Answer:
93.15 %
Explanation:
We have to start with the chemical reaction:
Now, we can balance the reaction:
Our initial data are the 15.71 g of 
, so we have to do the following steps:
1) <u>Convert from grams to moles of using the molar mass (110.98 g/mol).</u>
2) <u>Convert from moles of to moles of 
using the molar ratio. ( 1 mol = 1 mol of ).</u>
3) <u>Convert from moles of to grams of using the molar mass. (100 g/mol).</u>
Finally, we can calculate the yield percent:
I hope it helps!
Answer:
The answer is 6.25g.
Explanation:
First create your balanced equation. This will give you the stoich ratios needed to answer the question:
2C8H18 + 25O2 → 16CO2 + 18H2O
Remember, we need to work in terms of NUMBERS, but the question gives us MASS. Therefore the next step is to convert the mass of O2 into moles of O2 by dividing by the molar mass:
7.72 g / 16 g/mol = 0.482 mol
Now we can use the stoich ratio from the equation to determine how many moles of H2O are produced:
x mol H2O / 0.482 mol O2 = 18 H2O / 25 O2
x = 0.347 mol H2O
The question wants the mass of water, so convert moles back into mass by multiplying by the molar mass of water:
0.347 mol x 18 g/mol = 6.25g
The reaction of an acid and a base, which forms water and a salt
Answer:
The starch test also known as the iodine test is used to test for the presence of starch.
Sorry if this isn’t detailed enough. You can go on google/wikipedia and find much more detail if necessary. ^-^
Answer:
900 J/mol
Explanation:
Data provided:
Enthalpy of the pure liquid at 75° C = 100 J/mol
Enthalpy of the pure vapor at 75° C = 1000 J/mol
Now,
the heat of vaporization is the the change in enthalpy from the liquid state to the vapor stage.
Thus, mathematically,
The heat of vaporization at 75° C
= Enthalpy of the pure vapor at 75° C - Enthalpy of the pure liquid at 75° C
on substituting the values, we get
The heat of vaporization at 75° C = 1000 J/mol - 100 J/mol
or
The heat of vaporization at 75° C = 900 J/mol
