C) Sliver Carbonate AgCO3
The reaction for the combustion of methane can be expressed as follows.
CH4 + 2O2 --> CO2 + 2H2O
We solve first for the amount of carbon dioxide in moles by dividing the given volume by 22.4L which is the volume of 1 mole of gas at STP.
moles of CO2 = (5.6 L) / (22.4 L/1 mole)
moles of CO2 = 0.25 moles
Then, we can see that every mole of carbon dioxide will need 1 mole of methane
moles methane = (0.25 moles CO2) x (1 moles O2/1 mole CO2)
= 0.25 moles CH4
Then, multiply this by the molar mass of methane which is 16 g/mole. Thus, the answer is 4 grams methane.
13.48 is your answer.
M=moles/liters we have M=.115 and moles=1.55
Input those values into the equation
.115=1.55/L
Solve for L
1.55/.115=13.48
Answer:
Aluminum atoms are in 3.78 g of aluminum
02214076x10^23)8.4367659370640769254888x10^22 atoms, therefore there are 8.4367659370640769254888x10^22 atoms of Aluminum(Al) present in 3.78 grams of Aluminum(Al).
Explanation:
