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3 years ago

Which of the following gases is the most abundant in a volcanic eruption?

Chemistry

2 answers:

expeople1 [14]3 years ago

8 0

Option C: Sulfur Dioxide is the answer

Hope this helps

Karolina [17]3 years ago

6 0

Answer:

is the answer.

Sulfur Dioxide

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How many moles of water are in 1.23x10^18 water molecules

Harlamova29_29 [7]

By 1.23 x 1024 you mean 10 to the power of 24 molecules? If so all you need to do is divide the number of molecules you have by Avagadros number, 6.022 x 10^23. This will give you the mols of water, or the mols of anything, since there is always 6.022 x 10^23 molecules in 1 mol of substance.

1.23x10^24 atoms/6.022x10^23 atom/mol = 2.04 mol H20

6 0

3 years ago

Read 2 more answers

Treatment of (S)-( )-5-methyl-2-cyclohexenone with lithium dimethylcuprate gives, after protonolysis, a good yield of a mixture

tekilochka [14]

Answer:

use google and use the first link

Explanation:

6 0

3 years ago

17) A 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with what volume? HI

galina1969 [7]

Answer: A 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with 3 mL volume.

Explanation:

Given: $M_{1}$ = 0.20 M, $V_{1}$ = 15.0 mL

$M_{2}$ = 0.10 M, $V_{2}$ = ?

Formula used is as follows.

$M_{1}V_{1} = M_{2}V_{2}$

Substitute the values into above formula s follows.

$M_{1}V_{1} = M_{2}V_{2}\\0.20 M ]times 15.0 mL = 0.10 M ]times V_{2}\\V_{2} = 30 mL$

Thus, we can conclude that a 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with 3 mL volume.

4 0

3 years ago

Construct a three-step synthesis of trans-2-pentene from acetylene by dragging the appropriate formulas into the bins. Note that

adelina 88 [10]

Answer:

The three-step synthesis of trans-2-pentene from acetylene is as follows.

Step -1: Formation of higher order terminal alkyne on reaction with sodium acetylides with haloalkanes.

Step -2: Formation terminal alkyne to nonterminal alkynes.

Step -3: Formation of trans-pent - 2-pent-ene by reduction.

Explanation:

Synthesis of trans-pent-2-yne from ethyne takes place is mainly a three step synthesis which involves formation of higher order terminal alkyne on reaction with sodium acetylides with haloalkane. Second step involves the further alkylation of terminal alkynes to higher order nonterminal alkynes and the third step involves the formation of trans-2-ene by dissolving reduction method.

The chemical reaction of each step of chemical reactions is as follows.