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yan [13]
2 years ago
8

How many electrons in an atom can have each of the following quantum number or sublevel designations? (a) 4p (b) n = 3, l = 1, m

1 = +1 (c) n = 5, l = 3
Chemistry
1 answer:
leonid [27]2 years ago
6 0

Answer:

a) Six electrons

b) Two electrons

c) Fourteen electrons

Explanation:

n is the principal quantum number and defines the energy level of orbital. The shape of the orbital is described by azimuthal quantum number (l) and it also determine the angular momentum. It values give the following information

l = 0, define s orbital (single orbital)

l = 1, define p orbitals (three orbitals)

l = 2, define d orbitals (five orbitals)

l = 3, define f orbitals (seven orbitals)

These are further specified by magnetic quantum number (ml) which gives the orientation of the orbital. Its value ranges from +1 to -1, for example ml value of five d orbitals are +2, +1, 0, -1, -2. From this information we can predict the number of electrons that will have the given sub-level designations

a) n = 4 and orbital is p, there are three p orbitals as the ml is not defined, so six electrons will have this quantum number

b) In this part, the orbital is defined i.e. ml = +1. A single orbital can have only two electrons, so these electrons will have the given quantum number.

c) l = 3, is for f orbital, which have seven orbitals. The total number of electrons in it is fourteen. All of these electrons will have this quantum number.

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Vilka [71]

Answer:

58.61 grams

Explanation:

Taking The molecular weight of NaCl = 58.44 grams/mole

<u>Determine how many grams of NaCl to prepare the bath solution </u>

first we will calculate the moles of NaCl that is contained in 6L of 170 mM of NaCI solution

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= 1020 / 1000 = 1.020 moles

next

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4 0
2 years ago
Hafnium has six naturally occurring isotopes: 0.16% of 174Hf, with an atomic weight of 173.940 amu; 5.26% of 176Hf, with an atom
ser-zykov [4K]

Answer:

177.277amu

Explanation:

the total occuring isotopes for Hafnium is =6.

First isotope had an atomic weight of 173.940amu

Second isotope =175.941amu

Third isotope =176.943amu

Fourth isotope=177.944amu

Fifth isotope. =178.946amu

sixth isotope .179.947amu

<em>Avera</em><em>ge</em><em> </em><em>ato</em><em>mic</em><em> </em><em>wei</em><em>ght</em><em> </em><em>of</em><em> </em><em>Haf</em><em>nium</em><em>=</em><em> </em><em>sum</em><em> </em><em>of</em><em> </em><em>all</em><em> </em><em>the </em><em>atomi</em><em>c</em><em> </em><em>weights</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>iso</em><em>topes</em><em>/</em><em> </em><em>Tota</em><em>l</em><em> </em><em>occu</em><em>ring</em><em> </em><em>isotopes</em>

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6 0
2 years ago
A solution is prepared by condensing 4.00 l of a gas, measured at 27°c and 748 mmhg pressure, into 58.0 g of benzene. calculate
fgiga [73]
First, we are using the ideal gas law to get n the number of moles:

PV = nRT

when P is the pressure = 748 mmHg/760 = 0.984 atm

V is the volume = 4 L

R is ideal gas constant = 0.0821

T is the temperature in Kelvin = 300 K

∴ n =  0.984atm*4L/0.0821*300

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when the concentration = moles * (1000g / mass)

                                         = 0.1598 * (1000g / 58 g )

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when the freezing point = 5.5 °C

and Kf = - 5.12 °C/m

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                                                            = 5.5 °C - (5.12°C/m * 2.755m)

                                                            = -8.6 °C

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c2 is the concentration of the diluted solution to be prepared
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7 0
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