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3 years ago

ANSWER ASAP ITS CHEMISTRY HIGH SCHOOL 10 POINTS!

Chemistry

1 answer:

loris [4]3 years ago

5 0

Answer:

3rd option. 1–butanamine

Explanation:

To name the compound above, the following must be observed:

1. Locate the functional group in the chain. In this case the functional group is amine.

2. Locate the longest continuous carbon chain. This gives the parent name of the compound. In this case, the longest chain has 4 carbon i.e butane.

3. Since the functional group is amine, the parent name becomes butanamine i.e replacing the –e at the end in butane with –amine

4. Indicate the position of the functional group in the chain. In this case the functional group is at carbon 1

5. Name the compound by putting the above together.

The name of the compound is:

1–butanamine or butan–1–amine

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Answer:

ΔH⁰(11.4g NH₄NO₃) = -30.59Kj (4 sig. figs. ~mass of NH₄NO₃(s) given) (exothermic)

Explanation:

3NH₄NO₃(s) + C₁₀H₂₂(l) + 14O₂(g) => 3N₂(g) + 17H₂O(g) + 10CO₂(g)

ΔH⁰(f): 3(-365.6)Kj 1(-301)Kj 14(0)Kj 3(0)Kj 17(-241.8)Kj 10(-393.5)Kj

= -1096.8Kj = -301Kj = 0Kj = 0Kj = -4110.6Kj = -3930.5Kj

ΔHₙ°(rxn) = ∑ (ΔH˚(f)products) - ∑(ΔH˚(f)reactants)

= [3(0)Kj + 17(-241.8)Kj + (-393.5)Kj] - [(-(1096.8)Kj + (-301)Kj + (0)Kj]

= [-(8041.1) - (-1397.8)]Kj

= -6643.3Kj (for 3 moles NH₄NO₃ used in above equation)

∴ Standard Heat of Rxn = -6643.3Kj/3moles = -214.8Kj/mole NH₄NO₃(s)

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3 years ago

Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidit

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Answer:

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Volume of NaOH = 1.7 ml = 0.0017 L

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$0.0811 M=\frac{n}{0.0017 L}$

n = 0.0001378 mol

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According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

$\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol$ of sulfuric acid.

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$x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L$

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