Answer:
Explanation: It is already known that 1 mole of the gas( or 32g of O2) is equivalent to 22.4 Litres of the oxygen gas. So, 8g is equivalent to = (22.4/32) × 8 = 5.6 L of the gas.
Explanation:
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Answer:
decomposition reaction.
Explanation:
It is a decomposition reaction as potassium chlorate compound breaks to form potassium chloride and oxygen. This reaction requires heat as source of energy to break down the compound so it is endothermic in nature.
Answer:
The age of the sample is 4224 years.
Explanation:
Let the age of the sample be t years old.
Initial mass percentage of carbon-14 in an artifact = 100%
Initial mass of carbon-14 in an artifact = ![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
Final mass percentage of carbon-14 in an artifact t years = 60%
Final mass of carbon-14 in an artifact = ![[A]=0.06[A_o]](https://tex.z-dn.net/?f=%5BA%5D%3D0.06%5BA_o%5D)
Half life of the carbon-14 = 
![[A]=[A_o]\times e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-kt%7D)
![[A]=[A_o]\times e^{-\frac{0.693}{t_{1/2}}\times t}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-%5Cfrac%7B0.693%7D%7Bt_%7B1%2F2%7D%7D%5Ctimes%20t%7D)
![0.60[A_o]=[A_o]\times e^{-\frac{0.693}{5730 year}\times t}](https://tex.z-dn.net/?f=0.60%5BA_o%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-%5Cfrac%7B0.693%7D%7B5730%20year%7D%5Ctimes%20t%7D)
Solving for t:
t = 4223.71 years ≈ 4224 years
The age of the sample is 4224 years.
Answer:
1120 gm
Explanation:
6. Consider the reaction: CzHo (g) + 02 (8) - 4 CO2(g) + 6H2O (1)
(a) Balance the equation.
(b) How many grams of oxygen are required to react with 10 moles of ethane for a complete
combustion reaction?
FIRST, CORRECT THE EQUATION THEN BALANCE
2C2H6(G) + 7O2------------> 4CO2 + 6H2O
so for 10 moles of ethane, we need
7 X 5 = 35 MOLES O2
=35 MOLES O2
O2 HAS A MOLAR MASS OF 2X16 = 32 gm
35 MOLES OF O2 HAS A MASS OF 35 X 32 =1120 gm
