Kp= (COCl2)/[(CO)(Cl2)]= 1.49 x 10^8
1.49 x 10^8= (COCl2/((2.22x10-4)(2.22x10-4))
COCl2= 1.49x10^8 x ((2.22x10-4)(2.22x10-4))= 7.34 atm
Actually Rb or Rubidium in zero state has the following
electron configuration:
<span>1s22s2</span><span>2p6</span><span>3s2</span><span>3p63d10</span><span>4s2</span><span>4p65s1</span>
However we can see that the ion has a 1 positive charge,
which means that it lacks 1 electron, therefore the answer from the choices is:
<span>d.
rb+: 1s22s22p63s23p64s23d104p6</span>
Answer:
There are 0.0267 moles of oxygen in 0.658 grams of magnesium nitrate.
Explanation:
Mass of magnesium nitrate = m = 0.658 g
Molar mass of magnesium nitrate = M = 148 g/mol
Moles of magnesium nitrate = n
1 mole of magnesium nitrate has 6 moles of oxygen atoms. Then 0.004446 moles magnesium nitrate will have :
There are 0.0267 moles of oxygen in 0.658 grams of magnesium nitrate.
Answer:
The net ionic equation will be MgCl₂ + 2 NaOH → Mg(OH)₂ + 2 NaCl
Explanation:
Ionization of MgCl₂ is as follows
MgCl₂ → Mg²⁺ + 2 Cl⁻
Ionization of NaOH is as follows
NaOH → Na⁺ + OH⁻
It is a one type of substitution reaction where OH⁻ combined with Mg²⁺ to give magnesium hydroxide .
On the other hand Cl⁻ combined with Na⁺ to give sodium chloride as product.
Using proper stoichiometry to balanced the number of atoms in both side .
