Answer:
nBACO3=m/M=9,83/197=0,05(mol) ->nHCl=0,05.2/1=0,1(mol) =>VHCl=n/CM=0,1/0,44=0,227(lít)
Explanation:
Answer: pH of HCl =5, HNO3 = 1,
NaOH = 9, KOH = 12
Explanation:
pH = -log [H+ ]
1. 1.0 x 10^-5 M HCl
pH = - log (1.0 x 10^-5)
= 5 - log 1 = 5
2. 0.1 M HNO3
pH = - log (1.0 x 10 ^ -1)
pH = 1 - log 1 = 1
3. 1.0 x 10^-5 NaOH
pOH = - log (1.0 x 10^-5)
pOH = 5 - log 1 = 5
pH + pOH = 14
Therefore , pH = 14 - 5 = 9
4. 0.01 M KOH
pOH = - log ( 1.0 x 10^ -2)
= 2 - log 1 = 2
pH + pOH = 14
Therefore, pH = 14 - 2 = 12
(1) activation energy of the reaction to decrease.
Answer:
The new total body concentration would be 300 mOsM
Explanation:
In order to do this, we need to convert all concentrations to moles.
First, with the total body concentration, we have the initial volume of 3 liters and the concentration of 300 M (I will omit til the end the part of mOs)
The moles of the body concentration in this volume is:
moles = M * V
moles = 300 * 3 = 900 moles
To this moles, we add 150 moles of NaCL so, the total moles now is:
moles = 900 + 150 = 1050 moles
Finally, we can calculate the concentration with the new volume of 3.5 L (the sum of 3 and 0.5 liters added):
M = 1050 / 3.5
<em>M = 300 mOsM</em>
<em>So the concentration remains the same as initial</em>
