Answer:
1.5
Explanation:
It is given that:
Compound A and B originate from Sulfur + Oxygen.
Compound A:
6g sulfur + 5.99g Oxygen
Compound B:
8.6g sulfur + 12.88g oxygen
By comparing the ratios:
Compound A:
S: O = 6.00: 5.99
S/0 = 6.0g S / 5.99g O
Compound B:
S: O = 8.60: 12.88
S / O = 8.60g S / 12.88g O
The mass ratio of A and that of B
(6.0g S / 5.99g O) ÷ (8.60g S / 12.88g O)
(6.0 g S / 5.99g O) × (12.88g O / 8.60g S)
(6 × 12.88) / (5.99 × 8.60)
= 77.28 / 51.514
= 1.50017
= 1.5
Answer:
Calcium's atomic radius is roughly 175 pm.
Explanation:
We know that magnesium has an atomic radius of 150 pm.
The atomic radius of strontium measures 200 pm.
Since calcium's position is between magnesium and strontium in group 2 of the periodic table, its atomic radius should be roughly averaged between magnesium's and strontium's atomic radii because atomic radius is not constant.
Thus;
Calcium's atomic radius is approximately calculated as follows;
The average atomic radius is (200 + 150)/2 = 175 pm.
Given:
Mass of the ionic compound = 10.00 g
Mass of water = 75.0 g
Initial temperature of water T1= 23.2 C
Final temperature of water T2 = 31.8 C
Specific heat of water c = 4.18 J/gC
To determine:
Enthalpy of dissolution of the ionic compound
Heat gained by water equation:
Q = mcΔT
m = mass of water
c = specific heat
ΔT = change in temperature (T2-T1)
Q = 75.0 g * 4.18 J/gC * (31.8-23.2)C = 2696 J
Thus, the heat gained by water equals heat lost by the ionic compound (enthalpy of dissolution)
Therefore, q(ionic) = 2696 J
ΔH = q(ionic)/mass of ionic compound = 2696 J/10.00 g = 2.7 *10² J/g
Answer: A) enthalpy change = 2.7*10² J/g
Utilize the ideal gas law:
n = PV / RT
P = 100kPa = 100 x 1000 x (9.8 x 10^{-6}) = 0.98 atm
Convert kPa to atm, where 1 Pa = 9.8 x 10^{-6} atm.
T = 293 K
V = 6.8 L
R = 1/12
Substituting all values leads to:
n = 0.272
