Answer:
The answer is B- LiNO3(aq)+H2O(I)
Explanation:
I just did the assignment also UNUS ANNUS
Answer:
The answer to your question is pH = 1.45
Explanation:
Data
pH = ?
Volume 1 = 200 ml
[HCl] 1 = 0.025 M
Volume 2 = 150 ml
[HCl] 2 = 0.050 M
Process
1.- Calculate the number of moles of each solution
Solution 1
Molarity = moles / volume
-Solve for moles
moles = 0.025 x 0.2
result
moles = 0.005
Solution 2
moles = 0.050 x 0.15
-result
moles = 0.0075
2.- Sum up the number of moles
Total moles = 0.005 + 0.0075
= 0.0125
3.- Sum up the volume
total volume = 200 + 150
350 ml or 0.35 l
4.- Calculate the final concentration
Molarity = 0.0125 / 0.35
= 0.0357
5.- Calculate the pH
pH = -log [H⁺]
-Substitution
pH = -log[0.0357]
-Result
pH = 1.45
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:
ΔT = 0.78 °C
Explanation:
Given data:
Mass of Al = 9.5 g
Specific heat capacity of Al = 0.9 J/g.°C
Temperature change = ?
Heat added = 67 J
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
67 J = 9.5 g × 0.9 j/g.°C × ΔT
67 J = 85.5 j/°C × ΔT
ΔT = 67 J / 85.5 j/°C
ΔT = 0.78 °C
