Answer:
Solubility of O₂(g) in 4L water = 3.42 x 10⁻² grams O₂(g)
Explanation:
Graham's Law => Solubility(S) ∝ Applied Pressure(P) => S =k·P
Given P = 0.209Atm & k = 1.28 x 10⁻³mol/L·Atm
=> S = k·P = (1.28 x 10⁻³ mole/L·Atm)0.209Atm = 2.68 x 10⁻³ mol O₂/L water.
∴Solubility of O₂(g) in 4L water at 0.209Atm = (2.68 x 10⁻³mole O₂(g)/L)(4L)(32 g O₂(g)/mol O₂(g)) = <u>3.45 x 10⁻² grams O₂(g) in 4L water. </u>
Explanation:
Reaction rate is a measure of speed of a chemical reaction.
Usually, based on rates, reactions can be divided into two parts:
- Slow reactions which usually takes a long time.
- Fast reactions which takes place completely in seconds or microseconds.
Reaction rate can be defined as a change in the concentration of a reactant or product per unit time:
The unit is mol dm⁻³ s⁻¹
Reaction rate =
The measurement of the reaction rate is based on the disappearance of a reactant or appearance of a product with increasing reaction time.
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Reaction rate brainly.com/question/9642960
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Answer:
Explanation:
Hello there!
In this case, for the reaction by which water is decomposed to molecular hydrogen and oxygen:
It is necessary to perform the inspection balance process since there is a dissimilar number of atoms of oxygen on both sides; therefore, by putting a 2 on water we balance oxygen:
But now, there are four hydrogens on the left; therefore, we put a 2 on hydrogen to finally balance it:
And obviously, the coefficient in oxygen is an unwritten 1.
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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 %