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tatyana61 [14]
3 years ago
12

A solution has a pOH of 8.20. What is the concentration of H3O+?

Chemistry
1 answer:
Sedbober [7]3 years ago
7 0

Answer:

.000001585

Explanation:

14-8.20=5.8

10^-5.8 = H3O+

- Hope this helped! Please let me know if you need further explanation.

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What will be the pH of 1.0 mol dm-3 of NH4OH, which is 1% dissociated
Zielflug [23.3K]
Reaction is NH4OH <-> NH4+ OH- (note this is reversible)

Draw up an ICE table

Let x be equilibrium conc of OH- assume init conc of OH is 0M and init conc of NH4+ is 0M also. Init conc of NH4OH is 0.1M so equilibrium conc will be 0.1-x.

%dissociation = x/0.1-x * 100%

1 = 100x/0.1-x

0.1-x = 100x

101x = 0.1

x = 0.0009901

pOH = -log(0.0009901) = 3.00
5 0
3 years ago
Question 8
castortr0y [4]

Answer:

The 5 chemical changes are. 1) Tarnish, 2) Rust, 3) Dissolve, 4) Burn, 5) Bake.

5 0
3 years ago
An isotope has 15 protons, 16 neutrons, and 15 electrons. What is the isotopic notation of the isotope
VikaD [51]

Answer:

^{31}_{15}P is the isotopic notation of the atom

Explanation:

The isotope notation is:

^a_bX

<em>Where a is the mass number = Number of protons + Number of neutrons</em>

<em>b is atomic number = Number of protons</em>

The atomic number define the nature of the atom, the element with atomic number = 15 is phosphorus, P:

^a_bP

a = 15 protons + 16 neutrons = 31

b = 15

^{31}_{15}P is the isotopic notation of the atom.

3 0
3 years ago
Use the periodic table to determine the electron configuration for iodine (i). express your answer in condensed form.
Dovator [93]
Iodine electron configuration is:

1S^2 2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 5S^2 4d^10  5P^5
when Krypton is the noble gas in the row above iodine in the periodic table,
we can change 1S^2  2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 by the symbol
[Kr] of Krypton.

So we can write the electron configuration of Iodine:
[Kr] 5S^2 4d^10 5P^5

8 0
3 years ago
Read 2 more answers
A 0.04358 g sample of gas occupies 10.0-ml at 292.0 k and 1.10 atm. upon further analysis, the compound is found to be 25.305% c
kirill [66]
<span>Answer: A 0.04403 g sample of gas occupies 10.0-mL at 289.0 K and 1.10 atm. Upon further analysis, the compound is found to be 25.305% C and 74.695% Cl. What is the molecular formula of the compound? --------------------------------------... Seems like I did a problem very similar to this--this must be the "B" test. But the halogen was different. 25.305% C/12 = 2.108 74.695% Cl/35.5 = 2.104 So the empirical formula would be CH. However, there are many compounds which fit this bill, so we have to use the gas data. (And I made, in the previous problem, the simplifying assumption that 289C and 1.10 atm would offset each other, so I'll do that, too.) 0.044 grams/10 ml = x/22.4 liters 0.044g/0.010 liters = x/22.4 liters 22.4 liters/0.010 liters = 2240 (ratio) 2240 x .044 = 98.56 (actual atomic weight) CCl = 35.5+12 or 47.5, so two of those is 95 grams/mole. This is sufficiient to distinguish C2CL2, (dichloroacetylene) from C6CL6 (hexachlorobenzene) which would mass 3 times as much.</span>
3 0
3 years ago
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