Answer:
Mass = 36 g
Explanation:
Given data:
Mass of water formed = ?
Mass of hydrogen = 4.04 g
Mass of oxygen = 31.98 g
Solution:
Chemical equation:
2H₂ + O₂ → 2H₂O
Number of moles of hydrogen:
Number of moles = mass/molar mass
Number of moles = 4.04 g/ 2 g/mol
Number of moles = 2.02 mol
Number of moles of oxygen:
Number of moles = mass/molar mass
Number of moles = 31.98 g/ 32 g/mol
Number of moles = 1.0 mol
Now we will compare the moles of water with hydrogen and oxygen.
O₂ : H₂O
1 : 2
H₂ : H₂O
2 : 2
2.02 : 2.02
Number of moles of water formed by oxygen are less thus oxygen will limiting reactant.
Mass of water:
Mass = number of moles × molar mass
Mass = 2 mol × 18 g/mol
Mass = 36 g
The substance has a higher density than water
Use PV = nRT
(2 atm)(.3 liters) = n(8.314 mol*K)(303°K)
.6 = n(2519.142)
Divide by 2519.142
n = .00023818 mols of HCl * 36.46g of HCl/ 1 mol of HCl
Grams of HCl = 0.00868
Critical Thinking Questions
1. Why do you think forensic scientists are so careful that the tests they do are sensitive, reproducible, and specific? What might happen if they were less careful about this?
They have to be careful to ensure as much accuracy as possible.
2.Which type of evidence do you think is most useful in an investigation? Why?
Physical evidence would probably be most important because it is the best way to connect someone directly with that crime.
3.Why do you think that forensic scientists continue to look for class characteristics given their limitations?
Class characteristics are good in court because it provides details of different aspects of the crime.
Answer:
Attached in the photo.
Explanation:
Hello,
The answers in the attached photo. Just three things:
- In the second point a parenthesis is missing to properly understand the molecule (after the oxygen), nevertheless, I assumed it was an ether.
- In the sixth point, there's a missing hydrogen for it to be an ether as well.
- In the tenth point the second parenthesis is not clear, it seems there's a missing subscript, nevertheless I draw it assuming complete octates.
Best regards.
