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Jlenok [28]
4 years ago
12

If a base is added to water, what will occur?

Chemistry
1 answer:
viktelen [127]4 years ago
6 0
This will increase the concentration of [OH-] ions.
For example:
- Ammonia NH3 (base):
NH3(aq) + H2O (l) → NH4+  + OH-

and 
-NaOH:
NaOH(s) + H2O(l) → Na+ + OH-
and KOH:
KOH(s) → K+(aq) + OH-(aq)

So we can see that the base increase the [OH-] & increase the characteristic anion [OH-] of the solvent.
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Which of the following would most likely be reduced when combined with
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Answer:

This is based on most likely so it's A.Fe

5 0
3 years ago
What change in volume results if 170.0 mL of gas is cooled from 30.0 °C to 20.0 °C? (CHARLES LAW)
Inessa05 [86]

Answer:

164.4 L

Explanation:

use charles' law formula Volume 1 over Temp. 1 equaled to Volume 2 over Temp. 2

5 0
3 years ago
The symbols used on a map are explained in the
Nady [450]
Answer - LEGEND

The legend of a map is also called the map key. It is the part included in a map to, as it were, unlock the map. The legend (or map key) gives explanation to what each symbol and color used in the map represents.
5 0
4 years ago
A solid ball has a density of 2.5 g/cm and a volume of 20 cm3. What is the
Svetach [21]

Answer:

50 g

Explanation:

d= m/v

rearranging the above equation

m = d x v

m = 2.5 g x 20 g/cm3

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7 0
3 years ago
What is the solubility (in g/L) of calcium fluoride at 25°C? The solubility product constant for calcium fluoride is 3.4 × 10–11
PilotLPTM [1.2K]

<u>Answer:</u> The correct answer is Option b.

<u>Explanation:</u>

The balanced equilibrium reaction for the ionization of calcium fluoride follows:

CaF_2\rightleftharpoons Ca^{2+}+2F^-

                s       2s

The expression for solubility constant for this reaction will be:

K_{sp}=[Ca^{2+}][F^-]^2

We are given:

K_{sp}=3.4\times 10^{-11}

Putting values in above equation, we get:

3.4\times 10^{-11}=(s)\times (2s)^2\\\\3.4\times 10^{-11}=4s^3\\\\s=2.04\times 10^{-4}mol/L

To calculate the solubility in g/L, we will multiply the calculated solubility with the molar mass of calcium fluoride:

Molar mass of calcium fluoride = 78 g/mol

Multiplying the solubility product, we get:

s=2.04\times 10^{-4}mol/L\times 78g/mol=159.12\times 10^{-4}g/L=0.015g/L

Hence, the correct answer is Option b.

3 0
3 years ago
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