The correct answer to this question is this one:
Assuming all the barium bromide dissolved (which it should), the concentration of BaBr2 in solution should be zero: it should all dissociate into Ba+2 and 2Br- ions.
Turn those grams of BaBr2 into moles of BaBr2, then divide by the volume to get the concentration.
Recognize that every formula unit of BaBr2 has one ion of Ba+2, and 2 ions of Br-1. That means that when this substance dissociates, you'll get one concentration of Ba+2 ions, and a concentration of Br- ions TWICE as large. Whatever the concentration of Ba+2 ions is that you calculate, double it for the conentration of the Br-1 ion.
The correct answer of the given question above would be option B. IRON 0.449. Based on the given details above about an unknown substance that has a mass of 14.7 g and the substance absorbs 1.323×102 J of heat, the temperature of the substance is raised from 25.0 ∘C to45.0 ∘C, most likely, the substance is IRON. Hope this answers the question.
Answer:
2 moles NaClO3
Explanation:
2NaClO3 --> 2NaCl + 3O2
moles NaClO3 = 3 moles O2 x (2 moles NaClO3/3 moles O2) = 2 moles
conversion factor: 2 moles NaClO3 = 3 moles O2
Answer:
i. n = 5
ii. ΔE = 7.61 ×
KJ/mole
Explanation:
1. ΔE = (1/λ) = -2.178 ×
(
-
)
(1/434 ×
) = -2.178 ×
(
)
⇒ 434 ×
= (1/-2.178 ×
)![\frac{n^{2}_{final} *n^{2}_{initial} }{n^{2}_{initial} - n^{2}_{final} }](https://tex.z-dn.net/?f=%5Cfrac%7Bn%5E%7B2%7D_%7Bfinal%7D%20%2An%5E%7B2%7D_%7Binitial%7D%20%20%20%7D%7Bn%5E%7B2%7D_%7Binitial%7D%20-%20n%5E%7B2%7D_%7Bfinal%7D%20%20%20%20%7D)
But,
= 2
434 ×
= (1/2.178 ×
)![\frac{2^{2} n^{2}_{initial} }{n^{2}_{initial} - 2^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7B2%5E%7B2%7D%20n%5E%7B2%7D_%7Binitial%7D%20%20%7D%7Bn%5E%7B2%7D_%7Binitial%7D%20-%202%5E%7B2%7D%20%20%7D)
434 ×
× 2.178 ×
= ![(\frac{4n^{2}_{initial} }{n^{2}_{initial} - 4 })](https://tex.z-dn.net/?f=%28%5Cfrac%7B4n%5E%7B2%7D_%7Binitial%7D%20%20%7D%7Bn%5E%7B2%7D_%7Binitial%7D%20-%204%20%7D%29)
⇒
= 5
Therefore, the initial energy level where transition occurred is from 5.
2. ΔE = hf
= (hc) ÷ λ
= (6.626 × 10−34 × 3.0 ×
) ÷ (434 ×
)
= (1.9878 ×
) ÷ (434 ×
)
= 4.58 ×
J
= 4.58 ×
KJ
But 1 mole = 6.02×
, then;
energy in KJ/mole = (4.58 ×
KJ) ÷ (6.02×
)
= 7.61 ×
KJ/mole