Answer: B) 2 (as indicated by electron distribution shown), but taking into account the real properties of this element, 4,7,8 also occur (see below).
Explanation:
This is the electron complement/atomic number of ruthenium, which actually has the structure [Kr] 5s1 4d7
Nevertheless, Ru does not form Ru(I) compounds and few Ru(II) compounds (RuCl2, RuBr2, RuI2). It also forms Ru(III)Cl3 and a larger number of Ru(IV) compounds, e.g. RuO2, RuS2. It also forms RuO4
Answer:
I am not really sure, but I think Fr.
Explanation:
Answer:
Cd(s) + AgNO₃(aq) → Cd(NO₃)₂ (aq) + Ag(s)
Oxidized: Cd
Reduced: Ag
Explanation:
Cd(s) + AgNO₃(aq) → Cd(NO₃)₂ (aq) + Ag(s)
Cd → Cd²⁺ + 2e⁻ Half reaction oxidation
1e⁻ + Ag⁺ → Ag Half reaction reduction
Ag changed oxidation number from +1 to 0
Cd changed oxidation number from 0 to +2
Let's ballance the electrons
( Cd → Cd²⁺ + 2e⁻ ) .1
( 1e⁻ + Ag⁺ → Ag ) .2
Cd + 2e⁻ + 2Ag⁺ → 2Ag + Cd²⁺ + 2e⁻
Finally the ballance equation is:
Cd(s) + 2AgNO₃(aq) → Cd(NO₃)₂ (aq) + 2Ag(s)
There are 34 g of oxygen in the container.
We can use the<em> Ideal Gas Law</em> to solve this problem.
But
, so
and

STP is 0 °C and 1 bar, so

CH4+2O2–>CO2+2H2O
4Fe+3O2–>2Fe2O3