Mitochondria because is the power house for cell,site for respiration or energy release.
Ummmmmmmmmmmmmmmmmmmmmmmmm, 21
setup 1 : to the right
setup 2 : equilibrium
setup 3 : to the left
<h3>Further explanation</h3>
The reaction quotient (Q) : determine a reaction has reached equilibrium
For reaction :
aA+bB⇔cC+dD
![\tt Q=\dfrac{C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=%5Ctt%20Q%3D%5Cdfrac%7BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
Comparing Q with K( the equilibrium constant) :
K is the product of ions in an equilibrium saturated state
Q is the product of the ion ions from the reacting substance
Q <K = solution has not occurred precipitation, the ratio of the products to reactants is less than the ratio at equilibrium. The reaction moved to the right (products)
Q = Ksp = saturated solution, exactly the precipitate will occur, the system at equilibrium
Q> K = sediment solution, the ratio of the products to reactants is greater than the ratio at equilibrium. The reaction moved to the left (reactants)
Keq = 6.16 x 10⁻³
Q for reaction N₂O₄(0) ⇒ 2NO₂(g)
![\tt Q=\dfrac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=%5Ctt%20Q%3D%5Cdfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
Setup 1 :
Q<K⇒The reaction moved to the right (products)
Setup 2 :
Q=K⇒the system at equilibrium
Setup 3 :
Q>K⇒The reaction moved to the left (reactants)
Answer:
Mass = 76986 g
Explanation:
Given data:
Dimensions of tank = 126 cm× 47 cm× 13 cm
Mass of water required to filled = ?
Solution:
First of all we will calculate the volume of tank which is equal to the volume of water required to fill it.
Volume = length ×height ×width
Volume = 126 cm × 13 cm× 47 cm
Volume = 76986 cm³
Mass of water:
Mass = density × volume
density of water is 1 g/cm³
Mass = 1 g/cm³× 76986 cm³
Mass = 76986 g
Answer:
1.146 x 10⁴ year.
Explanation:
- The decay of carbon-14 is a first order reaction.
- The rate constant of the reaction (k) in a first order reaction = ln (2)/half-life = 0.693/(5730 year) = 1.21 x 10⁻⁴ year⁻¹.
- The integration law of a first order reaction is:
<em>kt = ln [A₀]/[A]</em>
<em></em>
k is the rate constant = 1.21 x 10⁻⁴ year⁻¹.
t is the time = ??? years.
[A₀] is the initial percentage of carbon-14 = 100.0 %.
[A] is the remaining percentage of carbon-14 = 1/4[A₀] = 25.0 %.
∵ kt = ln [Ao]/[A]
∴ (1.21 x 10⁻⁴ year⁻¹)(t) = ln (100.0%)/[25.0 %]
(1.21 x 10⁻⁴ year⁻¹)(t) = 1.386.
∴ <em>t </em>= 1.386/
(1.21 x 10⁻⁴ year⁻¹) = <em>1.146 x 10⁴ year.</em>
