The physical method that can be used for obtaining a sample of salt from a small beaker of salt and water would be evaporation.
<h3>Separation of salt and water</h3>
A mixture of salt and water can be separated by a method known as evaporation. This is based on the assumption that the salt in question is a water-soluble salt.
In order to separate the salt/water mixture:
- Place the mixture in a suitable evaporating dish
- Boil the mixture until all the water evaporates.
- The remaining residue would be the salt
Care should be taken not to overheat the residue in order to avoid melting. Evaporation is generally used to separate a mixture of water and soluble salt. If the salt is insoluble, filtration using a suitable filter paper will filter off the salt while the water is collected as the filtrate.
More on evaporation can be found here: brainly.com/question/1097783
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Which physical method can be used for obtaining a sample of salt from a small beaker of salt water?
<span>Kind of substance besides water:
The best example of hydrogen bonding excluding water is DNA. The two strands of polymers are connected by hydrogen bonds between the nucleotide bases</span>.
To convert Celsius to kelvin, we need to know the formula. Kelvin = Celsius + 273.15. Now, we can figure out how many degrees kelvin -12 degrees Celsius is.
Kelvin = -12 + 273.15
Kelvin = 261.15
The answer is B, early in the morning water covered areas (lakes,ponds,puddles,etc.) will vaporize a little bit because of the heat from the sun and it will continue all day, vapors rise towards the atmosphere and since it's a lot cooler there it will condense into a cloud which is full of tiny frozen water particles. Hope this helps <span />
Answer:
6626 g
Explanation:
Given that:
Density of water = 1.00 g/ml, volume of water = 42800 ml.
Since density = mass/ volume
mass of water = volume of water * density of water = 42800 ml * 1 g/ml = 42800 g
Initial temperature of water = 22°C and final temperature of water = 45°C.
specific heat capacity for water = 4.184 J/g°C
ΔT water = 45 - 22 = 23°C
For iron:
mass = m,
specific heat capacity for iron = 0.444 J/g°C
Initial temperature of iron = 1445°C and final temperature of water = 45°C.
ΔT iron = 45 - 1445 = -1400°C
Quantity of heat (Q) to raised the temperature of a body is given as:
Q = mCΔT
The quantity of heat required to raise the temperature of water is equal to the temperature loss by the iron.
Q water (gain) + Q iron (loss) = 0
Q water = - Q iron
42800 g × 4.184 J/g°C × 23°C = -m × 0.444 J/g°C × -1400°C
m = 4118729.6/621.6
m = 6626 g
