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
Answer:
OPTION D,
AS AT THIS PLACE ONLY THIS FACTIR CAN OCCUR..
Answer:
0.095 moles of O₂ are left over.
Explanation:
First of all, state the balanced reaction:
2NO + O₂ → 2NO₂
We determine moles of each reactant:
20.2 g . 1mol / 30g = 0.673 moles of NO
13.8g . 1mol / 32g = 0.431 moles of oxygen
Oxygen is the excess reactant. Let's see.
For 2 moles of NO I need 1 mol of O₂
Then, for 0.673 moles of NO I may use (0.673 .1) /2 = 0.336 moles
I have 0.431 moles of O₂ and I only need 0.336 mol. According to reaction, stoichiometry is 2:1.
In conclussion, the moles of excess reactant that will be left over:
0.431 - 0.336 = 0.095 moles
Answer:
The answer is in the problem
Explanation:
As general rule of number of oxygen is -2:
O → -2
Alkali metals (Li, Na, K) are always +1
Na → +1
Alkali earth methals (Be, Mg, Ca...) are always +2
Ca → +2
The halogen group (F, Cl, Br...) is always -1
F → -1
The oxidation number of Si (+/- 4)
Aluminium is, usually +3
And to complete the octet rule in nitrogen, 3 electrons are required. That means:
N → -3
