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postnew [5]
3 years ago
5

Consider a solution that contains 60.0% r isomer and 40.0% s isomer. if the observed specific rotation of the mixture is –43.0°,

what is the specific rotation of the pure r isomer?
Chemistry
1 answer:
pychu [463]3 years ago
5 0
The solution for this problem is:
Let x denote the specific rotation, R; andLet y denote the specific rotation, S = -x 
Solution:60 x - 40 x/100 = - 43
20x = - 4300Divide both sides by 20The answer is:x = - 215 is the specific rotation of the pure r isomer.
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Using standard heats of formation, calculate the standard enthalpy change for the following reaction. 2H2S(g) 3O2(g)2H2O(l) 2SO2
Marizza181 [45]

Answer:

\Delta _rH=-1124.14kJ/mol

Explanation:

Hello!

In this case, since the standard enthalpy change for a chemical reaction is stood for the enthalpy of reaction, for the given reaction:

2H_2S(g) +3O_2(g)\rightarrow 2H_2O(l) +2SO_2(g)

We set up the enthalpy of reaction considering the enthalpy of formation of each species in the reaction at the specified phase and the stoichiometric coefficient:

\Delta _rH=2\Delta _fH_{H_2O,liq}+2\Delta _fH_{SO_2,gas}-2\Delta _fH_{H_2S,gas}-3\Delta _fH_{O_2,gas}

In such a way, by using the NIST database, we find that:

\Delta _fH_{H_2O, liq}=-285.83kJ/mol\\\\\Delta _fH_{SO_2, gas}=-296.84kJ/mol\\\\\Delta _fH_{O_2,gas}=0kJ/mol\\\\\Delta _fH_{H_2S,gas}=-20.50kJ/mol

Thus, we plug in the enthalpies of formation to obtain:

\Delta _rH=2(-285.73kJ/mol)+2(-296.84kJ/mol)-2(-20.50kJ/mol)-3(0kJ/mol)\\\\\Delta _rH=-1124.14kJ/mol

Best regards!

8 0
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Which of the following best describes an example of applied chemistry?
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Answer:

Explanation:

Examples of applied chemistry include creation of the variety of laundry detergents on the market and development of oil refineries.

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Learn more about ethyl alcohol: brainly.com/question/1049383

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