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4 years ago

Make the following conversion. 120 mL = _____ cm³ 12.0 1200 120 1.20

Chemistry

2 answers:

r-ruslan [8.4K]4 years ago

5 0

Answer: 120

Explanation:

Volume of the gas is defined as the space occupied by a substance. It is expressed in units like $cm^3$ , $m^3$ , L and ml.

All these units of volume are inter convertible.

We are given:

Volume of the gas = $120ml$

Converting this unit of volume into $cm^3$ by using conversion factor:

$1cm^3=1ml$

$120ml=\frac{1}{1}\times 120=120cm^3$

Thus the volume in $cm^3$ would be 120.

Lina20 [59]4 years ago

4 0

The correct answer would be 120.

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Below are shown, for five metals, reduction reactions and standard electrode potential values. which of these metals is the leas

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........................................................................................................................
Solution:
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On other hand, reduction potential of Na is minimum, hence it is most prone to undergo oxidation. Hence, it is most reactive.

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3 years ago

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Ow does the weight of the water displaced compare to the buoyant force?

Ghella [55]

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Expatiation:

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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

Is there any way the pandemic helps this society?<br><img src="https://tex.z-dn.net/?f=22%20%5Ctimes%2012" id="TexFormula1" titl

Alexxandr [17]

Answer:

u meant the answer to this ques

Explanation:

×12

------

22 ×

----------

264

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or the answer can be 22×12=264

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4 years ago