<span>BaCl2+Na2SO4---->BaSO4+2NaCl
There is 1.0g of BaCl2 and 1.0g of Na2SO4, which is the limiting reagent?
"First convert grams into moles"
1.0g BaCl2 * (1 mol BaCl2 / 208.2g BaCl2) = 4.8 x 10^-3 mol BaCl2
1.0g Na2SO4 * (1 mol Na2SO4 / 142.04g Na2SO4) = 7.0 x 10^-3 mol Na2SO4
(7.0 x 10^-3 mol Na2SO4 / 4.8 x 10^-3 mol BaCl2 ) = 1.5 mol Na2SO4 / mol BaCl2
"From this ratio compare it to the equation, BaCl2+Na2SO4---->BaSO4+2NaCl"
The equation shows that for every mol of BaCl2 requires 1 mol of Na2SO4. But we found that there is 1.5 mol of Na2SO4 per mol of BaCl2. Therefore, BaCl2 is the limiting reagent.</span>
Answer:1
Explanation:i know cuz I got it right
The answer is:
carboxylate
and protonated amines
The explanation:
when
The Carboxylic acids are polar . and the functional group carboxyl is formed from the hydroxyl and carbonyl group.
and about protonated amines:
-alkyl groups improve the basicity, aryl groups diminish it.
-Primary amides have an aromatic group or alkyl attached to an amino-carbonyl function.
- Primary amines are basic functions which could be protonated to the ammonium ion
From Boyle's law, we can use the equation
P1V1 = P2V2
to find the volume when the pressure is increased to 760 kPa:
V2 = P1V1 / P2
V2 = (210 kPa)(15.0 L) / 760 kPa
V2 = 4.14 Liters
Therefore, the volume is 3.99 Liters after the pressure increased to 760 kPa.
But if the pressure is increased to 790 kPa, the volume of the krypton will decrease to
V2 = P1V1 / P2
V2 = (210 kPa)(15.0 L) / 790 kPa
V2 = 3.99 Liters
We should apply Boyle's Law here given initial pressure, initial volume and final volume.
P1V1= P2V2
(6.5 atm) (13 L) = P2 (3.3 L)
Solve for P2 on your calculator and that should get you to the answer.
