Answer:
1 mole of iron =6.023×10^23 particles
1 particles of iron=1/6.023×10^23 mole
7.46×10^25 particles =1/6.023×10^23×7.46×10^25
=1.238×10^48 mole is a required answer.
Answer:
The final temperature of the water is 28.0 °C
Explanation:
Step 1: Data given
Mass of liquid water = 10.0 grams
Temperature = 23.0 °C
Heat absorbed = 209 Joules
Since heat was absorbed by the water, you must have a positive value for
Δ
T
Step 2: Calculate final temperature
q = m*c* ΔT
⇒ with m = the mass of the water = 10.0 grams
⇒ with c = the specific heat of water = 4.184 J/g°C
⇒ with ΔT = The change in temperature = T2 - T1 = T2 - 23.0 °C
⇒ with q = the heat absorbed = 209 Joule
209 = 10.0 * 4.184 * ΔT
ΔT = 5
ΔT = 5 = T2 - 23
T2 = 28 °C
The final temperature of the water is 28.0 °C
