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adelina 88 [10]
3 years ago
8

Write the conventional, total ionic, and net ionic equations for Mg(s)+H2O(l)

Chemistry
1 answer:
mojhsa [17]3 years ago
5 0

Answer:

2Mg(s) + 2H2O(l) =======> 2Mg2+(aq) + 4OH-(aq)

Explanation:

2Mg(s) + 2H2O(l) =======>2Mg(OH)2(aq)

But

2Mg(OH)2(aq) ========> 2Mg2+(aq) + 4OH-(aq)

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ycow [4]

Mole is base unit of substance. The number of particles in one mole is equal to 6.022 × 10²³. The number of particle in a mole is called as Avogadro's number.

The mole of substance is the amount of substance that contain elementary particles as the number of atoms. the symbol of mole is mol. it is the SI unit of amount of substance. The number particles are ion one mole of substances is 6.022 × 10²³. The number of particles in one mole of substance is called as Avogadro's number.

Thus, Mole is base unit of substance. The number of particles in one mole is equal to 6.022 × 10²³. The number of particle in a mole is called as Avogadro's number.

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!!!!HELP PLEASE!!!!!Which of the following molecules has a trigonal pyramidal shape?
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Why knowing the density of water important?? explain why please so I can understand it don’t copy for the internet I won’t under
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it allows us to determine whether objects will float or sink when placed in a liquid or even a gas.

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Fusion reactors are more desirable than fission because:
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An unknown gaseous substance has a density of 1.06 g/L at 31 °C and 371 torr. If the substance has the following percent composi
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Answer:

C) C4H6 - Right answer

Explanation:

Let's combine the Ideal Gases Law with density to get the molecular formula for the unknown gas.

Density = mass / volume

1.06 g /L means that 1.06 grams of compound occupy 1 liter of volume.

P . V = n . R . T

Pressure in Torr must be converted to atm

760 Torr are 1 atm

371 Torr  are __ (371 .1)/760 = 0.488 atm

0.488 atm . 1L = 1.06g/MM . 0.082 . 304K

(0.488 atm . 1L) / 0.082 . 304K = 1.06g/ MM

Mass / Molar mass = Moles → That's why the 1.06 g / MM

0.0195 mol = 1.06g / MM

1.06g/0.0195 mol = MM →  54.3 g/m

Now, let's use the composition

100 g of compound have 88.8 g of C

54.3 g of compound have ___ (54.3  . 88.8) /100 = 48 g of C

100 g of compound have 11.2 g of H

54.3 g of compound have __ (54.3  .  11.2)/100 = 6 g of H

48 g of C are included un 4 atoms

6 g of H are included in 6 atoms

4 0
3 years ago
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