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Dovator [93]
3 years ago
5

What would be the result if Earth's unique atmosphere did not exist?

Chemistry
2 answers:
Goshia [24]3 years ago
6 0

Answer:

D) Life will no longer be possible

Explanation:

Odyssey ware

egoroff_w [7]3 years ago
5 0
Then the earth would most likely not be able to support life. Our earth would be similar to mars. In that it might have supported life. It the thin atmosphere would over all be unable to sustain more complex species. The earth would also be much colder because the green house effect is what maintains our relatively warm atmosphere.
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35g of HF was prepared by reacting 112g of CaF2 with an excess of H2SO4 calculate the percentage yield
Tom [10]

Answer: 61%


The reaction equation should be

CaF2 + H2SO4 → 2HF + CaSO4

For every 1 molecule CaF2 used, there will be 2 molecules of HF formed. The molecular mass of CaF2 is 78/mol while the molecular mass of HF is 20g/mol. If the yield is 100%, the amount of HF formed by 112g CaF2 would be: 112g/(78g/mol) * 2 * (20g/mol)=57.43g

The percentage yield of the reaction would be: 35g/57.43g= 60.94%

7 0
4 years ago
Sally's teacher tells her to find the masses of a sugar cube and a glass of water. Sally finds the masses to be 10 g for the sug
Kisachek [45]
110 grams, the law of conservation of mass states that matter can not be created nor destroyed so putting sugar into water does not make it disappear
6 0
3 years ago
Read 2 more answers
7. How many significant figures are in the following numbers?
Vlad1618 [11]
There are
A. 6 significant figures
B.5 sig figs
C.2 sig figs
D.5 sig figs
7 0
3 years ago
Determine the empirical formula of a compound containing 40.6 grams of carbon, 5.1 grams of hydrogen, and 54.2 grams of oxygen.
Lilit [14]

The Empirical formula of compound is C₁H₂O₁. The Molecular Formula of the compound is 4 (C₁H₂O₁).

<h3>What is Empirical Formula ?</h3>

Empirical formula is the simplest whole number ratio of atoms present in given compound.

Element   %   Atomic mass   Relative no. of atoms  Simplest whole ratio

C          40.6       12                   \frac{40.6}{12} = 3.3                        \frac{3.3}{3.3} = 1

H          5.1          1                      \frac{5.1}{1} = 5.1                          \frac{5.1}{3.1} = 2

O         54.2       16                    \frac{54.2}{16} = 3.3                        \frac{3.3}{3.3} = 1

The Empirical formula of compound is C₁H₂O₁ or CH₂O

<h3>How to find the Molecular formula of compound ?</h3>

Molecular formula = Empirical formula × n

n = \frac{\text{Molecular weight}}{\text{Empirical Formula weight}}

   = \frac{118.084}{30}

   = 4

Molecular formula = n × Empirical formula

                              = 4 (C₁H₂O₁)

Thus from the above conclusion we can say that The Empirical formula of compound is C₁H₂O₁. The Molecular Formula of the compound is 4 (C₁H₂O₁).

Learn more about the Empirical Formula here: brainly.com/question/1603500

#SPJ1

6 0
2 years ago
After completing an experiment to determine gravimetrically the percentage of water in a hydrate, a student reported a value of
zzz [600]

Answer:

A. Strong initial heating caused some of the hydrate sample to spatter out of the crucible.

Explanation:

Hi

The percentage of water in the sample is lower than expected.

A. Strong initial heating caused some of the hydrate sample to spatter out of the crucible:

If part of the sample is splashed from the crucible the mass of water detected will be less.

B. The dehydrated sample absorbed moisture after heating:

If the sample absorbs water after heating the percentage of water would be higher than expected.

C. The amount of the hydrate sample used was too small:

Depending on the sample size, different procedures can be chosen for analysis.

D. The crucible was not heated to constant mass before use:

In many occasions the crucible is heated next to the sample and not in previous form.

E. Excess heating caused the dehydration sample to decompose:

If the sample decomposes during heating, the analysis should be discarded.

success with your homework

4 0
3 years ago
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