Answer:
Qm = -55.8Kj/mole
Explanation:
NaOH(aq) + HNO₃(aq) => NaNO₃(aq) + H₂O(l)
Qm = (mc∆T)water /moles acid
Given => 100ml(0.300M) NaOH(aq) + 100ml(0.300M)HNO₃(aq)
=> 0.03mole NaOH(aq) + 0.03mole HNO₃(aq)
=> 0.03mole NaNO₃(aq) + 0.03mole H₂O(l)
ΔH⁰rxn = [(200ml)(1.00cal/g∙°C)(37 – 35)°C]water / 0.03mole HNO₃
= 13,333 cal/mole x 4.184J/cal = 55,787J/mol = 55.8Kj/mole (exothermic)*
Heat of reactions comes from formation of H-Oxy bonds on formation of water of reaction and heats the 200ml of solvent water from 35⁰C to 37⁰C.
Answer:
A.
Explanation:
The lower the pH the greater the concentration of H+ ions. Hydrochloric acid with a pH of 1 is a strong acid and is highly ionised in solution. It has the most H+ ions
10 is mixture of elements and compounds
11 is compound
12-15 are mixture of compounds
Q = mcΔT = (4.00 g)(0.129 J/g•°C)(40.85 °C - 0.85 °C)
Q = 20.6 J of energy was involved (more specifically, 20.6 J of heat energy was absorbed from the surroundings by the sample of solid gold).
Answer:
c = 0.377 J/g.°C
c = 0.2350 J/g.°C
J = 27.3 J
Explanation:
We can calculate the heat (Q) absorbed or released by a substance using the following expression.
Q = c × m × ΔT
where,
c: specific heat
m: mass
ΔT: change in the temperature
<em>It takes 49.0J to raise the temperature of an 11.5g piece of unknown metal from 13.0°C to 24.3°C. What is the specific heat for the metal? Express your answer numerically, in J/g.°C</em>
Q = c × m × ΔT
49.0 J = c × 11.5 g × (24.3°C - 13.0°C)
c = 0.377 J/g.°C
<em>The molar heat capacity of silver is 25.35 J/mol.°C. How much energy would it take to raise the temperature of 11.5g of silver by 10.1°C? Express your answer numerically, in Joules. What is the specific heat of silver?</em>
<em />
The molar mass of silver is 107.87 g/mol. The specific heat of silver is:

Q = c × m × ΔT
Q = (0.2350 J/g.°C) × 11.5 g × 10.1°C = 27.3 J