x21 +ANSWER
(i) Ne-22
(ii)1s2s22p6
(iii)21.3
An element X exists in three forms A, B and C in the ratio 1:2:3. If C has 10 protons and the number of neutrons in A, B and C are 10, 11 and 12 respectively,Give the following:(i) Representation of form C of the element X(ii) Electronic configuration of form B of the element(iii) Calculate the average atomic mass.
(i)C has 10 protons and 12 neutrons so a mass of 10 +12 =22
element 10 is Neon (Ne) so this isotope is Ne-22
(ii) they all have the sane atomic number so the same number of electrons
with an electronic structure of 1s2s22p6
(iii) A weighs 20, B weighs 21, C weighs 22
the ratio is 1:2:3
weighted average weight is therefor
(1X20 +2X21 +3X22)/6 =21.3
Answer:
The number of atoms in the outermost shell
Explanation:
For example, the electron shells in the alkali metals contain the following numbers of electrons:
Li: 2, 1
Na: 2, 8, 1
K: 2, 8, 8, 1
They all have one electron in their outermost shell, and they have similar chemical properties.
Answer:
1) 90.0 mL
2) 11.25 M
3) 0.477 M
4) 144 mL
Explanation:
The main formula that will be used for all these calculations is:
C₁V₁ = C₂V₂
C stands for concentration and V stands for volume and the subscripts 1 and 2 indicate an initial concentration or volume and a final concentration or volume.
For each problem, it's best to start by figuring out what you have and what you need to find. Figure out if you're looking for an initial value or a final value.
1) We need to find the initial volume. So, take what values you have and plug them in and then solve for whatever variable:
5.00 M · V₁ = 500.0mL · 0.900 M - divide by 5.00
C₁ = 90.0 mL
2) This time we're finding the initial concentration:
20.0mL · C₁ = 150.0mL · 1.50 M - divide by 20.0mL
C₂ = 11.25 M
3) Now we're finding the final concentration:
12.00mL · 3.50 M = 88.0mL · C₂ - divide by 88.0mL
C₂ = 0.477 M
4) Finally, we're looking for the final volume:
9.0mL · 8.0 M = 0.50 M · V₂ - divide by 0.50 M
V₂ = 144mL
A) heating a pan of water until the water is all gone because then it would change from a liquid top a gas.
Answer:
1120 L.
Explanation:
Hello!
In this case, as no conditions of pressure of temperature are given for this problem, we can assume that the scuba diver dives at STP (1 atm and 273.15 K), which means that 1 mole of air would occupy a volume of 22.4 L.
In such a way, since she needs 50.0 moles of air, the following ratio is useful to compute the size (volume) of the tank she needs:
Thereby, we plug in to obtain:
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