<span>An insect would have an easier time walking on the surface of water than on the surface of ethanol. Water's stronger intermolecular forces lead to higher surface tension. Higher surface tension allows water to support the insect. I hope this helps.</span>
It’s basically that’s any system that’s closed to all transfers of matter and energy the mass of the system has to remain constant over time because they can’t change meaning you can’t add or remove from it
Answer:
109.7178g of H2O
Explanation:
First let us generate a balanced equation for the reaction. This is illustrated below:
2C3H8O + 9O2 —> 6CO2 + 8H2O
Next we will calculate the molar mass and masses of C3H8O and H20. This is illustrated below:
Molar Mass of C3H8O = (3x12.011) + (8x1.00794) + 15.9994 = 36.033 + 8.06352 + 15.9994 = 60.09592g/mol.
Mass of C3H8O from the balanced equation = 2 x 60.09592 = 120.19184g
Molar Mass of H2O = (2x1.00794) + 15.9994 = 2.01588 + 15.9994 = 18.01528g/mol
Mass of H2O from the balanced equation = 8 x 18.01528 = 144.12224g
From the equation,
120.19184g of C3H8O produced 144.12224g of H20.
Therefore, 91.5g of C3H8O will produce = (91.5 x 144.12224) /120.19184 = 109.7178g of H2O
Answer:
2.2 °C/m
Explanation:
It seems the question is incomplete. However, this problem has been found in a web search, with values as follow:
" A certain substance X melts at a temperature of -9.9 °C. But if a 350 g sample of X is prepared with 31.8 g of urea (CH₄N₂O) dissolved in it, the sample is found to have a melting point of -13.2°C instead. Calculate the molal freezing point depression constant of X. Round your answer to 2 significant digits. "
So we use the formula for <em>freezing point depression</em>:
In this case, ΔTf = 13.2 - 9.9 = 3.3°C
m is the molality (moles solute/kg solvent)
- 350 g X ⇒ 350/1000 = 0.35 kg X
- 31.8 g Urea ÷ 60 g/mol = 0.53 mol Urea
Molality = 0.53 / 0.35 = 1.51 m
So now we have all the required data to <u>solve for Kf</u>:
