This is an incomplete question, here is a complete question.
A certain first-order reaction has a rate constant of 5.50 × 10⁻³ s⁻¹. How long will it take for the reactant concentration to drop to 1/8 of its initial value?
Answer : The time taken will be, 378.1 s
Explanation :
Expression for rate law for first order kinetics is given by:
where,
k = rate constant =
t = time passed by the sample = ?
a = let initial amount of the reactant = x
a - x = amount left after decay process =
Now put all the given values in above equation, we get
Thus, the time taken will be, 378.1 s
Graph 4. Graph 4 is the answer but I need 20 characters
Answer:
They are fused into helium and create energy. But it takes 4
hydrogen atoms to make 1 atom of helium. The Hydrogen is first
converted to Deuterium (heavy hydrogen), and the two deuterium
atoms fuse to make the Helium atom. This process releases a lot of
energy, not the least because of the neutrons released. If this helps please rank Brainliest. Thanks!
The answer for this one is B, you are adding more of the product, making K