The change in energy of the system : -63 J
<h3>Further explanation</h3>
Given
279 J work
216 J heat
Required
The change in energy
Solution
Laws of thermodynamics 1
ΔU=Q+W
Rules :
- receives heat, Q +
- releases heat, Q -
- work is done by a system, W -
- work is done on a system, W +
a gas work on the surrounding : W =-279 J
a gas absorb heat from surrounding : Q = +216 J
Internal energy :
= -279+216
= -63 J
Answer:
a) 72 °F= 22.22 °C
b) 213.8 °C= 416.84°F
c) 180 °C= 453.15 °K
d) 315 °K= 107.33 °F
e) 1750 °F= 1227.594 °K
f) 0 °K= -459.67 °F
Explanation:
Para realizar el intercambio de unidades debes tener en cuenta las siguientes conversiones:
- Fahrenheit a Celsius:

- Celsius a Fahrenheit: °F= °C*1.8 + 32
- Celsius a Kelvin: °K= °C + 273.15
- Kelvin a Fahrenheit: F= (K -273.15)*1.8 + 32
- Fahrenheit a Kelvin:

Entonces se obtiene:
a) 72 °F=
=22.22 °C
b) 213.8 °C= 213.8*1.8 + 32= 416.84°F
c) 180 °C= 180°C + 273.15= 453.15 °K
d) 315 °K= (315 -273.15)*1.8 + 32= 107.33 °F
e) 1750 °F=
= 1227.594 °K
f) 0 °K= (0 -273.15)*1.8 + 32= -459.67 °F
The answer is D because large hydrocarbons have a higher number of carbon which means an increase in boiling point and also rate of flow decreases, so they become solids.
Answer:
Trimethylacetaldehyde
Explanation:
For the <u>unknow compound</u> we have a molar mass of 86 g/mol. We have an even value for the mass, so the compound does <u>not have nitrogen</u> and we can have several posibilities:
A) 
B) 
C) 
D) 
If we check the IR info a signal in 1730 cm-1 appears, this indicates that we have an <u>oxo group</u> (C=O). So, the D option can be discarded. The groups that can have the oxo group are: Carboxylic acids, <u>Ketones and aldehydes</u>.
We don have a signal in 3000 cm-1, so the carboxylic acid can be discarded. Now, is we check the info for the 1H NMR we only have 2 signals. If we only have 2 we will have a very<u> symmetric compound</u>.
By trial an error the find the compound <u>Trimethylacetaldehyde</u> (Figure 1).