Answer:
1.42 L
Explanation:
Step 1:
The following data were obtained from the question :
Molarity of KBr = 2.40 M
Mole of KBr = 3.40 moles
Volume of solution =?
Step 2:
Determination of the volume of the solution.
Molarity of solution is simply the mole of the solute per unit volume the of solution. It is given as :
Molarity = mole /Volume
Volume = mole /Molarity
Volume = 3.4/2.4
Volume = 1.42 L
Therefore, the volume of the solution is 1.42 L
Answer:
- <u>Cadmium has larger atomic radius than sulfur.</u>
Explanation:
Down a period, atomic radii decrease from left to right due to the increase in the number of protons and electrons across a period: when a proton is added the pull of the electrons towards the nucleus is larger, so the size of the atom decreases.
Hence, you can compare the elements that belong to a same period and predict that the atom with lower atomic number (number of protons) will haver larger atomic radius. With that:
- Oxygen and fluorine are in the period 3, being oxygen to the left of fluorine, so oxygen is larger than fluorine.
- Sulfur and chlorine are in the period 4, being sulfur to the left of chlorine, so sulfur is larger than chlorine.
Now see whan happens down a group. Atomic radius increases from top to bottom within a group due to electron shielding. That permits you to compare the size of the elements in a group:
- Fluorine and chlorine are in the same group (17), with chlorine directly below fluorine, so the atomic radius of chlorine is larger than the atomic radius of fluorine.
- Sulfur and oxygen are in the same group (16), with sulfur directlly below oxygen, so sulfur the atomic radius of sulfur is larger than the atocmi radius of oxygen.
So far, you can rank the atomic radius of sulfur, chlorine, fluorine, and oxygen, in increasing order as:
- O < F < Cl < S, concluding that O, F, and Cl have smaller atomic radius than S.
Cadmiun, Cd, is to the left and below sulfur, so both electron shielding (down a group) and increase of the number of protons (down a period) lead to predict the cadmium has a larger atomic radius than sulfur.
<span>Africa was more south and west and South America was more south and east of their current positions. I would say this would be because because South America and Africa used to be together in Gondwanaland millions of years ago and then apparently drifted apart as Wegener thought and then with the advent of plate tectonics it became apparent that the mid-Atlantic ridge runs between them and due to spreading along it Africa and S America got separated.</span>
Answer:
NAD+, FAD.
Explanation:
The citric acid cycle is popularly known as the Kreb's cycle. The cycle involve the oxidation of acetyl-CoA to produce energy. The Kreb's cycle is a chemical process that produces produces two carbon dioxide molecules,NADH,FADH2 and one ATP.
When oxygen is depleted, the citric acid cycle stops, apart from oxygen NAD+ and FAD could be added to the system to restore citric acid cycle activity. NAD+ acts as an electron acceptor.
Citric acid cycle/Kreb's cycle is an aerobic process that occurs in the mitochondria and produces thirty-six(36) ATPs.
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)
