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harkovskaia [24]
3 years ago
11

Please help! I don't understand electron configurations..

Chemistry
2 answers:
Eddi Din [679]3 years ago
8 0
1. [Ne] 3s² 3p² ( simple formula )

2. Na: 1 valence electron
F: 7 valence electrons
Ga: 3 valence electrons
Bi : 5 valence electrons
Lelechka [254]3 years ago
7 0
You want me to explain it as well?
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Explain how the fire spreading from tree to tree provides a model for conduction.
Inessa05 [86]

the fire spreading is what represents the form of heat transferring from one atom to another within an object and direct contact

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3 years ago
I need help with this!
Natali [406]

\boxed{ \rm{ \red2Cs + Sr(CrO4) → Cs2(CrO4) + Sr}}

8 0
2 years ago
If an equal number of moles of reactants are used, do the following equilibrium mixtures contain primarily reactants or products
puteri [66]

<u>Answer:</u>

<u>For a:</u> The equilibrium mixture contains primarily reactants.

<u>For b:</u> The equilibrium mixture contains primarily products.

<u>Explanation:</u>

There are 3 conditions:

  • When K_{eq}>1; the reaction is product favored.
  • When K_{eq}; the reaction is reactant favored.
  • When K_{e}=1; the reaction is in equilibrium.

For the given chemical reactions:

  • <u>For a:</u>

The chemical equation follows:

HCN(aq.)+H_2O(l)\rightleftharpoons CN^-(aq.)+H_3O^+(aq.);K_{eq}=6.2\times 10^{-10}

The expression of K_{eq} for above reaction follows:

K_{eq}=\frac{[CN^-][H_3O^+]}{[HCN][H_2O]}=6.2\times 10^{-10}

As, K_{eq}, the reaction will be favored on the reactant side.

Hence, the equilibrium mixture contains primarily reactants.

  • <u>For b:</u>

The chemical equation follows:

H_2(g)+Cl_2(g)\rightleftharpoons 2HCl(g);K_{eq}=2.51\times 10^{4}

The expression of K_{eq} for above reaction follows:

K_{eq}=\frac{[HCl]^2}{[H_2][Cl_2]}=2.51\times 10^{4}

As, K_{eq}>1, the reaction will be favored on the product side.

Hence, the equilibrium mixture contains primarily products.

4 0
3 years ago
Please help/show work!
Keith_Richards [23]

Answer:

7.00335g

Explanation:

n=\frac{V}{V_m} \\n = \frac{11.2}{22.4} \\n= 0.5 mol

n=\frac{m}{M} \\m=nM\\m=(0.5)(14.0067)\\m=7.00335g

3 0
1 year ago
BRAINLIESTTT ASAP!! <br><br> Who take FLVS Chemistry ( Seg 1) ? Please only answer if you do. :)
kirza4 [7]

Hi again what do you need help on this time lol?

6 0
3 years ago
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