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Alekssandra [29.7K]
3 years ago
12

19 pts - How do you determine if something is isoelectronic?

Chemistry
1 answer:
Anna71 [15]3 years ago
5 0
Isoelectronic means equal number of electrons.

The method to determine if elements are sioelectronic is:

1) Determine the atomic number of the element (search the element in the periodic table).

2) The atomic number is equal to the number of protons (always, by definition).

3) If the atom is neutral, the number of electrons is equal to the number of protons.

3) If the species is not neutral then:

     - if it is a negative ion, add as many electrons as the magnitude of the charge of the ion

    - if iit is a positive ion, subtract as many electrons as the magnitude of the charge of the ion.

4) One by one
<span>
species          atomic number              charge          number of electrons
                      number of protons

O₂,                    8                                     0                 8 (each atom)
F-                     9                                      -1               9 + 1 = 10    
Ne                  10                                      0                10
Na                  11                                       0               11

So, in that group only F- and Ne are isoelectronic, each with 10 electrons

Se2-               34                                      -2                34 + 2 = 36
Br-                  35                                      -1                35 + 1 = 36
Kr                   36                                       0                36
Rb+                37                                      +1               37 - 1 = 36

So, in that group all are isoelectronic.
   
I-                     53                                     -1                 53 + 1 = 54
Xe                   54                                     0                  54
Cs                   55                                     0                  55  
Ba2+               56                                    +2                 55 - 2 = 53

So, in that group only I- and Xe are isoelectronic.
</span>
You might be interested in
A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.
makvit [3.9K]

Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H

What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> <u>smallest</u> whole number ratio of elements in a compound

Molecular Formula => <u>actual</u> whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4'  to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

              %     =>         grams                 =>                 moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H =>  7.14 : 14.29

<u>To reduce mole values to the smallest whole number ratio,  divide all mole values by the smaller mole value of the set.</u>

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

3 0
3 years ago
Having energy in the reactant lead to an exothermic reaction.<br>true<br>false​
Ostrovityanka [42]

Answer:

false

Explanation:

first of all;-energy lead to an indotermic reaction.

indotermic is a reaction that absorbs energy \

*it has positive enthalpy of reaction

*Heat content of product is greater than that of reactant

*Heat is added to reactant side

example;- CO^2+2H^2+891kj --------- CH4 +2O2

6 0
3 years ago
What is the concentration of h+ ions in a 2.20 m solution of hno3?
kvv77 [185]

The question is improperly formatted.

What is the concentration of H+ ions in a 2.2 M solution of HNO3.

Answer:-

2.2 moles of H+ per litre

Explanation:-

M stands for molarity. 2.2 M means 2.2 moles of HNO3 is present per litre of the solution.

Now HNO3 has just 1 H in it's formula. HNO3 would give H+. So 2.2 moles of HNO3 would mean 2.2 moles of H+ per litre.

7 0
3 years ago
Which is the best practice recommended in the safety video to mix and acid or a base with a solvent?
Vesna [10]

Answer:

Never pour water into acid but acid into water

Explanation:

If water is poured into extremely concentrated acid/bases, the rate of volatility and exothermic reaction is too rapid and might cause a chemical eruption, leading to acid burns.

Safety precautions hence dictate the reverse is practiced.

I believe this is a clear answer.

4 0
3 years ago
A 200.0 mL solution of 0.40 M ammonium chloride was titrated with 0.80 M sodium hydroxide. What was the pH of the solution after
choli [55]

Answer:

9.25

Explanation:

Let first find the moles of NH_4Cl and NaOH

number of moles of NH_4Cl = 0.40  mol/L × 200 ×  10⁻³L

= 0.08 mole

number of moles of NaOH = 0.80  mol/L × 50 ×  10⁻³L

= 0.04 mole

The equation for the reaction is expressed as:

NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}

The ICE Table is shown below as follows:

                            NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}

Initial (M)              0.08            0.04                            0

Change (M)         - 0.04          -0.04                          + 0.04

Equilibrium (M)      0.04             0                                0.04

K_a*K_b = 10^{-14} \ at \ 25^0C

K_a = \frac{10^{-14}}{K_b}

K_a = \frac{10^{-14}}{1.76*10^{-5}}

K_a= 5.68*10^{10}

pK_a = - log \ (K_a)

pK_a = - log \ (5.68*10^{-10})

pK_a = 9.25

pH = pKa + \ log (\frac{HB}{HA} )   for buffer solutions

pH = pKa + \ log (\frac{moles \ of \ base }{ moles\ of \ acid} ) since they are in the same solution

pH = 9.25 + \ log (\frac{0.04 }{ 0.04} )

pH = 9.25

8 0
3 years ago
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