What is the total amount of heat absorbed by 100.0
grams of water when the temperature of the water is
increased from 30.0°C to 45.0°C?
A) 418 J
B) 6270 J
C) 12 500 J
D) 18 800 J

Option (B) 6270J

Explanation:

The following were obtained from the question:

M = 100g

T1 = 30°C

T2 = 45°C

ΔT = 45 —30 = 15°C

C = 4.18J/g°C

Q=?

Q = MCΔT

Q = 100 x 4.18 x 15

Q = 6270J

Therefore, the total amount of heat absorbed is 6270J

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.