Answer:
(II) only correctly rank the bonds in terms of increasing polarity.
Explanation:
Bond polarity is proportional to difference in electronegativity between bonded atoms.
Atoms Electronegativity Bond Electronegativity difference
Cl 3.0 Cl-F 1.0
Br 2.8 Br-Cl 0.2
F 4.0 Cl-Cl 0
H 2.1 H-C 0.4
C 2.5 H-N 0.9
N 3.0 H-O 1.4
O 3.5 Br-F 1.2
I 2.7 I-F 1.3
Si 1.9 Cl-F 1.0
P 2.2 Si-Cl 1.1
Si-P 0.3
Si-C 0.6
Si-F 2.1
So, clearly, order of increasing polarity : O-H > N-H > C-H
So, (II) only correctly rank the bonds in terms of increasing polarity
There are:
3.41 moles of C
4.54 moles of H
3.40 moles of O.
Why?
To solve the problem, the first thing that we need to do is to write the chemical formula of the ascorbic acid.

Now, we know that there are 100 grams of the compound, so, the masses of each element will represent the percent in the compound.
We have that:

To know the percent of each element, we need to to the following:

So, we know that for the 100 grams of the compound, there are:
40.92 grams of C
4.58 grams of H
54.49 grams of O
We know the molecular masses of each element:

Now, to calculate the number of moles of each element, we need to divide the mass of each element by the molecular mass of each element:

Hence, we have that there are 3.41 moles of C, 4.54 moles of H, and 3.40 moles of O.
Have a nice day!
Let's assume that the gas has ideal gas behavior.
Then we can use ideal gas equation,
PV = nRT
Where, P is Pressure of the gas (Pa), V is volume of the gas (m³), n is the number of moles of gas (mol), R is the Universal gas constant (8.314 J mol⁻¹ K⁻¹) and T is the temperature in Kelvin (K)
The given data for the gas is,
P = 2.8 atm = 283710 Pa
V = 98 L = 98 x 10⁻³ m³
T = 292 K
R = 8.314 J mol⁻¹ K⁻¹
n = ?
By applying the formula,
283710 Pa x 98 x 10⁻³ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 292 K
n = 11.45 mol
Hence,moles of gas is 11.45 mol.
Answer: 317 joules
Explanation:
The quantity of heat energy (Q) gained by aluminium depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
In this case,
Q = ?
Mass of aluminium = 50.32g
C = 0.90J/g°C
Φ = (Final temperature - Initial temperature)
= 16°C - 9°C = 7°C
Then, Q = MCΦ
Q = 50.32g x 0.90J/g°C x 7°C
Q = 317 joules
Thus, 317 joules of heat is gained.