To solve this problem we will use the following equation:
w = (m of solute) / (m of solution)
w - percentage
It is necessary to mention here that mass of solution is a sum of the mass of solute and mass of water.
<span>w = mass CaCl2/(mass of water + mass of CaCl2)
</span>
mass of water = x
0.35 = 36 / (x + 36)
0.35 × (x + 36) = 36
0.35x + 12.6 = 36
0.35x = 23.4
x = 66.86 g of water is necessary
Answer:
This question is incomplete
Explanation:
This question is incomplete as the volume of the base that was used during the titration was not provided. However, the completed question is in the attachment below.
The formula to be used here is CₐVₐ/CbVb = nₐ/nb
where Cₐ is the concentration of the acid = unknown
Vₐ is the volume of the acid used = 25 cm³ (as seen in the question)
Cb is the concentration of the base = 0.105 mol/dm³ (as seen in the question)
Vb is the volume of the base = 22.13 cm³ (22.1 + 22.15 + 22.15/3)
nₐ is the number of moles of acid = 1 (from the chemical equation)
nb is the number of moles of base = 2 (from the chemical equation)
Note that the Vb was based on the concordant results (values within the range of 0.1 cm³ of each other on the table) of the student
Cₐ x 25/0.105 x 22.13 = 1/2
Cₐ x 25 x 2 = 0.105 x 22.13 x 1
Cₐ x 50 = 0.105 x 22.13
Cₐ = 0.105 x 22.13/50
Cₐ = 0.047 mol/dm³
The concentration of the sulfuric acid is 0.047 mol/dm³
It is a geothermal power plant
The reaction formula of this is C3H8 + 5O2 --> 3CO2 + 4H2O. The ratio of mole number of C3H8 and O2 is 1:5. 0.025g equals to 0.025/44.1=0.00057 mole. So the mass of O2 is 0.00057*5*32=0.0912 g.
<em>Answer:</em>
The equlibrium concentration sof Ca+2 ion willl be 4.9×10∧-3 M
<em>Data Given:</em>
Ksp of CaSO4 = 2.4 × 10∧-5
CaSO4 ⇔ Ca+2 + SO4∧-2
<em>Solution:</em>
Ksp = [Ca+2].[ SO4∧-2]
2.4 × 10∧-5 = [x].[x]= x²
x = 4.9×10∧-3 M
<em>Result:</em>
- The conc. of Ca+2 ion is 4.9×10∧-3 M