Answer:
Mole fraction N₂ = 0.336
Explanation:
Mole fraction of a gas can be determined in order to know the partial pressure of the gas, and the total pressure, in the mixture.
Total pressure in the mixture: Sum of partial pressure from all the gases
Total pressure = 183 mmHg + 443 mmHg + 693 mmHg =1319 mmHg
Mole fraction N₂ = Partial pressure N₂ / Total pressure
443 mmHg / 1319 mmHg = 0.336
Remember that mole fraction does not carry units
The location of the valence electron or the outermost electron is expressed in quantum numbers. There are five quantum numbers: prinicipal (n), angular momentum (l), magnetic (ms) and magnetic spin (ms) quantum numbers. This is based on Bohr's atomic model where electrons orbit around the nucleus. These electrons are in the orbitals with specific energy levels. Starting from energy level 1 that is closest to the nucleus, the energy level decreases to 2, 3, 4, 5, 6, and 7. These energy level numbers represent the principal quantum number. Within each orbital also contains subshell. From increasing to decreasing order, these subshells are the s, p, d and f subshells. These subshells represent the angular momentum quantum numer. Specifically, s=0, p=1, d=2 and f=3. Therefore, if the electron is in the orbital 5p, the quantum number would be: 5, 1. Applying these to the choices, the correct pairing would be:
2p: n=2. l=1
3d: n=3, l=2
2s: n=2. l=0
4f: n=4. l=3
1s: n=1, l=0
<h2>Halogens.</h2>
Explanation:
All of the halogens form salts with alkali metals with the formula MX.
Halogens are the members of group 17 in the periodic table. They have seven electron in their valence shell and tend to form negative ion with minus one charge. They get there name because of their property to form salts. Halogens are very reactive and form salts with the metals.
Answer:
A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)
B - Increase(P), Increase(q), Decrease (R)
C - Triple (P) and reduce (q) to one third
Explanation:
<em>According to Le Chatelier principle, when a system is in equilibrium and one of the constraints that affect the rate of reaction is applied, the equilibrium will shift so as to annul the effects of the constraint.</em>
P and Q are reactants, an increase in either or both without an equally measurable increase in R (a product) will shift the equilibrium to the right. Also, any decrease in R without a corresponding decrease in either or both of P and Q will shift the equilibrium to the right. Hence, Increase(P), Increase(q), and Decrease (R) will shift the equilibrium to the right.
In the same vein, any increase in R without a corresponding increase in P and Q will shift the equilibrium to the left. The same goes for any decrease in either or both of P and Q without a counter-decrease in R will shift the equilibrium to the left. Hence, Increase (R), Decrease (P), Decrease(q), and Triple both (Q) and (R) will shift the equilibrium to the left.
Any increase or decrease in P with a commensurable decrease or increase in Q (or vice versa) with R remaining constant will create no shift in the equilibrium. Hence, Triple (P) and reduce (q) to one third will create no shift in the equilibrium.
The reaction between LiOH and HCl is;
LiOH + HCl → LiCl + H₂O
The stoichiometric ratio between LiOH and HCl is 1 : 1
moles of LiOH added = moles of HCl needed to neutralize.
Molar mass of LiOH = 24 g/mol
moles = mass / molar mass
LIOH moles = 1.65 / 24 = 0.06875 mol
Hence needed HCl moles = 0.06875 mol
Molarity = moles (mol) / Volume (L)
Hence needed HCl volume = moles / molarity
= 0.06875 mol / 0.150 mol/L
= 0.458 L = 458 mL
