All categories

3 years ago

Given following electron configurations, 1s22s22p63s23p64s23d104p65s24d9 and 1s22s22p63s23p64s23d104p65s14d10.

1 answer:

8 0

Explanations:- Part 1: We could count the total number of electrons by looking at the electron configurations. Both of these electrons configurations have 47 electrons. If we look at the periodic table then 47 is the atomic number of silver. So, the name of the element is silver and its represented as Ag.

Part 2: As per the rule, Completely filled and half filled orbitals are more stable. First electron configuration has 9 electrons in 4d and we know that d is more stable if it has 5 electrons(half filled) or it has 10 electrons(full filled).

For stability reasons, one of the electron from 5s goes to 4d and for this reason the second electron configuration is found most often in nature for silver.

Few other examples are Cr and Cu.

You might be interested in

When 1.2383 g of an organic iron compound containing Fe, C, H, and O was burned in O2, 2.3162 g of CO2 and 0.66285 g of H2O were

uysha [10]

Answer:

$\boxed{\text{C$_{15}$H$_{21}$FeO$_{6}$}}$

Explanation:

Let's call the unknown compound X.

1. Calculate the mass of each element in 1.23383 g of X.

(a) Mass of C

$\text{Mass of C} = \text{2.3162 g } \text{CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{0.632 07 g C}$

(b) Mass of H

$\text{Mass of H} = \text{0.66285 g }\text{H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g } \text{{H$_{2}$O}}} = \text{0.074 157 g H}$

(c)Mass of Fe

(i)In 0.4131g of X

$\text{Mass of Fe} = \text{0.093 33 g Fe$_{2}$O$_{3}$}\times \dfrac{\text{111.69 g Fe}}{\text{159.69 g g Fe$_{2}$O$_{3}$}} = \text{0.065 277 g Fe}$

(ii) In 1.2383 g of X

$\text{Mass of Fe} = \text{0.065277 g Fe}\times \dfrac{1.2383}{0.4131} = \text{0.195 67 g Fe}$

(d)Mass of O

Mass of O = 1.2383 - 0.632 07 - 0.074 157 - 0.195 67 = 0.336 40 g

2. Calculate the moles of each element

$\text{Moles of C = 0.63207 g C}\times\dfrac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.052 629 mol C}\\\\\text{Moles of H = 0.074157 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.073 568 mol H}\\\\\text{Moles of Fe = 0.195 67 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.845 g Fe}} = \text{0.003 5038 mol Fe}\\\\\text{Moles of O = 0.336 40} \times \dfrac{\text{1 mol O}}{\text{16.00 g O}} = \text{0.021 025 mol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{0.052629}{0.003 5038}= 15.021\\\\\text{H: } \dfrac{0.073568}{0.003 5038} = 20.997\\\\\text{Fe: } \dfrac{0.003 5038}{0.003 5038} = 1\\\\\text{O: } \dfrac{0.021025}{0.003 5038} = 6.0006$

4. Round the ratios to the nearest integer

C:H:O:Fe = 15:21:1:6

5. Write the empirical formula

$\text{The empirical formula is } \boxed{\textbf{C$_{15}$H$_{21}$FeO$_{6}$}}$

5 0

3 years ago

Anthracite is so hard and pure it is also referred to as a _____.

Serga [27]

<span>Anthracite is also referred to as hard coal because it is so hard and pure. It is very hard and highly compacted with variety of coal that has submetallic luster. It has highest carbon content, the lowest impurities, and highest density of energy of all the coal deposits.</span>

6 0

3 years ago

If 120 g of potassium chloride are mixed with 100 g of water at 80°C, how much will not dissolve?

djverab [1.8K]

Answer:

yes

Explanation:

How many grams of KCl will dissolve in 1 liter of H2O at 50 °C? 5. 58.0 g of K2Cr2O7 is added to 100 g H2O at. 0 °C. With constant stirring, to what temp-.

4 0

3 years ago

A bond where electrons are transferred from a cation to an anion is called a (n) ___________bond.\

emmasim [6.3K]

When elements transfer electrons from a cation and an anion it called ionic bond.

6 0

3 years ago

Gas a (L^2. atm/mol^2) b(L/mol)

Amiraneli [1.4K]

Answer:

Explanation:

solve it urself idiot

7 0

2 years ago