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VLD [36.1K]
2 years ago
8

4. What is the main hydrocarbon component of natural gas? A. ethane

Chemistry
2 answers:
KIM [24]2 years ago
7 0

Answer:

Q.4) C. Methane

Q.7) A. False

Q.8) A. True

777dan777 [17]2 years ago
4 0
Tryna boost my score for college chemistry could you give me the brainiest and a thanks? Hope you find your answer your looking for!
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Compare and contrast of cotton balls and pine cone
Alex
<span>Cotton balls and pine cones are similar in that they both grow on plants and they contain the seeds of that plant. However, that is where the similarities end. By sight and feel alone, cotton balls and pine cones are vastly different. Cotton balls are soft and light, while pine cones have the potential to be a little dry and pri.ckly and have a bit of heft to them.</span>
3 0
2 years ago
Please help me it’s due today at 7:40pm please help me please please help me
snow_tiger [21]

Answer:

inertia

Explanation:

5 0
3 years ago
Which are examples of biotic factors? Select three options.
alina1380 [7]

The biotic factors are bacteria soil, dead leaves, and stream water.

<h3>What is an abiotic factor?</h3>

An abiotic factor is a non-living part of an ecosystem that shapes its environment.

Biotic and abiotic factors make up a community via interaction.

Biotic factors are considered living things (having "life") while abiotic factors are simply non-living things.

The dead leaves of plants are an abiotic factor, the bacteria in soil are living matter and stream flowing, etc.

Hence, the biotic factors are bacteria soil, dead leaves, and stream water.

Learn more about the abiotic factor here:

brainly.com/question/12689972

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6 0
1 year ago
Write the correct name for each compound. a. N2O5 b. Si2F4 c. S3F d. OF2 e. H4P6 f. C2O4 g. HF3
PSYCHO15rus [73]

Answer:

a. Boron trifluoride

b. Propane

c. Dinitrogen pentoxide

d. Carbon Dioxide

e. Silicon Octafluroride?

Explanation:

Glad to help :)

3 0
3 years ago
A student wants to prepare 1.00 L of a 1.00 M solution of NaOH (molar mass 40.00 g/mol). If solid NaOH is available, how would t
Serga [27]

Explanation:

1)

Molarity=\frac{\text{Mass of substance}}{\text{Molar mass of substance}\times \text{Volume of solution(L)}}

Mass of NaOH = m

MOlar mass of NaOH = 40 g/mol

Volume of NaOH solution = 1.00 L

Molarity of the solution= 1.00 M

1.00 M=\frac{m}{40 g/mol\times 1.00 L}

m=1.00 M\times 40 g/mol\times 1.00 L = 40. g

A student can prepare the solution by dissolving the 40. grams of NaOH in is small volume of water and making that whole volume of solution to volume of 1 L.

Upto two significant figures mass should be determined.

2)

M_1V_1=M_2V_2 (dilution equation)

Molarity of the NaOH solution = M_1=2.00 M

Volume of the solution = V_1=?

Molarity of the NaOH solution after dilution = M_2=1.00 M

Volume of NaOH solution after dilution= V_2=1 L

M_1V_1=M_2V_2

V_1=\frac{1.00 M\times 1.00 L}{2.00 M}=0.500 L

A student can prepare NaOH solution of 1.00 M by diluting the 0.500 L of 2.00 M solution of NaOH with water to 1.00 L volume.

Upto three significant figures volume should be determined.

8 0
3 years ago
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