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Andreyy89
3 years ago
13

Enter the cycle number, n, in the x column and the number of radioactive atoms in the y column. When you finish, resize the wind

ow.
Chemistry
1 answer:
kirza4 [7]3 years ago
7 0
Answer Answer: I am not too sure about the X or Y, but they didn't seem necessary in my quiz. However, after that, you'll want to click "Exponential" Explanation: I just took the test, that's about it!
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Can you please explain to me what mole concept is<br><br>​
dybincka [34]

a mole is defined as the amount of substance that contains as many atoms, molecules, icons, electron or any other elementary entities as there are exactly 12 gm of. of carbon atoms. the number of 12 gm of. Is called Avogadro's number

3 0
4 years ago
How many grams of H2 will be produced from 12 grams of Mg according to the following equation?
LUCKY_DIMON [66]

Answer:

0.98 g of H₂

Explanation:

the balanced equation for the reaction is

Mg + 2HCl ---> MgCl₂ + H₂

molar ratio of Mg to H₂ is 1:1

number of Mg moles reacted = 12 g/ 24.3 g/mol = 0.49 mol

according to molar ratio

when 1 mol of Mg reacts 1 mol of H₂ is formed

therefore when 0.49 mol of Mg reacts - 0.49 mol of H₂ forms

therefore mass of H₂ formed = 0.49 mol x 2 g/mol = 0.98 g

mass of H₂ formed is 0.98 g

7 0
3 years ago
if you are told to get 100 mL of stock solution to use to prepare smaller size sample for an experiment, which piece of glasswar
fgiga [73]

Answer:

A beaker  

Step-by-step explanation:

Specifically, I would use a 250 mL graduated beaker.

A beaker is appropriate to measure 100 mL of stock solution, because it's easy to pour into itscwide mouth from a large stock bottle.

You don't need precisely 100 mL solution.

If the beaker is graduated, you can easily measure 100 mL of the stock solution.

Even if it isn't graduated, 100 mL is just under half the volume of the beaker, and that should be good enough for your purposes (you will be using more precise measuring tools during the experiment).

6 0
3 years ago
Which term describes the part of the periodic table that is arranged by increasing atomic number and repeats periodically?
erastovalidia [21]

Answer:

period

Explanation:

8 0
3 years ago
Calculate E o , E, and ΔG for the following cell reactions (a) Mg(s) + Sn2+(aq) ⇌ Mg2+(aq) + Sn(s) where [Mg2+] = 0.025 M and [S
Rudik [331]

E⁰(cell) = 2.24V

E(cell) = 2.246V

∆G = -433 KJ/mol

<u>Explanation:</u>

Mg(s) + Sn²⁺(aq) ⇌ Mg²⁺(aq) + Sn(s)

[Mg2+] = 0.025 M

[Sn2+] = 0.040 M

First we need the standard reduction potentials:

. . . . . . . . . . . . . . . . . E°(V)

Mg²⁺ + 2 e⁻ ⇌ Mg(s). . .−2.372

Sn²⁺ + 2 e⁻ ⇌ Sn(s) . . . −0.13

Take the more negative (or less positive in other cases) one, and write it as an oxidation:

Mg(s) ⇌ Mg²⁺ + 2 e⁻. . .+2.372 V

Combine them,

Mg(s) + Sn²⁺ ⇌ Mg²⁺ + Sn(s)

E°(cell) = +2.372 – 0.13 V = 2.24 V

To get the cell potential under the conditions given, use the Nernst Equation:

E(cell) = E°(cell) – [(0.059)/n]•logQ = 2.24 V – 0.0295 V • log [Mg²⁺]/[Sn²⁺]

Note that the solids don't appear in Q, only the concs. of the dissolved ions.

E(cell) = 2.24 V – 0.0295 V X log (0.025)/(0.040)

          = 2.24 + 0.006 V ≈ 2.246 V

The concentration ratio in Q (Sn²⁺ and Mg²⁺) is too close to 1 to shift E(cell) significantly from E°(cell) given the precision I have for the Sn reduction potential.

∆G = –nFE(cell) = –2(96.485 kJ/mol•V)(2.246 V) = –433 kJ/mol

E⁰(cell) = 2.24V

E(cell) = 2.246V

∆G = -433 KJ/mol

4 0
4 years ago
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