Answer:
Molality = 0.0862 mole/kg
Explanation:
Molality = (number of moles of solute)/(mass of solvent in kg)
Number of moles of solute = (mass of Creatinine in the blood sample)/(Molar mass of Creatinine)
To obtain the mass of creatinine in 10 mL of blood. We're told that 1 mg of Creatinine is contained in 1 decilitre of blood.
1 decilitre = 100 mL
1 mg of Creatinine is contained in 100 mL of blood
x mg of Creatinine is contained in 10 mL of blood.
x = (1×10/100) = 0.1 mg = 0.0001 g
Molar mass of Creatinine (C₄H₇N₃O) = 113.12 g/mol
Number of moles of Creatinine in the 10 mL blood sample = (0.0001/113.12) = 0.000000884 moles
Mass of 10 mL of blood = density × volume = 1.025 × 10 = 10.25 mg = 0.01025 g = 0.00001025 kg
Molality of normal creatinine level in a 10.0-ml blood sample = (0.000000884/0.00001025)
Molality = 0.0862 moles of Creatinine per kg of blood.
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Answer:
that results in an alcohol and a carboxylic acid.
Explanation:
Saponification is a chemical reaction process of alkaline hydrolysis of esters(R'COOR group) by which soap is obtained.
For Example, when a base such as sodium hydroxide [NaOH] is used to hydrolyze an ester, the products are a carboxylate salt and an alcohol. Because soaps are prepared by the alkaline hydrolysis of fats and oils.
In a saponification reaction, alkaline hydrolysis of fats and oils with sodium hydroxide yields propane-1,2,3-triol and the corresponding sodium salts of the component fatty acids.
i.e Fat or oil + caustic alkali ⇒ Soap + propane-1,2,3-triol
As a specific example, ethyl acetate and NaOH react to form sodium acetate and ethanol:
The reaction goes to completion in the image below:
Answer:
Explanation:
Hello!
In this case, since the phosphoric acid is a triprotic acid, we infer it has three stepwise ionization reactions in which one hydrogen ion is released at each step, considering they are undergone due to the presence of water, thus, we proceed as follows:
Moreover, notice each step has a different acid dissociation constant, which are quantified in the following order:
Ka1 > Ka2 > Ka3
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Answer:
5.00g
Explanation:
Molarity is moles per liter
Therefore:
x moles/.250L = .5 moles/1L
Solve for x: .125 moles required
Question asks for grams of NaOH so multiply the moles by the molar mass of NaOH
.125(39.997) = 4.999625g
Rounds to 5.00g
For the compound B the following statement is correct-
B. It is an ether because it is unable to form a hydrogen bond, so it is less soluble in water.
The solubility of alcohol in water depends upon the capability of formation of hydrogen bond in the solute. Now in alcohol the -OH group is polar in nature which enhance the possibility of hydrogen bond formation and it is more soluble in water.
On the other hand although there presence a -O- functional group in ether. It is less soluble in water due to non polarity of the functional group.
From the given data it is seen that compound A is more soluble in water than compound B. Thus it may be predicted that compound A is alcohol and B is ether.
Henceforth, for the compound B the following statement is correct-
B. It is an ether because it is unable to form a hydrogen bond, so it is less soluble in water.
The reason of incorrect options:
A. compound B cannot be an alcohol as it is less soluble in water.
C. In ether the functional group is -O-, thus electronegative atom (O) is present.
D. As both the compound (alcohol and ether) has equal molecular mass thus the organic chain will be same in alcohol and the hydrogen bond interaction will be more prominent than the dispersion force between the -OH group.
