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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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What is the concentration of a phosphoric acid solution of a 25.00 mL sample if the acid requires 42.24 mL of 0.135 M NaOH for n
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0.0760 m

do this by:

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8 0
4 years ago
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The periodic table of the elements lists the elements in order of increasing atomic number. Each element has its own unique squa
Vesna [10]

Answer:

A) They all have 12 protons.

Explanation:

Magnesium has an atomic number of 12 which means it has 12 protons. All Magnesium atoms have 12 protons, the neutrons however may differ which produces things called isotopes where the atoms have same protons but neutrons change.

Keep in mind the atomic number is unique to each element, so 12 atomic number will always be Magnesium, 1 will always be Hydrogen and so on.....

6 0
3 years ago
If 28.0 grams of Pb(NO3)2 react with 18.0 grams of NaI, what mass of PbI2 can be produced? Pb(NO3)2 + NaI → PbI2 + NaNO3
ss7ja [257]

Answer:- 27.7 grams of PbI_2 are produced.

Solution:- The balanced equation is:

Pb(NO_3)_2+2NaI\rightarrow PbI_2+2NaNO_3

let's convert the grams of each reactant to moles and calculate the grams of the product and see which one gives least amount of the product. This least amount would be the answer as the least amount we get is from the limiting reactant.

Molar mass of Pb(NO_3)_2 = 207.2+2(14.01)+6(16)  = 331.22 gram per molmolar mass of NaI = 22.99+126.90 = 149.89 gram per molMolar mass of [tex]PbI_2 = 207.2+2(126.90) = 461 gram per mol

let's do the calculations for the grams of the product for the given grams of each of the reactant:

28.0gPb(NO_3)_2(\frac{1molPb(NO_3)_2}{331.22gPb(NO_3)_2})(\frac{1molPbI_2}{1molPb(NO_3)_2})(\frac{461gPbI_2}{1molPbI_2})

= 39.0gPbI_2

18.0gNaI(\frac{1molNaI}{149.89gNaI})(\frac{1molPbI_2}{2molNaI})(\frac{461gPbI_2}{1molPbI_2})

= 27.7gPbI_2

From above calculations, NaI gives least amount of PbI_2, so the answer is, 27.7 g of PbI_2 are produced.

8 0
3 years ago
Can someone please explain this to me? I need help ​
Vladimir [108]

Answer:

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Explanation:

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Here is a list of all the types of waves in order from shortest wavelength to longest wavelengths:

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6 0
4 years ago
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