Answer:
8740 joules are required to convert 20 grams of ice to liquid water.
Explanation:
The amount of heat required (
), measured in joules, to convert ice at -50.0 ºC to liquid water at 0.0 ºC is the sum of sensible heat associated with ice and latent heat of fussion. That is:
(1)
Where:
- Mass, measured in grams.
- Specific heat of ice, measured in joules per gram-degree Celsius.
,
- Temperature, measured in degrees Celsius.
- Latent heat of fussion, measured in joules per gram.
If we know that
,
,
,
and
, then the amount of heat is:
![Q = (20\,g)\cdot \left\{\left(2.06\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot [0\,^{\circ}C-(-50\,^{\circ}C)]+334\,\frac{J}{g} \right\}](https://tex.z-dn.net/?f=Q%20%3D%20%2820%5C%2Cg%29%5Ccdot%20%5Cleft%5C%7B%5Cleft%282.06%5C%2C%5Cfrac%7BJ%7D%7Bg%5Ccdot%20%5E%7B%5Ccirc%7DC%7D%20%5Cright%29%5Ccdot%20%5B0%5C%2C%5E%7B%5Ccirc%7DC-%28-50%5C%2C%5E%7B%5Ccirc%7DC%29%5D%2B334%5C%2C%5Cfrac%7BJ%7D%7Bg%7D%20%5Cright%5C%7D)

8740 joules are required to convert 20 grams of ice to liquid water.
Answer:
The answer to your question is V2 = 434.7 l
Explanation:
Data
Volume 1 = V1 = 240 l Volume 2 = ?
Temperature 1 = T1 = 479°K Temperature 2 = T2 = 293°K
Pressure 1 = P1 = 300 KPa Pressure 2 = P2 = 101.325 Kpa
Process
1.- Use the combined gas law to solve this problem
P1V1/T1 = P2V2/t2
-Solve for V2
V2 = P1V1T2 / T1P2
2.- Substitution
V2 = (300)(240)(293) / (479)(101.325)
3.- Simplification
V2 = 21096000 / 48534.675
4.- Result
V2 = 434.7 l
Answer:
In chemical nomenclature, the IUPAC nomenclature of organic chemistry is a systematic method of naming organic chemical compounds as recommended by the International Union of Pure and Applied Chemistry (IUPAC). ... IUPAC names can sometimes be simpler than older names, as with ethanol, instead of ethyl alcohol.
Explanation:
In chemical nomenclature, the IUPAC nomenclature of organic chemistry is a systematic method of naming organic chemical compounds as recommended by the International Union of Pure and Applied Chemistry (IUPAC). ... IUPAC names can sometimes be simpler than older names, as with ethanol, instead of ethyl alcohol.
Answer: The laboratory value of potassium (3.0 mmol / L) is consistent with the client's symptoms of hypokalemia.
Explanation:
Hello!
Let's solve this!
Hypokalemia is a disorder in the body's electrolyte balance, when the decrease in blood potassium (K) ion levels is below 3.5 mmol / L. Potassium losses can occur through the digestive tract: such as vomiting and
diarrhea The most frequent symptoms of potassium loss include: tiredness, muscle weakness and cramping.
In conclusion, the laboratory value of potassium (3.0 mmol / L) is consistent with the client's symptoms of hypokalemia.