The moles that are present in 150 g of ammonium hydrogen phosphite is 1.136 moles
<u><em>calculation</em></u>
<em>The formula of ammonium hydrogen phosphite = (NH4)2HPO4</em>
moles = mass/molar mass
mass= 150 g
molar mass of (NH4)2HPO4 = [ ( 14+ 1x4)2) +1 + 31 + ( 4 x16)] =132 g/mol
moles is therefore = 150 g/ 132 g/mol =1.136 moles
Answer:
Rb > K > Na > Li
Explanation:
Ionization energy increases up the period and up the group.
From the periodic table, we can see that Li is the going to have the highest ionization energy and Rb is going to have the lowest.
Therefore, the correct answer is the first choice
Answer:
Option b.) Potassium ion, is written correctly.
Explanation:
Hope this helps! :)
if i am correct it shall be 12. because i am thinking, 1 mole = 1 ar.
CBr4 is a symmetric tetrahedral molecule so it will be non-polar.
