The three particles found in an atom are the protons, neutrons and electrons. Protons have a positive charge. Electrons have negative charge. Lastly, neutrons have no net electrical charge. Protons and neutrons are much heavier than electrons and are located in the center of the atom.
Answer:
C. Carbon dioxide
Explanation:
Carbon dioxide is one of the end-product of combustion reactions involving many fuels today.
With the rapid increase in urbanization and technological development, man demand for energy increased tremendously. The discovery of fossil fuels paved the way for the astronomical increase in the concentration of carbon dioxide in the atmosphere. The burning of fossil fuels like coal and oil invovles the process where the carbon atoms present in these fuels combine with oxygen in the air to make CO2. This has resulted in an increase in the concentration of atmospheric carbon dioxide (CO2).
The burning fossil fuels for electricity, industry, heat, and transportation are the major sources of the emossion of carbon dioxide.
Also, the cutting down of trees for paper production, building construction and for the establishment of settlements also increase the concentration of carbon dioxide in the atmosphere. Trees are help remove carbon dioxide from the atmosphere through the process of photosynthesis. However, when these trees are cut down, carbon dioxide accumulates in the atmosphere.
Answer:
3. V = 0.2673 L
4. V = 2.4314 L
5. V = 0.262 L
6. V = 2.224 L
Explanation:
3. assuming ideal gas:
∴ R = 0.082 atm.L/K.mol
∴ V1 = 225 L
∴ T1 = 175 K
∴ P1 = 150 KPa = 1.48038 atm
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(175 K))/((1.48038 atm)(225 L))
⇒ n = 0.043 mol
∴ T2 = 112 K
∴ P2 = P1 = 150 KPa = 1.48038 atm
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(112 K)(0.043 mol))/(1.48038 atm)
⇒ V2 = 0.2673 L
4. gas is heated at a constant pressure
∴ T1 = 180 K
∴ P = 1 atm
∴ V1 = 44.8 L
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(180 K))/((1 atm)(44.8 L))
⇒ n = 0.3295 mol
∴ T2 = 90 K
⇒ V2 = RT2n/P
⇒ V2 = ((0.082 atm.L/K.mol)(90 K)(0.3295 mol))/(1 atm)
⇒ V2 = 2.4314 L
5. V1 = 200 L
∴ P1 = 50 KPa = 0.4935 atm
∴ T1 = 271 K
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(271 K))/((0.4935 atm)(200 L))
⇒ n = 0.2251 mol
∴ P2 = 100 Kpa = 0.9869 atm
∴ T2 = 14 K
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(14 K)(0.2251 mol))/(0.9869 atm)
⇒ V2 = 0.262 L
6.a) ∴ V1 = 24.6 L
∴ P1 = 10 atm
∴ T1 = 25°C = 298 K
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(298 K))/((10 atm)(24.6 L))
⇒ n = 0.0993 mol
∴ T2 = 273 K
∴ P2 = 101.3 KPa = 0.9997 atm
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(273 K)(0.0993 mol))/(0.9997 atm)
⇒ V2 = 2.224 L
Two moles of SO2 means 1.204x10^24 molecules, since there are 3 atoms in one molecule, multiply 1.204 x10^24 by 3 and you get 3.612x10^24.
Explanation:
Deliquescent substances are solids that absorb moisture from the atmosphere until they dissolve in the absorbed water and form solutions. Efflorescent: Efflorescent substances are solids that can undergo spontaneous loss of water from hydrated salts.
