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Ratling [72]
3 years ago
9

Jules Verne wrote the book Twenty Thousand Leagues Under the Sea. If one league - 5.556 km and one furlong - 660.0 feet, how man

y furlongs did the Nautilus travel? (assume the twenty thousand leagues is good for 3 sig. fig.) ball park-550,000 furlong T
Chemistry
1 answer:
Vlad [161]3 years ago
7 0

Answer:

The Nautilus travel 5.52\times 10^5 furlongs.

Explanation:

Given: 1 league = 5.556 km

1 furlong = 660.0 feet

To find: 20,000 leagues = ? furlongs

Solution:

20,000 leagues = 20,000\times 5.556 km =111,120 km

1 km = 3280.84 feet

111,120 km= 111,120\times 3280.84 feet=364,566,940.8 feet

If ,1 furlong = 660.0 feet.

Then, 1 foot = \frac{1}{660.0} furlong

364,566,940.8 feet=364,566,940.8 \times \frac{1}{660.0}

=552,374.1524 furlongs\approx 5.52\times 10^5 furlongs

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Ira Lisetskai [31]
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3 years ago
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Consider the following reaction: 2Mg(s)+O2(g)-->2MgO(s) delta H=-1204kJ
Pachacha [2.7K]

Answer:

a. The reaction is exothermic.

b. -87,9 kJ

c. 9,60g of Mg(s)

d. 602kJ are absorbed

Explanation:

Based on the reaction:

2Mg(s) + O₂(g) → 2MgO(s) ΔH = -1204kJ

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c. If -238 kJ of heat were transferred. The moles of Mg(s) that react must be:

-238kJ × ( 2mol Mg / -1204kJ) = 0,395 moles of Mg(s). In grams:

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2MgO(s) → 2Mg(s) + O₂(g)  ΔH = +1204kJ

40,5g of MgO(s) are:

40,5g MgO × ( 1mol MgO / 40,3044g) = 1,00 moles of MgO(s)

As 2 moles of MgO absorbe 1204kJ of energy:

1,00 moles of MgO(s) × ( +1204 kJ / 2mol MgO) = <em>602kJ are absorbed</em>

<em></em>

I hope it helps!

7 0
3 years ago
Determine the theoretical maximum moles of ethyl acetate, , that could be produced in this experiment. The reactant, acetic acid
kherson [118]

Answer:

0.1832 moles of ethyl acetate (C_{4}H_{8}O_{2})

Explanation:

1. Find the balanced chemical equation:

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CH_{3}COOH+C_{2}H_{5}OH=C_{4}H_{8}O_{2}+H_{2}O

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As the problem says that the acetic acid CH_{3}COOH is the limiting reagent, use stoichiometry to find the moles of ethyl acetate produced:

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Let me know if you need anything else. 

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