Answer: 7.98 grams of 
are produced if 10.7 grams of 
are reacted.
Explanation:
To calculate the number of moles, we use the equation:
.....(1)
Putting values in equation 1, we get:
The chemical equation for the reaction is
By Stoichiometry of the reaction:
2 moles of produce = 1 mole of
So, 0.100 moles of produce= 
of
Mass of =
Hence 7.98 grams of are produced if 10.7 grams of are reacted.
Answer:
Anita did not make the track team this year, and she feels really disappointed and upset. What question would most likely redirect her thinking and provide a solution?
Why did this terrible thing happen to me?
What other sports can I try this spring?
Why does everything I attempt go wrong?
Why did Karen get on the team but I didn't?
Answer - What other sports can I try this spring?
Explanation:
Failures and successes are part of our daily lives as humans. It can be very challenging for one to accept his or her failures and bring oneself above it.
when failure occur, the best way to reverse the negative impacts it brings is to move forward and decide on what next to do.
The actions that must be taken in the processes should be quite productive enough to bring back the results one desired.
When the effects of failures prove irreversible, one would then give the mind a diversion and help his or herself think their way though the failure that was encountered.
Answer:
The answer to your question is: 2.20 x 10 ²³ molecules
Explanation:
Data
mass = 45.7 g
molecules of CF₂Cl₂ = ?
Process
1.- Calculate the mass number of CF₂Cl₂
C = 12 F =2 x 19 Cl = 2 x 35.5
total = 12 + 38 + 71
total = 121 g
2.- Use the Avogradro's number to solve the problem
121 g ------------------- 6.023 x 10²³ molecules of CF₂Cl₂
45.7 g --------------- x
x = (45.7 x 6.023 x 10²³) / 121
x = 2.75 x 10²⁵ / 121
x = 2.20 x 10 ²³ molecules
Explanation:
Here I send you all the 3 elements that you are asking for.
Notice that first on the left you will find lewis structure, then condensed and finally chemical formula of each of the compound you enlisted.
Limiting reactant : O₂
Mass of N₂O₄ produced = 95.83 g
<h3>Further explanation</h3>
Given
50g nitrous oxide
50g oxygen
Reaction
2N20 + 302 - 2N204
Required
Limiting reactant
mass of N204 produced
Solution
mol N₂O :
mol O₂ :
2N₂O+3O₂⇒ 2N₂O₄
ICE method
1.136 1.5625
1.0416 1.5625 1.0416
0.0944 0 1.0416
Limiting reactant : Oxygen-O₂
Mass N₂O₄(MW=92 g/mol) :
