<u>Answer:</u> No crystals of potassium sulfate will be seen at 0°C for the given amount.
<u>Explanation:</u>
We are given:
Mass of potassium nitrate = 47.6 g
Mass of potassium sulfate = 8.4 g
Mass of water = 130. g
Solubility of potassium sulfate in water at 0°C = 7.4 g/100 g
This means that 7.4 grams of potassium sulfate is soluble in 100 grams of water
Applying unitary method:
In 100 grams of water, the amount of potassium sulfate dissolved is 7.4 grams
So, in 130 grams of water, the amount of potassium sulfate dissolved will be
As, the soluble amount is greater than the given amount of potassium sulfate
This means that, all of potassium sulfate will be dissolved.
Hence, no crystals of potassium sulfate will be seen at 0°C for the given amount.
Answer:
when concentration increases, particles colide more often
Answer
is: activation energy of this reaction is 212,01975 kJ/mol.<span>
Arrhenius equation: ln(k</span>₁/k₂) =
Ea/R (1/T₂ - 1/T₁).<span>
k</span>₁
= 0,000643 1/s.<span>
k</span>₂
= 0,00828 1/s.
T₁ = 622 K.
T₂ = 666 K.
R = 8,3145 J/Kmol.
<span>
1/T</span>₁ =
1/622 K = 0,0016 1/K.<span>
1/T</span>₂ =
1/666 K = 0,0015 1/K.<span>
ln(0,000643/0,00828) = Ea/8,3145 J/Kmol ·
(-0,0001 1/K).
-2,55 = Ea/8,3145 J/Kmol </span>· (-0,0001 1/K).<span>
Ea = 212019,75 J/mol = 212,01975 kJ/mol.</span>
Answer :
(a) The dimensions in are,
(b) The distance in centimeters is, 40.64 cm
Explanation :
(a) The conversion used :
and,
As we are given the thickness, width, length dimensions in inches, inches and feet respectively.
Now we have to determine the thickness, width, length dimensions in centimeter, centimeter and meter respectively.
The dimensions in are,
(b) The conversion used :
As we are given the distance in inches. Now we have to determine the distance in centimeter.
The distance in centimeters is, 40.64 cm