Answer:
THE MOLAR MASS OF THE UNKNOWN MOLECULAR SUBSTANCE IS 200 G/MOL.
Explanation:
Mass of the unknown substance = 0.50 g
Freezing point of the solution = 3.9 °C
Freezing point of pure benzene = 5.5 °C
Freezing point dissociation constant Kf = 5.12°C/m
First, calculate the temperature difference between the freezing point of pure benzene and the final solution freezing point.
Change in temperature = 5.5 -3.9 = 1.6 °C
Next is to calculate the number of moles or molarity of the compound that dissolved.
Using the formula:
Δt = i Kf m
Assume i = 1
So,
1.6 °C = 1 * 5.12 * x/ 0.005 kg of benzene
x = 1.6 * 0.008 / 5.12
x = 0.0128 / 5.12
x = 0.0025 moles.
Next is to calculate the molar mass using the formula, molarity = mass / molar mass
Molar mass = mass / molarity
Molar mass = 0.50 g /0.0025
Molar mass = 200 g/mol
Hence, the molar mass of the unknown compound is 200 g/mol
My Friday is going good and I might get to go over to my BFF house!!!
Hi there! Let's solve this problem shall we!
⠀Volume = 10g
Mass = 2 mL
In this specific problem, they are asking us to find the <u><em>density </em></u>of the object. So,<u><em> using the information given to us</em></u> (volume and mass), let's solve the problem!
Now, if you remember, D = M ÷ V
So, let's fill in the blanks!
D = Our unknown value
M = 2mL
V = 10g
Here is the filled out formula:
D = M ÷ V
D = 2mL ÷ 10g
D = 5 g/mL
*Make sure you put the units for your final solution!*
1) Hydrocarbon: CH3 - CH2 - CH2 - CH2 - CH3
2) Only single bonds => alkane => sufix ane
3) no substitutions
4) 5 carbons = > prefix penta.
Therefore, the name is pentane.
Answer:
At 430.34 K the reaction will be at equilibrium, at T > 430.34 the
reaction will be spontaneous, and at T < 430.4K the reaction will not
occur spontaneously.
Explanation:
1) Variables:
G = Gibbs energy
H = enthalpy
S = entropy
2) Formula (definition)
G = H + TS
=> ΔG = ΔH - TΔS
3) conditions
ΔG < 0 => spontaneous reaction
ΔG = 0 => equilibrium
ΔG > 0 non espontaneous reaction
4) Assuming the data given correspond to ΔH and ΔS
ΔG = ΔH - T ΔS = 62.4 kJ/mol + T 0.145 kJ / mol * K
=> T = [ΔH - ΔG] / ΔS
ΔG = 0 => T = [ 62.4 kJ/mol - 0 ] / 0.145 kJ/mol*K = 430.34K
This is, at 430.34 K the reaction will be at equilibrium, at T > 430.34 the reaction will be spontaneous, and at T < 430.4K the reaction will not occur spontaneously.