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Goryan [66]
3 years ago
10

2. The combustion of heptane (C7H16) in oxygen spontaneously occurs. The products of the reaction are carbon dioxide and water v

apor. The enthalpy of this reaction is –2877.5 kJ. Using the given information, write the complete thermochemical equation for this reaction. Explain how you developed this equation, and specify what the equation tells us about the reaction?
Chemistry
1 answer:
masya89 [10]3 years ago
8 0

The combustion of heptane (C7H16) in oxygen spontaneously occurs. The products of the reaction are carbon dioxide and water vapor. The balanced chemical equation is 7H16 + 11O2 → 7CO2 + 8H2O H = –2877.5 kJ.

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Pavel [41]
NH₃:

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N=7*10²²
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4NH₃                   <span>+                        3O</span>₂                      ⇒<span>          2N</span>₂<span> + 6H</span>₂<span>O
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6 0
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3 years ago
A diploid somatic ("body") cell has 2n = 20 chromosomes. At the end of mitosis, each daughter cell would have ______ chromosomes
kozerog [31]

Answer:

At the end of mitosis, 2n = 20

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Explanation:

Mitosis is a type of cell division in which daughter cell produced are genetically identical to their mother cell. So, no. of chromosome does not change after mitosis.

So, at the end of mitosis, each daughter cell would have <u>20</u> chromosome.

Meiosis is a type of cell division in which mother cell produces two haploid cells ones with a single set of chromosomes.

Meiosis is a two step cell division, Meiosis I and Meiosis II.

In meiosis I, homologous pair separates, so no. of chromosomes becomes half.

In meiosis II, sister chromatids separates. So, the number of chromosomes remains same (i.e. Have same no. of chromosome as present in cell produced after meiosis I).

So, at the end of mitosis, each daughter cell would have <u>20</u> chromosome.

At the end of meiosis I, each daughter cell would have n = 10 chromosomes. At the end of meiosis II, each daughter cell would have n = 10 chromosomes.

7 0
3 years ago
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Answer:

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5 0
2 years ago
Read 2 more answers
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madreJ [45]

Answer:

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Explanation:

Hello,

In this case, for the dissociation of calcium fluoride:

CaF_2(s)\rightleftharpoons Ca^{2+}+2F^-

The equilibrium expression is:

Ksp=[Ca^{2+}][F^-]^2

In such a way, via the ICE procedure, including an initial concentration of calcium of 0.01 M (due to the calcium nitrate solution), the reaction extent x is computed as follows:

3.9x10^{-11}=(0.01+x)(2*x)^2\\\\x=0.0000312M

Thus, the molar solubility equals the reaction extent x, therefore:

Molar \ solubility=3.12x10^{-5}M

Regards.

4 0
3 years ago
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