When moist air cools below its dew point on a cold surface it forms dew. It is in the liquid state.
Rain is again a form of precipitation in which water is in the liquid state.
When air temperature is below freezing, the precipitation that results is referred to as frost. Hail is a form of precipitation in the ice balls whereas as sleet is a mixture of rain and snow.
Ans: Forms of frozen water- frost, hail and sleet
Answer:
Formation of intermolecular hydrogen bonding between water molecules and molecules of n-butanol
Explanation:
Low molecular weight alcohols are miscible with water in all proportions. The reason for this is that, when a low molecular weight alcohol is dissolved in water, intermolecular hydrogen bonds are formed between the low molecular weight alcohol and water molecules.
Low molecular weight alcohols such as n-butanol contain the polar -OH group which interacts with water via hydrogen bonding.
Answer:
it will be classical as gas
Answer:
Explanation:
Sn(WC)2
if it is tungsten carbide this should be correct but there are many versions of carbide
Sn(MC2)2
could also be possible
the 2 next to MC should be a subscript
<span>C2H5
First, you need to figure out the relative ratios of moles of carbon and hydrogen. You do this by first looking up the atomic weight of carbon, hydrogen, and oxygen. Then you use those atomic weights to calculate the molar masses of H2O and CO2.
Carbon = 12.0107
Hydrogen = 1.00794
Oxygen = 15.999
Molar mass of H2O = 2 * 1.00794 + 15.999 = 18.01488
Molar mass of CO2 = 12.0107 + 2 * 15.999 = 44.0087
Now using the calculated molar masses, determine how many moles of each product was generated. You do this by dividing the given mass by the molar mass.
moles H2O = 11.5 g / 18.01488 g/mole = 0.638361 moles
moles CO2 = 22.4 g / 44.0087 g/mole = 0.50899 moles
The number of moles of carbon is the same as the number of moles of CO2 since there's just 1 carbon atom per CO2 molecule.
Since there's 2 hydrogen atoms per molecule of H2O, you need to multiply the number of moles of H2O by 2 to get the number of moles of hydrogen.
moles C = 0.50899
moles H = 0.638361 * 2 = 1.276722
We can double check our math by multiplying the calculated number of moles of carbon and hydrogen by their respective atomic weights and see if we get the original mass of the hydrocarbon.
total mass = 0.50899 * 12.0107 + 1.276722 * 1.00794 = 7.400185
7.400185 is more than close enough to 7.40 given rounding errors, so the double check worked.
Now to find the empirical formula we need to find a ratio of small integers that comes close to the ratio of moles of carbon and hydrogen.
0.50899 / 1.276722 = 0.398669
0.398669 is extremely close to 4/10, so let's reduce that ratio by dividing both top and bottom by 2 giving 2/5.
Since the number of moles of carbon was on top, that ratio implies that the empirical formula for this unknown hydrocarbon is
C2H5</span>
