You must burn 1.17 g C to obtain 2.21 L CO2 at
STP.
The balanced chemical equation is
C+02+ CO2.
Step 1. Convert litres of CO, to moles of CO2.
STP is 0 °C and 1 bar. At STP the volume of 1 mol
of an ideal gas is 22.71 L.
Moles of CO2= 2.21 L CO2 × (1 mol CO2/22.71 L
CO2) = 0.097 31 mol CO2
Step 2. Use the molar ratio of C:CO2 to convert
moles of CO to moles of C
Moles of C= 0.097 31mol CO2 × (1 mol C/1 mol
CO2) = 0.097 31mol C
Step 3. Use the molar mass of C to calculate the
mass of C
Mass of C= 0.097 31mol C × (12.01 g C/1 mol C) =
1.17 g C
It looks as if you are using the old (pre-1982)
definition of STP. That definition gives a value of
1.18 g C.
Answer:
hope it helps...
Explanation:
Along the third type of plate boundary, two plates move laterally and pass each other along giant fractures in Earth's crust. Transform faults are so named because they are linked to other types of plate boundaries. The majority of transform faults link the offset segments of oceanic ridges.
Answer:
yes
Explanation:
Usually, it would not affect the crucible, but depending on the temperature of the flame the enamel of the crucible may begin to melt and stick to the metal object being used to handle the crucible. This tiny amount that is melted off can cause very small changes in the original mass of the crucible, which although it is almost unnoticeable it is still there. Therefore, the answer to this question would be yes.
Answer:Acids taste sour, react with metals, react with carbonates, and turn blue litmus paper red. Bases taste bitter, feel slippery, do not react with carbonates and turn red litmus paper blue.
Explanation:
- Sour taste (though you should never use this characteristic to identify an acid in the lab)
- Reacts with a metal to form hydrogen gas.
- Increases the H+ concentration in water.
Answer: 18.65L
Explanation:
Given that,
Original volume of oxygen (V1) = 30.0L
Original temperature of oxygen (T1) = 200°C
[Convert temperature in Celsius to Kelvin by adding 273.
So, (200°C + 273 = 473K)]
New volume of oxygen V2 = ?
New temperature of oxygen T2 = 1°C
(1°C + 273 = 274K)
Since volume and temperature are given while pressure is held constant, apply the formula for Charle's law
V1/T1 = V2/T2
30.0L/473K = V2/294K
To get the value of V2, cross multiply
30.0L x 294K = 473K x V2
8820L•K = 473K•V2
Divide both sides by 473K
8820L•K / 473K = 473K•V2/473K
18.65L = V2
Thus, the new volume of oxygen is 18.65 liters.