After 3 half life periods you would have 5 grams of krypton left because half of 40 is 20 half of 20 is 10 and half of 10 is 5
1. electrostatic interactions
<span>3. van de waals interactions </span>
<span>4. hydrogen bonding </span>
Answer:
Explanation:
(NH4)3 PO4 +NaOH arrow Na3PO4 +3NH3 +3H2O
Start by seeing what happens with the Na. You need 3 on the left, so put a 3 in front of NaOH
(NH4)3 PO4 +3NaOH arrow Na3PO4 +3NH3 +3H2O Next work with the nitrogens. YOu have 3 on the left and 3 on the right, so they are OK. Next Go to the stray oxygens.
You have 3 on left in (NaOH) and three on the right in 3H2O so they are fine as well. The last thing you should look at are hydrogens.
There are 12 + 3 on the left which is 15. There are 9 (in 3NH3) and 6 more in the water. They seem fine.
Why didn't I do something with the PO4^(-3)? The reason is a deliberately stayed away from them and balanced everything else. Since they were untouched with 1 on the left and 1 on the right, they are balanced.
Species Na H O N PO4
Left 3 15 3 3 1
Right 3 15 3 3 1
The mass of chlorine that react with 9.00 g of Al to form AlCl3 is 35.465 grams
Explanation
write the equation for reaction
that is
2 Al + 3 Cl2 = 2 Al CL3
find the moles of Al reacted
moles = mass/molar mass
9 g/ 27 g/mol = 0.333 moles of Al
by use of mole ratio between Al to Cl2 which is 2:3 find the moles of Cl2
mole of cl2 = 0.333 x3/2 = 0.4995 moles
mass of Cl2 is therefore = moles x molar mass
= 0.4995 x71 = 35.465 moles
