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attashe74 [19]
2 years ago
15

An element X exists in three forms A, B and C in the ratio 1:2:3. If C has 10 protonsand the number of neutrons in A, B and C ar

e 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.
Chemistry
1 answer:
Strike441 [17]2 years ago
4 0

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

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77°F

Explanation:

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Explain why ethics and skepticism are integral to science.
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8 0
3 years ago
Consider the single‑step, bimolecular reaction. CH3Br+NaOH⟶CH3OH+NaBr When the concentrations of CH3Br and NaOH are both 0.120 M
GarryVolchara [31]

Answer:

Explanation:

CH₃Br+NaOH⟶CH₃OH+NaBr

It is a single step bimolecular reaction so  order of reaction is 2 , one for CH₃Br and one for NaOH .

rate of reaction = k x [CH₃Br] [ NaOH]

.008 = k x .12 x .12

k = .55555

when concentration of CH₃Br is doubled

rate of reaction = .555555 x [.24] [ .12 ]

= .016 M/s

when concentration of NaOH  is halved

rate of reaction = .555555 x [.12] [ .06 ]

= .004  M/s

when concentration of both CH₃Br and Na OH is made 5 times

rate of reaction = .555555 x .6 x .6

= 0.2 M/s

8 0
3 years ago
What is the mass of 4.38 x 10^24 formula units of sodium chloride (NaCl)?
lana66690 [7]

Answer:

Mass = 427.05 g

Explanation:

Given data:

Formula units of sodium chloride = 4.38 ×10²⁴

Mass of NaCl = ?

Solution:

One mole contain formula units = 6.022 ×10²³

4.38×10²⁴ formula units × 1mol /6.022 ×10²³ formula units

0.73 ×10¹ mol

7.3 mol

Mass of NaCl:

Mass = number of moles × molar mass

Mass = 7.3 mol × 58.5 g/mol

Mass = 427.05 g

5 0
3 years ago
Question 1 of 3
AysviL [449]

Answer:

The volume of the solution is 133 ml

Explanation:

Having a 0.250 molar solution means 0.250 moles in 1000 ml of solution (1 liter)

We calculate the weight of 1 mole of CuNO3, and use 3 simple rules for calculations:

CuNo3---> Weight of 1 mole= 63, 546g+ 14, 0067g+ 3X 15, 9999g= 125, 55g/mol

1 mol----125, 55g

0,250mol---X= (0,250molx 125,55g)/1mol= 31,3 g

31,3g----1000ml

4,16g---x= (4,16gx1000ml)/31,3g = 133ml

6 0
3 years ago
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