Answer:
Covalent substances have weaker intermolecular attractions.
Explanation:
i just took the test
Answer:
9 : 8
Explanation:
Aluminum oxide has the following formula Al₂O₃.
Next, we shall determine the mass of Al and O₂ in Al₂O₃. This can be obtained as follow:
Mass of Al in Al₂O₃ = 2 × 27 = 54 g
Mass of O₂ in Al₂O₃ = 3 × 16 = 48 g
Finally, we shall determine the mass ratio of Al and O₂. This can be obtained as follow:
Mass of Al = 54 g
Mass of O₂ = 48 g
Mass of Al : Mass of O₂ = 54 : 48
Mass of Al : Mass of O₂ = 54 / 48
Mass of Al : Mass of O₂ = 9 / 8
Mass of Al : Mass of O₂ = 9 : 8
Therefore, the mass ratio of Al and O₂ in Al₂O₃ is 9 : 8
Answer:
P₂= 116.7 atm
Explanation:
Here apply the Boyle's law equations that states :at constant temperature the volume of a dry mass of a gas is inversely proportional to its pressure.
This is simplified as;
P₁V₁=P₂V₂ where P is pressure and V is volume
Given that;
P₁=1
V₁=1.81 m³
P₂=?
V₂=1.55*10^-2 m³
Apply the formula
1*1.81 =P₂*1.55*10^-2 m³
1.81/1.55*10^-2 =P₂
P₂= 116.7 atm
Answer:
T2 = 2843.1 oK. This is a huge temperature. Check it for errors.
Explanation:
Remark
This is the same question as the other one I've answered. Only the numbers have been altered.
Givens
v1 = 56 mL
P1 = 1 atm
T1 = 273o K
v2 = 162
P2 = 3.6 atm
T2 = ?
Formula
Vi * P1 / T1 = V2 * P2/T2
Solution
Rearrange the formula so T2 is on the left
T2 = V2 P2 * T1 / (V1 * P1) Now just put the numbers in.
T2 = 162 * 3.6* 273 / (56 *1)
T2 = 159213.6/56
T2 = 2843.1