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Finger [1]
3 years ago
12

What mass of KCl will produce a saturated solution in 500.0 g of water at 20°C? The solubility of KCl at 20°C is 34.0 g/100 g H2

O.
Chemistry
2 answers:
IgorLugansk [536]3 years ago
5 0
34g -------- 100g H₂O
Xg ---------- 500g H₂O
X = (500×34)/100
<u>X = 170g KCl

</u>
:)
Anna11 [10]3 years ago
3 0

Answer:

170 g of KCl will produce a saturated solution in 500 g of water at 20 ° C

Explanation:

The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them. That is, what is intended with it is to find the fourth term of a proportion knowing the other three. Remember that proportionality is a constant relationship or ratio between different magnitudes.

If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other) , the direct rule of three must be applied. To solve a direct rule of three, the following formula must be followed:

a ⇒ b

c ⇒ x

Then

x=\frac{c*b}{a}

Solubility is the measure of the ability of a certain substance to dissolve in another.  In this case the solubility of KCl at 20 ° C indicates that 34.0 g dissolve in 100 g H2O at the mentioned temperature. Then it is possible to apply a rule of three to know the mass of KCl will produce a saturated solution in 500 g of water at 20 ° C: if in 100 g of H2O 34 g of KCl will dissolve, in 500 g of H2O how much mass of KCl will dissolve ?

mass of KCl=\frac{500 g H2O*34 g KCl}{100 g H2O}

mass of KCl=170 g

Then, <u><em>170 g of KCl will produce a saturated solution in 500 g of water at 20 ° C</em></u>

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c) Light of frequency less than v₀ does not possess enough energy to cause electrons to break free from the metal surface. The energy of light with frequency less than v₀ is less than the work function of the metal (which is the minimum amount of energy of light required to excite electrons on metal surface enough to break free).

As evident from the graph, electron kinetic energy remains at zero as long as the frequency of incident light is less than v₀.

d) When frequency of the light exceeds v₀, there is an increase of electron kinetic energy from zero steadily upwards with a constant slope. This is because, once light frequency exceeds, v₀, its energy too exceeds the work function of the metal and the electrons instantaneously gain the energy of incident light and convert this energy to kinetic energy by breaking free and going into motion. The energy keeps increasing as the energy and frequency of incident light increases and electrons gain more speed.

e) The slope of the line segment gives the Planck's constant. From the mathematical relationship, E = hv₀,

And the slope of the line segment is Energy of ejected electrons/frequency of incident light, E/v₀, which adequately matches the Planck's constant, h = 6.63 × 10⁻³⁴ J.s

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