In this problem, we need to use the ideal gas law. The following is the formula used in ideal gas law: PV = nRT, where n refers to the moles and R is the gas constant.
Given
P = 10130.0 kPa
V = 50 L
T = 300 degree celcius + 273.15 = 573.15 K
R = 8.314 L. kPa/K.mol
Solution
To get the moles which represent the "n" in the formula, we need to rearrange the equation.
PV = nRT PV
---- ------ ---> n = --------
RT RT RT
10130.0 kPa x 50 L
n= ---------------------------------------------
8.314 L. kPa/K.mol x 573.15 K
506,500
= ----------------------------
4,765.17 mol K
=106.29 mol Ar
So the moles of argon gas is 106.29 moles
The answer you need is Volume.
Answer:
1, just I) color.
Explanation:
Physical properties are the properties that can be observed without changing the composition of a substance, such as color, temperature, density, and boiling point.
A physical change is a change in the substance that only modifies its aggregation state, such as solidification, and boiling.
Chemical properties are the properties that need a reaction to being observed, such as the combustibility, which needs a combustion reaction to being quantified.
When a chemical reaction occurs, and the composition of the substance change, it's a chemical change.
So, the heating copper with carbon is a chemical reaction, and purification by electrolysis is too. Color is the only physical property.
Answer:
a) 88.48%
b) 0.05625 mol
Explanation:
2CH₃CH₂OH(l) → CH₃CH₂OCH₂CH₃(l) + H₂O(g) Reaction 1
CH₃CH₂OH(l) → CH₂═CH₂(g) + H₂O(g) Reaction 2
a) CH₃CH₂OH = 46.0684 g/mol
CH₃CH₂OCH₂CH₃ = 74.12 g/mol
1 mol CH₃CH₂OH ______ 46.0684 g
x ______ 50.0 g
x = 1.085 mol CH₃CH₂OH
1 mol CH₃CH₂OCH₂CH₃ ______ 74.12 g g
y ______ 35.9 g
y = 0.48 mol CH₃CH₂OCH₂CH₃
100% yield _____ 0.5425 mol CH₃CH₂OCH₂CH₃
w _____ 0.48 mol CH₃CH₂OCH₂CH₃
w = 88.48%
b) Only 0.96 mol of ethanol reacted to form diethyl ether. This means that 0.125 mol of ethanol did not react. 45% of 0.125 mol reacted to form ethylene. Therefore, 0.05625 mol of ethanol reacted by the side reaction (reaction 2). Since 1 mol of ethanol leads to 1 mol of ethylene, 0.05625 mol of ethanol produces 0.05625 mol of ethylene.
