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Igoryamba
2 years ago
11

What is the electron arrangement of K

Chemistry
1 answer:
OverLord2011 [107]2 years ago
5 0
A scale factor is it
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Can someone please explain this i will give brainliest
IrinaVladis [17]
1) 2
2) 2
3) there are 2 atoms of hydrogen and 1 atom of oxygen, altogether is 3
4) subscript tells how many atoms of each element are present in the molecule
5) No, it is not balanced because the oxygen atom is not equal on both sides of the reactants and products. This wouldn’t be considered balanced as there are two oxygen atoms in the reactants section and only one oxygen on the products section.

It is:
2 H2 + O2 -> 2 H2O
4 0
2 years ago
Determine the molecular formula of a compound that has a molar mass of 183.2 g/mol and an empirical formula of c2h5o2.
Tpy6a [65]
C6H15O6

Good luck and don't forget to rate or mark Brainliest :)
6 0
2 years ago
Read 2 more answers
What is the charge of Fe(OH)3?
klasskru [66]
The charge for this compound is positive. For Fe, it's charge is positive 3, and for OH, it's charge is negative 1. You would then criss cross the charges of each and come out with Fe(OH)3. I hope this helped!! :)
6 0
3 years ago
Which of the following is true of the relationships between heat of fusion and heat of vaporization?
klio [65]

Answer:

The heat of vaporization is typically larger than the heat of fusion

Next question answer:

The liquid water absorbs heat from the skin surface and is transferred to the air when the water evaporates.

Explanation:

6 0
2 years ago
A 1.0l buffer solution contains 0.100 mol of hc2h3o2 and 0.100 mol of nac2h3o2. the value of ka for hc2h3o2 is 1.8×10−5. part a
Mandarinka [93]
There are two ways to solve this problem. We can use the ICE method which is tedious and lengthy or use the Henderson–Hasselbalch equation. This equation relates pH and the concentration of the ions in the solution. It is expressed as

pH = pKa + log [A]/[HA] 

 where pKa = - log [Ka]
[A] is the concentration of the conjugate base
[HA] is the concentration of the acid

Given:
Ka = 1.8x10^-5
NaOH added = 0.015 mol
HC2H3O2 = 0.1 mol
NaC2H3O2 = 0.1 mol

Solution:
pKa = - log ( 1.8x10^-5) = 4.74

[A] = 0.015 mol + 0.100 mol = .115 moles
[HA] = .1 - 0.015 = 0.085 moles

pH = 4.74 + log (.115/0.085)
pH = 4.87
6 0
2 years ago
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