Answer:
Mole percent of water in azeotrope is 69%
Explanation:
Clearly, the azeotrope consists of water and perchloric acid.
So, 72% perchloric acid by mass means 100 g of azeotrope contains 72 g of perchloric acid and 28 g of water.
Molar mass of water = 18.02 g/mol and molar mass of perchloric acid = 100.46 g/mol
So, 72 g of perchloric acid = 
of perchloric acid = 0.72 mol of perchloric acid
Also, 28 g of water = 
of water = 1.6 mol of water
Hence total number of mol in azeotrope = (1.6+0.72) mol = 2.32 mol of azeotrope
So, mole percent of water in azeotrope = [(moles of water)/(total no of moles)]
%
Mole percent of water = 
% = 69%
Explanation:
Experiment Initial [CS2] (mol/L) Initial Rate (mol/L·s)
1 0.100 2.7 × 10−7
2 0.080 2.2 × 10−7
3 0.055 1.5 × 10−7
4 0.044 1.2 × 10−7
a) Choose the rate law for the decomposition of CS2.
Comparing equations 1 and 3, reducing the initial concentration by almost half (from 0.100 to 0.055) leads too the rate of reaction to be reduced by almost half (from 2.7 × 10−7 to 1.5 × 10−7).
This signifies that the reaction is a first order reaction.
Rate = k [CS2]
(b) Calculate the average value of the rate constant.
Taking equation 1.
Rate = k [CS2]
k = Rate / [CS2]
k = 0.100 / (2.7 × 10−7) = 0.037 x 10^8 = 3.7 x 10^6s-1
Answer:
7
Explanation:
cuz it is trust me hope it helps though
They differ in their molecular structures and properties.
Answer:
0.0305 moles of MgCO₃
Explanation:
In order to solve this problem, we first need to calculate the molecular weight of MgCO₃:
- MgCO₃ MW = Atomic mass Mg + Atomic mass C + (Atomic mass O)*3
- MgCO₃ MW = 24.3 + 12 + 16*3 = 84.3 g/mol
Finally we <u>divide the mass by the molecular weight</u>, to calculate the <em>number of moles</em>:
- 2.57 g MgCO₃ ÷ 84.3 g/mol = 0.0305 moles.
