Answer:
• One mole of oxygen is equivalent to 16 grams.
→ But at STP, 22.4 dm³ are occupied by 1 mole.
Knowing the ratio between atoms we can write an empirical formula:
<span>C4H6O </span>
<span>we compute the molar mass of this single formula: </span>
<span>4x12 + 6 x 1 + 16 x1 = 70 g / mol </span>
<span>Now, as we know the actual molar mas being 280 g/mol, we divide this number by 70 and we get the ratio between empirical formula and molecular actual formula: </span>
<span>280 / 70 = 4 </span>
<span>This means that actual molecular formula is: </span>
<span>(C4H6O)4 or </span>
<span>C16H24O4 </span>
When you heat up most substances it gives them more Kinetic energy and the substance becomes less arranged in an ordered state, further apart and move faster. therefore the answer is the first: They gain a higher average kinetic energy
Hope that helps :)
Answer:
15.70mg would remain
Explanation:
Partition coefficient is used to extract or purify a solute from a solvent selectively to avoid interference from other substances. For the problem, formula is:
Kp = Concentration 9-fluorenone in ether / Concentration of solute in H₂O
After the solute, 9-fluorenone, is extracted with water, the mass that remains in ether is:
(19mg - X)
<em>Where X is the mass that now is in the aqueous phase</em>
Replacing in Kp formula:
9.5 = (19mg - X) / 5mL / (X /10mL)
0.95X = 19mg - X / 5mL
4.75X = 19 - X
5.75X = 19
X = 19 / 5.75
X = 3.30mg
That means 9-fluorenone that remain in the ether layer is:
19mg - 3.30mg =
<h3>15.70mg would remain</h3>
Answer:
10.23 grams of sucrose should be added.
Explanation:
1.15 m means molality (moles of solute in 1kg of solvent)
1.15 moles of sucrose are contained in 1 kg of solvent (1000 g)
Let's determine the moles in our mass of solvent.
Firstly we convert the g to kg → 26 g . 1kg/1000g = 0.026 kg
m . mass (kg) = 1.15 mol/kg . 0.026kg → 0.0299 moles.
Finally we convert the moles to mass (mol . molar mass)
0.0299 mol . 342.3 g/mol = 10.23 g
