Answer:
i think it will increase the rate of chemical reaction as pressure is directly proportional to the reactivity of gas.
True because it is warmer closer to the equator
Yes, the law of conservation of mass holds.
Explanation:
In every chemical reaction, mass is always conserved. This implies that in chemical reaction, the process proceeds maintaining the same set of atom in the same proportion without new ones forming.
- From the description given, decomposition of the Silver carbonate will produce a silver residue with the given mass.
- The other mass that seems lost can be examined to be given off as carbon dioxide gas which is the other product of the reaction and oxygen.
- Therefore, since the products are silver, carbon dioxide and oxygen, the remaining mass is that of the carbon dioxide and oxygen. It is not lost.
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Explanation:
Rain is when water falls from clouds in droplets that are bigger than 0.5 mm. Droplets of water that are about 0.2mm to 0.45mm big are called drizzle. Rain is a kind of precipitation. Precipitation is any kind of water that falls from clouds in the sky, like rain, hail, sleet and snow.
Other hand, Water vapor, water vapour or aqueous vapor is the gaseous phase of water. It is one state of water within the hydrosphere. Water vapor can be produced from the evaporation or boiling of liquid water or from the sublimation of ice. Water vapor is transparent, like most constituents of the atmosphere.
Answer:
T° freezing solution → -11.3°C
T° boiling solution → 103.1 °C
Explanation:
Assuming 100 % dissociation, we must find the i, Van't Hoff factor which means "the ions that are dissolved in solution"
This salt dissociates as this:
SnCl₄ (aq) → 1Sn⁴⁺ (aq) + 4Cl⁻ (aq) (so i =5)
The formula for the colligative property of freezing point depression and boiling point elevation are:
ΔT = Kf . m . i
where ΔT = T° freezing pure solvent - T° freezing solution
ΔT = Kb . m . i
where ΔT = T° boiling solution - T° boiling pure solvent
Freezing point depression:
0° - T° freezing solution = 1.86°C/m . 1.22 m . 5
T° freezing solution = - (1.86°C/m . 1.22 m . 5) → -11.3°C
Boiling point elevation:
T° boiling solution - 100°C = 0.512 °C/m . 1.22 m . 5
T° boiling solution = (0.512 °C/m . 1.22 m . 5) + 100°C → 103.1 °C
