Water is the answer out of all the other options
The percent of O in Cr₂O₃ : 31.58%
<h3>Further explanation</h3>
Given
Cr = 52.00 amu, O = 16.00 amu
Required
The percent of O
Solution
MW Cr₂O₃ = 2 x Ar Cr + 3 x Ar O
MW Cr₂O₃ = 2.52+3.16
MW Cr₂O₃ =152 amu
Answer:
Option B is correct = 1,3
Explanation:
Chemical equation:
C₂H₄ + O₂ → CO₂ + H₂O
Balanced chemical equation:
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O
Step 1:
Left side Right side
C = 2 C = 1
H = 4 O = 3
O = 2 H = 2
Step 2:
C₂H₄ + O₂ → 2CO₂ + H₂O
Left side Right side
C = 2 C = 2
H = 4 O = 5
O = 2 H = 2
Step 3:
C₂H₄ + O₂ → 2CO₂ + 2H₂O
Left side Right side
C = 2 C = 2
H = 4 O = 6
O = 2 H = 4
Step 4:
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O
Left side Right side
C = 2 C = 2
H = 4 O = 6
O = 6 H = 4
Answer:
Explanation:
Hello,
In this case, given that the same temperature and pressure is given for all the gases, we can notice that 16.0 mL are related with two moles of carbon monoxide by means of the Avogadro's law which allows us to understand the volume-moles relationship as a directly proportional relationship. In such a way, since in the chemical reaction:
We notice two moles of carbon monoxide yield two moles of carbon dioxide, therefore we have the relationship:
Thus, solving for the yielded volume of carbon dioxide we obtain:
Best regards.
Problem 2
You start out with 216 ugrams of Fermium - 253. After 3 days, you will have 1/2 as much. 108 ugrams is what you have.
Another 3 days goes by. You started with 108 ugrams. That gets cut in 1/2 again. Now you have 54 ugrams.
Finally another 3 days goes by. You started with 54 ugrams. you now have 1/2 as much which would be 27 ugrams
#days Amount in micrograms
0 216
3 108
6 54
9 27
Problem One
You are using Nitrogen as your base example. The first thing you should do is fill in the table. Then you should try and make some rules. You need the rules in case the exam you are preparing for picks a different element to talk about these bond tendencies. In any event, it's handy to think this way.
<em><u>Table</u></em>
Bond Energy Kj/Mol Bond Length pico meters
N - N 167 145
N=N 418 125
N≡N 942 110
<em><u>Rules</u></em>
As the number of bonds INCREASES, the energy contained in the bond goes UP
As the number of bonds INCREASES, the length of the bond goes DOWN.
