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Blababa [14]
4 years ago
7

assume that you used a 0.660 g piece of pure aluminum in the synthesis of alum and obtained a 63.5% yield of alum. How many gram

s of alum did you obtain?
Chemistry
1 answer:
attashe74 [19]4 years ago
8 0

Answer:

1.015g

Explanation:

From the question,

Mass of Al used = 0.660g

The percentage yield = 63.5%

Mass of Alum used = y

To find the mass of Alum used, do the following:

%yield = Mass of Al /Mass and Alum

65% = 0.660/y

65/100 = 0.660/y

Cross multiply

65 x y = 0.660x100

Divide both side by 65

y = (0.660x100) / 65

y = 1.015g

The mass of alum obtained is 1.015g

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3 years ago
Which statement is true about a liquid but not a gas?
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Answer:

I think its D

Explanation:

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3 years ago
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A chemist must dilute 34.3mL of 1.72mM aqueous calcium sulfate solution until the concentration falls to 1.00mM. He'll do this b
EastWind [94]

Answer:

58.9mL

Explanation:

Given parameters:

Initial volume  = 34.3mL    = 0.0343dm³

Initial concentration  = 1.72mM   = 1.72 x 10⁻³moldm⁻³

Final concentration  = 1.00mM = 1 x 10⁻³ moldm⁻³

Unknown:

Final volume  =?

Solution:

Often times, the concentration of a standard solution may have to be diluted to a lower one by adding distilled water. To find the find the final volume, we must recognize that the number of moles of the substance in initial and final solutions are the same.

   Therefore;

              C₁V₁  =  C₂V₂

where C and V are concentration and 1 and 2 are initial and final states.

        now input the variables;

                      1.72 x 10⁻³ x  0.0343 = 1 x 10⁻³  x V₂

                        V₂ = 0.0589dm³ = 58.9mL

         

4 0
3 years ago
What is the volume of 0.80 grams of o2 gas at stp? (5 points) group of answer choices 0.59 liters 0.56 liters 0.50 liters 0.47 l
Vladimir [108]

Answer:

0.56L

Explanation:

This question requires the Ideal Gas Law:  PV=nRT where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the Ideal Gas constant, and T is the Temperature of the gas.

Since all of the answer choices are given in units of Liters, it will be convenient to use a value for R that contains "Liters" in its units:R=0.0821\frac{L\cdot atm}{mol\cdot K}

Since the conditions are stated to be STP, we must remember that STP is Standard Temperature Pressure, which means T=273.15K and P=1atm

Lastly, we must calculate the number of moles of O_2(g) there are.  Given 0.80g of O_2(g), we will need to convert with the molar mass of O_2(g).  Noting that there are 2 oxygen atoms, we find the atomic mass of O from the periodic table (16g/mol) and multiply by 2:  32g\text{ }O_2=1mol\text{ }O_2

Thus, \frac{0.80g \text{ }O_2}{1} \frac{1mol\text{ }O_2}{32g\text{ }O_2}=0.25mol\text{ }O_2=n

Isolating V in the Ideal Gas Law:

PV=nRT

V=\frac{nRT}{P}

...substituting the known values, and simplifying...

V=\frac{(0.025 mol \text{ }O_2)(0.0821\frac{L\cdot atm}{mol \cdot K} )(273.15K)}{(1atm)}

V=0.56L \text{ } O_2

So, 0.80g of O_2(g) would occupy 0.56L at STP.

5 0
2 years ago
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A certain amount of hydrogen peroxide was dissolved in 100. mL of water and then titrated with 1.68 M KMnO4. What mass of H2O2 w
Stels [109]

Answer:

see explanation below

Explanation:

First to all, this is a redox reaction, and the reaction taking place is the following:

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According to this reaction, we can see that the mole ratio between the peroxide and the permangante is 5:2. Therefore, if the titration required 21.3 mL to reach the equivalence point, then, the moles would be:

MhVh = MpVp

h would be the hydrogen peroxide, and p the permanganate.

But like it was stated before, the mole ratio is 5:2 so:

5MhVh = 2MpVp

Replacing moles:

5nh = 2MpVp

Now, we just have to replace the given data:

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Now to get the mass, we just need the molecular mass of the peroxide:

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Finally the mass:

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8 0
3 years ago
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