Answer:
Q = -3980.9 j
Explanation:
Given data:
Mass of sample = 30 g
Initial temperature = 56.7 °C
Final temperature = 25 °C
Specific heat of water = 4.186 j/g.°C
Amount of heat released = ?
Formula:
Q = m.c.ΔT
Q = heat released
m = mass of sample
c = specific heat of given sample
ΔT = change in temperature
Solution:
ΔT = T2 -T1
ΔT = 25 °C - 56.7 °C = - 31.7°C
Q = m.c.ΔT
Q = 30 g × 4.186 j/g.°C × - 31.7°C
Q = -3980.9 j
Answer:
(a) Homogeneous. 4.7 g of MgCl₂.
(b) 9.1 g
Explanation:
(a)
At 200°C, we can dissolve 54.6g of MgCl₂ in 100 g of water. The mass that we could dissolve in 38.2 g of water is:

Since we can dissolve up to 20.9 g of MgCl₂ and we added only 16.2 g, the mixture is homogeneous and we could add 20.9 g -16.2 g = 4.7 g of solute to make it saturated.
(b)
At 800°C, we can dissolve 66.1 g of MgCl₂ in 100 g of water. The mass that we could dissolve in 38.2 g of water is:

Since we can dissolve up to 25.3 g of MgCl₂ and we added only 16.2 g, we could add 25.3 g - 16.2 g = 9.1 g of solute to make it saturated.
They were natural philosophers that used experimentation to establish the cause effect relations and explanations, in an epoque when many "theories' were not supported by experiments but by some authority (antique philosophers like Aristoteles or the Catholic Church).
The question is incomplete, here is the complete question:
Calculate the volume in liters of a 0.13 M potassium dichromate solution that contains 200. g of potassium dichromate . Round your answer to 2 significant digits.
<u>Answer:</u> The volume of solution is 5.2 L
<u>Explanation:</u>
To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

We are given:
Molarity of solution = 0.13 M
Given mass of potassium dichromate = 200. g
Molar mass of potassium dichromate = 294.15 g/mol
Putting values in above equation, we get:

Hence, the volume of solution is 5.2 L
Your answer is C. Both gasoline and litter would need to be physically separated from the water, because neither bonds with the water.