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3 years ago

Why do you think the "super-hot" Earth's core is important in this

1 answer:

7 0

The "super-hot" earth's core is important because it generates the magnetic field around the earth. it also helps control evolution.

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Which of the following is NOT an

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The answer is C. Light because light is a form of energy

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A stream leaving a mountain range deposits a large part of its load in a(n): alluvial fan sandbar meander sand dune

pshichka [43]

the answer to you question is alluvial fan

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What do you call a circuit with only one path

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If I'm correct the answer should be a series circuit :) Hopefully this helps you out

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3 years ago

For the following reaction, 22.9 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas. nitrogen monox

ololo11 [35]

Answer:

1) Maximun ammount of nitrogen gas: $m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent: $m_{N2}=4.274 g$

Explanation:

The reaction

$2 NO (g) + 2 H_2 (g) \longrightarrow N_2 (g) + 2 H_2O (g)$

Moles of nitrogen monoxide

Molecular weight: $M_{NO}=30 g/mol$

$n_{NO}=\frac{m_{NO}}{M_{NO}}$

$n_{NO}=\frac{22.9 g}{30 g/mol}=0.763 mol$

Moles of hydrogen

Molecular weight: $M_{H2}=2 g/mol$

$n_{H2}=\frac{m_{H2}}{M_{H2}}$

$n_{H2}=\frac{.5.8 g}{2 g/mol}=2.9 mol$

Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess

1) Maximun ammount of nitrogen gas => when all NO reacted

$m_{N2}=0.763 mol NO* \frac{1 mol N_2}{2 mol NO}*\frac{28 g N_2}{mol N_2}$

$m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent:

$m_{N2}=(2.9 mol - 0.763 mol NO* \frac{1 mol H_2}{1 mol NO})*\frac{2 g H_2}{mol H_2}$

$m_{N2}=4.274 g$

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3 years ago

What is the percent by mass of carbon in acetone, c 3 h 6 o?

Tatiana [17]

In the problem, we are tasked to solved for the amount of carbon (C) in the acetone having a molecular formula of C 3 H 6 O. We need to find first the molecular weight if Carbon (C), Hydrogen (H), Oxygen (O).

Molecular Weight:
C=12 g/mol
H=1 g/mol
O=16 g/mol

To calculate for the percent by mass of acetone, we assume 1 mol of acetone.
%C= $( \frac{3(12 g/mol)}{3(12 g/mol)+6(1 g/mol)+16 g/mol} ) x 100%$
%C=62.07%
Therefore, the percent by mass of carbon in acetone is 62.07%

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3 years ago

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