Answer:
6.17 g/cm³
Explanation:
Data given:
one side of cube = 0.53 cm
mass of the cube is 0.92 g
density of the cube = ?
Solution:
First we will calculate for volume the cube
As we know all the sides or edges of a cube are equal so volume equation will be
So,
V = length x width x height
V = e³
as on side = 0.53 cm
then
V = (0.53 cm)³
V = 0.149 cm³
Now we will calculate density of cube
To calculate density, formula will be used
d = m/v . . . . . (1)
where
d = density
m = mass
v = volume
put values in above formula 1
d = 0.92 g / 0.149 cm³
d = 6.17 g/cm³
so. the density of cube = 6.17 g/cm³
(i) We start by calculating the mass of sugar in the solution:
mass of sugar = concentration × solution mass
mass of sugar = 2.5/100 × 500 = 12.5 g
Then now we can calculate the amount of water:
solution mass = mass of sugar + mass of water
mass of water = solution mass - mass of sugar
mass of water = 500 - 12.5 = 487.5 g
(ii) We use the following reasoning:
If 500 g solution contains 12.5 g sugar
Then X g solution contains 75 g sugar
X=(500×75)/12.5 = 3000 g solution
Now to get the amount of solution in liters we use density (we assume that is equal to 1):
Density = mass / volume
Volume = mass / density
Volume = 3000 / 1 = 3000 liters of sugar solution
Answer:
[NaOH} = 0.4 M
Explanation:
In a reaction of neutralization, we determine the equivalence point of the titration. In this case, we have a strong base and a strong acid.
(H₂SO₄, is considered strong, but the first deprotonation is weak)
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
As we have 2 protons in the acid, we need 2 OH⁻ from the base to form 2 molecules of water.
In the equivalence point we know mmoles of base = mmoles of acid
Let's finish the excersise with the formula
25 mL . M NaOH = 28.2 mL . 0.355M
M NaOH = (28.2 mL . 0.355M) / 25 mL → 0.400
Answer:
to fall towards 7, making the solution less alkaline as more water is added.
