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Phantasy [73]
3 years ago
6

Which substance is a primary source of hydrocarbon

Chemistry
1 answer:
IceJOKER [234]3 years ago
6 0

Answer:

Hydrocarbons are molecules consisting of both hydrogen and carbon. They are most famous for being the primary constituent of fossil fuels, namely natural gas, petroleum, and coal. For this reason, fossil fuel resources are often referred to as hydrocarbon resources.Jun 25, 2018

https://energyeducation.ca ›

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Relative formula mass of glucose? (C6H12O6)
Rom4ik [11]

To find the mass of glucose, you must multiply the atomic weight of each of the elements in the molecule by the subscripts in the formula:

C_{6}=6*(12.01g)=72.06g

H_{12}=12*(1.008g)=12.096g

O_{6}=6*(15.999g)=95.994g

Then you add all of them together:

72.06g+12.096g+95.994g=180.15g

Therefore, the molar weight of glucose is 180.15 grams.

3 0
3 years ago
For an aqueous solution of sodium chloride (NaCl) .Determine the molarity of 3.45L of a solution that contains 145g of sodium
worty [1.4K]
Molar mass NaCl = 58.44 g/mol

number of moles:

mass NaCl / molar mass

145 / 58.44 => 2.4811 moles of NaCl

Volume = 3.45 L

Therefore :

M = moles / volume in liters:

M = 2.4811 / 3.45

M = 0.719 mol/L⁻¹

hope this helps!
6 0
3 years ago
How would the electron configuration of nitrogen change to make a stable configuration?
lesantik [10]

Answer: The electronic configuration of nitrogen is 1s2 2s2 2p3. Thus nitrogen has a half-filled p orbital, which is comparatively more stable. Thus the p orbital is the outermost orbital. To achieve a stable gas configuration, nitrogen needs to have a fulfilled p orbital.

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5 0
2 years ago
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Can someone help me please :)
Fudgin [204]
I think it’s the third option but I’m not entirely sure
3 0
2 years ago
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Identify the oxidizing agent and the reducing agent for 4Li(s) + O_2 (g) to 2Li_2O(s).
wolverine [178]

Answer: Li is the reducing agentg and O is the oxidizing agent.

Explanation:

1) The oxidizing agent is the one that is reduced and the reducing agent is the one that is oxidized.

2) The given reaction is:

4Li(s) + O₂ (g) → 2 Li₂O(s)

3) Determine the oxidation states of each atom:

Li(s): oxidation state = 0 (since it is alone)

O₂ (g): oxidation state = 0 (since it is alone)

Li in Li₂O (s) +1

O in Li₂O -2

That because 2× (+1) - 2 = 0.

4) Determine the changes:

Li went from 0 to + 1, therefore it got oxidized and it is the reducing agent.

O went from 0 to - 2, therefore it got reduced and it is the oxidizing agent.

7 0
3 years ago
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