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fomenos
3 years ago
12

How do surface waves form?

Chemistry
1 answer:
ivann1987 [24]3 years ago
7 0
Ocean surface waves are surface waves that occur at the surface of an ocean. They usually result from distant winds or geologic effects and may travel thousands of miles before striking land. They range in size from small ripples to huge tsunamis. There is surprisingly little actual forward motion of individual water particles in a wave, despite the large amount of forward energy it may carry.
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Which is the best term to use when describing the energy of position?
Minchanka [31]
Answer: potential.

Chemical energy is the energy provided by a chemical reaction.

Kinetic energy is the energy due to the speed.

Potential energy is the energy due to the position. For example, an object on the top of a mountain, has the possibility to perform work if it falls.

Electromagnetic energy. is propagated by waves: radio waves, infrared radiation, microwaves, etc.
4 0
4 years ago
Read 2 more answers
The combustion of gasoline produces carbon dioxide and water. Assume gasoline to be pure octane (C8H18) and calculate the mass (
Mnenie [13.5K]

Answer:

3.1 kg

Explanation:

Step 1: Write the balanced combustion equation

C₈H₁₈ + 12.5 O₂ ⇒ 8 CO₂ + 9 H₂O

Step 2: Calculate the moles corresponding to 1.0 kg of C₈H₁₈.

The molar mass of C₈H₁₈ is 114.23 g/mol.

1.0 × 10³ g × 1 mol/114.23 g = 8.8 mol

Step 3: Calculate the moles of CO₂ produced from 8.8 moles of C₈H₁₈

The molar ratio of C₈H₁₈ to CO₂ is 1:8. The moles of CO₂ produced are 8/1 × 8.8 mol = 70 mol.

Step 4: Calculate the mass corresponding to 70 moles of CO₂

The molar mass of CO₂ is 44.01 g/mol.

70 mol × 44.01 g/mol = 3.1 × 10³ g = 3.1 kg

5 0
3 years ago
Consider the following reaction:
iren [92.7K]

Answer:

A. ΔG° = 132.5 kJ

B. ΔG° = 13.69 kJ

C. ΔG° = -58.59 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.

ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)

where,

n: moles

ΔH°f: standard enthalpy of formation

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CaCO₃(s))

ΔH° = 1 mol × (-635.1 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1206.9 kJ/mol)

ΔH° = 178.3 kJ

We can calculate the standard entropy of the reaction (ΔS°) using the following expression.

ΔS° = ∑np . S°p - ∑nr . S°r

where,

S: standard entropy

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)) - 1 mol × S°(CaCO₃(s))

ΔS° = 1 mol × (39.75 J/K.mol) + 1 mol × (213.74 J/K.mol) - 1 mol × (92.9 J/K.mol)

ΔS° = 160.6 J/K. = 0.1606 kJ/K.

We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.

ΔG° = ΔH° - T.ΔS°

where,

T: absolute temperature

<h3>A. 285 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 285K × 0.1606 kJ/K = 132.5 kJ

<h3>B. 1025 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1025K × 0.1606 kJ/K = 13.69 kJ

<h3>C. 1475 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1475K × 0.1606 kJ/K = -58.59 kJ

5 0
3 years ago
100 points
Mariana [72]

Answer:

Where is the puzzle!??

I surely can help you

5 0
3 years ago
Read 2 more answers
Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask. be sure your answer has the correct number
Vera_Pavlovna [14]
<span>A chemist adds 155.0ml of a 4.10 X 10^-5 mmol/L of a zinc oxalate (ZnC2O4)solution to a reaction flask. Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask. 1mmol = 10^-3 mol Therefore 4.10*10^-5mmol = 4.10*10^-8mol molar mass ZnC2O4 = 65.39+(2*12.011)+(4*15.99) = 153.372g/mol You have 4.10*10^-8 mol/litre =153.372 * 4.10*10^-8 = 6.29*10^-6 grams / litre (* see below) But you have 155ml. Mass of ZnC2O4 = 155/1000*6.29*10^-6 g Mass is = 9.75*10^-7 grams 1µg = 10^-6 g You then have 9.75*10^-7/10^-6 = 0.975µg ZnC2O4 (*see below) at this point you could have said: 1µg = 10^-6 g therefore you have a solution of 6.29µg per litre, 155ml = 6.29*155/1000 = 0.975µg ZnC2O4</span>
3 0
3 years ago
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