1. 3.0% ----> 3.0 kg fat= 100 kg body weigh
also remember that 1 kg= 2.20 lbs
![65 kg \frac{3.0 kg}{100 kg} x \frac{2.20 lb}{1 kg} = 4.29 lbs](https://tex.z-dn.net/?f=65%20kg%20%20%5Cfrac%7B3.0%20kg%7D%7B100%20kg%7D%20x%20%20%5Cfrac%7B2.20%20lb%7D%7B1%20kg%7D%20%3D%204.29%20lbs)
2. 0.94 g/mL----> 0.94 grams= 1 mL
1 Liters= 1000 mL
1kg= 1000 grams
Answer:
c = 0.898 J/g.°C
Explanation:
1) Given data:
Mass of water = 23.0 g
Initial temperature = 25.4°C
Final temperature = 42.8° C
Heat absorbed = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Specific heat capacity of water is 4.18 J/g°C
ΔT = 42.8°C - 25.4°C
ΔT = 17.4°C
Q = 23.0 g × × 4.18 J/g°C × 17.4°C
Q = 1672.84 j
2) Given data:
Mass of metal = 120.7 g
Initial temperature = 90.5°C
Final temperature = 25.7 ° C
Heat released = 7020 J
Specific heat capacity of metal = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 25.7°C - 90.5°C
ΔT = -64.8°C
7020 J = 120.7 g × c × -64.8°C
7020 J = -7821.36 g.°C × c
c = 7020 J / -7821.36 g.°C
c = 0.898 J/g.°C
Negative sign shows heat is released.
Answer:
![\% Fe^{+2}=70%](https://tex.z-dn.net/?f=%5C%25%20Fe%5E%7B%2B2%7D%3D70%25)
Explanation:
Hello,
In this case, we could considering this as a redox titration:
![Fe^{+2}+(Cr_2O_7)^{-2}\rightarrow Fe^{+3}+Cr^{+3}](https://tex.z-dn.net/?f=Fe%5E%7B%2B2%7D%2B%28Cr_2O_7%29%5E%7B-2%7D%5Crightarrow%20Fe%5E%7B%2B3%7D%2BCr%5E%7B%2B3%7D)
Thus, the balance turns out (by adding both hydrogen ions and water):
![Fe^{+2}\rightarrow Fe^{+3}+1e^-\\(Cr_2^{+6}O_7)^{-2}+14H^++6e^-\rightarrow 2Cr^{+3}+7H_2O\\\\6Fe^{+2}+(Cr_2^{+6}O_7)^{-2}+14H^++6e^-\rightarrow Cr^{+3}+7H_2O+6Fe^{+3}+6e^-](https://tex.z-dn.net/?f=Fe%5E%7B%2B2%7D%5Crightarrow%20Fe%5E%7B%2B3%7D%2B1e%5E-%5C%5C%28Cr_2%5E%7B%2B6%7DO_7%29%5E%7B-2%7D%2B14H%5E%2B%2B6e%5E-%5Crightarrow%202Cr%5E%7B%2B3%7D%2B7H_2O%5C%5C%5C%5C6Fe%5E%7B%2B2%7D%2B%28Cr_2%5E%7B%2B6%7DO_7%29%5E%7B-2%7D%2B14H%5E%2B%2B6e%5E-%5Crightarrow%20Cr%5E%7B%2B3%7D%2B7H_2O%2B6Fe%5E%7B%2B3%7D%2B6e%5E-)
Thus, by stoichiometry, the grams of Fe+2 ions result:
![m_{Fe^{+2}}=0.04021\frac{molK_2Cr_2O_7}{L}*0.02872L*\frac{6molFe^{+2}}{1molK_2Cr_2O_7}*\frac{56gFe^{+2}}{1molFe^{+2}}=0.388gFe^{+2}](https://tex.z-dn.net/?f=m_%7BFe%5E%7B%2B2%7D%7D%3D0.04021%5Cfrac%7BmolK_2Cr_2O_7%7D%7BL%7D%2A0.02872L%2A%5Cfrac%7B6molFe%5E%7B%2B2%7D%7D%7B1molK_2Cr_2O_7%7D%2A%5Cfrac%7B56gFe%5E%7B%2B2%7D%7D%7B1molFe%5E%7B%2B2%7D%7D%3D0.388gFe%5E%7B%2B2%7D)
Finally, the mass percent is:
![\% Fe^{+2}=\frac{0.388g}{0.5562g}*100\%\\ \% Fe^{+2}=70%](https://tex.z-dn.net/?f=%5C%25%20Fe%5E%7B%2B2%7D%3D%5Cfrac%7B0.388g%7D%7B0.5562g%7D%2A100%5C%25%5C%5C%20%20%5C%25%20Fe%5E%7B%2B2%7D%3D70%25)
Best regards.