2) amine is the correct answer
First, we need to get the concentration of [NaH2PO4]:
[NaH2PO4] =( mass / molar mass ) * volume L
when we have mass NaH2PO4 = 6.6 g & molar mass = 120g/mol & V = 0.355 L
So by substitution:
[NaH2PO4] = (6.6g / 120g/mol) * 0.355 L = 0.0195 M
then, we need to get the concentration of [Na2HPO4]:
[Na2HPO4]= (mass / molar mass ) * volume L
So by substitution:
[Na2HPO4] = (8g/ 142g/mol) * 0.355 L = 0.02 M
and when Pka of the 2nd ionization of phosphoric acid = 7.21
So by substitution in the following formula, we can get the PH:
PH = Pka + ㏒[A]/[AH]
∴PH = 7.21 + ㏒[0.02]/[0.0195]
∴ PH = 7.2
Answer:
0.5m/s2
Explanation:
Data obtained from the question include:
u (initial velocity) = 0m/s
v (final velocity) = 30m/s
t (time) = 1min = 60secs
a (Acceleration) =?
a = v — u/t
a = 30—0 / 60
a = 30/60
a = 0.5m/s2
Answer:
....,................................
Explanation:
1= A
2=D
3=C
4=C
