Answer:
pH = 1.6
Explanation:
0 mL NaOH:
⇒ [ H3O+] = M HBr = 0.1 M
⇒pH = -log [H3O+] = 1
30 mL NaOH:
⇒ mol NaOH = 0.1 mol / L * 0.03 L = 3 E-3 mol
⇒ mol HBr = 0.05 L * 0.1 mol/L = 5 E-3 mol
⇒ M HBr = ( 5 E-3 mol - 3 E-3 mol) / 0.08 L = 0.025 M
⇒ pH = - log (0.025) = 1.6
Remembering the equation Q=MCdeltaT where
q=is the amount of heat energy
M=mass
C=specific heat
deltaT= change in temperature
Therefore, using the equation we can substitute values and solve for q.
Q= (15 grams) (0.129 J/(gx°C))(85-22)
Q=(15) ((0.129 J/(gx°C)) (63)
Q=121.9 Joules
The energy needed to raise the temperature of 15 grams of gold from 22 degrees Celsius to 85 degrees Celsius is then 121.9 Joules or 122 Joules (if rounded up).
Ho123 right I’m sorry of I’m wrong
Answer:
2NaOH + (NH4) 2SO4 = Na2SO4(s) + 2NH3(g) + 2H2O(l)
Two moles of sodium hydroxide reacts with 1 Mike of ammonium sulphate to give 1 mole of Sodium sulphate, 2 moles of ammonia gas and 2 moles of water
Explanation:
Answer:
76.03 °C.
Explanation:
Equation:
C2H5OH(l) --> C2H5OH(g)
ΔHvaporization = ΔH(products) - ΔH (reactants)
= (-235.1 kJ/mol) - (-277.7 kK/mol)
= 42.6 kJ/mol.
ΔSvaporization = ΔS(products) - ΔS(reactants)
= 282.6 J/K.mol - 160.6 J/K.mol
= 122 J/K.mol
= 0.122 kJ/K.mol
Using gibbs free energy equation,
ΔG = ΔH - TΔS
ΔG = 0,
T = ΔH/ΔS
T = 42.6/0.122
= 349.18 K.
Coverting Kelvin to °C,
= 349.18 - 273.15
= 76.03 °C.
