Answer:
a) 210.3 g/mol
b) 210.2 g/mol
c) 384.5 g/mol
Explanation:
First step we will calculate the molar masses of ; carbon atom, hydrogen atom and oxygen atom in each .
<u> Molar mass of dibenzyl ketone</u>
Molar mass of dibenzyl ketone = ∑ molar masses of atoms in dibenzyl ketone
= carbon( 15 ) = 15 ( 12.0107 ) + oxygen ( 14 ) = 1 ( 15.999 ) + hydrogen(14) =14(1.00784)
= 210.26926 ≈ 210.3 g/mol
<u> Molar mass of benzil</u>
Molar mass of Benzil = ∑ molar masses of atoms in Benzil
= carbon( 14) = 14(12.0107) + oxygen(2) = 2 ( 15.999) + hydrogen(10) =10(1.00784)
= 210.2262 ≈ 210.2 g/mol
<u>Molar mass of 2,3,4,5-tetraphenylcyclopentadienone</u>
Molar mass = ∑ molar masses of atoms
= carbon ( 29) = 29(12.0107) + oxygen (1) = 1( 15.999 ) + hydrogen(20) = 20(1.00784 )
≈ 384.5 g/mol
1.
The balanced chemical reaction is:
N2 +3 I2 = 2NI3
We are given the amount of product formed.
This will be the starting point of our calculations.
3.58 g NI3 ( 1 mol NI3 / 394.71 g NI3 ) ( 3
mol I2 / 2 mol NI3 ) = 0.014 mol I2.
Thus, 0.014 mol of I2 is needed to form the
given amount of NI3.
1×10^-4 = 0,0001M
pH = -log[H+]
pH = -log0,0001
pH = 4
Answer:
0.576M and 0.655m
Explanation:
<em>...Dissolves 15.0g of styrene (C₈H₈) in 250.mL of a solvent with a density of 0.88g/mL...</em>
<em />
Molarity is defined as moles of solute (Styrene in this case) per liter of solution whereas molality is the moles of solute per kg of solvent. Thus, we need to find the moles of styrene, the volume in liters of the solution and the mass in kg of the solvent as follows:
<em>Moles styrene:</em>
Molar mass C₈H₈:
8C = 12.01g/mol*8 = 96.08g/mol
8H = 1.005g/mol* 8 = 8.04g/mol
96.08g/mol + 8.04g/mol = 104.12g/mol
Moles of 15.0g of styrene are:
15.0g * (1mol / 104.12g) = 0.144 moles of styrene
<em>Liters solution:</em>
250mL * (1L / 1000mL) = 0.250L
<em>kg solvent:</em>
250mL * (0.88g/mL) * (1kg / 1000g) = 0.220kg
Molarity is:
0.144 moles / 0.250L =
<h3>0.576M</h3>
Molality is:
0.144 moles / 0.220kg =
<h3>0.655m</h3>
To solve this problem, we establish the general energy balance:
ΔE = ΔU + ΔKE + ΔPE
ΔE = Q + W
Q + W = ΔU + ΔKE + ΔPE
In this case, ΔKE and ΔPE are both zero or negligible.
Given:
m = 33.0 grams of CO2
Tsub = 77 K
P = 1 atm
ΔE = Q + W
ΔE = mCpΔT + ΔPV
solve for mCpΔT, find the value of Cp for CO2, then solve for Q. Next, solve for W using the ideal gas law. Add the two values and that will be the value of the delta E.
