Answer:
as follow
Explanation:
2
a acceptor
b Donor electron
3
A. MgBr2 B. Na2O
C. HCl D. AlCl3
E. K3N F. CaS
* the number is under the charactor/atoms
Answer:
a) Hydrogen and Fluorine
b) Lithium and Oxygen
c) Potassium and Phosphorus
d) Nickel and Oxygen
e) Mercury and Chlorine
Explanation:
The capital letters in the compounds represent a single element, while some elements have one capital letter and one small letter after it.
a) H & F
b) Li & O
c) K & P
d) Ni & O
e) Hg & Cl
<span>How many moles of acetic acid and of sodium acetate are present in 50.0 mL of solution?
We calculate it as follows:
moles of acetic acid = .120 mol/L (.050 L ) = 0.006 mol
moles of sodium acetate = 0.150 mol/L (0.050 L) = 0.008 mol
Hope this answers the question. Have a nice day.</span>
Answer:
<em><u>Percent volume-volume (%(v/v)) = 100 x (volume of solute / volume of solution)</u></em>
<em><u>Percent volume-volume (%(v/v)) = 100 x (volume of solute / volume of solution)ex. 20 ml of methanol dissolved in enough water to make 200 ml of solution would result in a 10 % methanol solution</u></em>
<em><u>Percent volume-volume (%(v/v)) = 100 x (volume of solute / volume of solution)ex. 20 ml of methanol dissolved in enough water to make 200 ml of solution would result in a 10 % methanol solution-the units may be any units of volume you chose - as long as they are consistent</u></em>
<em><u>Percent volume-volume (%(v/v)) = 100 x (volume of solute / volume of solution)ex. 20 ml of methanol dissolved in enough water to make 200 ml of solution would result in a 10 % methanol solution-the units may be any units of volume you chose - as long as they are consistent-this concentration unit is most often used when mixing two liquids</u></em>
It’s 0 because if u add them and divide 0 it’s 0
