Calculate the amount of heat required to raise the temperature of a 65-g sample of water from 32 ∘c to 65 ∘c. (the specific heat capacity of water is 4.184 j/(g⋅∘c).

The amount of heat required is 9.0 kJ.

q = mCΔT

ΔT = T_f – T_i = 65 °C – 32 °C = 33 °C

q = 65 g × 4.184 J·°C⁻¹g⁻¹ × 33 °C = 9000 J = 9.0 kJ

8974.68 J.

Explanation:

Q = m.C.ΔT, where,

Q is the amount of heat required to raise the temperature of water sample,

m is the mass of water sample (m = 65.0 g),

C is the specific heat of water (C = 4.184 J/g.°C),

ΔT is the temperature difference (the final temperature - the initial temperature (ΔT = 65.0 - 32.0 = 33.0 °C).

∴ The amount of heat required to raise the temperature of water sample Q = m.C.ΔT = (65.0 g) (4.184 J/g.°C) (33.0 °C) = 8974.68 J.

chemistry sucks a**

explanation:

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.