Unit of electro negativity
1 answer:
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<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:- .
Solution:- Mass of Iron added to water is 93.3 g. Initial temperature of iron metal is 65.58 degree C and final temperature of the system is 19.68 degree C.
temperature change, for iron metal = 65.58 - 19.68 = 45.9 degree C
specific heat for the metal is given as 0.444 J per g per degree C.
let's calculate the heat lost by iron metal using the equation:
where, q is the heat energy, m is mass, c is specific heat and delta T is change in temperature. let's plug in the values and calculate q for iron metal:
q = 1901.42 J
Using same equation we will calculate the heat gained by water.
mass of water is 75.0 g.
for water = 19.68 - 16.95 = 2.73 degree C
specific heat for water is 4.184 J pr g per degree C. Let's plug in the values:
q = 856.674 J
Total heat lost by iron metal is the sum of heat gained by water and calorimeter.
So, heat gained by calorimeter = heat lost by iron metal - geat gained by water
heat gained by calorimeter = 1901.42 J - 856.674 J = 1044.746 J
Change in temperature for calrimeter is same as for water that is 2.73 degree C
For calorimeter,
So, the heat capacity of calorimeter is .