4. The half-life of C-14 is 30 years. If a scientist had 100.0 g at the beginning, how
many grams would be left after 120 years has elapsed?
Half Lives
0
1
2
Time (years)
Amount (grams)
Show table
Answer:
6,25 gm
Explanation:
4. The half-life of C-14 is 30 years. If a scientist had 100.0 g at the beginning, how
many grams would be left after 120 years has elapsed?
10 years would be 120/30 = 4 half lives0
s0 the fraction of c-14 left would be four (1/2) multiplied together
1/2 X 1/2 X 1/2 X 1/2 = 1/16
SO 1/16 OF THE 100.0 gm would be left
100.0/16 = 6,25 gm
help me. correct answer will be marked as brnlst
Answer:
PCI3
Explanation:
It does not obey the octet rule on the nitrogen atom.
Methane (CH4) is used in laboratory burner. When 1 mole of methane burns at constant pressure, it produces 804 kJ of heat and does 3 kJ of work.Methane (CH4) is used in laboratory burner. When 1 mole of methane burns at constant pressure, it produces 804 kJ of heat and does 3 kJ of work.
The enthalpy ΔH of combustion is - 804 kJ. The change in internal energy ΔE of the reaction process is -801 kJ
The objective of this question is to determine the value of ΔH and ΔE of combustion is for 1 mole of methane.
If we look closely at the question, we will realize that at constant pressure, methane burns and produces heat of 804 kJ into the surroundings.
Since heat is released;
ΔH = -804 kJ
It also does a work of 3kJ, since work is done on the system
W = - 3kJ
According to the first law of thermodynamics;
ΔE = ΔH - W
ΔE = (-804 - (-3 )) kJ
ΔE = -801 kJ
Therefore, we can conclude that the in 1 mole of methane, the enthalpy ΔH of combustion is - 804 kJ. The change in internal energy ΔE of the reaction process is -801 kJ
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A sample of carbon dioxide gas has a pressure of 8.33 bar and a volume of 10.3 mL at a temperature of 29.1 °C. How many
molecules of carbon dioxide gas are in the sample?
molecules:
Using the metric "stairs" convert the following:
The heat of combustion of propane, C3H8 (g) is -2057 kJ/mol. What would be the enthalpy change if enough propane was burned to give off 12 moles of carbon dioxide gas
Considering the reaction stoichiometry, the enthalpy change if enough propane was burned to give off 12 moles of carbon dioxide gas is 8228 kJ.
The balanced reaction is:
C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l)
The heat of combustion of propane, C₃H₈, is -2057 kJ/mol. This is, 2057 kJ is released for every 1 mol C₃H₈.
So to determine the enthalpy change if enough propane was burned to emit 12 moles of carbon dioxide, you must take into account the stoichiometry of the reaction.
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
C₃H₈: 1 mole O₂: 5 moles CO₂: 3 moles H₂O: 4 molesThen you can apply the following rule of three: if by stoichiometry 3 moles of CO₂ are produced by 1 mole of C₃H₈, 12 moles of CO₂ are produced by how many moles of C₃H₈?
[tex]amount of moles of C_{3} H_{8} =\frac{12 moles of CO_{2}x1 mole of C_{3} H_{8} }{3 moles of CO_{2}}[/tex]
amount of moles of C₃H₈= 4 moles
So to determine the enthalpy change, you can apply the following rule of three: If for each mole of C₃H₈ 2057 kJ are released, for 4 moles of C₃H₈ how much heat is released?
[tex]Heat released=\frac{4 molesx2057 kJ}{1 mole}[/tex]
Heat released= 8228 kJ
The enthalpy change if enough propane was burned to give off 12 moles of carbon dioxide gas is 8228 kJ.
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