give some practical examples where prefixes are used

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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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

Answer:

PCI3

Explanation:

It does not obey the octet rule on the nitrogen atom.

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  

the total pressure of the bottle is 1.024 atm, what is the partial

pressure of O2 in the bottle? is PV

Answer:

for better understanding and visual understanding

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:

Then 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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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

The 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]

The 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]

Learn more about concentration units here: https://brainly.com/question/14594475

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

Answer:

B) Sterile droppers and petri dishes