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Mademuasel [1]
3 years ago
8

The energy on the earth

Chemistry
1 answer:
Andreas93 [3]3 years ago
3 0

Answer:

<em>The earth-atmosphere energy balance is the balance between incoming energy from the Sun and outgoing energy from the Earth. When it reaches the Earth, some is reflected back to space by clouds, some are absorbed by the atmosphere, and some is absorbed in the Earth's surface. ...</em>

Explanation:

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When adding or subtracting deelmals, how many digits should the answer contain?
Alexus [3.1K]

Answer:

Depends, but in most cases, 2.

It's best to use as many digits as possible to keep it accurate.

Explanation:

This varies between teachers, as most schools go with 2 decimal places.

This is something that depends in your situation.

You technically want as many decimals as possible to keep it as accurate, but most people stick with 2.

I personally do 3, and commonly do 5 sometimes.

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What happens to the length of a year for each planet as you get farther away from the sun
Bond [772]

Answer:

they get colder and darker, with less light

Explanation:

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Which model best represents an element?
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2 years ago
A compound contains 1.2 g of carbon, 3.2 g of oxygen and 0.2g of hydrogen. Find the formula of the compound
Karolina [17]

Answer:

The empirical formula of the compound is C_{0.504}HO_{1.008}.

Explanation:

We need to determine the empirical formula in its simplest form, where hydrogen (H) is scaled up to a mole, since it has the molar mass, and both carbon (C) and oxygen (O) are also scaled up in the same magnitude. The empirical formula is of the form:

C_{x}HO_{y}

Where x, y are the number of moles of the carbon and oxygen, respectively.

The scale factor (r), no unit, is calculated by the following formula:

r = \frac{M_{H}}{m_{H}} (1)

Where:

m_{H} - Mass of hydrogen, in grams.

M_{H} - Molar mass of hydrogen, in grams per mole.

If we know that  M_{H} = 1.008\,\frac{g}{mol} and m_{H} = 0.2\,g, then the scale factor is:

r = \frac{1.008}{0.2}

r = 5.04

The molar masses of carbon (M_{C}) and oxygen (M_{O}) are 12.011\,\frac{g}{mol} and 15.999\,\frac{g}{mol}, then, the respective numbers of moles are: (r = 5.04, m_{C} = 1.2\,g, m_{O} = 3.2\,g)

Carbon

n_{C} = \frac{r\cdot m_{C}}{M_{C}} (2)

n_{C} = \frac{(5.04)\cdot (1.2\,g)}{12.011\,\frac{g}{mol} }

n_{C} = 0.504\,moles

Oxygen

n_{O} = \frac{r\cdot m_{O}}{M_{O}} (3)

n_{O} = \frac{(5.04)\cdot (3.2\,g)}{15.999\,\frac{g}{mol} }

n_{O} = 1.008\,moles

Hence, the empirical formula of the compound is C_{0.504}HO_{1.008}.

3 0
2 years ago
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