The question is incomplete. The complete question is:
A certain molecular compound M has a solubility in acetone of 0.737g/mL at 10.°C
.Calculate the mass of M that's dissolved in 9.0 L of a saturated solution of M in acetone at this temperature.
Be sure your answer has the correct unit symbol and number of significant digits.
Answer: 6633 g of M is dissolved in 9.0 L of saturated solution in hexane at this temperature.
Explanation:
Given : 0.737g of solute is present in 1ml of solution
i.e 1ml of solution contains = 0.737 g of solute
Thus 9L or 9000ml of solution contains=
Thus 6633 g of M is dissolved in 9.0 L of saturated solution in hexane at this temperature.
In order to gauge the feedback mechanism of the way your hormone responds, you need at least three hours to monitor the situation.
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Answer:
The elements in each group have the same number of electrons in the outer orbital. Those outer electrons are also called valence electrons. They are the electrons involved in chemical bonds with other elements.
Explanation:
Answer:
The electrons are found on the atomic nucleus.
Explanation:
Answer:
The human body when metabolizing meat, milk or eggs, being high in protein, manufactures uric acid as a metabolite, and releases more urea in the urine.
Uric acid generates an increased retention of liquid or water, that is why people who ingest excessively meat suffer from GOTA disease.
Gout disease is systemic and life-threatening puffiness.
Furthermore uric acid is an organic compound of carbon, nitrogen, oxygen and hydrogen. Its chemical formula is C5H4N4O3. It is found in urine in small amounts.
And urea is defined as a renal parameter from the quantitative point of view; It is a colorless, crystalline chemical compound of formula CO (NH2) 2. It is found abundantly in urine and faeces.
Explanation:
To have an accurate idea of the function of the kidneys, it is sufficient to carry out a simple blood test with determination of Urea and creatinine and an analysis of a urine sample in which the presence of cells (red blood cells and leukocytes) is assessed. and / or proteins (albumin). Today these 2 types of analysis are part of the routine of any analytical study.
