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Why does the moon and the earth have the same mass

Evgesh-ka [11]

Answer:

Moon rocks contain few volatile substances (e.g. water), which implies extra baking of the lunar surface relative to that of Earth. The relative abundance of oxygen isotopes on Earth and on the Moon are identical, which suggests that the Earth and Moon formed at the same distance from the Sun.

Explanation:

3 0

2 years ago

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All rocks can form into other types of rocks

natka813 [3]

Answer:false

Explanation:

4 0

3 years ago

1.) is H2SO4 a sulfuric acid or a sulfurous acid?

Lorico [155]

1) is sulfuric acid
2)is nitrous acid
3)is hydrochlorous acid
4)is hydrobromous acid
5)is hydrophosphoric acid
6)is fluoric acid
7)is sulfuric acid
8)is chlorous acid
9)is nitric acid
10)is iodic acid
11)is acetous acid
12)is chlorous acid.

3 0

3 years ago

Some kayakers coat the bottom of their kayaks with wax to make them more slippery. Kayakers wax their boats to

Sindrei [870]

The answer is : C. reduce their input force
they put wax on their kayak to make their boat cut smoothly through the water, which make it able to obtain further distance with less input force, which will increase it's overall speed

8 0

3 years ago

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3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15

shutvik [7]

Answer:

C₅H₁₀O₅

Explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

$\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}$

(b) Mass of H

$\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}$

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

$\text{Moles of C = 1400 mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{116.6}{116.6}= 1\\\\\text{H: } \dfrac{233.1}{116.6} = 1.999\\\\\text{O: } \dfrac{116}{116.6} = 1.00$

4. Round the ratios to the nearest integer

C:H:O = 1:2:1

5. Write the empirical formula

The empirical formula is CH₂O.

6. Calculate the molecular formula.

EF Mass = (12.01 + 2.016 + 16.00) u = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

$n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{150 u}}{\text{30.03 u}} = 5.00 \approx 5$

MF = (CH₂O)₅ = C₅H₁₀O₅

The molecular formula of X is C₅H₁₀O₅.

8 0

4 years ago