1AlBr3+ 3K ---> 3KBr + 1Al
<u>Answer:</u> The moles of oxygen and carbon dioxide in air is
and
respectively
<u>Explanation:</u>
To calculate the number of moles, we use the equation:

Given mass of atmosphere = 
Average molar mass of atmosphere = 28.96 g/mol
Putting values in above equation, we get:

We know that:
Percent of oxygen in air = 21 %
Percent of carbon dioxide in air = 0.0415 %
Moles of oxygen in air = 
Moles of carbon dioxide in air = 
Hence, the moles of oxygen and carbon dioxide in air is
and
respectively
Answer:
Option B. 4 moles of the gaseous product
Explanation:
Data obtained from the question include:
Initial volume (V1) = V
Initial number of mole (n1) = 2 moles
Final volume (V2) = 2V
Final number of mole (n2) =..?
Applying the Avogadro's law equation, we can obtain the number of mole of the gaseous product as follow:
V1/n1 = V2/n2
V/2 = 2V/n2
Cross multiply
V x n2 = 2 x 2V
Divide both side by V
n2 = (2 x 2V)/V
n2 = 2 x 2
n2 = 4 moles
Therefore, 4 moles of the gaseous product were produced.
Answer:
Sodium Bicarbonate on decomposition produces Carbon dioxide gas and Water vapors.
<span> 2 NaHCO</span>₂<span> </span> →<span> Na</span>₂<span>CO</span>₃<span> (s) </span>+ <span> CO</span>₂<span> (g) + H</span>₂<span>O (g)
</span>
Explanation:
Let suppose you burn 168 g ( 2 moles ) of NaHCO₃, a gas will produced and product is left behind. On measuring the product formed it will be almost equal to 105 g. This shows that the product is Na₂CO₃ and 1 mole of it is being produced after decomposition of sodium bicarbonate.
Answer:
0.48
Explanation:
all you need is to decide 12% with 100% then you multiply it by 4L.