The number of electrons in outer ring for cesium would be ONE 1.
Answer:
2C4H10 + 13O2 → 8CO2 + 10H2O
Hope this helps brainleist??
Explanation:
From the chemical equation given:
H2SO4+2KOH--->K2SO4+2H2O
the two reactants, H2SO4 and KOH, are in 1:2 stoichiometric ratio.
No. of moles of KOH = 2* no. of moles of H2SO4
=2*0.1*0.033
The concentration of KOH = no. of moles / volume
=2*0.1*0.033/0.05
=0.132M
Answer:
3p^2
Explanation:
after filling 3s^2 only two electrons left out of 14 so the next sub shell is 3p therefore ,X represents 3p^2
