The first and Third graph
Answer:
1.023 J / g °C
Explanation:
m = 37.9 grams
ΔT = 25.0*C
H = 969 J
c = ?
The equation relating these equation is;
H = mcΔT
making c subject of formulae;
c = H / mΔT
c = 969 J / (37.9 g * 25.0*C)
Upon solving;
c = 1.023 J / g °C
Explanation:
Some Rules Regarding Oxidation Numbers:
- Hydrogen has oxidation number of + 1 except in hydrides where it is -1
- Oxygen has oxidation number of -2 except in peroxides where it is -1
- Some elements have fixed oxidation numbers. E.g Halogen group elements has oxidation number of -1
- Oxidation number of a compound is the sum total of the individual elements and a neutral compound has oxidation number of 0.
A. HI
Hydrogen has oxidation of + 1
Oxidation number of I:
1 + x = 0
x = -1
B. PBr3
Br has oxidation number of - 1
Oxidation number of Pb:
x + 3 (-1) = 0
x = + 3
C. KH
Hydrogen has oxidation of + 1
Oxidation number of K:
1 + x = 0
x = -1
D. H3PO4
Hydrogen has oxidation number of + 1
Oxygen has oxidation number of -2
Oxidation number of P:
3(1) + x + 4(-2) = 0
3 + x - 8 =0
x = 5
Answer : The maximum amount of nickel(II) cyanide is 
Explanation :
The solubility equilibrium reaction will be:

Initial conc. 0.220 0
At eqm. (0.220+s) 2s
The expression for solubility constant for this reaction will be,
![K_{sp}=[Ni^{2+}][CN^-]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BNi%5E%7B2%2B%7D%5D%5BCN%5E-%5D%5E2)
Now put all the given values in this expression, we get:


Therefore, the maximum amount of nickel(II) cyanide is 
Answer:
D.
Explanation:
Carbon atoms are not strongly electronegative and tend to form covalent bonds.