Answer:
31395 J
Explanation:
Given data:
mass of water = 150 g
Initial temperature = 25 °C
Final temperature = 75 °C
Energy absorbed = ?
Solution:
Formula:
q = m . c . ΔT
we know that specific heat of water is 4.186 J/g.°C
ΔT = final temperature - initial temperature
ΔT = 75 °C - 25 °C
ΔT = 50 °C
now we will put the values in formula
q = m . c . ΔT
q = 150 g × 4.186 J/g.°C × 50 °C
q = 31395 J
so, 150 g of water need to absorb 31395 J of energy to raise the temperature from 25°C to 75 °C .
<span>You use the Henderson - Hasselbalch equation
pH = pKa + log ([salt]/[acid])
pKa = -log (8.2*10^-5) = 4.081
pH = 4.081 + (0.590/0.190)
pH = 4.081 + log 3.105
pH = 4.081 + 0.49206
pH = 4.573</span>
Answer:
The answer is "0.00172172603".
Explanation:
Given:

Formula:


El arroz si produce electricidad:)