Answer:
Q = 8.8 kJ
Explanation:
Step 1: Data given
The specific heat of a solution = 4.18 J/g°C
Volume = 296 mL
Density = 1.03 g/mL
The temperature increases with 6.9 °C
Step 2: Calculate the mass of the solution
mass = density * volume
mass = 1.03 g/mL * 296 mL
mass = 304.88 grams
Step 3: Calculate the heat
Q = m*c*ΔT
⇒ with Q = the heat in Joules = TO BE DETERMINED
⇒ with m = the mass of the solution = 304.88 grams
⇒ with c = the specific heat of the solution = 4.18 J/g°C
⇒ with ΔT = the change in temperature = 6.9 °C
Q = 304.88 g * 4.18 J/g°c * 6.9 °C
Q = 8793.3 J = 8.8 kJ
Q = 8.8 kJ
%(NaHCO3)= ((mass NaHCO3)/(mass NaHCO3 + mass water))*100%
m=Volume*Density
Density of water =1 g/ml
m(water) = Volume(water)*Density(water) = 600.0 ml * 1g/ml=600g water
%(NaHCO3)= ((20.0 g)/(20.0 g + 600 g))*100%=0.0323*100%=32.3%
I think it is 11% I read it on a article on msn.
Amplitude is the awnser glad I could help you
