The mole fraction of KBr in the solution is 0.0001
<h3>How to determine the mole of water</h3>
We'll begin by calculating the mass of the water. This can be obtained as follow:
- Volume of water = 0.4 L = 0.4 × 1000 = 400 mL
- Density of water = 1 g/mL
- Mass of water =?
Density = mass / volume
1 = Mass of water / 400
Croiss multiply
Mass of water = 1 × 400
Mass of water = 400 g
Finally, we shall determine the mole of the water
- Mass of water = 400 g
- Molar mass of water = 18.02 g/mol
- Mole of water = ?
Mole = mass / molar mass
Mole of water = 400 / 18.02
Mole of water = 22.2 moles
<h3>How to de terminethe mole of KBr</h3>
- Mass of KBr = 0.3 g
- Molar mass of KBr = 119 g/mol
- Mole of KBr = ?
Mole = mass / molar mass
Mole of KBr = 0.3 / 119
Mole of KBr = 0.0025 mole
<h3>How to determine the mole fraction of KBr</h3>
- Mole of KBr = 0.0025 mole
- Mole of water = 22.2 moles
- Total mole = 0.0025 + 22.2 = 22.2025 moles
- Mole fraction of KBr =?
Mole fraction = mole / total mole
Mole fraction of KBr = 0.0025 / 22.2025
Mole fraction of KBr = 0.0001
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The molecular weight of hemoglobin can be calculated using osmotic pressure
Osmotic pressure is a colligative property and it depends on molarity as
πV = nRT
where
π = osmotic pressure
V = volume = 1mL = 0.001 L
n = moles
R = gas constant = 0.0821 L atm / mol K
T = temperature = 25°C = 25 + 273 K = 298 K
Putting values we will get value of moles

we know that

Therefore

The balanced equation that shows the reaction between oxalic acid and permanganate ion in an acidic medium is: 2MnO4- + 5H2C2O4 + 6H+ -> 2Mn(2+) + 10CO2 + 8H2O. Thus, 1 mole of oxalic acid reacts with 0.4 mole of permanganate ion. This was obtained using stoichiometry:
1 mol H2C2O4 x (2 mol MnO4-/ 5 mol H2C2O4) = 0.4 mol MnO4-
In this redox reaction, the permanganate is reduced to manganese(II) ion.
2.47 moles are produce if 85g of N2 react with 180g of Mg
Answer:
Long bones contain yellow bone marrow and red bone marrow, which produce blood cells.
Explanation:
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