The independent variable is the variable being changed. In this case, the independent variable is the calculators. The dependent variable is essentially what you are looking for that <u>depends</u> on the independent variable. In this case it would be time. The constant variable or controlled variable are something that doesn't change and would skew the results. One may be the exact same problem for both groups. Try to come up with two more.
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
![\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.
When an atom emits a beta
particle from the nucleus, the nucleus only has one more proton and one less
neutron and this will make the atomic mass number remains unchanged while the
atomic number increases by 1.
<span>I hope this answer helps you!</span>
hi im breanna
Answer:
The mole is simply a very large number that is used by chemists as a unit of measurement.
Explanation:
The mole is simply a very large number,
6.022
×
10
23
, that has a special property. If I have
6.022
×
10
23
hydrogen atoms, I have a mass of 1 gram of hydrogen atoms . If I have
6.022
×
10
23
H
2
molecules, I have a mass of 2 gram of hydrogen molecules. If I have
6.022
×
10
23
C
atoms, I have (approximately!) 12 grams.
The mole is thus the link between the micro world of atoms and molecules, and the macro world of grams and litres, the which we can easily measure by mass or volume. The masses for a mole of each element are given on the periodic table as the atomic weight. So, if have 12 g of
C
, I know, fairly precisely, how many atoms of carbon I have. Given this quantity, I know how many molecules of
O
2
are required to react with the
C
, which I could measure by mass or by volume.
