<span>When M(OH)2 dissolves we have
M(OH)2 which produces M2+ and 2OHâ’
pH + pOH=14
At ph =7; we have
7+pOH=14
pOH=14â’7 = 7
Then [OHâ’]=10^(â’pOH)
[OH-] = 10^(-7) = 1* 10^(-7)
At ph = 10. We have,
pOH = 4. And [OH-] = 10^(-4) = 1 * 10^(-4)
Finally ph = 14. We have, pOH = 0
And then [OH-] = 10^(-0) -----anything raised to zero power is 1, but (-0)...
So [OH-] = 1</span>
Answer:
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Explanation:
Answer:
The electronic configuration that are incorrectly written is 1s²2s³2p⁶, 4s²3d¹⁰4p⁷, 3s¹ and 2s²2p⁴.
Explanation:
The electronic configuration of the elements corresponds to how all the electrons of an element are arranged in energy levels and sub-levels.
There are 7 energy levels —from 1 to 7— whose sublevels are described as s, p, d and f.
All electronic configurations begin with the term "1s" —corresponding to the sublevel s of level 1— so 4s²3d¹⁰4p⁷, 3s¹ and 2s²2p⁴ are incorrectly written. In addition, 4s²3d¹⁰4p⁷ is written incorrectly because is impossible to jump from the sublevel "s" to the sublevel "d" —which is found from level 3 and up— without passing through the sublevel "p".
In the case of 1s²2s³2p⁶, the wrong thing is that the sublevel "s" can only hold two electrons, not three.
The other options are correctly written.
Answer:
4) transferred from the valence shell of one atom to the valence shell of another atom
Explanation:
Electrons are located outside of the nucleus which contains the protons and the neutrons.
For bonds to form, valence electrons located in the outermost shell electrons are involved. These are the valence electrons. These outer shell electrons can be shared or transferred between two combining atoms to form stable atoms.
In ionic bonds, the electrons are transferred from one specie to another. The atom that loses the electrons becomes positively charged and the receiving atom becomes negatively charged. This is the crux of ionic bonds.
