Answer:
4.56 X 10^ -4 g/mL
Explanation:
A solution is prepared by diluting 6.0 mL of a 7.6 x 10-4 g/mL solution to a total volume of 10.0 mL. Calculate the concentration of the dilute solution.
(7.6 X10^-4 gm/m L) x( 6.0 m L ) = 45.6 X 10^-4 g
this is dissolved )in 10 m L=45.6 X 10^-4 g/ 10
4.56 X 10^ -4 g/mL
check
6/10 =0.6
4.56/7.6 = o.,6
Answer:
a i think
Explanation:
because it was the only one that compared something to another
Answer:
the mass of the air in the room is 4.96512 kg ( in 0°C)
Answer:
pKa = 3.675
Explanation:
∴ <em>C</em> X-281 = 0.079 M
∴ pH = 2.40
let X-281 a weak acid ( HA ):
∴ HA ↔ H+ + A-
⇒ Ka = [H+] * [A-] / [HA]
mass balance:
⇒<em> C</em> HA = 0.079 M = [HA] + [A-]
⇒ [HA] = 0.079 - [A-]
charge balance:
⇒ [H+] = [A-] + [OH-]... [OH-] is negligible; it comes from to water
⇒ [H+] = [A-]
∴ pH = - log [H+] = 2.40
⇒ [H+] = 3.981 E-3 M
replacing in Ka:
⇒ Ka = [H+]² / ( 0.079 - [H+] )
⇒ Ka = ( 3.981 E-3 )² / ( 0.079 - 3.981 E-3 )
⇒ Ka = 2.113 E-4
⇒ pKa = - Log ( 2.113 E-4 )
⇒ pKa = 3.675
Answer:
1552.83J Released
Explanation:
1. mass/m=225
Initial temp:86C, final:32.5C
Changed Temp: 32.5-86= -53.5C
s=0.129 J/gC
Formula: q= m times s times changed Temp.
q=(225)(0.129)(-53.5)
q= -1552.83 J
q=1552.83 J Released
