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IgorC [24]
3 years ago
15

Non example of electrical energy

Chemistry
1 answer:
valentinak56 [21]3 years ago
4 0

Answer:

Non-example – creating energy, losing energy. Energy sources - The places we get energy from for heat and electricity. Examples – coal, sun, wind. Non-example – socket, batteries. Energy transformation - The movement of energy where the energy changes forms.

Explanation:

Hope this helps :)

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polet [3.4K]
The answer is 1. A sample of 0.50 moles of gas is placed in a container of volume of 2.5 L. What is the pressure of the gas in torr if the gas is at 25oC? P = 4.89 atm = 3719 torr

2. A sample of gas is placed in a container at 25oC and 2 atm of pressure. If the temperature is raised to 50oC, what is the new pressure? P = 2.17 atm

3. At 1 atm of pressure water boils at 100oC, if the sample was placed under 2 atm of pressure, what would be the temperature? (This would be like a pressure cooker).
T = 746 K = 473oC = 883oF

4. At what temperature would water boil if the pressure is 600 torr? (Use information from problem 3: this shows why food doesn't cook well at higher elevations)
T = 294 K = 21.5oC = 70.7oF

5. Calculate the volume of 40.6 g of F2 at STP. V = 23.9 L

6. A sample of 2.0 moles of hydrogen gas is placed in a container with a volume of 10.4 L. What is the pressure of the gas in torr if the gas is at 25oC? P = 4.70 atm = 3576

7. The tire pressure is 32 psi. What is the pressure in torr if 1 atm = 14.7 psi?
P = 1654 torr

8. A gas is placed in a balloon with a volume of 3.0 L at 28oC and 900 torr. What would be the new volume for the gas if placed under STP? V = 3.2 L

9. How many moles of gas would occupy a volume of 14 L at a pressure of 700 torr and a temperature of 30oC? n = 0.52 mol

10. Calculate the volume of 24.0 g of HCl at STP. V = 14.8 L

11. What is the volume of one mole of acetylene gas at STP? V =22.414 L

12. What is the volume of 0.75 mol of gas at 72oC and 2 atm? V = 10.6 L

13. After eating beans, a student collects a sample of gas at 0.97 atm and 26oC which occupies a volume of 3.5 L, calculate its volume at STP. V = 3.1 L

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15. A mixture of Ar and CO gases is collected over water at 28oC and an atmospheric pressure of 1.05 atm. If the partial pressure of Ar is 600 torr, what is the partial pressure of CO? (vapor pressure of water at 28oC is 28.3 mmHg) PCO = 0.223 atm

16. Determine the partial pressures of each of the gases in the following mixture: 17.04 g NH3, 40.36 g Ne and 19.00 g F2. The gases are at 1.5 atm of pressure.
PNH3 = 0.428 atm; PNe = 0.857 atm; PF2 = 0.2124 atm

17. Potassium chlorate decomposes under heat as follows:

2 KClO3 (s) -------> 2 KCl (s) + 3 O2 (g)

The oxygen gas is collected over water at 25oC. The volume of gas is 560 mL measured at 1 atm. Calculate the number of grams of KClO3 used in the reaction. (vapor pressure of water = 0.0313 atm) nO2 = 0.022 mol; 1.81 g KClO3
5 0
2 years ago
Is water a compound or mixture
enyata [817]

Answer:

compiound

Explanation:

8 0
3 years ago
Read 2 more answers
How does temperature, pressure and surface area affect the dissolving process
larisa [96]
If I am correct, yes. As I was told in chemistry, the surface area affects the dissolving of the "sugar". If you put regular Surat in a hot cup of coffee, it will dissolve at a quick pace, but what if you put the same amount of sugar in the same amour of coffee, but the sugar was fine powder? It would dissolve even faster since it has more surface area. So temperature does affect the dissolving. Hope this helps!
8 0
3 years ago
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How much energy is required when a 142.1 gram sample of aluminum goes from 25.5°C to 46°C? (the specific heat of aluminum is 0.9
vfiekz [6]

Answer:

2621.75 j heat is required to increase the temperature 25.5°C to 46°C.

Explanation:

Given data:

Mass of sample = 142.1 g

Initial temperature = 25.5°C

Final temperature = 46°C

Specific heat capacity of Al = 0.90 J/g.°C

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 46°C - 25.5°C

ΔT = 20.5°C

Q = 142.1 × 0.90 J/g.°C × 20.5°C

Q = 2621.75 j

Thus,  2621.75 j heat is required to increase the temperature 25.5°C to 46°C.

4 0
3 years ago
A 54.2 L sample of gas at 115 K is heated to 345 K, at constant pressure. What volume does the gas now occupy?
sweet [91]
V1/T1 = V2/T2, so 
<span>V2 = V1 * T2 / T1 </span>
<span>V2 = 54.2 L * 345 K / 115 K </span>
<span>V2 = 163 L</span>
7 0
4 years ago
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