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stellarik [79]
3 years ago
14

Molar mass of K2SO4 = 174.0 g/mol

Chemistry
1 answer:
Nastasia [14]3 years ago
8 0
Not sure what you need to know, but if it’s true or false it would be false. it’s actually 174.26
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For the reaction NH4NO3 (s) → N2O (g) + 2H2O (l), you decompose 1 mole NH4NO3 and only get 0.75 moles of N2O. The percent yield
attashe74 [19]

Answer:

The answer to your question is False.

Explanation:

Data

1 mole of NH₄NO₃

0.75 moles of N₂O

Percent yield = 25%

Chemical reaction

               NH₄NO₃   ⇒   N₂O  +  2H₂O

Process

1.- Determine the theoretical yield

              1 mol NH₄NO₃ ------------- 1 mol of N₂O

2.- Calculate the percent yield

             Percent yield = Actual yield / Theoretical yield  x 100

-Substitution

              Percent yield = 0.75 / 1 x 100

-Simplification

              Percent yield = 0.75 x 100

-Result

              Percent yield = 75%

Conclusion

False, the actual percent yield is 75%

5 0
3 years ago
What medium can your voice travel through?
Nookie1986 [14]

Sound waves are most effective when traveling through solids. Less effective when in liquids and least effective in air.

3 0
3 years ago
Read 2 more answers
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
What is the most striking part of the Rutherford scattering stimulation?
Sati [7]

Answer:

None of the alpha particles fired at the foil are being repelled back, like they were in the Rutherford atom simulation.I hope this is correct.

3 0
4 years ago
According to the below equation, how many moles of SO2 are required to generate 1.43×1024 water molecules?
podryga [215]
Make sure the equation is always balanced first. (It is balanced for this question already) 6.022 x 10^23 is Avogadro’s number. In one mole of anything there is always 6.022 x 10^23 molecules, formula units, atoms. For one mol of an element/ compound use molar mass (grams).

Multiply everything on the top = 8.61x10^47
Multiple everything on bottom= 1.20x10^24
Divide top and bottom = 7.15x10^23

Answer: 7.15x10^23 mol SO2
3 0
4 years ago
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