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There are 9 polyatomic ions so which one and the compound “Sn(NO2)2 is Tin(ll) Nitrite
Answer:
The pH of the buffer solution = 8.05
Explanation:
Using the Henderson - Hasselbalch equation;
pH = pKa₂ + log ( [HPO₄²-]/[H₂PO4⁻]
where pKa₂ = -log (Ka₂) = -log ( 6.1 * 10⁻⁸) = 7.21
Concentration of OH⁻ added = 0.069 M (i.e. 0.069 mol/L)
[H₂PO4⁻] after addition of OH⁻ = 0.165 - 0.069 = 0.096 M
[HPO₄²-] after addition of OH⁻ = 0.594 + 0.069 = 0.663 M
Therefore,
pH = 7.21 + log (0.663 / 0.096)
pH = 7.21 + 0.84
pH = 8.05
Answer: 1.27 bar
Explanation:
1 atm = 1.01325 bar
1.25 atm = Z (let Z be the unknown value)
To get the value of Z, cross multiply
Z x 1 atm = 1.25 atm x 1.01325 bar
1 atm•Z = 1.2665625 atm•bar
To get the value of Z, divide both sides by 1 atm
1 atm•Z/1 atm = 1.2665625 atm•bar/1atm
Z = 1.2665625 bar
(Round up Z to the nearest hundredth as 1.27 bar)
Thus, 1.25 atm when coverted gives 1.27 bar
Answer:
Groups 14, 15, and 16 have 2,3, and 4 electrons in the p sublevel (p sublevel has 3 "spaces" AKA orbitals), because Hunds says one in each orbital before doubling up if you had 2 electrons, group 14, they would both be in the first orbital, with 3 electrons, group 15, two in the first orbital one in the 2nd none in the 3rd. With 4 electrons, group 16, then you would have 2 in the first 2 orbitals and NONE in the 3rd.
Explanation:
If you are in group 13 you only have 1 electron so it can only be in one orbital. with group 17, you have 5 electrons, so 2 in the first 2 in the second and 1 in the 3rd, correct for Hunds rule anyway. Noble gasses, group 18, have 6 elecctrons, so every orbital is full any way you look at it.
