1) in periodic acid (HIO₄), iodine has oxidation number +7, hydrogen has oxidation number +1, oxygen has -2, compound has neutral charge:
+1 + x + 4 · (-2) = 0.
x = +7.
2) in molecule of iodine (I₂), iodine has oxidation number 0, because iodine is nonpolar molecule.
3) in sodium iodide (NaI), iodine has oxidation number -1, sodium has oxidation number +1:
+1 + x = 0.
x = -1.
4) in iodic acid (HIO₃), iodine has oxidation number +5, hydrogen has oxidation number +1, oxygen has -2, compound has neutral charge:
+1 + x + 3 · (-2) = 0.
x = +5.
The answer would be either b or c because they are both correct
Answer:
Not doubled
Explanation:
The equation below represent the ideal gases relationship
PV ÷ T = constant
Here
P denotes pressure,
V denotes volume,
T denotes temperature in degrees Kelvin
Now
20 ° c = 273 + 20
= 293 K
And,
40 ° c = 313 K
So,
V = Vo. 313 K ÷ 293 K = 1.07 Vo
So, the volume is NOT doubled.
In the case when the temperature would be determined in degrees celsius at 0 degrees so the volume would be zero
We have to know final temperature of the gas after it has done 2.40 X 10³ Joule of work.
The final temperature is: 75.11 °C.
The work done at constant pressure, W=nR(T₂-T₁)
n= number of moles of gases=6 (Given), R=Molar gas constant, T₂= Final temperature in Kelvin, T₁= Initial temperature in Kelvin =27°C or 300 K (Given).
W=2.4 × 10³ Joule (Given)
From the expression,
(T₂-T₁)=
(T₂-T₁)= 
(T₂-T₁)= 48.11
T₂=300+48.11=348.11 K= 75.11 °C
Final temperature is 75.11 °C.
