Answer:
Twice as much.
Explanation:
That's because the freezing point depression depends on the total number of solute particles.
C₆H₁₂O₆(s) ⟶ C₆H₁₂O₆(aq)
0.01 mol of C₆H₁₂O₆ gives 0.01 mol of solute particles.
NaCl(s) ⟶ Na⁺(aq) + Cl⁻(aq)
1 mol of NaCl gives 0.01 mol of Na⁺(aq) and 0.01 mol of Cl⁻(aq).
That's 0.02 mol of particles, so the freezing point depression of 0.01 mol·L⁻¹ NaCl will be twice that of 0.01 mol·L⁻¹ C₆H₁₂O₆.
Using the specific heat capacity formula:
q = mc ∆ t
60.0 J = (6g)(x)(11*C)
x = 0.9 J/g*C
Aluminum
Answer:
V = 57.39 L
Explanation:
Given that,
Temperature, T = 300 K
Pressure, P = 0.987 atm
No. of moles of Ne, n = 2.30 mol
We need to find the volume of Ne. We know that, the ideal gas law is as follows :
PV = nRT
Where
P is pressure and R is gas constant
So, the volume of the Ne is 57.39 L.
Answer:
The answer is b
Explanation:
have common properties and are good conductors of heat and electricity. They reflect light and are malleable, and ductile.
None of the alpha particles fired at the foil are being repelled back, like they were in the Rutherford atom simulation.I hope this correct.