Answer:
B. 13 electrons
Explanation:
Bcoz protons are the same with electrons and the neutrons too.
Answer:
The sun would appear to move more slowly across Mercury's sky.
Explanation:
This is because, the time it takes to do one spin or revolution on Mercury is 176 days (which is its period), whereas, the time it takes to do one spin or revolution on the Earth is 1 day.
Since the angular speed ω = 2π/T where T = period
So on Mercury, T' = 176days = 176 days × 24 hr/day × 60 min/hr × 60 s/min = 15,206,400 s
So, ω' = 2π/T'
= 2π/15,206,400 s
= 4.132 × 10⁻⁷ rad/s
So on Earth, T" = 1 day = 1 day × 24 hr/day × 60 min/hr × 60 s/min = 86,400 s
So, ω" = 2π/T"
= 2π/86,400 s
= 7.272 × 10⁻⁵ rad/s
Since ω' = 4.132 × 10⁻⁷ rad/s << ω" = 7.272 × 10⁻⁵ rad/s, <u>the sun would appear to move more slowly across Mercury's sky.</u>
I think it’s microbiology, since it would help the student understand the process of composting the best. Botany is plants- not for composting. Geophysics- he’s not researching tectonic plates. Oceanography- he isn’t interested in the marina trench
Mass 1 + %abundance of first isotope + Mass 2 + %abundance of second isotope
/ 100
This is RAM.
Answer:
3,964 years.
Explanation:
- It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
- Half-life time is the time needed for the reactants to be in its half concentration.
- If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).
- Also, it is clear that in first order decay the half-life time is independent of the initial concentration.
- The half-life of the element is 5,730 years.
- For, first order reactions:
<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>
Where, k is the rate constant of the reaction.
t1/2 is the half-life of the reaction.
∴ k =0.693/(t1/2) = 0.693/(5,730 years) = 1.21 x 10⁻⁴ year⁻¹.
- Also, we have the integral law of first order reaction:
<em>kt = ln([A₀]/[A]),</em>
where, k is the rate constant of the reaction (k = 1.21 x 10⁻⁴ year⁻¹).
t is the time of the reaction (t = ??? year).
[A₀] is the initial concentration of the sample ([A₀] = 100%).
[A] is the remaining concentration of the sample ([A] = 61.9%).
∴ t = (1/k) ln([A₀]/[A]) = (1/1.21 x 10⁻⁴ year⁻¹) ln(100%/61.9%) = 3,964 years.
