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Tanya [424]
3 years ago
13

The accepted value for the percent by mass of water in a hydrate is 36.0%. In a laboratory activity, a student determined the pe

rcent by mass of water in the hydrate to be 37.8%. What is the percent error for the student's measured value?
(1) 5.0% (3) 1.8%

(2) 4.8% (4) 0.05%
Chemistry
1 answer:
valentina_108 [34]3 years ago
4 0
The original or accepted value for the percent by mass of water in a hydrate = 36%
Percen by mass of water in the hydrate determined by the student
in the laboratory = 37.8%
So the difference between the actual and the percent by mass in water determined by student = (37.8 - 36.0)%
             = 1.8%
So the percentage of error made by the student = (1.8/36) * 100 percent
                                                                             = (18/360) * 100 percent
                                                                             = ( 1/20) * 100 percent
                                                                              = 5 percent
So the student makes an error of 5%. Option "1" is the correct option.
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Calculate the standard molar enthalpy for the complete combustion of liquid ethanol (C2H5OH) using the standard enthalpies of fo
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Answer:

Explanation:

For the reaction

C2H5OH (l) + 3 O2(g) = 2CO2(g) + 3 H2O

We can calculate the  standard molar enthalpy of combustion using the standard enthalpies of formation of the species involved in the reaction according to Hess law:

ΔHºc =  2ΔHºf CO2 (g) + 3ΔHºfH2O(l)  - ( ΔHºf C2H5OH (l) - 3ΔHºfO2 (g) )

( we were not give the water state but we know we are at standard conditions so it is in its liquid state )

The ΔHºfs can be found in appropiate reference or texts.

ΔHºc =  2ΔHºf CO2 (g)+ 3ΔHºfH2O(l)  - (  ΔHºf C2H5OH (l) -+3ΔHºfO2 (g) )

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5 0
3 years ago
0.53g of acetanilide was subjected to kjeldahl determination and the ammonia produced was collected in 50cm3 of 0.50M of h2so4.o
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Answer:

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The acid in excess in titrated with Na₂CO₃ as follows:

Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂

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<em>Moles Na₂CO₃ anf Moles H₂SO₄ in excess:</em>

0.025L * (0.05mol / L) = 1.25x10⁻³ moles Na₂CO₃ / 0.01360L =

0.09191M * 0.250L = 0.0230 moles H₂SO₄ in excess.

<em>Moles H₂SO₄ added:</em>

0.050L * (0.50mol / L) = 0.0250 moles H₂SO₄ added

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0.0250 moles - 0.0230 moles = 0.0020 moles H₂SO₄

<em>Moles of NH₃ = Moles N:</em>

0.0020 moles H₂SO₄ * (2mol NH₃ / 1mol H₂SO₄) = 0.0040 moles NH₃ = Moles N

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<h3>10.57% of N in acetanilide</h3>
7 0
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7 0
3 years ago
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