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Levart [38]
3 years ago
10

What is the mass of 14.10 moles of Li2O?

Chemistry
1 answer:
galina1969 [7]3 years ago
7 0

Answer:

421.32774 grams

Explanation:

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Jill then wrote the formulas for all the possible compounds that sodium can form with the other ions. What formulas did she writ
sp2606 [1]

Explanation:

Sodium has atomic number of 11 and its electronic configuration is given by:

[Na]=1s^22s^22p^63s^1

The nearest stable electronic configuration to sodium is of the neon. So, in order to attain stability of noble gas it will loose its single electron.

Na\rightarrow Na^++1e^-

[Na^+]=1s^22s^22p^63s^0

Sodium has single valency that is 1.

Let nbe the valency of the ion 'X'

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

Na_nX_1

So, the formulas for all the possible compounds that sodium can form with the other ions will be:

Na_nX_1

4 0
3 years ago
Which of the following is the balanced equation for the
NNADVOKAT [17]
A.) is the answer =)
6 0
3 years ago
Which compounds are lewis acids? select all that apply. (ch3ch2)2hc+ cl3c− ch3nh2 bbr3?
alexandr402 [8]

The Lewis acids are electron pair acceptors due to the availability of vacant orbitals they accept electrons from the donor compounds.

  • (CH_3CH_2)_2CH^{+} is a Lewis acid due to the unfilled valence shell on carbon atom (possessing positive charge) it accepts electron readily.
  • CCl_{3}^{-} is not a Lewis acid due to the presence of negative charge which represents that there are electrons available for donation.
  • CH_3NH_2 is not a Lewis acid due to the presence of lone pair on nitrogen which is easily donated to the acceptors.
  • BBr_3 is a Lewis acid due to the availability of vacant p-orbital on boron atom.

Hence, (CH_3CH_2)_2CH^{+} and BBr_3 are Lewis acids.

3 0
3 years ago
Particles in a liquid move around just as freely as particles in a solid.
andre [41]
It is false :D I am 95% sure
5 0
3 years ago
Read 2 more answers
How many milliliters of a 0.211 M HI solution are needed to reduce 24.0 mL of a 0.354 M KMnO4 solution according to the followin
Novay_Z [31]

Answer:

The answer to your question is 242 ml

Explanation:

Data

HI 0.211 M   Volume = x

KMnO₄ 0.354 M   Volume = 24 ml

Balanced Chemical reaction

     12HI + 2KMnO₄ + 2H₂SO₄ → 6I₂ + Mn₂SO₄ + K₂SO₄ + 8H₂O

Process

1.- Calculate the moles of KMnO₄  0.354 M in 24 ml

Molarity = moles / volume (L)

moles = Molarity x volume (L)

moles = 0.354 x 0.024

moles = 0.0085

2.- From the balanced chemical reaction we know that HI and KMnO₄ react in the proportion 12 to 2. Then,

              12 moles of HI --------------- 2 moles of KMnO₄

                x                     --------------- 0.0085 moles of KMnO₄

             x = (0.0085 x 12)/2

             x = 0.051 moles of HI

3.- Calculate the milliliters of HI 0.211 M

Molarity = moles/volume

Volume = moles/molarity

Volume = 0.051/0.211

Volume = 0.242 L or Volume = 242 ml

8 0
3 years ago
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