Answer:
13598 J
Explanation:
Q = m × c × ∆T
Where;
Q = amount of energy (J)
m = mass (grams)
c = specific heat capacity
∆T = change in temperature
m = 65g, specific heat capacity of water = 4.184J/g°C, initial temperature= 100°C, final temperature = 150°C
Q = 65 × 4.184 × (150 - 100)
Q = 271.96 × 50
Q = 13598 J
Hence, 13598 J of energy is required to boil 65 grams of 100°C water and then heat the steam to 150°C.
Using the relationship M1V1 = M2V2 where M1 and M2 are the molar concentrations (mol/L or mmol/ml) and V1 and V2 are the volumes of the solutions, we can arrive at the following answer for the given problem:
<span>15.0M (L of stock solution) = 2.35M (0.25L) *all volumes were converted to liters.
L of stock solution = (2.35*0.25)/15.0
Therefore, 0.0392L or 39.17 ml of stock solution is needed. </span>
Answer:
pH = 2 and pOH = 12
Explanation:
Given [OH⁻] = 1 x 10⁻¹²M
pOH = -log[OH⁻] = -log(1 x 10⁻¹²) = - ( -12 ) = 12
pH + pOH = 14 => pH = 14 - pOH = 14 - 12 = 2
