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aleksandrvk [35]
3 years ago
6

Which correctly lists the three gases that are part of Earth's cycles in both the atmosphere and biosphere?

Chemistry
2 answers:
tiny-mole [99]3 years ago
8 0

Answer:

The correct answer is actually, B) carbon dioxide, nitrogen, oxygen

zubka84 [21]3 years ago
4 0

Answer:

B

Explanation:

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In the following reaction, which element or compound is the oxidizing agent?
8_murik_8 [283]

Answer:

O_{2} is the oxidizing agent

Explanation:

An oxidizing agent is an element in a reaction that accepts the electrons of another element. It is typically hydrogen, oxide, or any halogen. In this case, it is oxygen. The answer is 02.

3 0
3 years ago
Read 2 more answers
A 10.0 cm3 sample of copper has a mass of 89.6 g. What is the density of copper
natali 33 [55]

Answer:

<h2>Density = 8.96 g/cm³</h2>

Explanation:

The density of a substance can be found by using the formula

Density =  \frac{mass}{volume}

From the question

mass of copper = 89.6 g

volume = 10 cm³

Substitute the values into the above formula and solve

That's

Density =  \frac{89.6}{10}

We have the final answer as

<h3>Density = 8.96 g/cm³</h3>

Hope this helps you

7 0
3 years ago
Why is a chemical equilibrium described as dynamic?
Katen [24]
Because when equilibrium is reached, the reaction is still occurring in both directions, it's just that rate(forward) =rate(reverse) so there is no net change in the concentrations of the reactants or products.
8 0
3 years ago
Read 2 more answers
I need it please !!!
anyanavicka [17]

Answer:

im pretty sure its 2...

Explanation:

if its wrong im sorry

if its right brainliest pls?

6 0
3 years ago
Glucose, C 6 H 12 O 6 , is used as an energy source by the human body. The overall reaction in the body is described by the equa
MA_775_DIABLO [31]

Answer:

m_{O_2}=61.87gO_2

m_{CO_2}=85.07gCO_2

Explanation:

Hello,

Considering the given reaction's stoichiometry, grams of oxygen result:

m_{O_2}=58.0gC_6H_{12}O_6*\frac{1molC_6H_{12}O_6}{180gC_6H_{12}O_6}*\frac{6molO_2}{1molC_6H_{12}O_6}*\frac{32gO_2}{1molO_2}\\m_{O_2}=61.87gO_2

Moreover, the mass of produced carbon dioxide turns out:

m_{CO_2}=58.0gC_6H_{12}O_6*\frac{1molC_6H_{12}O_6}{180gC_6H_{12}O_6}*\frac{6molCO_2}{1molC_6H_{12}O_6}*\frac{44gCO_2}{1molCO_2}\\m_{O_2}=85.07gCO_2

Best regards.

6 0
3 years ago
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