Answer:
it is easier for them to have an octet of electrons(8e)
Each ion will: obtain the noble gas structure, each atom has high ionization energy
Explanation:
.
A. Cuz it contains more salt
Hope this helps good luck on ur examsss :))))
Answer:
H₂O is the limiting reactant
Theoretical yield of 240 g Al₂O₃ and 14 g H₂
Explanation:
Find how many moles of one reactant is needed to completely react with the other.
6.5 mol Al × (3 mol H₂O / 2 mol Al) = 9.75 mol H₂O
We need 9.75 mol of H₂O to completely react with 6.5 mol of Al. But we only have 7.2 mol of H₂O. Therefore, H₂O is the limiting reactant.
Now find the theoretical yield:
7.2 mol H₂O × (1 mol Al₂O₃ / 3 mol H₂O) × (102 g Al₂O₃ / mol Al₂O₃) ≈ 240 g Al₂O₃
7.2 mol H₂O × (3 mol H₂ / 3 mol H₂O) × (2 g H₂ / mol H₂) ≈ 14 g H₂
Since the data was given to two significant figures, we must round our answer to two significant figures as well.
Answer:
The new pressure at constant volume is 1066.56 kPa
Explanation:
Assuming constant volume, the pressure is diectly proportional to the temperature of a gas.
Mathematically, P1/T1 = P2 /T2
P1 = 880 kPA= 880 *10^3 Pa
T1 = 250 K
T2 = 303 K
P2 =?
Substituting for P2
P2 = P1 T2/ T1
P2 = 880 kPa * 303 / 250
P2 = 266,640 kPa/ 250
P2 = 1066.56 kPa.
The new pressure of the gas is 1066.56 kPa
Answer:
1.88 × 10²³ particles
Explanation:
Given data:
Volume of H₂ = 0.7 L
Number of particles at STP = ?
Solution:
First of all we will calculate the number of moles of H₂.
PV = nRT
n = PV / RT
n = 1 atm . 7 L / 0.0821 L. atm. mol⁻¹. k⁻¹ × 273.15 K
n = 7 atm. L / 22.43 L. atm. mol⁻¹
n = 0.312 mol
it is known that,
2 g H₂ = 1 mol = 6.022 × 10²³ particles
For 0.312 mol
0.312 × 6.022 × 10²³ particles
1.88 × 10²³ particles
