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GarryVolchara [31]
3 years ago
5

1. Which kingdom is made up of only autotrophs?

Chemistry
1 answer:
Reika [66]3 years ago
3 0

Answer: I believe it's C

Hope this helped<3

Can you please make my answer brainly

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C should be the answer to the following formula
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4. Methane, CH4, is one carbon atom bonded to four hydrogen atoms. In order to show
oee [108]

The answer is C. the carbon symbol C with one dot on each

the top, bottom, left, and right sides

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HELP ASAP!!! How many grams of lead are in a 41.2 gram sample of lead (II) oxide? With work please :)
Afina-wow [57]
That would be 58.6 oxide because if the 41.2 plus the oxide would be 48 plus the other particcles inside it so it equaled 58.6 oxide

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3 years ago
the mass of a single potassium atom is 6.50×10-23 grams. How many potassium atoms would there be in 114 milligrams of potassium?
Nataly [62]

Answer:LOL

42

Explanation:

8 0
2 years ago
A student ran the following reaction in the laboratory at 311 K:CH4(g) + CCl4(g) 2CH2Cl2(g)When she introduced 4.10×10-2 moles o
statuscvo [17]

<u>Answer:</u> The value of K_{eq} is 0.044

<u>Explanation:</u>

We are given:

Initial moles of methane = 4.10\times 10^{-2}mol=0.0410moles

Initial moles of carbon tetrachloride = 6.51\times 10^{-2}mol=0.0651moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}

So, concentration of methane = \frac{0.0410}{1.00}=0.0410M

Concentration of carbon tetrachloride = \frac{0.0651}{1.00}=0.0651M

The given chemical equation follows:

                    CH_4(g)+CCl_4(g)\rightleftharpoons 2CH_2Cl_2(g)

<u>Initial:</u>          0.0410    0.0651

<u>At eqllm:</u>     0.0410-x   0.0651-x       2x

We are given:

Equilibrium concentration of carbon tetrachloride = 6.02\times 10^{-2}M=0.0602M

Evaluating the value of 'x', we get:

\Rightarrow (0.0651-x)=0.0602\\\\\Rightarrow x=0.0049M

Now, equilibrium concentration of methane = 0.0410-x=[0.0410-0.0049]=0.0361M

Equilibrium concentration of CH_2Cl_2=2x=[2\times 0.0049]=0.0098M

The expression of K_{eq} for the above reaction follows:

K_{eq}=\frac{[CH_2Cl_2]^2}{[CH_4]\times [CCl_4]}

Putting values in above expression, we get:

K_{eq}=\frac{(0.0098)^2}{0.0361\times 0.0603}\\\\K_{eq}=0.044

Hence, the value of K_{eq} is 0.044

8 0
3 years ago
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