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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Why do we need to build models of molecules to study atomic composition?
Answer:
for better understanding and visual understanding
what are the strongest acids and bases in chemstry
Consider the following chemical equation:
C2H8 + O2 --> CO2 + H2O
Which of the following would be the best practice(s) for balancing the equation? There may or may not be more than one correct answer.
A.Reduce the coefficients if possible.
B.Put a 2 in front of the hydrocarbon.
C.Split the hydrocarbon into individual elements.
D.Split the water into HOH.
E.Keep the hydrocarbon together.
The best way to balancing a combustion equation of an organic compound, like the one in the question, is to start balancing the carbon, then the hydrogen and finish with the oxygen. For that you have to split the hydrocarbon into individual elements. And once you have finished you shall always try to reduce the coefficient if possible. Therefore, the answers are A. and C.
Using the metric "stairs" convert the following:
Recreational drug users often obtain the drug as a white powder with unknown purity. Some users snort the power, whereas others make a solution and take it orally or inject it. To achieve a ketamine high, users usually need to take between 0.1 mg/kg and 1 mg/kg of ketamine. However, taking 2 mg/kg can lead the user to experience a K-hole or to enter an unconscious state.
If the user dissolves exactly 1 g of ketamine in 1/4 cup of water, what is the concentration of ketamine in milligrams per milliliter (mg/mL)? 1 cup=236.5 mL
What volume, in milliliters, of the recreational ketamine in the previous question would a 59.0 kg user need to take to experience a high at 0.600 mg/kg?
How many milliliters would that same user need to take to become unconscious at 2.00 mg/kg?
A user dissolves exactly 1 g of ketamine in 1/4 cup of water.
Given 1 cup = 236.5 mL, the volume of 1/4 cup of water is:
1/4 cup × (236.5 mL/1 cup) = 59.13 mL
1 g of ketamine is dissolved in 59.13 mL of water. The concentration of ketamine, in mg/mL, is:
C = 1000 mg / 59.13 mL = 16.91 mg/mL
Part 2The volume of this solution that a 59.0 kg user (body mass) would have to consume to experience a high at 0.600 mg/Kg is:
[tex]59.0 kgBM \times \frac{0.600mgKetamine}{1kgBM} \times \frac{1mL Soluiton}{16.91mgKetamine} = 2.09 mL[/tex]
Part 3The volume of this solution that a 59.0 kg user (body mass) would have to consume to become unconscious at 2.00 mg/kg is:
[tex]59.0 kgBM \times \frac{2.00mgKetamine}{1kgBM} \times \frac{1mL Soluiton}{16.91mgKetamine} = 6.98 mL[/tex]
If a user dissolves exactly 1 g of ketamine in 1/4 cup of water, the resulting concentration is 16.91 mg/mL. A 59.0 kg user would have to consume 2.09 mL of this solution to experience a high of 0.600 mg/kg. The same user would have to consume 6.98 mL of this solution to become unconscious at 2.00 mg/kg.Learn more about concentration units here: https://brainly.com/question/14594475
What equipment is generally used to make lyophilized medications suitable for administering to the patient? a) Test tubes or sterile ampules b) Petri dishes and sterile droppers c) Sterile syringes or graduated cylinders d) Measuring cups and clean, warm water
Answer:
B) Sterile droppers and petri dishes
Calculate the freezing point of a solution of 3.46 g of a compound, X, in 160 g of benzene. When a separate sample of X was vaporised, its density was found to be 3.27 g/Lat 116°C and 773 torr. The freezing point of pure benzene is 5.45°C, and Kis 5.12°C kg/mol.
Answer:
Calculate the freezing point of a solution of 3.46 g of a compound,X,in 160 g of benzene. When separate sample of X was vaporised, its density was found 3.27 g/L at 116 c and 773 torr. The freezing point of pure benzen is 5.45 c, and kf is 5 .12 c/m.
Explanation:
First determine molar mass from
P*molar mass = density*R*T
Solve for molar mass. Then
moles = grams/molar mass
Solve for moles
molality = moles/kg solvent
Solve for molality
delta T = Kf*m
Solve for delta T, then convert to freezing point
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.
give some practical examples where prefixes are used
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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1. A closed bottle contains a mixture of Ar and O2 at 25.3°C. If the partial pressure of Ar in
the bottle is 0.447 atm and the total pressure of the bottle is 1.024 atm, what is the partial
pressure of O2 in the bottle?
Answer: 0.79 atm
Explanation:
The Ideal Gas Law is PV=nRT.
since both Ar and O2 have the same volume and temperature, the only variable causing the differnce in pressure betwwen the 2 gases is the number of moles
the total pressure = the partial pressure of Ar + the partial pressure of O2
1.024 atm = 0.447 amt + ?
so
? = 1.24 -0,447 = 1.24 - 0.45= 0.79 atmthe total pressure of the bottle is 1.024 atm, what is the partial
pressure of O2 in the bottle? is PV
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:
Calculate energy output (ATP production) of full oxidation of saturated fatty acid with 24 carbons in the chain?
..
Explain