First you figure out how many grams of sulfur has been used.
2.080g - 1.659g = 0.421g of S used
Now change all of your grams to mols.
1.659g of Cu / 63.546g/mol = 0.02611mol of Cu
0.421g of S / 32.065g/mol = 0.01313 mol of S
From this you can see that S has less so we divide both numbers by the number of moles of S so we can get a ratio of S to Cu.
0.01313/0.01313 = 1 for S
0.02611/0.01313 = 2 for Cu
So the empirical formula would be Cu2S or copper sulfide.
Answer:
Both oil and gasoline molecules are nonpolar, while water is polar. Nonpolar solvents have a tendency to dissolve other nonpolar molecules.
Explanation:
Molecules may be categorized as "polar" or "nonpolar" according to <em>difference in the atom's electronegativity.</em>
<u>Water is polar</u> because it consists of two types of atoms that<em> do not cancel out each other.</em> It is made of two atoms of Hydrogen and only one atom of Oxygen. This makes the Oxygen<u> partially negative</u> and the Hydrogen <u>partially positive.</u> This allows them to readily bond with other polar molecules like sugar. However, it cannot mix freely with oil and gasoline because<em> both of these are nonpolar. </em>Nonpolar molecules do not have much difference when it comes to their atoms' electronegativity. <em>Therefore, they have the tendency to dissolve molecules which are nonpolar as well. </em>This explains why oil molecules can mix freely with gasoline.
By Gayle-Lussac's law, the pressure and temperature of a fixed volume and amount of gas is directly proportional.
Thus,
P/T = constant
So if the temperature is increased four times, the pressure is also increased four times.
Answer:
Q = -18118.5KJ
W = -18118.5KJ
∆U = 0
∆H = 0
∆S = -60.80KJ/KgK
Explanation:
W = RTln(P1/P2)
P1 = 1bar = 100KN/m^2, P2 = 1500bar = 1500×100 = 150000KN/m^2, T = 23°C = 23 + 273K = 298K
W = 8.314×298ln(100/150000) = 8.314×298×-7.313 = -18118.5KJ ( work is negative because the isothermal process involves compression)
∆U = Cv(T2 - T1)
For an isothermal process, temperature is constant, so T2 = T1
∆U = Cv(T1 - T1) = Cv × 0 = 0
Q = ∆U + W = 0 + (-18118.5) = 0 - 18118.5 = -18118.5KJ
∆H = Cp(T2 - T1)
T2 = T1
∆H = Cp(T1 - T1) = Cp × 0 = 0
∆S = Q/T
Mass of water = 1kg
Heat transferred (Q) per kilogram of water = -18118.5KJ/Kg
∆S = (-18118.5KJ/Kg)/298K = -60.80KJ/KgK
Answer:
Explanation:the best guess I will go with is H but I might be wrong
