We can use the following formula when making diluted solutions from concentrated solutions
c1v1 = c2v2
where c1 is concentration and v1 is volume of the concentrated solution
and c2 is concentration and v2 is final volume of diluted solution
substituting the values in the equation
18.4 M x V = 0.1 M x 600 mL
V = 3.26 mL
volume of 3.26 mL of 18.4 M sulfuric acid solution should be taken and diluted upto 600 mL
Answer:
It helps microorganisms grow
Explanation:
I think this is the correct answer because I tried searching it up and you sterilise the petri dish before adding the agar, I've used agar in petri dishes a year ago and im pretty sure they don't make cells glow or make bacteria appear larger.
if I had to have a second guess I'd go with the first one but considering the fact that bacteria is usually to small to see with the naked eye the last option seems correct
Your answer would be A for your homework
Complete question:
The decomposition of SO2Cl2 is first order in SO2Cl2 and has a rate constant of 1.44×10⁻⁴ s⁻¹ at a certain temperature.
If the initial concentration of SO2Cl2 is 0.125 M , what is the concentration of SO2Cl2 after 210 s ?
Answer:
After 210 s the concentration of SO2Cl2 will be 0.121 M
Explanation:
![ln\frac{[A_t]}{[A_0]} =-kt](https://tex.z-dn.net/?f=ln%5Cfrac%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D%20%3D-kt)
where;
At is the concentration of A at a time t
A₀ is the initial concentration of A
k is rate constant = 1.44×10⁻⁴ s⁻¹
t is time
ln(At/A₀) = -( 1.44×10⁻⁴)t
ln(At/0.125) = -( 1.44×10⁻⁴)210
ln(At/0.125) = -0.03024
At/0.125 = 0.9702
At = 0.125*0.9702
At = 0.121 M
Therefore, after 210 s the concentration of SO2Cl2 will be 0.121 M
