Answer:
0.036 M
Explanation:
To do this, let's mark the dye as D and bleach as B.
We have the concentrations of both, and we already know that they react in a 1:1 mole ratio. The total volume of reaction is 9 + 1 = 10 mL or 0.010 L, and we hava both concentrations.
The problem already states that the dye reacts completely, so this is the limiting reagent, while bleach is the excess.
To know the remaining amount of bleach, we need to do this with the moles. First, let's calculate the initial moles of D and B:
moles D = 3.4x10⁻⁵ * 0.009 = 3.06x10⁻⁷ moles
moles B = 0.36 * 0.001 = 3.6x10⁻⁴ moles
Now that we have the moles, and that we know that all the dye reacts completely, let's see how many moles of bleach are left:
moles of B remaining = 3.6x10⁻⁴ - 3.06x10⁻⁷ = 3.597x10⁻⁴ moles
These are the moles presents of B after the reaction has been made. The concentration of the same will be:
[B] = 3.597x10⁻⁴ / 0.010
[B] = 0.0357
With 2 SF it would be:
[B] = 3.6x10⁻² M
Lets take 100 g of this compound,
so it is going to be 2.00 g H, 32.7 g S and 65.3 g O.
2.00 g H *1 mol H/1.01 g H ≈ 1.98 mol H
32.7 g S *1 mol S/ 32.1 g S ≈ 1.02 mol S
65.3 g O * 1 mol O/16.0 g O ≈ 4.08 mol O
1.98 mol H : 1.02 mol S : 4.08 mol O = 2 mol H : 1 mol S : 4 mol O
Empirical formula
H2SO4
Answer:
4.34.
Explanation:
<em>∵ pH = pKa + log [salt]/[Acid]</em>
∴ pH = - log(Ka) + log [salt]/[Acid]
∴ pH = - log(6.8 x 10⁻⁵) + log(0.75)/(0.50)
<em>∴ pH = 4.167 + 0.176 = 4.343 ≅ 4.34.</em>
<em></em>
Answer: Option (b) is the correct answer.
Explanation:
It is given that mass of Mg is 97.22 g and it is known that molar mass of Mg is 24.305 g/mol.
So, calculate the number of moles as follows.
No. of moles = 
= 
= 4 mol
Also, it is known that 1 mole has 
atoms/mol. Therefore, calculate the number of atoms in 4 mol as follows.
= 
atoms
or, = 
atoms
Thus, we can conclude that there are atoms in 97.22 grams of Mg.
