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:
1.37x10²⁵atoms of carbon
2.74x10²⁵ atoms of oxygen.
33.7g of KNO₃
Explanation:
To answer this question you must use molar mass of carbon dioxide (44g/mol) and 1 mole are 6.022x10²³atoms.
1.00kg are 1000g of CO₂. Moles are:
1000g CO₂ * (1mol / 44g) = 22.73 moles of CO₂ = 22.73 moles of carbon.
In atoms:
22.73 moles C * (6.022x10²³atoms / 1mole) = 1.37x10²⁵atoms of carbon
There are 22.73 moles of CO₂ * 2 = 45.45 moles of oxygen are present in the carbon dioxide. In atoms:
45.45 moles Oxygen * (6.022x10²³atoms / 1mole) = 2.74x10²⁵ atoms of oxygen.
1 mole of Potassium nitrate, KNO₃, contains 3 moles of oxygen. 1 mol of oxygen are:
1.00 mol O * (1mol KNO₃ / 3 moles O) = 0.33 moles of KNO₃
As molar mass of KNO₃ is 101.1g/mol:
0.33 moles of KNO₃ * (101.1g / mol) = 33.7g of KNO₃
Answer:
38.7%
41.3%
20%
Explanation:
The percentage composition helps to know the what percent of the total mass of a compound is made up of each of the constituent elements or groups.
To solve this problem:
- find the formula mass by adding the atomic masses of the atoms that makes up the compound.
- place the mass contribution of the element or group to the formula mas and multiply by 100;
Compound:
Ca₃(PO₄)₂
Formula mass = 3(40) + 2[31 + 4(16)]
= 120 + 2(95)
= 120 + 190
= 310
%C = 
x 100 = 38.7%
%P = 
x 100 = 41.3%
%O = 
x 200 = 20%
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