Answer:
a. x = + 2
b. x = + 3
c. x = + 2
d. x = + 3
Explanation:
The oxidation number is a formal charge assigned to an atom present in a molecule or formula unit or ion based on some arbitrary rules.
a.
<u>Hg</u>O
The oxidation number of Hg in HgO is:
x + (-2) = 0
x = +2
b.
<u>Al₄</u>C₃
The oxidation number of Al in Al<u>₄</u>C₃ is:
4x +(3 × -4) = 0
4x - 12 = 0
4x = +12
x = 12/4
x = +3
c. CrF₂
x + ( 2 × - 1) = 0
x - 2 = 0
x = + 2
d. Fe₂S₃
2x + ( 3 × - 2) = 0
2x + (-6) = 0
2x = 6
x = 6/2
x = +3
Answer:
4
Explanation:
If I remember correctly, the atomic # will always be the same as the # of electrons
A monobromination reaction of an alkane involves an alkane and bromine. The position of the hydrogen atom that will be substituted by the bromine free radical will depend on the order of the alkane. The bromine will attach to the carbon that has the most substituents.
<span>4 I</span>₂<span>+ 9 O</span>₂<span> = 2 I</span>₄<span>O</span>₉
Reactants :
I₂ , O₂
Products :
I₄O₉
hope this helps!
Answer:
The answer is 1.15m.
Since molality is defined as moles of solute divided by kg of solvent, we need to calculated the moles of H2SO4 and the mass of the solvent, which I presume is water.
We can find the number of H2SO4 moles by using its molarity
C=nV→nH2SO4=C⋅VH2SO4=6.00molesL⋅48.0⋅10−3L=0.288
Since water has a density of 1.00kgL, the mass of solvent is
m=ρ⋅Vwater=1.00kgL⋅0.250L=0.250 kg
Therefore, molality is
m=nmass.solvent=0.288moles0.250kg=1.15m