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exis [7]
2 years ago
10

15. What does an indicator do if placed in an acidic or alkaline solution?

Chemistry
1 answer:
dedylja [7]2 years ago
5 0

Answer: 15. When an acid is dissolved in water we get an acidic solution. Indicators are substances that change color when they are added to acidic or alkaline solutions. Litmus, phenolphthalein, and methyl orange are all indicators that are commonly used in the laboratory.

17. In a solution that decreases in acidity, methyl orange moves from the color red to orange and finally to yellow with the opposite occurring for a solution increasing in acidity. In an acid, it is reddish and in alkali, it is yellow.

18. The pH scale measures whether there is more hydronium or hydroxide in a solution. In other words, it tells us how basic or acidic the solution is. A lower pH means something is more acidic, also known as a stronger acid. A higher pH means it is more alkaline or a stronger base.

21. A base or alkali accepts hydrogen ions, and when added to water, it soaks up the hydrogen ions formed by the dissociation of water so that the balance shifts in favor of the hydroxyl ion concentration, making the solution alkaline or basic.

I don't know most of the question i only answered the ones i know i hope it helped you.

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duplet-A set of two things or people
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The property of metals to be made into wire is called
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hope this helps, please mark brainliest. :)

6 0
2 years ago
10 POINTS
Snowcat [4.5K]
E=hc/λ =6.626×10^-34×3 ×10^8 / 3×10^7 × 10^-9 = 6.626×10 ^-24J.
7 0
3 years ago
The salt formed by the reaction of the weak acid hydrocyanic acid, HCN, with the strong base potassium hydroxide is
iogann1982 [59]

Answer:

2.28 × 10^-3 mol/L

Explanation:

The equation for the equilibrium is

CN^- + H2O ⇌ HCN + OH^-

                    Ka = 4.9 × 10^-10

               KaKb = Kw

4.9 × 10^-10 Kb = 1.00 × 10^-14

                   Kb = (1.00 × 10^-14)/(4.9 × 10^-10) = 2.05 × 10^-5

Now, we can set up an ICE table

                     CN^- + H2O ⇌ HCN + OH^-

I/(mol/L)      0.255                     0         0

C/(mol/L)       -x                        +x        +x

E/(mol/L)  0.255 - x                   x         x

Ka = x^2/(0.255 - x) = 2.05 × 10^-5

Check for negligibility

0.255/(2.05 × 10^-5) = 12 000 > 400. ∴ x ≪ 0.255

    x^2 = 0.255(2.05 × 10^-5) = 5.20 × 10^-6

        x = sqrt(5.20 × 10^-6)    = 2.28 × 10^-3

[OH^-] = x mol/L                     = 2.28 × 10^-3 mol/L

5 0
2 years ago
98 POINTS!!! WILL AWARD BRAINLIST!! EXTRA POINTS!! PLS HELP!!
BabaBlast [244]

Answers are at the pdf. 

Since there are 16 questions, all answers cannot be uploaded because number of characters is beyond the limit. 

Download pdf
4 0
3 years ago
Read 2 more answers
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