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natulia [17]
3 years ago
11

The density of pure water is 1.0 g/ml. Four students measured and calculated the density of pure water samples. The results are

as follows: student 1 got 0.85
g/ml, student 2 got 0.94 g/ml, student 3 got 0.95 g/ml, and student 4 got 1.3 g/ml. Which student had the smallest error in their measurement?

A. Student 3
B. Student 2
C. Student 4
D. Student 1
Chemistry
1 answer:
balandron [24]3 years ago
5 0

Answer:

Student 4

Explanation:

Student 1: 15

student 2 : 6

Student 3: 5

Student 4: 3

P.S: Just guessed

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<u>Answer:</u> The heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H

The equation used to calculate enthalpy change is of a reaction is:

\Delta H_{rxn}=\sum [n\times \Delta H_{(product)}]-\sum [n\times \Delta H_{(reactant)}]

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C_4H_6(g)+2H_2(g)\rightarrow C_4H_{10}(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(1\times \Delta H_{(C_4H_{10})})]-[(1\times \Delta H_{(C_4H_6)})+(2\times \Delta H_{(H_2)})]

We are given:

\Delta H_{(C_4H_{10})}=-2877.6kJ/mol\\\Delta H_{(C_4H_6)}=-2540.2kJ/mol\\\Delta H_{(H_2)}=-285.8kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(1\times (-2877.6))]-[(1\times (-2540.2))+(2\times (-285.8))]\\\\\Delta H_{rxn}=234.2J

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3 years ago
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In the equilibrium constant expression for the reaction below what is the correct exponent for N2O4?
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5 0
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