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Serjik [45]
2 years ago
10

List the four different sublevels and (given that only a maximum of two electrons can occupy an orbital) determine the maximum n

umber of electrons that can exist in each sublevel. Drag the appropriate items to their respective bins. ResetHelp l = 0 l = 1 l = 2
Chemistry
1 answer:
timurjin [86]2 years ago
8 0

Answer:

l = 0 → s = 2 electrons;

l = 1 → p = 6 electrons;

l = 2 → d = 10 electrons;

l= 3  → f = 14 electrons.

Explanation:

For the quantum theory, the probability to find an electron is higher in the space region called orbital. It's impossible to determine where the electron is and his velocity at the same time (uncertainty principle). So, the theory determines four quantum numbers to characterize an electron, so it's easy to identify it:

  • n is the principal quantum number and identify the shell where the electron is. It varies from 1 to 7 and is represented by the letters K, L, M, N, O, P, and Q;
  • l is the azimuthal quantum number and identify the subshell (or sublevel) where the electron is. It varies from 0 to 3 and is represented by the letters s, p, d, and f;
  • ml is the magnetic quantum number, and it represents the orbital. It varies from -l to +l, passing by 0. Each orbital can have at least 2 electrons;
  • ms is the spin number and represents the spin of the electrons. It can be +1/2 or - 1/2.

Then, the sublevel s (l= 0) only has 1 orbital (ml = 0) so, it can have ate least 2 electrons; the sublevel p (l= 1) has 3 orbitals (ml = -1, ml= 0, ml = +1), so it can have at least 6 electrons; the sublevel d (l = 2) has 5 orbitals (ml = -2, ml = -1, ml = 0, ml = +1, ml = +2), so it can have at least 10 electrons; and the sublevel f (l = 3) has 7 orbitals (ml = -3, ml = -2, ml = -1, ml = 0, ml = +1, ml = +2, ml = +3), so it can have at least 14 electrons.

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If the tip of the syringe, "The Titrator", was not filled with NaOH before the initial volume reading was recorded, would the co
Ostrovityanka [42]

Answer:

The concentration of acetic acid in vinegar of that trial would be <u><em>greater than</em></u> the actual concentration.

Explanation:

"The titrator" contains the base solution (NaOH) with which the soution of vinegar (acetic acid) is being titrated.

Under the assumption that the tip of the syringe was not filled before the initial volume reading was recorded, part of the volume of the base that you release will be retained in the tip of the syringe, and, consequently, the actual volume of base added to the acetic acid will be less than what you will calculate by the difference of readings.

So,  in your calculations you will use a larger volume of the base than what was actually used, yielding a fake larger number of moles of base than the actual amount added.

So, as at the neutralization point the number of equivalents of the base equals the number of acid equivalents, you will be reporting a greater number of acid equivalents, which in turn will result in a greater concentration than the actual one. This means that the concentration of acetic acid in vinegar of that trial would be greater than the actual concentration.

6 0
2 years ago
Magnesium has an atomic mass of 24.3. there are two isotopes of magnesium - one contains 12 neutrons and the other contains 13 n
Juli2301 [7.4K]
So what we know:
-Atomic Mass = Protons + Neutrons
-Atomic Number is the number of protons

Magnesium's atomic number is 12, so the natural occurring isotope for magnesium is Mg-12 (12 protons and 12 neutrons). Added up we have an atomic mass of 24 amu. Which means if we added one neutron in Mg-13, our atomic mass would be 25 amu.

We can use the equation:
(amu of isotope 1)x + (amu of isotop 2)(x-1) = Average atomic mass
where isotope 1 is the fractional abundance we're solving for.

Plugged in it looks like this:
24x + 25(1-x) = 24.3

Now to solve for x:
24x + 25 - 25x = 24.3
   -x + 25 = 24.3
        -x = -.7
         x = .7

So in this case, the fractional abundance of Mg-12 would be .7, or 70%.<span />
3 0
2 years ago
Acetylene burns in air according to the following equation: C2H2(g) + 5 2 O2(g) → 2 CO2(g) + H2O(g) ΔH o rxn = −1255.8 kJ Given
professor190 [17]

Answer:  -227 kJ

Explanation:

The balanced chemical reaction is,

C_2H_2(g)+\frac{5}{2}O_2(g)\rightarrow 2CO_2(g)+H_2O(g)

The expression for enthalpy change is,

\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+ n_{H_2O}\times \Delta H_{H_2O})]-[(n_{C_2H_2}\times \Delta H_{C_2H_2})+(n_{O_2}\times \Delta H_{O_2})]

where,

n = number of moles

\Delta H_{O_2}=0 (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

-1255.8=[(2\times -393.5)+(1\times -241.8)]-[(1\times \Delta H_{C_2H_2})+(\frac{5}{2}\times 0)]

-1255.8=[(-787)+(-241.8)]-[(1\times \Delta H_{C_2H_2})+(0)]

\Delta H_{C_2H_2}=-227kJ

Therefore, the enthalpy change for C_2H_2 is -227 kJ.

7 0
2 years ago
Solving applied density problems Mr. Auric Goldfinger, criminal mastermind, intends to smuggle several tons of gold across Inter
Margarita [4]

Answer:

thickness = 0.29 cm

Explanation:

In order to make fake iron ball [made of gold] we should get mass of fake ball

should be equal to that of Iron ball.So for that we should calculate

volume of iron ball using diameter given ;formula is 4/3 pi r^3

given d= 6 cm;so radius r= 6/2 = 3 cm

then volume of Iron ball = 4/3 *3.14* 3^3 = 113.04 cm^3

So mass of iron ball = volume x density = 113.04 * 5.15 g/cm^3 = 582.156 g

This must be the mass of gold ball ;now calculate volume of gold ball using its

density

volume of gold ball = mass of gold ball/density of gold ball = 582.156 g/19.3 g/cm^3

= 30.1635 cm^3

Now this must be the volume of hollow sphere whose outer radius R = 3cm

and inner radius r= ??

Volume of hollow ball = 4/3pi[R3-r^3]

30.1635 cm^3 = 4/3 pi [3^3-r^3]

30.1635* 3/4*3.14 = 27 - r^3

7.2046 = 27- r^3

r^3 = 19.7954

r= 2.7051 cm

So the thick ness = outer radius- inner radius = 3 - 2.7051 = 0.2949 cm

rounding to 2 significant figures

we get thickness = 0.29 cm

8 0
2 years ago
A student requires 2.00 L of 0.100 M NH4NO3 from a 1.75 M NH4NO3 stock solution. What is the correct way to get the solution?
AysviL [449]

Answer:

not sure, but B

Explanation:

7 0
2 years ago
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