Water boils at 100 Degrees Celsius or 212 degrees Fahrenheit
|F| = 112 - 67 = 45 N
F = 45 N ( to the left)
Tomato juice or acid rain
Answer:
13.8072 kj
Explanation:
Given data:
Mass of water = 100.0 g
Initial temperature = 4.0 °C
Final temperature = 37.0°C
Specific heat capacity = 4.184 j/g.°C
Heat absorbed = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 37.0°C - 4.0 °C
ΔT = 33.0°C
Q = 100.0 g ×4.184 j/g.°C × 33.0°C
Q = 13807.2 j
Joule to KJ:
13807.2 j × 1kj /1000 j
13.8072 kj
Answer: When 20 grams of potassium chlorate, KClO3, is dissolved in 100 grams of water at 80 ºC, the solution can be correctly described as:, unsaturated
At approximately what temperature does the solubility of sodium chloride, NaCl, match the solubility of potassium dichromate, K2Cr2O7?, 60 ºC
Explanation:
