C). Viscosity
hope this helps
Answer:
2914 J
Explanation:
Step 1: Given data
- Mass of the copper tubing (m): 665.0 g
- Initial temperature: 15.71 °C
- Final temperature: 27.09 °C
- Specific heat of copper (c): 0.3850 J/g.°C
Step 2: Calculate the temperature change
ΔT = 27.09 °C - 15.71 °C = 11.38 °C
Step 3: Calculate the energy required (Q)
We will use the following expression.
Q = c × m × ΔT
Q = 0.3850 J/g.°C × 665.0 g × 11.38 °C
Q = 2914 J
The value of current solubility product is calculated as below
K = (Ag+)( Cl-)
Ag+ = 0.01 M
Cl-= 1 x10^-5M
K is therefore = 1 x10^-5 x 0.01 = 1 x10 ^ -7 M
The K obtained is greater than Ksp
that is K> KSp
1x10^-7 > 1.7 x10 ^-10
will precipitation of AgCl form?
yes the precipitation of AgCl will be formed since K> KSP
Answer:
13.5 years
Explanation:
Initial Concentration [Ao] = 10g
Final Concentration [A] = 0.768g
Time t= 50 years
Half life t1/2 = ?
These quantities are related by the following equations;
ln[A] = ln[Ao] - kt ......(i)
t1/2 = ln(2) / k ...........(ii)
where k = rate constant
Inserting the values in eqn (i) and solving for k, we have;
ln(0.768) = ln(10) - k (50)
-0.2640 = 2.3026 - 50k
50k = 2.3026 + 0.2640
k = 2.5666 / 50 = 0.051332
Insert the value of k in eqn (ii);
t1/2 = ln(2) / k
t1/2 = 0.693 / 0.051332 = 13.5 years
