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bezimeni [28]
3 years ago
11

How many moles of oxygen are needed to completely react with 9.5 g of sodium

Chemistry
1 answer:
maria [59]3 years ago
8 0
<span>4 Na + O</span>₂<span> = 2 Na</span>₂<span>O
</span>
4* 23 g Na --------> 16 g O₂
9.5 g Na ------------> ?

Mass of O₂ = 9.5 * 16 / 4 * 23

Mass = 152 / 92

Mass = 1.6521 g of O₂

Molar mass O₂ = 16.0 g/mol

1 mole O₂ ------------ 16.0 g 
? mole O₂ ------------ 1.6521 g

mole O₂ = 1.6521 * 1 / 16.0

≈ 0.10325 moles of O₂

hope this helps!

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QUICK! Which best explains how the collisions of materials in space contribute to the formation of layers in protoplanets?
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3 years ago
Among the following options, a valid Lewis structure of __________ cannot be drawn without violating the octet rule.
Murrr4er [49]

Answer:

d. IF3

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From the given options, a valid Lewis structure that cannot be drawn without violating the octet rule is IF3

8 0
3 years ago
How many moles of Cu(OH)2 are soluble in 1L of sodium hydroxide (NaOH) when the pH is 8.23?
Morgarella [4.7K]

Answer:

4.96E-8 moles of Cu(OH)2

Explanation:

Kps es the constant referring to how much a substance can be dissolved in water. Using Kps, it is possible to know the concentration of weak electrolytes. Then, pKps is the minus logarithm of Kps.

Now, we know that sodium hydroxide (NaOH) is a strong electrolyte, who is completely dissolved in water. Therefore the pH depends only on OH concentration originating from NaOH. Let us to figure out how much is that OH concentration.

pH= -log[H]\\pH= -log (\frac{kw}{[OH]})

8.23 = - log(\frac{Kw}{[OH]} \\10^{-8.23} = Kw/[OH]\\ [OH] = Kw/10^{-8.23}

[OH]=1.69E-6

This concentration of OH affects the disociation of Cu(OH)2. Let us see the dissociation reaction:

Cu(OH)_2 -> Cu^{2+} + 2OH^-

In the equilibrum, exist a concentration of OH already, that we knew, and it will be added that from dissociation, called "s":

The expression for Kps is:

Kps= [Cu^{2+}] [OH]^2

The moles of (CuOH)2 soluble are limitated for the concentration of OH present, according to the next equation.

Kps= s*(2s+1.69E-6)^2

"s" is the soluble quantity of Cu(OH)2.

The solution for this third grade equation is s=4.96E-8 mol/L

Now, let us calculate the moles in 1 L:

moles Cu(OH)_2 = 4.96E-8 mol/L * 1 L = 4.96E-8 moles

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