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storchak [24]
3 years ago
14

If 30g of HCl is reacted with excess NaOH, and 10g of NaCl is produced, what is the theoretical yield of the experiment?

Chemistry
1 answer:
KATRIN_1 [288]3 years ago
4 0
1. %yield = actual yield / theoretical yield x 100
               = 3 g / 4. 46 g x 100
               = 62.11 % percent yield

Hope This Helps! :3
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Lunar Eclipe's do not happen every month because the moon, earth and sun must be lined up in a certain way.
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To protect yourself from consuming too much _____, you should limit your consumption of swordfish, king mackerel, and shark. a.
Serggg [28]

Answer:

b. mercury

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How many mole of ZnCl2 will be produced from 55.0 g of Zn, assuming HCL is available in excess
Ilia_Sergeevich [38]

Answer:

0.84 mol

Explanation:

Given data:

Moles  of ZnCl₂ produced = ?

Mass of Zn = 55.0 g

Solution:

Chemical equation:

2HCl + Zn  →  ZnCl₂ + H₂

Number of moles of Zn:

Number of moles = mass / molar mass

Number of moles = 55.0 g/ 65.38 g/mol

Number of moles = 0.84 mol

Now we will compare the moles of Zn with ZnCl₂ from balance chemical equation.

                                      Zn          :             ZnCl₂

                                         1          :               1

                                      0.84       :           0.84

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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

7 0
3 years ago
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