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mestny [16]
2 years ago
6

How many PO43? ions are in a mole of K3PO4?

Chemistry
1 answer:
adoni [48]2 years ago
7 0

K₃PO₄ → 3K⁺ (aq) + PO₄³⁻(aq)

One mole of PO₄³⁻ ion gets dissociated from one mole of K₃PO₄

As per the definition of Avogadro's number, 1 mole = 6.022 x 10²³ ions

One mole of PO₄³⁻ ions x  (6.022 x 10²³ ions/ 1 mole of PO₄³⁻ ions )

= 6.022 x 10²³ ions

Therefore , there are 6.022 x 10²³ PO₄³⁻ ions in a mole of K₃PO₄.

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How many grams are in 1.6 moles of potassium bromide?​
slava [35]

Answer:

190.4g

Explanation:

1.6mol of KBr (119.002g KBr/1 mol) = 190.4g

since you want to find grams, take the molar mass of KBr (119.002) per 1 mol and use it as your conversion factor (119.002g KBr/1 mol) which will then cancel out mols and leave you with grams.

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3 years ago
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7 0
3 years ago
Explain why the atomic radius of aluminium is smaller than that of sodium.
garik1379 [7]
Okay, to explain this you might have to grab a periodic table.

Do you have one? Good. Look at the most left side of the periodic table. The first group is the largest atoms in the periodic table. If you go to the right of the periodic table, the atoms get progressively smaller and smaller.

Why is this? Don't atoms get more electrons, and so become significantly bigger as they move to the right?

Although atoms do get more electrons as they go to the right, they also get more protons too. Protons pull on electrons and make atoms smaller. Because of this, going from left to right in a periodic table makes the atoms smaller and smaller, since more and more protons are added.

In this scenario, Aluminum is more to the right than Sodium, which means that it has more protons. Because of this, the protons in Aluminum pull more strongly on electrons than sodium, thus making aluminum smaller.

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6 0
3 years ago
The pH of a 0.29 M solution of carbonic acid (H2CO3) is measured to be 3.44. Calculate the acid dissociation constant Ka of carb
arsen [322]

Answer: 4.55x10^-7

Explanation:

H2CO3 + H20 <==> H3O+ + HCO3-

Desigining an ICE table, we have:

Initial conc. of H2CO3 = 0.29 M

Initial conc. of H3O+ = 0

Initial conc. of HCO3- = 0

Change in conc. of H2CO3 = - x

Change in conc. of H3O+ = x

Change in conc. of HCO3- = x

Equilibrium conc. of H2CO3 = 0.29 - x

Equilibrium conc. of H3O+ = x

Equilibrium conc. of HCO3- = x

but pH = 3.44

pH = - log[H30+]

[H30+] = 10^-3.44 = 3.63x10^-4

[H30+] = 3.63x10^-4

Ka = [H3O+].[HCO3-] / [H2CO3]

[H30+] = x = 3.63x10^-4

[HCO3-] = 3.63x10^-4

[H2CO3] = 0.29 - x = 0.29 - 3.63x10^-4 = 0.289637

Ka = (3.63x10^-4)(3.63x10^-4)/0.289637 = 4.55x10^-7

6 0
3 years ago
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