I believe it would be c: 3 because you can have the numbers -1,0,1.
Answer:
2HNO3+Ca(OH)2 = Ca(NO3)2+2H2O
Explanation:
The reaction between Nitric acid(HNO3)and Calcium hydroxide(Ca(OH)2) gives Calcium Nitrate( Ca(NO3)2 and Water( H2O)
Complete question:
ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J. calculate the magnitude of q for the process.
Answer:
The magnitude of q for the process 568 J.
Explanation:
Given;
change in internal energy of the gas, ΔU = 475 J
work done by the gas, w = 93 J
heat added to the system, = q
During gas expansion process, heat is added to the gas.
Apply the first law of thermodynamic to determine the magnitude of heat added to the gas.
ΔU = q - w
q = ΔU + w
q = 475 J + 93 J
q = 568 J
Therefore, the magnitude of q for the process 568 J.
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
