The concentration of AlCl3 solution if 150 ml of the solution contains 550 mg of cl- ion is 0.0344 M
calculation
concentration = moles /volume in liters
volume in liters = 150 /1000= 0.15 L
number of moles calculation
write the equation for dissociation of Al2Cl3
that is AlCl3 ⇔ Al^3+ + 3 Cl ^-
find the moles of Cl^- formed
moles =mass/molar mass
mass in grams= 550/ 1000 =0.55 grams
molar mass of Cl^- =35.5 g/mol
moles is therefore= 0.55/35.5 =0.0155 moles
by use of mole ration betweem AlCl3 to Cl^- which is 1:3 the moles of AlCl3 is =0.0155 x 1/3= 5.167 x10^-3 moles
concentration of AlCl3 is therefore= 5.167 x10^-3/ 0.15 =0.0344 M
<u>Answer:</u> The volume of HBr solution required is 130.16 mL
<u>Explanation:</u>
To calculate the concentration of base, we use the equation given by neutralization reaction:
![n_1M_1V_1=n_2M_2V_2](https://tex.z-dn.net/?f=n_1M_1V_1%3Dn_2M_2V_2)
where,
are the n-factor, molarity and volume of acid which is ![H_2SO_4](https://tex.z-dn.net/?f=H_2SO_4)
are the n-factor, molarity and volume of base which is ![KOH](https://tex.z-dn.net/?f=KOH)
We are given:
![n_1=2\\M_1=123\times 10^{-4}M=0.0123M\\V_1=50mL\\n_2=1\\M_2=0.00945M\\V_2=?mL](https://tex.z-dn.net/?f=n_1%3D2%5C%5CM_1%3D123%5Ctimes%2010%5E%7B-4%7DM%3D0.0123M%5C%5CV_1%3D50mL%5C%5Cn_2%3D1%5C%5CM_2%3D0.00945M%5C%5CV_2%3D%3FmL)
Putting values in above equation, we get:
![2\times 0.0123\times 50=1\times 0.00945\times V_2\\\\V_2=130.16mL](https://tex.z-dn.net/?f=2%5Ctimes%200.0123%5Ctimes%2050%3D1%5Ctimes%200.00945%5Ctimes%20V_2%5C%5C%5C%5CV_2%3D130.16mL)
Hence, the volume of KOH solution required is 130.16 mL
Answer:
Value of
is 0.090.
Explanation:
Initial molarity of
=
= 0.0700 M
Construct an ICE table corresponding to the combustion reaction of carbon to determine ![K_{c}](https://tex.z-dn.net/?f=K_%7Bc%7D)
![C(s)+O_{2}(g)\rightarrow 2CO(g)](https://tex.z-dn.net/?f=C%28s%29%2BO_%7B2%7D%28g%29%5Crightarrow%202CO%28g%29)
I (M): - 0.0700 0
C (M): - -x +2x
E (M): - 0.0700-x 2x
So,
, where [CO] and
represents equilibrium concentration of CO and
respectively.
Here, ![[CO]=2x=0.060](https://tex.z-dn.net/?f=%5BCO%5D%3D2x%3D0.060)
⇒x = 0.030
So,
= 0.0700-x = (0.0700-0.030) = 0.040
Hence, ![K_{c}=\frac{(0.060)^{2}}{0.040}=0.090](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%280.060%29%5E%7B2%7D%7D%7B0.040%7D%3D0.090)
Answer:
30L
Explanation:
The following data were obtained from the question:
Volume of stock solution (V1) = 1L
Molarity of stock solution (M1) = 6M
Molarity of diluted solution (M2) = 0.2M
Volume of diluted solution (V2) =..?
The volume of the diluted solution can be obtained as follow:
M1V1 = M2V2
6 x 1 = 0.2 x V2
Divide both side by 0.2
V2 = 6/0.2
V2 = 30L
Therefore, the solution must be diluted to 30L in order to obtain 0.2M of the solution.