The density of pure water is 1 g/cm^3.
Its density is 0.98 g cm 3 at room temperature, in comparison with the handiest zero.92 g cm 3 for ice, a reality that has to be defined through atomic, and molecular concepts. If ice has been no longer much less dense than water, it might sink, having a devastating impact on lake backside ecosystems. believe it or now not, ice is honestly about 9% much less dense than water. for the reason that water is heavier, it displaces the lighter ice, causing the ice to glide to the pinnacle.
The density of ice is about 90 percent that of water, but that could range because ice can contain air, too. meaning that about 10 percent of an ice cube or iceberg will be above the water line. The density of water is maximum at four∘C, and the density of the ice is much less than the water due to its susceptible intermolecular pressure of attraction. as the density of water is more, it's miles heavier than ice. therefore ice floats on the floor of the water. Ice continually floats due to the fact it's far less dense than everyday water. because frozen water molecules shape a crystal.
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Answer:
1.56 g of water was involved in the reaction
Explanation:
From the stoichiometric equation
2Na + 2H2O = 2NaOH + H2
NB : Mm Na= 23, Mm H2O = ( 2+16)= 18
2(23) of Na requires 2(18) of water
Hence 1.99g of Na will require 1.99×2×18/2(23) of water = 1.56 g of water
The question is incomplete, here is the complete question.
A chemist prepares a solution of copper(II) fluoride by measuring out 0.0498 g of copper(II) fluoride into a 100.0mL volumetric flask and filling the flask to the mark with water.
Calculate the concentration in mol/L of the chemist's copper(II) fluoride solution. Round your answer to 3 significant digits.
<u>Answer:</u> The concentration of copper fluoride in the solution is 
<u>Explanation:</u>
To calculate the molarity of solute, we use the equation:
We are given:
Given mass of copper (II) fluoride = 0.0498 g
Molar mass of copper (II) fluoride = 101.54 g/mol
Volume of solution = 100.0 mL
Putting values in above equation, we get:
Hence, the concentration of copper fluoride in the solution is
The activation energy Ea can be related to rate constant (k) at temperature (T) through the equation:
ln(k2/k1) = Ea/R[1/T1 - 1/T2]
where :
k1 is the rate constant at temperature T1
k2 is the rate constant at temperature T2
R = gas constant = 8.314 J/K-mol
Given data:
k1 = 0.543 s-1; T1 = 25 C = 25+273 = 298 K
k2 = 6.47 s-1; T = 47 C = 47+273 = 320 K
ln(6.47/0.543) = Ea/8.314 [1/298 - 1/320]
2.478 = 2.774 *10^-5 Ea
Ea = 0.8934*10^5 J = 89.3 kJ
