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Mamont248 [21]
3 years ago
13

The process of nuclear division that produces haploid sex cells is mitosis. Please select the best answer from the choices provi

ded T F
Chemistry
1 answer:
Ivanshal [37]3 years ago
4 0

Answer:

Explanation:

meiosis

Haploid sperm and eggs are produced via meiosis.

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Can someone please quickly answer this question?
dimulka [17.4K]

Answer:

The difference between the two is:

[Kr]4d^1^05s^25p^2 is the Orbital Occupancy

[Kr]5s^24d^1^05p^2 is the Orbital Filling Order

Both are correct, I don't think your teacher will be so nit-picky to care.

3 0
3 years ago
Could you help answer these two question
Katen [24]

Answer:

number 29 is D

Explanation: When the moon orbits the earth some of the light reflects off the moon, making it visible on the Earth.

I dont know how to answer 28 fully, is there a picture?

4 0
3 years ago
In an experiment magnesium ribbon was heated in air. The product was found to be heavier than the original ribbon. Potassium man
KatRina [158]

Burning a magnesium ribbon in the air is an addition reaction while heating potassium manganate 7 is a decomposition reaction.

<h3>Addition and decomposition reactions</h3>

Magnesium burns in air to produce magnesium oxide as follows:

2Mg + O_2 --- > 2MgO

Potassium manganate 7 burns to produce multiple products as follows:

2 KMnO_4 --- > K_2MnO_4 + MnO_2(s) + O_2

Thus, the MgO will be heavier than Mg. On the other hand, MnO_2 will be less heavy than KMnO_4.

More on reactions can be found here: brainly.com/question/17434463

#SPJ1

7 0
2 years ago
The measured voltage of an electrochemical cell consisting of pure nickel immersed in a solution of Ni2+ ions of unknown concent
Verizon [17]

Answer:

[Ni²⁺] = 1.33 M

Explanation:

To do this, we need to use the Nernst equation which (in standard conditions of temperature)

E = E° - RT/nF lnQ

However R and F are constant, and the reaction is taking place in 25 °C so we can assume the nernst equation like this:

E = E° - 0.05916/n logQ

As the nickel is in the cathode, this means that this element is being reducted while Cadmium is being oxidized, therefore the REDOX reaction would be:

Cd(s) + Ni²⁺(aq) --------> Cd²⁺(aq) + Ni(s)

With this, Q:

Q = [Cd²⁺] / [Ni²⁺]

Now, we need to know the value of the standard reduction potentials, which can be calculated with the semi equations of reduction and oxidation:

Cd(s) ------------> Cd²⁺ + 2e⁻       E°₁ = 0.40 V

Ni²⁺ + 2e⁻ -------------> Ni(s)        E°₂ = -0.25 V

E° = E°₁ + E°₂

E° = 0.40 - 0.25 = 0.15 V

Now that we have all the data, we can solve for the [Ni²⁺]:

0.133 = 0.15 - 0.05916/2 log(5/[Ni²⁺])

0.133 - 0.15 = -0.05916/2 log(5/[Ni²⁺])

-0.017 = -0.02958 log(5/[Ni²⁺])

-0.017/-0.02958 = log(5/[Ni²⁺])

0.5747 = log(5/[Ni²⁺])

10^(0.5747) = 5/[Ni²⁺]

[Ni²⁺] = 5/3.7558

<h2>[Ni²⁺] = 1.33 M</h2>
7 0
3 years ago
Plz answer plzplzpzlpl​
astraxan [27]
Whats the question tho?
8 0
3 years ago
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