Answer:
[KOH] = 0.10M in KOH
Explanation:
Molar Concentration [M] = moles solute/volume solution in liters
moles KOH = 0.56g/56g/mole = 0.01mole
Volume of solution = 100cm³ = 100ml = 0.10 liter
[KOH] = 0.01 mole KOH / 0.10 liter solution = 0.10M in KOH
Answer:
81 °C
Explanation:
This is a calorimetry question so a few things you will need for this. The calorimetry equation q=mcΔT & the specific heat of water (4.2J/g•°C). Other definitions are:
q = heat added/released by a sample
m = mass of sample
c=specific heat of sample
ΔT = change in temperature
from here we can rearrange the equation to state:
q/(mc) = ΔT
1200J/((20.0g)(4.2J/g•°C)) = ΔT
14°C = ΔT
If the starting temperature was 95.0°C and we know that the temperature was cooled by 14°C then the final temperature of the water would be 81.
Answer:
Rn] 5f14 6d4 7s2 mp: none d: none. Seaborgium ..... Element. Density. Boiling or Melting Point in °C. Electron Configuration. Symbol.
Explanation:
