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
What effect does removing predators have on prey?
:It means that as they are removed the predators go hungry and they could become extinct.
Hope I helped!
so you can see those fluorine atoms have really spread out around the central phosphorus atom. this gives us a trigonal bi-pyramidal molecular geometry for pf5.
H. The atom would no longer be aluminum
If you added a proton to an atom of aluminum it would become a silicone ion
Answer:
c = 0.377 J/g.°C
c = 0.2350 J/g.°C
J = 27.3 J
Explanation:
We can calculate the heat (Q) absorbed or released by a substance using the following expression.
Q = c × m × ΔT
where,
c: specific heat
m: mass
ΔT: change in the temperature
<em>It takes 49.0J to raise the temperature of an 11.5g piece of unknown metal from 13.0°C to 24.3°C. What is the specific heat for the metal? Express your answer numerically, in J/g.°C</em>
Q = c × m × ΔT
49.0 J = c × 11.5 g × (24.3°C - 13.0°C)
c = 0.377 J/g.°C
<em>The molar heat capacity of silver is 25.35 J/mol.°C. How much energy would it take to raise the temperature of 11.5g of silver by 10.1°C? Express your answer numerically, in Joules. What is the specific heat of silver?</em>
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The molar mass of silver is 107.87 g/mol. The specific heat of silver is:

Q = c × m × ΔT
Q = (0.2350 J/g.°C) × 11.5 g × 10.1°C = 27.3 J