Explanation:
this is your answer hope it will help you
I would say D, because the object's speed moves up, therefore accelerating.
Answer:
A compound has atoms of different elements chemically joined together They can't be separated without a chemical reaction.
Answer : The molarity after a reaction time of 5.00 days is, 0.109 M
Explanation :
The integrated rate law equation for second order reaction follows:
![k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%5BA%5D_o%7D%5Cright%29)
where,
k = rate constant = 
t = time taken = 5.00 days
[A] = concentration of substance after time 't' = ?
= Initial concentration = 0.110 M
Now put all the given values in above equation, we get:
![9.7\times 10^{-6}=\frac{1}{5.00}\left (\frac{1}{[A]}-\frac{1}{(0.110)}\right)](https://tex.z-dn.net/?f=9.7%5Ctimes%2010%5E%7B-6%7D%3D%5Cfrac%7B1%7D%7B5.00%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%280.110%29%7D%5Cright%29)
![[A]=0.109M](https://tex.z-dn.net/?f=%5BA%5D%3D0.109M)
Hence, the molarity after a reaction time of 5.00 days is, 0.109 M
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane