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saveliy_v [14]
3 years ago
11

Does absorbance increase or decrease as the solution concentration of the absorbing substance increases?

Chemistry
1 answer:
AleksAgata [21]3 years ago
5 0
I believe the answer is absorbance increases
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i am making a solution, and i want to increase the rate at which the solute dissolves. Describe two ways I can achieve this.
Paladinen [302]
Heat the solvent/solution with the solute over a Bunsen burner whilst mixing it with a glass storing rod or break up the solid into a powder to increase its surface area and then mix it with the solvent/solution.
4 0
4 years ago
A boat Pulls a 60Kg water-skier with a force of 105 N, What is the skier’s acceleration?
kicyunya [14]

Answer:1.75 acceleration

Explanation: because 105N divide by 60kg = 1.75

8 0
3 years ago
when metals react they ________________ (gain/lose) electrons. as you go down any group containing metals, it becomes easier to
balandron [24]

Answer:

When metals react they lose electrons.

Explanation:

An ion is defined as a molecule or atom made up of a different number of electrons and protons. This makes the electric charge of the molecule net, not neutral.

In other words, an ion is a molecule or atom that has a positive or negative electric charge, making it an atom whose electric charge is not neutral.

The characteristic property of a metallic atom is to lose one or more of its electrons to form a positive ion. That is, metals lose electrons and acquire a net positive charge, which is why they are electropositive.

4 0
3 years ago
What Phase is water at 120 degrees Celsius?
Rina8888 [55]
At atmospheric pressure water boils at 100 degC.
So the water would be gas/vapor/steam.
4 0
3 years ago
You are given a sulfuric acid solution of unknown concentration. You dispense 10.00 mL of the unknown solution into an Erlenmeye
dmitriy555 [2]

Answer:

"0.053457 M" of sulfuric acid.

Explanation:

The given values are:

V = 10 mL solution

V_{added} = 12.20 mL

V_{total} = 22.20 mL

then,

M 0.103 M of NaOH,

V_{rinsed} = experiment  will not be affected

V_{total \ base} = 10.38 mL

Now,

⇒  mol of NAOH = MV

                            = 0.103\times 10.38

                            =  1.06914  \ m

Whether Sulfuric acid, then

⇒  H_{2}SO_{4} + 2NaOH = Na_{2}SO_{4} + 2H_{2}O

⇒  mol \ of \ acid =\frac{1}{2}\times \ mol \   of  \ base

⇒  1.06914 \ m \ mol \ of \ base = \frac{1}{2}\times 1.06914 = 0.53457 \ m \ mol \ of \ acid

Before any dilution:

V_{sample} = 10  \ mL

⇒  M \ acid = \frac{m \ mol}{V}

                 =\frac{ 0.53457 }{10}

                 =0.053457 \ M (Sulfuric acid)

6 0
3 years ago
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