Today, natural sciences<span> are more </span>commonly divided<span> into life </span>sciences<span>, such as botany and zoology; and physical </span>sciences<span>, which include physics, chemistry, geology, astronomy and materials </span>science<span>.</span>
Answer:
% (COOK)2H2O = 37.826 %
Explanation:
mix: (COOK)2H2O + Ca(OH)2 → CaC2O4 + H2O
∴ mass mix = 4.00 g
∴ mass (CaC2O4)H2O = 1.20 g
∴ Mw (COOK)2H2O = 184.24 g/mol
∴ Mw (CaC2O4)H2O = 146.12 g/mol
∴ r = mol (COOK)2H2O / mol (CaC2O4)H2O = 1
- % (COOK)2H2O = (mass (COOK)2H2O / mass Mix) × 100
⇒ mass (COOK)2H2O = (1.20 g (CaC2O4)H2O)×(mol (CaC2O4)H2O / 146.12 g (CaC2O4)H2O)×(mol (COOK)2H2O/mol (CaC2O4)H2O)×(184.24 g (COOK)2H2O/mol (COOK)2H2O)
⇒ mass (COOK)2H2O = 1.513 g
⇒ % (COOK)2H2O = ( 1.513 g / 4 g )×100
⇒ % (COOK)2H2O = 37.826 %
Answer:
2.90 x 10⁻¹¹moldm⁻³
Explanation:
Given parameters:
[H⁺] = 3.5 x 10⁻⁴mol/dm³
Unknown
[OH⁻] = ?
Solution;
The ionic product of water can be used to solve this problem. It has been experimentally determined to be 1 x 10⁻¹⁴mol² dm⁻⁶
[H⁺] [OH⁻] = 1 x 10⁻¹⁴
Therefore;
[OH⁻] = 
= 
= 0.29 x 10⁻¹⁰moldm⁻³
= 2.90 x 10⁻¹¹moldm⁻³
You must add 45 mL of the 80 % alcohol to the 30 % alcohol to get a 35 % solution.
You can use a modified dilution formula to calculate the volume of 80 % alcohol
V1×C1 + V2×C2 = V3×C3
Let the volume of 80 % mixture 1 = <em>x</em> mL. Then the volume of the final 35 % mixture 3 = (405 + <em>x</em> ) mL
(<em>x</em> mL×80 % alc) + (405 mL×30 % alc) = (405 + <em>x</em>)mL × 35 % alc
80x + 12 150 = 14 175 + 35 x
45x = 2025
x = 2025/45 = 45
