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Mashcka [7]
3 years ago
8

What are the respective concentrations (moles/liter) of k+ and po43- afforded by dissolving 0.800 mol k3po4 in water and dilutin

g to 1.63l?
Chemistry
1 answer:
Kitty [74]3 years ago
8 0

<em>Answer: </em>

  •                      Concentration of K+  = 1.47 M
  •                      Concentration of Po4∧-2 = 0.4908 M

<em>Given Data:</em>

   No. of moles of K3PO4 = 0.800 mole

  Molarity of K3PO4 =  0.800÷1.63 = 0.4908

<em>Chemical equations:</em>

        K3PO4 ⇔3 K+   +  PO4∧-2

<em>Solution: </em>

         K3PO4 : K+                                                K3PO4 : PO4∧-2

            1  :    3                                                            1      :    1

        0.4908 = 0.4908 × 3 = 1.47 M                   0.4908 = 0.4908 × 1 = 0.4908 M

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Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of sodium sulfa
notka56 [123]

Answer:

27%

Explanation:

Hello,

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

2NaOH+H_2SO_4-->Na_2SO_4+2H_2O

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4

Finally, the percent yield turns out into:

Y=\frac{1.92g}{7.1g} *100

Y=27.0%

Best regards.

6 0
3 years ago
Question 24 (1 point)
aleksley [76]

Answer:

Explanation:

Idk

4 0
3 years ago
How can you read a determine the masses
krok68 [10]
I need you to explain the question
4 0
3 years ago
A compound is 2% H, 32.7% S, and 65.3% O by mass. What is the subscript on the O in the empirical formula for this compound?
IceJOKER [234]
Since all of those percents add up to 100, you can just directly convert that to grams. So now you can use 2 grams H, 32.7 grams S, and 65.3 grams O. Use that info and convert that to moles for an answer of 2mol H, 1mol S, and 4mol O. In every empirical question you need to divide each quantity of moles by the lowest number. In this case, that number is one, so they stay the same, but it's important to remember that step. You're final chemical formula would be H2SO4 and the answer to your question would be that the subscript for oxygen is 4. Hope this helped! 
4 0
3 years ago
A liquid dietary supplement is packaged in 10-ml dropper containers to deliver 2000 international units of vitamin d3 in each dr
RoseWind [281]

Answer : 37 drops are delivered per milliliter of the solution.

Explanation :

The problem gives us lot of extra information.

We want to find the number of drops delivered in 1 milliliter here.

We have been given that, one drop of the solution delivers 0.027 mL of solution.

Let us use this as a conversion factor, \frac{1 drop}{0.027 mL}

Let us find number of drops in 1 mL using this conversion factor.

1 mL \times \frac{1 drop}{0.027mL} = 37.0 drops

Therefore we can say that 37 drops are delivered per milliliter of the solution.

7 0
3 years ago
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