1 mole ----------- 22.4 L ( at STP )
3.5 mols --------- ?
V = 3.5 x 22.4 / 1
V = 78.4 L
hope this helps!
Through hemoglobin transport in blood, material exchange is carried out on the alveolar membrane.
The chemical could have more or less of a reaction to the other chemicals in the experiment
First we have to find moles of C:
Molar mass of CO2:
12*1+16*2 = 44g/mol
(18.8 g CO2) / (44.00964 g CO2/mol) x (1 mol C/ 1 mol CO2) =0.427 mol C
Molar mass of H2O:
2*1+16 = 18g/mol
As there is 2 moles of H in H2O,
So,
<span>(6.75 g H2O) / (18.01532 g H2O/mol) x (2 mol H / 1 mol H2O) = 0.74mol H </span>
<span>Divide both number of moles by the smaller number of moles: </span>
<span>As Smaaler no moles is 0.427:
So,
Dividing both number os moles by 0.427 :
(0.427 mol C) / 0.427 = 1.000 </span>
<span>(0.74 mol H) / 0.427 = 1.733 </span>
<span>To achieve integer coefficients, multiply by 2, then round to the nearest whole numbers to find the empirical formula:
C = 1 * 2 = 2
H = 1.733 * 2 =3.466
So , the empirical formula is C2H3</span>
The molality of the solution = 17.93 m
<h3>Further explanation</h3>
Given
6.00 L water with 6.00 L of ethylene glycol(ρ=1.1132 g/cm³= 1.1132 kg/L)
Required
The molality
Solution
molality = mol of solute/ 1 kg solvent
mol of solute = mol of ethylene glycol
- mass of ethylene glycol :
= volume x density
= 6 L x 1.1132 kg/L
= 6.6792 kg
= 6679.2 g
- mol of ethylene glycol (MW=62.07 g/mol)
=mass : MW
=6679.2 : 62.07
=107.608
6 L water = 6 kg water(ρ= 1 kg/L)
