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stich3 [128]
3 years ago
9

Classify the bonds formed between each pair as ionic, polar covalent, or nonpolar covalent based solely on each element's positi

on on the periodic table.
O-O, Fe-O, Al-N, O-I,S-I, Br-Br, P-Br, Be-Cl
Chemistry
1 answer:
Amanda [17]3 years ago
5 0
<span>you have to find the differences of electronegativity for the atoms joined by the bond. if their diff is <0.4,
then it's non-polar if it's between 0.4 to 1.7
then it's polar 1.7 and up it's ionic Electronegativity values can be found on most periodic tables.</span>
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A 1.0-L buffer solution is 0.10 M in HF and 0.050 M in NaF. Which action destroys the buffer? (a) adding 0.050 mol of HCl (b) ad
Volgvan

Answer:

(a) adding 0.050 mol of HCl

Explanation:

A buffer is defined as the mixture of a weak acid and its conjugate base -or vice versa-.

In the buffer:

1.0L × (0.10 mol / L) = 0.10 moles of HF -<em>Weak acid-</em>

1.0L × (0.050 mol / L) = 0.050 moles of NaF -<em>Conjugate base-</em>

-The weak acid reacts with bases as NaOH and the conjugate base reacts with acids as HCl-

Thus:

<em>(a) adding 0.050 mol of HCl:</em> The addition of 0.050moles of HCl produce the reaction of 0.050 moles of NaF producing HF. That means after the reaction, all NaF is consumed and you will have in solution just the weak acid <em>destroying the buffer</em>.

(b) adding 0.050 mol of NaOH: The NaOH reacts with HF producing more NaF. Would be consumed just 0.050 moles of HF -remaining 0.050 moles of HF-. Thus, the buffer <em>wouldn't be destroyed</em>.

(c) adding 0.050 mol of NaF: The addition of conjugate base <em>doesn't destroy the buffer</em>

3 0
3 years ago
BRAINLIESTTT ASAP!!!
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Answer:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 4p²

Explanation:

This atom will likely have 4 electron shells denotation of – 2.8.8.4

Orbitals shells show the probability, in space around the nucleus, where to find an electron. It is important to note that the 3rd shell has an additional d orbital (-in addition to s and p). However, because the d orbital has a higher energy state than the 4s and 4p orbitals, the d orbital only fills up when these latter ones are completely filled. In this case, the 4p does not completely fill (hence we don't see the d orbital in the notation).

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