidation</u>

The reduction potential values for each half-reaction are:

Ag_2S~+~e^-~->~Ag_(_s_)~+~S^-^2~_(_a_q_) - 0.69 V

Al^+^3~_(_a_q_)+~2e^-~->~Al_(_s_) -1.66 V

In the aluminum half-reaction, we have an oxidation reaction, therefore we have to <u>flip</u> the reduction potential value:

Al_(_s_)~->~Al^+^3~+~2e^-~ +1.66 V

Finally, to calculate the overall potential we have to <u>add</u> the two values:

1.66 V - 0.69 V = <u>0.97 V</u>

For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:

Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/~Ag_(_s_)

I hope it helps!

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