If 0.484J of heat is added to 0.1372g of water, how much will the temperature increase? 0.263 °C
0.927 °c
0.843 °C
0.460 °C

ΔT = 0.843 °C

Explanation:

Given data:

Heat added = 0.484 J

Mass of water = 0.1372 g

Temperature increase = ΔT = ?

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

Specific heat capacity of water is 4.184 j/g.°C

Now we will put the values in formula.

0.484 j = 0.1372 g × 4.184 j/g.°C× ΔT

0.484 j = 0.574j/°C× ΔT

ΔT = 0.484 j / 0.574j/°C

ΔT = 0.843 °C

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.

bcde is the incorrect answer it is a