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TEA [102]
3 years ago
5

In general what is the effect of increased temperature e on solubility of a gas

Chemistry
2 answers:
AnnyKZ [126]3 years ago
6 0
If theres choices its C i had the question before lol comment if im wrong then ill fix it

Yakvenalex [24]3 years ago
3 0
Solubility increases, considering the molecules when heated will spread out and become more active. I’m not sure how a gas can be considered soluble but my beginning remarks are standard laws
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Before of an results an experiment are accepted, which of these points must it fulfill?
jek_recluse [69]

Answer:

I think it will be (B) only because you have to do it repeatedly and analyze it and not noted, in order for the results are accepted. I May be wrong. Thanks for letting me help you.

8 0
4 years ago
Can anyone help me with these I’m really co fused on what to do it’s for chemistry?
liq [111]

Answer:

It's just asking you to rearrange to equations to get what you want, so no numbers involved it's just teaching you the basics.

Explanation:

3 0
4 years ago
Which of the following statements is true?
natulia [17]
The 1 one electrons have positive charge
6 0
3 years ago
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Calculate the number of miles of magnesium,chlorine, and oxygen atoms in 5.00 moles of magnesium perchlorate
Zigmanuir [339]

Answer:

5.00 mol Mg

10.0 mol Cl

40.0 mol O

Explanation:

Step 1: Given data

Moles of Mg(ClO₄)₂: 5.00 mol

Step 2: Calculate the number of moles of Mg

The molar ratio of Mg(ClO₄)₂ to Mg is 1:1.

5.00 mol Mg(ClO₄)₂ × 1 mol Mg/1 mol Mg(ClO₄)₂ = 5.00 mol Mg

Step 3: Calculate the number of moles of Cl

The molar ratio of Mg(ClO₄)₂ to Cl is 1:2.

5.00 mol Mg(ClO₄)₂ × 2 mol Cl/1 mol Mg(ClO₄)₂ = 10.0 mol Cl

Step 4: Calculate the number of moles of O

The molar ratio of Mg(ClO₄)₂ to Cl is 1:8.

5.00 mol Mg(ClO₄)₂ × 8 mol O/1 mol Mg(ClO₄)₂ = 40.0 mol O

4 0
3 years ago
What is the pOH of a 0.00037 M solution of barium hydroxide
ruslelena [56]

Answer: pOH = 3.13

Ba(OH)2 is a very basic substance. The general formula for pOH is - log(OH)

Barium Hydroxide produces 2 moles of OH for every mole of Ba(OH)2 present in the solution.


0.00037 M = 3.7 * 10^-4 Ba(OH)2 will produce 2 *0.00037 M of OH-

OH- = 2* 0.00037 = 0.00074

pOH = - log(0.00074) = 3.13


5 0
4 years ago
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