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asambeis [7]
2 years ago
8

1 point There are 118 known elements. How many elements have atoms, in their ground-state, with core electrons whose quantum num

bers are n
Chemistry
1 answer:
Diano4ka-milaya [45]2 years ago
8 0

The question is incomplete; the complete question is;

There are 118 known elements. How many elements have atoms, in their ground state, with core electrons whose quantum numbers are n=3 and l=1?

a. 10

b. 6

c. 100

d. 18

Answer:

c. 100

Explanation:

If we look closely at the periodic table, we will notice that there 100 elements that occur after argon whose electronic configuration is; 1s2 2s2 2p6 3s2 3p6.

This means that there every other element after argon must have a 3p sublevel as part of its core electrons. In the 3p sublevel, n=3 and l=1.

If we diligently count the number of elements from potassium (immediately after argon) to Oganesson (element 118) we have a total of 100 elements all having core electrons of quantum numbers are n=3 and l=1.

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A 23.0g sample of a compound contains 12.0g of C, 3.0g of H, and 8.0g of O.What the empirical formula of the compound
Kryger [21]

Answer:

The empirical formula of compound is C₂H₆O.

Explanation:

Given data:

Mass of carbon = 12 g

Mass of hydrogen = 3 g

Mass of oxygen = 8 g

Empirical formula of compound = ?

Solution:

First of all we will calculate the gram atom of each elements.

no of gram atom of carbon = 12 g / 12 g/mol = 1 g atoms

no of gram atom of hydrogen = 3 g / 1 g/mol = 3 g atoms

no of gram atom of oxygen = 8 g / 16 g/mol = 0.5 g atoms

Now we will calculate the atomic ratio by dividing the gram atoms with the 0.5 because it is the smallest number among these three.

          C:H:O  =     1/0.5  :   3/0.5  :   0.5/0.5

          C:H:O  =     2      :     6      :     1

The empirical formula of compound will be C₂H₆O

5 0
2 years ago
Of the elements: b, c, f, li, and na. the element with the highest ionization energy is
padilas [110]
Carbon has the highest ionization energy as its energy 1086KJ\Mol and the rest are between 500 and 800. 
6 0
2 years ago
A student wants to reclaim the iron from an 18.0-gram sample of iron(III) oxide, which
lilavasa [31]

Answer:

m_{Fe}=12.6gFe

Explanation:

Hello,

In this case, since we have grams of iron (III) oxide whose molar mass is 159.69 g/mol are able to compute the produced grams of iron by using its atomic mass that is 55.845 g/mol and their 2:4 molar ratio in the chemical reaction:

m_{Fe}=18.0gFe_2O_3*\frac{1molFe_2O_3}{159.69gFe_2O_3}*\frac{4molFe_2O_3}{2molFe_2O_3} *\frac{55.845gFe}{1molFe_2O_3} \\\\m_{Fe}=12.6gFe

Best regards.

3 0
2 years ago
Read 2 more answers
The interior of a refrigerator has a volume of 0.600 m3. the temperature inside the refrigerator in 282 k, and the pressure is 1
Dafna11 [192]
In this instance we can use the ideal gas law equation to find the number of moles of gas inside the refrigerator 
PV = nRT
where 
P - pressure - 101 000 Pa
V - volume - 0.600 m³
n - number of moles
R - universal gas constant - 8.314 J/mol.K
T - temperature - 282 K
substituting these values in the equation 
101 000 Pa x 0.600 m³ = n x  8.314 J/mol.K x 282 K
n = 25.8 mol
there are 25.8 mol of the gas
to find the mass of gas
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6 0
2 years ago
Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll
makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature T_1 = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

a)

The truncated viral equation is expressed as:

\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}

where; B = - 152.5 \ cm^3 /mol   C = -5800 cm^6/mol^2

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}

4.138*10^{-4}  \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}

Multiplying through with V² ; we have

4.138*10^4  \ V ^3 = V^2 - 152.5 V - 5800 = 0

4.138*10^4  \ V ^3 - V^2 + 152.5 V + 5800 = 0

V = 2250.06  cm³ mol⁻¹

Z = \frac{PV}{RT}

Z = \frac{1800*2250.06}{8.314*10^3*523.15}

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

T_c = 647.1 \ K \\ \\ P_c = 22055 \  kPa  \\ \\ \omega = 0.345

T__{\gamma}} = \frac{T}{T_c}

T__{\gamma}} = \frac{523.15}{647.1}

T__{\gamma}} = 0.808

P__{\gamma}} = \frac{P}{P_c}

P__{\gamma}} = \frac{1800}{22055}

P__{\gamma}} = 0.0816

B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}

B_o = 0.083 - \frac{0.422}{0.808^{1.6}}

B_o = 0.51

B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}

B_1 = -0.282

The compressibility is calculated as:

Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}

Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}

Z = 0.9386

V= \frac{ZRT}{P}

V= \frac{0.9386*8.314*10^3*523.15}{1800}

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At T_1 = 523.15 \  K \ and  \ P = 1800 \ k Pa

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V  =  0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = \frac{PV}{RT}

Z = \frac{1800*10^3 *0.1249}{729.77*523.15}

Z = 0.588

3 0
2 years ago
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