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.
Hope this Helps!!!
Answer:
The original concentration is "35 mg/ml".
Explanation:
According to the question,
The solution is diluted,
= 1:50
The initial volume,
V1 = 1 ml
Final concentration,
= 0.07 mg
then,
The final volume,
V2 = 500 ml
As we know,
⇒ 
or,
⇒ 
On substituting the values, we get
⇒ 
⇒ 
⇒ 
Answer:
The anwer is not D the anwer is A
Explanation:
10%
At each trophic level only small proportion of energy is transferred to the next level
Sodium acetate<span> contains </span>both ionic and covalent<span> bonds. </span> <span>In </span>sodium acetate<span>, there is </span>ionic<span> bond between sodium and acetate </span>ions<span> and there is </span>covalent<span> bond in the atoms that are bondeed to the carbon atoms. Hope this answers the question.</span>
