The rate of disappearance of chlorine gas : 0.2 mol/dm³
<h3>Further explanation</h3>
The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.
For reaction :

The rate reaction :
![\tt -\dfrac{1}{a}\dfrac{d[-A]}{dt}= -\dfrac{1}{b}\dfrac{d[-B]}{dt}=\dfrac{1}{c}\dfrac{d[C]}{dt}=\dfrac{1}{d}\dfrac{d[D]}{dt}](https://tex.z-dn.net/?f=%5Ctt%20-%5Cdfrac%7B1%7D%7Ba%7D%5Cdfrac%7Bd%5B-A%5D%7D%7Bdt%7D%3D%20-%5Cdfrac%7B1%7D%7Bb%7D%5Cdfrac%7Bd%5B-B%5D%7D%7Bdt%7D%3D%5Cdfrac%7B1%7D%7Bc%7D%5Cdfrac%7Bd%5BC%5D%7D%7Bdt%7D%3D%5Cdfrac%7B1%7D%7Bd%7D%5Cdfrac%7Bd%5BD%5D%7D%7Bdt%7D)
Reaction for formation CCl₄ :
<em>CH₄+4Cl₂⇒CCl₄+4HCl</em>
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From equation, rate of reaction = rate of formation CCl₄ = 0.05 mol/dm³
Rate of formation of CCl₄ = reaction rate x coefficient of CCCl₄
0.05 mol/dm³ = reaction rate x 1⇒reaction rate = 0.05 mol/dm³
The rate of disappearance of chlorine gas (Cl₂) :
Rate of disappearance of Cl₂ = reaction rate x coefficient of Cl₂
Rate of disappearance of Cl₂ = 0.05 x 4 = 0.2 mol/dm³
Answer:
c 18.0ml
Explanation:
The average mass of one H2O molecule is 18.02 amu. The number of atoms is an exact number, the number of mole is an exact number; they do not affect the number of significant figures. The average mass of one mole of H2O is 18.02 grams. This is stated: the molar mass of water is 18.02 g/mol.
Answer:
10 m
Explanation:
The mole fraction of FeCl₃ of 0.15, that is, per mole of solution, there are 0.15 moles of FeCl₃ and 1 - 0.15 = 0.85 moles of water.
The molar mass of water is 18.02 g/mol. The mass corresponding to 0.85 moles is:
0.85 mol × 18.02 g/mol = 15 g = 0.015 kg
The molality of FeCl₃ is:
m = moles of solute / kilogram of solvent
m = 0.15 mol / 0.015 kg
m = 10 m
Explanation:
Copper(II) sulfide reacts with oxygen gas to give solid copper(II) oxide and sulfur trioxide gas.
The reaction is given as:

When 1 mol copper(II) sulfide react with 2 moles of oxygen gas it gives 1 mol of solid copper(II) oxide and 1 mol of sulfur trioxide gas
The gas formed in above reaction that is sulfur trioxide reacts with water to give sulfuric acid or hydrogen sulfate.
The reaction is given as:

1 mol of sulfur trioxide gas reacts with 1 mol of liquid water to produce 1 molo of liquid hydrogen sulfate or sulfuric acid