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kondor19780726 [428]
3 years ago
15

Elements, Compounds,& Mixtures,

Chemistry
1 answer:
mrs_skeptik [129]3 years ago
3 0

Answer:

1. Elements: F, H, E

2. Compounds: A, D, I

3. Mixtures: B, C, G

Explanation:

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What is a chemical property of carbon?
MA_775_DIABLO [31]
I'll give you three just in case a couple are ones you aren't quite looking for.

-Carbon has several allotropes, or different forms in which it can exist. A couple of allotropes included are graphite and diamond, which both have very different properties.

-Despite carbon's ability to make 4 bonds and its presence in many compounds, it is highly nonreactive under normal conditions.

-Carbon exists in three main isotopes: ¹²C, ¹³C, and ¹⁴C in which ¹⁴C is radioactive and used in dating carbon-containing samples (known as radiometric dating)

Hope this helps!!
7 0
4 years ago
This is a PERCENTAGE COMPOSITION by mass problem.
Norma-Jean [14]

Answer:

I think it will be d 12% it will be true

4 0
3 years ago
Read 2 more answers
Someone please answer the question attached.
nikdorinn [45]

Answer:

Option ( 1 )

Explanation:

This atom has a cub close pack structure, and therefore - the number of edge center present in X atoms =  4,

Number of unit cells present in X atoms = 6

____________________________________________________

Now the 6th coordination number of X atom = 6 * 4 = 24,

So respectively, the 3rd coordination number of X = 8 -

And thus the ratio between the 6th coordination number and the 3rd coordination number = 24 / 8 = 3,

Option ( 1 )

<u><em>Hope that helps!</em></u>

5 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
Which of the following does not change the rate of collisions between particles in a reaction?
Naddik [55]

Answer: C

Explanation: trust me bro

4 0
3 years ago
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