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balu736 [363]
4 years ago
8

Consider these elements: p, ca, si, s, ga. a. write the electron configuration for each element

Chemistry
1 answer:
dimulka [17.4K]4 years ago
7 0
Method:

1) Find the atomic number in a periodic table: the number of electrons equal the atomic number

2) Use Aufbau rule

Element     atomic number       electron configuration
<span>
P                15                            1s2 2s2 2p6 3s2 3p3

Ca              20                            </span><span><span>1s2 2s2 2p6 3s2 3p6 4s2

</span>Si                14</span><span>                            1s2 2s2 2p6 3s2 3p2

S                 16</span><span><span>                            1s2 2s2 2p6 3s2 3p4

</span>Ga               31.                   </span><span><span>        1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p</span>       </span>
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What is a chemical change and what is a chemical property? Give Example.
nordsb [41]
Look it up on google
8 0
3 years ago
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State whether the following statements are true or false. If false, explain why. (a) A reaction stops when equilibrium is reache
Elden [556K]

Answer:

(a) False;

(b) False;

(c) False;

(d) True.

Explanation:

(a) When equilibrium is reached, the forward reaction rate becomes equal to the reverse reaction rate, that's why the molarity of each species remains constant, but reactions don't stop.

(b) According to the principle of Le Chatelier, an increase in molarity of either reactants or products would lead to a disturbance of equilibrium. This disturbance would lead to the shift of equilibrium towards the side which would minimize such a disturbance.

(c) Equilibrium constant is only temperature-dependent, it's independent of molarity, pressure, volume etc. of any species present in the reaction.

(d) The greater the initial molarity of reactants, the more products can be formed, e. g., since the ratio of products to reactants should be kept constant, the larger the amount of reactants, the greater the amount of products formed to keep a constant ratio.

7 0
3 years ago
Are the following combinations allowed? If not, show two ways to correct them:
mafiozo [28]

The following combination of n=3 ; l=1 ; ml=-2 is not allowed. One way to correct this would be by changing the azimuthal quantum number, l and the other way would be to change the magnetic quantum number, m.

<h3>Is the following combination n=3; l=1; ml=-2 allowed or not.? If not, suggest two ways through which it can be corrected.</h3>

The following combination of n=3 ; l=1 ; ml=-2 is not allowed.

There are several rules that need to be followed for assigning electron quantum numbers. They are:

1. Principal quantum number should be 1 ≤ n

2. Azimuthal quantum number, 0 ≤ l ≤ n − 1

3. Magnetic quantum number, -l ≤ ml ≤ l

4. Spin quantum number as either -1/2 or +1/2

For n = 3,

l should be n - 1 or n - 2 or n - 3 = 2, 1, 0 respectively.

If we choose l = 1 then ml should be -1, 0 and +1

Therefore, one way to correct the combination would be to change the magnetic quantum number to -1

If we choose l = 2 then ml would be -2, -1, 0, +1, +2

Thus, another way to correct the combination is to choose the azimuthal quantum number as 2.

Thus, the following combination of n=3; l=1; ml=-2 is not allowed. One way to correct this would be by changing the azimuthal quantum number, l and the other way would be to change the magnetic quantum number, m.

To learn more about quantum numbers refer:

brainly.com/question/5927165

#SPJ4

5 0
2 years ago
Calculate the pH in titration of a weak acid: What is the pH in titration of formic acid (HCHO2, 0.200 M, 100.0 mL) after the ad
ki77a [65]

Answer:

pH = 12.61

Explanation:

First of all, we determine, the milimoles of base:

0.120 M = mmoles / 300 mL

mmoles = 300 mL . 0120 M = 36 mmoles

Now, we determine the milimoles of acid:

0.200 M = mmoles / 100 mL

mmoles = 100 mL . 0.200M = 20 mmoles

This is the neutralization:

HCOOH    +     OH⁻         ⇄        HCOO⁻     +    H₂O

20 mmol       36 mmol             20 mmol

                    16 mmol

We have an excess of OH⁻, the ones from the NaOH and the ones that formed the salt NaHCOO, because this salt has this hydrolisis:

NaHCOO  →  Na⁺  +  HCOO⁻

HCOO⁻  +  H₂O  ⇄   HCOOH  +  OH⁻   Kb →  Kw / Ka = 5.55×10⁻¹¹

These contribution of OH⁻ to the solution is insignificant because the Kb is very small

So:  [OH⁻] =  16 mmol / 400 mL →  0.04 M

- log  [OH⁻]  = pOH →  1.39

pH = 14 - pOH → 12.61

6 0
3 years ago
Please help! Please try and give me a long answer. Thanks!
Morgarella [4.7K]
I found this for your question

5 0
3 years ago
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