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kondaur [170]
3 years ago
9

Under which conditions would the solubility of a gas be greatest? A. high pressure and high temperature B. high pressure and low

temperature C. low pressure and high temperature D. low pressure and low temperature
Chemistry
1 answer:
LiRa [457]3 years ago
8 0

Answer:

High pressure and low temperature.

Explanation:

You might be interested in
Part A
KonstantinChe [14]

Answer:

The answer to your question is below

Explanation:

A.

[H₃O⁺] = 2 x 10⁻¹⁴ M

pH = ?

Formula

                                    pH = - log [H₃O⁺]

Substitution

                                    pH = - log [2 x 10⁻¹⁴]

Result

                                    pH = 13.7          

B.

[H₃O⁺] = ?

pH = 3.12

Formula

                                   pH = - log [H₃O⁺]

Substitution

                                   3.12 = - log [H₃O⁺]

                                   10^{-3.12} = [H_{3} O^{+}]

Result

                                  [H₃O⁺] = 7.59 M

   

7 0
3 years ago
How many sulfur atoms are there in 3.90 mol of sulfur?
Ilya [14]
 <span>In a mole of anything, there are 6.023 x 10^23 units. So, in 3.9 moles of sulfur, there are 3.9 * 6.023 x 10^23 = 23 x 10^23 = 2.3 x 10^24 atoms (keeping only 2 sig figs).  Hope I help!!
</span>
6 0
3 years ago
All organisms need glucose or a source of________ to carry out basic life functions
hichkok12 [17]

Answer:

chlorophyll

Explanation:

7 0
3 years ago
Help please ASAP just with number 9
Crank
I don’t know I just need to ask a question
5 0
3 years ago
Read 2 more answers
If 185 g of KBr are dissolved in 1.2 kg of water, what would be the expected change in boiling point? The boiling point constant
Fudgin [204]

Answer:

ΔTb = 0.66 C

Explanation:

Given

Mass of KBr = 185 g

Mass of water = 1.2 kg

Kb = 0.51 C/m

Explanation:

The change in boiling point (ΔTb) is given by the product of molality (m) of the solution and the boiling point constant (Kb)

\Delta T_{b}= K_{b}* m

Molality = \frac{moles\ KBr}{Kg\ water} \\\\moles KBr = \frac{mass\ KBr}{Mol.wt\ KBr} = \frac{185}{119} = 1.555\\\\Molality (m) = \frac{1.555 }{1.2} =1.296 m\\

[tex]\Delta T_{b}= 0.51 C/m * 1.296 m = 0.66 C[\tex]

6 0
4 years ago
Read 2 more answers
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