Answer: Concentration of in the equilibrium mixture is 0.31 M
Explanation:
Equilibrium concentration of = 0.729 M
The given balanced equilibrium reaction is,
Initial conc. x 0 0
At eqm. conc. (x-2y) M (y) M (3y) M
The expression for equilibrium constant for this reaction will be:
3y = 0.729 M
y = 0.243 M
Now put all the given values in this expression, we get :
concentration of in the equilibrium mixture =
Thus concentration of in the equilibrium mixture is 0.31 M
Answer:
0.7g of HCl
Explanation:
First, let us write a balanced equation for the reaction between HCl and Al(OH)3.
This is illustrated below:
Al(OH)3 + 3HCl —> AlCl3 + 3H2O
Next, let us obtain the masses of Al(OH)3 and HCl that reacted together according to the equation. This can be achieved as shown below:
Molar Mass of Al(OH)3 = 27 + 3(16+1)
= 27 + 3(17) = 27 + 51 = 78g/mol.
Molar Mass of HCl = 1 + 35.5 = 36.5g/mol
Mass of HCl from the balanced equation = 3 x 36.5 = 109.5g
Now we can obtain the mass of HCl that would react with 0.5g of Al(OH)3. This can be achieved as follow:
Al(OH)3 + 3HCl —> AlCl3 + 3H2O
From the equation above,
78g of Al(OH)3 reacted with 109.5g of HCl.
Therefore, 0.5g of Al(OH)3 will react with = (0.5 x 109.5)/78 = 0.7g of HCl
Answer:
d≈ 1.15 g/cm^3
Explanation:
The density of an object can be found by dividing the mass over the volume.
d= m/v
The mass of the object is 54.3 grams and the volume is 47.18 cubic centimeters.
m= 54.3 g
v= 47.18 cm^3
Substitute the values into the formula.
d= 54.3 g/ 47.18 cm
Divide.
d= 1.150911403136922 g/cm^3
Let’s round to the nearest hundredth. The 0 in the thousandth place tells us to leave the 5 in the hundredth place.
d ≈ 1.15 g/cm^3
The density is about 1.15 grams per cubic centimeter.