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Korvikt [17]
3 years ago
8

Which is a set of valid quantum numbers?

Chemistry
2 answers:
Mariana [72]3 years ago
7 0

The last set which is n=4 l=3 m=3 is a valid set

Ad libitum [116K]3 years ago
5 0

The answer on Edgenuity2020 is n = 4, l = 3, m = 3

Which is the last option.

You might be interested in
What exactly is a proton? Is it a member of the periodic table?
Pavel [41]

The atomic number is the number of protons in the nucleus of an atom. The number of protons define the identity of an element (i.e., an element with 6 protons is a carbon atom, no matter how many neutrons may be present).

4 0
3 years ago
Calculate the pH of a 0.50 M HIO. The Ka of hypoiodic acid, HIO, is 2.3x10–11.0.305.325.479.474.80
never [62]

Answer:

pH = 5.47

Explanation:

The equilibrium that takes place is:

HIO ↔ H⁺ + IO⁻

Ka = \frac{[H+][IO-]}{[HIO]} = 2.3 * 10⁻¹¹

At equilibrium:

  • [HIO] = 0.5 M - x
  • [H⁺] = x
  • [IO⁻] = x

<u>Replacing those values in the equation for Ka and solving for x:</u>

Ka=\frac{x^2}{0.5-x}=2.3*10^{-11} \\x^2=(2.3*10^{-11})(0.5-x)\\x^2=1.15*10^{-11}-2.3*10^{-11}x\\x^2+2.3*10^{-11}x-1.15*10^{-11}=0\\x=3.39*10^{-6}

Then [H⁺]=3.39 * 10⁻⁶, thus pH = 5.47

7 0
3 years ago
When the following oxidation–reduction reaction in acidic solution is balanced, what is the lowest whole-number coefficient for
Svet_ta [14]

Answer:

c. 8, product side

Explanation:

In order to balance a redox reaction we use the ion-electron method, which has the following steps:

Step 1: identify oxidation and reduction half-reaction.

Oxidation: MnO₄⁻(aq) → Mn²⁺(aq)

Reduction: Br⁻(aq) → Br₂(l)

Step 2: perform the mass balance adding H⁺ and H₂O where necessary

8 H⁺(aq) + MnO₄⁻(aq) → Mn²⁺(aq) + 4 H₂O(l)

2 Br⁻(aq) → Br₂(l)

Step 3: perform the electrical balance adding electrons where necessary.

8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)

2 Br⁻(aq) → Br₂(l) + 2 e⁻

Step 4: multiply both half-reactions by numbers that secure that the number of electrons gained and lost are the same.

2 × (8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l))

5 × (2 Br⁻(aq) → Br₂(l) + 2 e⁻)

Step 5: add both half-reactions side to side.

16 H⁺(aq) + 2 MnO₄⁻(aq) + 10 e⁻ + 10 Br⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 Br₂(l) + 10 e⁻

16 H⁺(aq) + 2 MnO₄⁻(aq) + 10 Br⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 Br₂(l)

3 0
3 years ago
My swimming pool is rectangular (16 feet by 34 feet) and has a depth of 6 feet. Lets imagine that my pool water is full to the t
Reil [10]

Answer:

Number of moles of photons required = 5.04 × 10⁴ moles

Explanation:

The energy of a photon can be calculated from Planck's equation E = hc/λ

Where h = 6.63 × 10-³⁴ Js and c, the velocity of light = 3.0 × 10⁸ m/s

Energy of one mole of photons = N₀ × hc/λ

wavelength of photon, λ = 520 nm = 5.20 × 10-⁷ m

Energy of one mole of photons = 6.02 × 10²³ × 6.63 × 10−³⁴ × 3 × 10⁸/5.20 × 10-⁷

Energy of one mole of photons = 2.30 × 10⁵ J/mol

Energy required to raise the temperature of a given mass of a substance, E = mcΔT

Where m is mass of substance,  c is specific heat capacity,  ΔT is temperature difference

Mass ofnwternin the pool = volume × density

Volume of water = Volume of swimming pool

Volume of water = 16 × 34 × 6 ft³ = 3264 ft³

1 ft³ = 28316.8 cm³; 3264 ft³ = 28316.8 × 3264 = 92426035.2 cm³

Density of water = 1 g/cm³

Mass of water = 92426035.2 cm³ × 1 g/cm³ = 92426035.2g

ΔT = 80°C - 50°C = 30°C, c = 4.18 J/g/K

Energy required to raise 92426035.2 g water by 30° C = 92426035.2 × 4.18 × 30

Energy required = 1.16 × 10¹⁰ J

Hence, number of moles of photons required = 1.16 × 10¹⁰ J/2.30 × 10⁵ J/mol

Number of moles of photons required = 5.04 × 10⁴ moles

5 0
3 years ago
If i have 8 L of 0.25 M solution, how many miles of MgCl2 are present
qaws [65]

Answer:

2 moles

Explanation:

The following were obtained from the question:

Molarity = 0.25 M

Volume = 8L

Mole =?

Molarity is simply defined as the mole of solute per unit litre of solution. It is represented mathematically as:

Molarity = mole of solute/Volume of solution.

With the above equation, we can easily find the number of mole of MgCl2 present in 8 L of 0.25 M MgCl2 solution as follow:

Molarity = mole of solute/Volume of solution.

0.25 = mole of MgCl2 /8

Cross multiply to express in linear form

Mole of MgCl2 = 0.25 x 8

Mole of MgCl2 = 2 moles

Therefore, 2 moles of MgCl2 are present in 8 L of 0.25 M MgCl2 solution

4 0
3 years ago
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