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klio [65]
3 years ago
14

A 1-cubic-centimeter cube of sodium reacts more rapidly in water at 25°C than does a 1-cubic-centimeter cube of calcium at 25°C.

This difference in rate of reaction is most closely associated with the different
A) surface area of the metal cubes
B) nature of the metals
C) density of the metals
D) concentration of the metals
Chemistry
2 answers:
Tamiku [17]3 years ago
4 0

As given that the

a) temperature is same for both the metals

b) the volume of both the cubes is same (Same dimension)

The metals are sodium and calcium

Sodium has atomic number = 11

the electronic configuration is 1s2 2s2 2p6 3s1

It will attain noble gas configuration by easily losing one electron only

calcium has atomic number = 20

the electronic configuration is 1s2 2s2 2p6 3s2 3p6 4s2

It will attain noble gas configuration by donating two electrons so calcium is less reactive than sodium

hence the difference in the rate of reaction is most closely associated with the different nature of the metals

ki77a [65]3 years ago
3 0
B) nature of the metals
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When 2.0 mol of methanol is dissolved in 45 grams of water, what is the mole fraction of methanol
kaheart [24]

The mole fraction of methanol in the mixture is 0.444

We'll begin by calculating the number of mole of water.

  • Mass of water = 45 g
  • Molar mass of water = 18 g/mol
  • Mole of water =?

Mole = mass / molar mass

Mole of water = 45 / 18

Mole of water = 2.5 moles

Finally, we shall determine the mole fraction of methanol.

  • Mole of water = 2.5 moles
  • Mole of methanol = 2 moles
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Mole fraction of methanol =?

Mole fraction = mole / total mole

Mole fraction of methanol = 2 / 4.5

Mole fraction of methanol = 0.444

Thus, the mole fraction of methanol is 0.444

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The reaction of perchloric acid (HClO4) with lithium hydroxide (LiOH) is described by the equation: HClO4 + LiOH → LiClO4 + H2O
dem82 [27]

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A. 0.35 M

Explanation:

Hello,

In this case, given the volume and concentration of lithium hydroxide and the volume of chloric acid, we can compute the concentration of the neutralized acid by using the following equation:

n_{acid}=n_{base}\\\\V_{acid}M_{acid}=V_{base}M_{base}\\\\M_{acid}=\frac{V_{base}M_{base}}{V_{acid}} =\frac{46.9mL*0.75M}{100mL}\\ \\M_{acid}=0.35M

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Now, you have to have the same number of electrons in the two half-reactions, so multiply the second one by 2 to get:

2 Ag+ + 2 e- --> 2 Ag

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