Answer:
Volume = 98.5 L.
Explanation:
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Answer:
the initial temperature of the iron sample is Ti = 90,36 °C
Explanation:
Assuming the calorimeter has no heat loss to the surroundings:
Q w + Q iron = 0
Also when the T stops changing means an equilibrium has been reached and therefore, in that moment, the temperature of the water is the same that the iron ( final temperature of water= final temperature of iron = T )
Assuming Q= m*c*( T- Tir)
mc*cc*(T-Tc)+mir*cir*(T - Tir) = 0
Tir = 20.3 °C + 300 g * 4.186 J/g°C * (20.3 C - 19 °C) / ( 51.9 g * 0.449 J/g°C )
Tir = 90.36 °C
Note :
- The specific heat capacity of water is assumed 1 cal/g°C = 4.186 J/g°C
- We assume no reaction between iron and water
<span>We look at the end of the day:
n(HNO3) added = 0.500*17.0/1000 = 0.00850 mol
n(NH3) = 0.200*75.0/1000 - 0.00850 = 0.00650 mol
[NH3] left = 0.00650*1000/(17.0+75.0) = 0.070652
M [OH-] = Kb * [NH3] = 0.070652*1.8*10^(-5) = 1.27174 x 10^(-6)
pOH = -log[OH-] ≈ 5.8956 pH = 14 - pOH ≈ 8.10</span>
