Warmer air is less dense than cold air.As air warm it rises while the cold air sink. Warmer air masses forces the cooler air to move which causes wind. These is illustrated when you open a hot oven the hotter air inside the oven rises into cooler air outside the oven.
Answer:
0.082g
Explanation:
The following data were obtained from the question:
Heat (Q) = 0.092J
Change in temperature (ΔT) = 0.267°C
Specific heat capacity (C) of water = 4.184J/g°C
Mass (M) =..?
Thus, the mass of present can be obtained as follow:
Q = MCΔT
0.092 = M x 4.184 x 0.267
Divide both side by 4.184 x 0.267
M = 0.092 / (4.184 x 0.267)
M = 0.082g
Therefore, mass of water was present is 0.082.
The electron domain geometry is trigonal bipyramidal while the molecular geometry of the compound is seesaw.
The shapes of molecules is determined by the number of electron pairs on the valence shell of the central atom in the molecule. These electron domains include lone pairs and bond pairs.
The lone pairs only contribute towards the electron domain geometry and not the molecular geometry. SCl4 has five electron domains hence its electron domain geometry is trigonal bipyramidal. The molecular geometry of the compound is seesaw.
Learn more: brainly.com/question/6505878
Answer:
Electrolysis
Explanation:
The electrolysis of water is one such experiment that shows that water is made up of hydrogen and oxygen atoms only in the ratio of 2 to 1.
In the electrolysis of water, electricity is passed through acidified water to cause it to decompose.
The electrolysis of water is also known as the electrolysis of dilute tetraoxosulphate (VI) acid.
At the cathode, H⁺ ions are discharged and hydrogen gas is liberated:
2H⁺ + 2e⁻ → H₂
At the anode, both the sulfate ion and hydroxyl ions migrate to this electrode. Only the OH⁻ is selected for preferential discharge due to its lower position in that activity series.
4OH⁻ → 2H₂O + O₂ + 4e⁻
Oxygen gas is produced at the anode.
This electrolysis demonstrates the volumetric composition of water that is, 2 volumes of hydrogen at the cathode and 1 volume of oxygen at the anode.
C + H2O -> H2 + CO
n(C) = 15.9/12 = 1.325 (mol)
=> n(H2) = 1.325 mol
We have:
PV = nRT
=> V = (nRT)/P
(R = 22.4/273 = 0.082)
V = (1.325 x 0.082 x 360)/1 = 39.114 (L)
