Answer:
Therefore the concentration of the reactant after 4.00 minutes will be 0.686M.
Explanation:
The unit of k is s⁻¹.
The order of the reaction = first order.
First order reaction: A first order reaction is a reaction in which the rate of reaction depends only the value of the concentration of the reactant.
![-\frac{d[A]}{dt} =kt](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%20%3Dkt)
[A] = the concentration of the reactant at time t
k= rate constant
t= time
Here k= 4.70×10⁻³ s⁻¹
t= 4.00
[A₀] = initial concentration of reactant = 0.700 M
![-\frac{d[A]}{dt} =kt](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%20%3Dkt)
![\Rightarrow -\frac{d[A]}{[A]}=kdt](https://tex.z-dn.net/?f=%5CRightarrow%20-%5Cfrac%7Bd%5BA%5D%7D%7B%5BA%5D%7D%3Dkdt)
Integrating both sides
![\Rightarrow\int -\frac{d[A]}{[A]}=\int kdt](https://tex.z-dn.net/?f=%5CRightarrow%5Cint%20-%5Cfrac%7Bd%5BA%5D%7D%7B%5BA%5D%7D%3D%5Cint%20kdt)
⇒ -ln[A] = kt +c
When t=0 , [A] =[A₀]
-ln[A₀] = k.0 + c
⇒c= -ln[A₀]
Therefore
-ln[A] = kt - ln[A₀]
Putting the value of k, [A₀] and t
- ln[A] =4.70×10⁻³×4 -ln (0.70)
⇒-ln[A]= 0.375
⇒[A] = 0.686
Therefore the concentration of the reactant after 4.00 minutes will be 0.686M.
In a saturated solution, extra solid X would remain solid, dissolve in an unsaturated solution, and crystallize in a supersaturated one.
A solution is said to be saturated when there is a maximum amount of solute present that has been dissolved in the solvent. As a result, the system is in an equilibrium between the dissolved and undissolved solutes: A solution is considered to be unsaturated if the solute concentration is less than the equilibrium solubility. A supersaturated solution is one that has more solute than is necessary to generate a saturated solution at a given temperature.
Learn more about Supersaturated here-
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Answer;
C. unchanged rock and mineral fragments
Explanation;
A large number of landforms and features found in desert environments are formed as the result of weathering. Weathering is defined as the breakdown and deposition of rocks by weather acting in situ
The two main types of weathering which occur in deserts are Mechanical weathering, which is the disintegration of a rock by mechanical forces that do not change the rock's chemical composition and Chemical weathering, which is the decomposition of a rock by the alteration of its chemical composition.
By contrast much of the weathered debris in deserts has resulted from mechanical weathering. Chemical weathering, however, is not completely absent in deserts. Over long time spans,clays and thin soils do form.
Balanced Eqn
2
C
2
H
6
+
7
O
2
=
4
C
O
2
+
6
H
2
O
By the Balanced eqn
60g ethane requires 7x32= 224g oxygen
here ethane is in excess.oxygen will be fully consumed
hence
300g oxygen will consume
60
⋅
300
224
=
80.36
g
ethane
leaving (270-80.36)= 189.64 g ethane.
By the Balanced eqn
60g ethane produces 4x44 g CO2
hence amount of CO2 produced =
4
⋅
44
⋅
80.36
60
=
235.72
g
and its no. of moles will be
235.72
44
=5.36 where 44 is the molar mass of Carbon dioxide
hope this helps
one mole of P weights about 31 grams
in one mole there are 6.022*10^23 atoms
we use the rule of threes
6.022*10^23atoms......weight..........31 grams
3.45*10^23 atoms.........weight...........x grams
x=(3.45*10^23*31)/6.022*10^23
x=106.95/6.022=<u><em>17.76 grams</em></u>