Answer:
3.07 Cal/g
Explanation:
Step 1: Calculate the heat absorbed by the calorimeter
We will use the following expression.
Q = C × ΔT
where,
- C: heat capacity of the calorimeter (37.60 kJ/K = 37.60 kJ/°C)
- ΔT: temperature change (2.29 °C)
Q = 37.60 kJ/°C × 2.29 °C = 86.1 kJ
According to the law of conservation of energy, the heat released by the candy has the same magnitude as the heat absorbed by the calorimeter.
Step 2: Convert 86.1 kJ to Cal
We will use the conversion factor 1 Cal = 4.186 kJ.
86.1 kJ × 1 Cal/4.186 kJ = 20.6 Cal
Step 3: Calculate the number of Cal per gram of candy
20.6 Cal/6.70 g = 3.07 Cal/g
PH = -log([H+])
[H+] = 10^(-pH)
[H+] = 10^(-9)
[H+][OH-] = Kw
Kw = 1.0*10^-14 at 25 degrees celsius.
[OH-] = Kw/[H+] = (1.0*10^-14)/(1*10^-9) = 1.0*10^-5
The concentration of OH- ions is 1.0*10^-5 M.
1.785714286 moles
The number of moles (n) for nitrogen is: [ n=50.0÷28.0 ] = 1.785714286 moles.
Your answer is C
amino acids are created by a bond of two carbon atoms and each carbon is part of a different group, one negative and one positive. when the chemical reactions occurs, it creates the several diff amino acids
Answer:
Hydrofluoric acid.
Explanation:
To know which of the acid is the strongest, let us determine the pka of each acid. This is illustrated below:
1. Acetic acid
Ka = 1.8x10^-5
pKa =..?
pKa = –logKa
pKa = –Log 1.8x10^-5
pKa = 4.74
2. Benzoic acid
Ka = 6.5x10^-5
pKa =..?
pKa = –logKa
pKa = –Log 6.5x10^-5
pKa = 4.18
3. Hydrofluoric acid.
Ka = 6.8x10^-4
pKa =..?
pKa = –logKa
pKa = –Log 6.8x10^-4
pKa = 3.17
4. Hypochlorous acid
Ka = 3.0x10^-8
pKa =..?
pKa = –logKa
pKa = –Log 3.0x10^-8
pKa = 7.52
Note: the smaller the pKa value, the stronger the acid.
The pka of the various acids as calculated above is given below:
Acid >>>>>>>>>>>>>>>>>> pKa
1. Acetic acid >>>>>>>>>> 4.74
2. Benzoic acid >>>>>>>> 4.18
3. Hydrofluoric acid >>>> 3.17
4. Hypochlorous acid >> 7.52
From the above illustration, we can see that hydrofluoric acid has the lowest pKa value. Therefore, hydrofluoric acid is the strongest among them.