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3 years ago

Do you think the younger scientist had enough evidence for his prediction?

Chemistry

2 answers:

grin007 [14]3 years ago

5 0

You have no context to the problem

Sveta_85 [38]3 years ago

4 0

Answer:

yes

Explanation:

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Round the measurement to 3 significant figures. 13.06

erastovalidia [21]

Answer: 13.1

Explanation: just round the .06 to 1 creating three sig figs.

8 0

3 years ago

If the results of an experiment disprove a hypothesis, then the a. results should not be reported. b. hypothesis is just a theor

Alina [70]

B.Hypothesis is just a theory.

3 0

3 years ago

Which of the following equations represents the law of conservation of mass?

weeeeeb [17]

Equations represents the law of conservation of mass is 2H₂O → 2H₂ + O₂

Law of conservation of mass stated that energy neither be created nor destroyed called as law of conservation of mass and as the same no of each atom present in the both side of the reaction this reaction represents the law of conservation of mass so in the equation 2H₂O → 2H₂ + O₂ the energy cannot be created nor destroyed water can form 2 mole of hydrogen and 1 mole of oxygen which form water

Know more about law of conservation of mass

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5 0

1 year ago

2. Will calcite scratch feldspar? Why or why not?

ahrayia [7]

Answer: No

Explanation: The reason for this is because calcite has a hardness of 9 while feldspar has a hardness of 21. Calcite can scratch Gypsum which has a hardness of 3.

7 0

3 years ago

The total mass of the atmosphere is about 5.00 x 1018 kg. How many moles each of air, O2, and CO2 are present in the atmosphere?

n200080 [17]

<u>Answer:</u> The moles of oxygen and carbon dioxide in air is $3.63\times 10^{19}mol$ and $7.18\times 10^{16}mol$ respectively

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of atmosphere = $5.00\times 10^{18}kg=5.00\times 10^{21}g$

Average molar mass of atmosphere = 28.96 g/mol

Putting values in above equation, we get:

$\text{Moles of atmosphere}=\frac{5.00\times 10^{21}g}{28.96g/mol}=1.73\times 10^{20}mol$

We know that:

Percent of oxygen in air = 21 %

Percent of carbon dioxide in air = 0.0415 %

Moles of oxygen in air = $\frac{21}{100}\times 1.73\times 10^{20}=3.63\times 10^{19}mol$

Moles of carbon dioxide in air = $\frac{0.0415}{100}\times 1.73\times 10^{20}=7.18\times 10^{16}mol$

Hence, the moles of oxygen and carbon dioxide in air is $3.63\times 10^{19}mol$ and $7.18\times 10^{16}mol$ respectively

6 0

3 years ago