The correct answer would be 3.49 times 10^ minus 24 molecules
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:
Bombardment by asteroids
Explanation:
All of the other ones explain themselves
Answer:
The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion. (option E)
Explanation:
Step 1: Data given
Volume = 500 mL = 0.500 L
The concentration sodium sulfate = 2.104 M
Step 2: The equation
Na2SO4 → 2Na+ + SO4^2-
For 1 mol Na2SO4 we have 2 moles sodium ion (Na+) and 1 mol sulfate ion (SO4^2-)
Step 3: Calculate the concentration of the ions
[Na+] = 2*2.104 M = 4.208 M
[SO4^2-] = 1*2.104 M = 2.104 M
The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion. (option E)
When energy transforms into mass, the amount of energy does not remain the same. When mass transforms into energy, the amount of energy also does not remain the same. However, the amount of matter and energy remains the same. ... You would weigh much less on the Moon because it is only about one-sixth the mass of Earth. So the answer is D