How much heat energy is required to raise the temperature of 0.298 mole of water from 35.93°c to 79.84°c? the specific heat capacity of water = 4.18 j/g°c.

There is 985.63 J heat required

Explanation:

Step 1: Data given

Number of moles of water = 0.298 moles

Initial temperature = 35.93 °C

Final temperature = 79.84 °C

Specific heat capacity of water = 4.18 J/g°C

Molar mass water = 18.02 g/mol

Step 2: Calculate mass of water

Mass of H2O = Moles of H2O * molar mass H2O

Mass of H2O = 0.298 moles * 18.02 g/mol

Mass of H2O = 5.37 grams

Step 3: Calculate heat energy

Q = m*c*ΔT

⇒ Q = heat energy in J

⇒ m = mass of water = 5.37 grams

⇒ c = the specific heat capacity of water = 4.18 J/g°C

⇒ ΔT = The change in temperature = T2 - T1 = 79.84 - 35.93 = 43.91°C

Q = 5.37g * 4.18 J/g°C * 43.91 °C

Q = 985.63 J

There is 985.63 J heat required

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.