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maw [93]
2 years ago
7

You have 150. g of a bleach solution. The percent by mass of the solute, sodium hypochlorite (NaOCI), is 3.62%.

Chemistry
1 answer:
Taya2010 [7]2 years ago
4 0

Answer:

There are 5.43 grams of NaOCl

Explanation:

The given percent by mass of the solute tells us that out of the 150 g of the solution, 3.62% are due solely to the solute.

In other words, <u>the mass of the solute in the solution is</u>:

  • 150 g * 3.62/100 = 5.43 g

Thus, in 150 grams of the given bleach solution, there are 5.43 grams of sodium hypochlorite.

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"How much NH4Cl, when present in 2.00 liters of 0.200 M ammonia, will give a solution with pH = 8.20? For NH3, Kb = 1.8 x 10-5"
Andru [333]

Answer:

245.66g of NH₄Cl is the mass we need to add to obtain the desire pH

Explanation:

The mixture of NH3/NH4Cl produce a buffer. We can find the pH of a buffer using H-H equation:

pH = pKa + log [A⁻] / [HA]

<em>Where [A⁻] is the molar concentration of the base, NH₃, and [HA] molar concentration of the acid, NH₄⁺. This molar concentration can be taken as the moles of each chemical</em>

<em />

First, we need to find pKa of NH₃ using Kb. Then, the moles of NH₃ and finally replace these values in H-H equation to solve moles of NH₄Cl we need to obtain the desire pH.

  • <em>pKa NH₃/NH₄⁺</em>

pKb = - log Kb

pKb = -log 1.8x10⁻⁵ = <em>4.74</em>

pKa = 14 - pKb

pKa = 14 - 4.74

pKa = 9.26

  • <em>Moles NH₃</em>

<em>2.00L ₓ (0.200mol NH₃ / L) = 0.400 moles NH₃</em>

  • <em>H-H equation:</em>

pH = pKa + log [NH₃] / [NH₄Cl]

8.20 = 9.26 + log [0.400 moles] / [NH₄Cl]

-1.06 =  log [0.400 moles] / [NH₄Cl]

0.0087 =  [0.400 moles] / [NH₄Cl]

[NH₄Cl] = 0.400 moles / 0.0087

[NH₄Cl] = 4.59 moles of NH₄Cl we need to add to original solution to obtain a pH of 8.20. In grams (Using molar mass NH₄Cl=53.491g/mol):

4.59 moles NH₄Cl ₓ (53.491g / mol) =

<h3>245.66g of NH₄Cl is the mass we need to add to obtain the desire pH</h3>

<em />

3 0
3 years ago
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Answer:

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Explanation:

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3 0
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Wewaii [24]
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6 0
3 years ago
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denpristay [2]
The reaction is as follows:
2 H₂(g) + O₂(g) → 2 H₂O(g) . ΔH = - 483.5 kJ
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7 0
3 years ago
if you have 3.50 moles of hydrogen and 5.00moles of nitrogen to produce ammonia, witch element is the reactant in excess? 3H2+N2
hram777 [196]
Nitrogen is the N and that is what is the element is the reactant in excess.
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