False because the kenetic energy is non proficient
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:Locate the row corresponding to known unit of torque along the left of the table. Multiply by the factor under the column for the desired units. For example, to convert 2 oz-in torque to n-m, locate oz-in row at table left. Locate 7.062 x 10-3 at intersection of desired n-m units column. Multiply 2 oz-in x (7.062 x 10-3 ) = 14.12 x 10-3 n-m.
Converting between units is easy if you have a set of equivalencies to work with. Suppose we wanted to convert an energy quantity of 2500 calories into watt-hours. What we would need to do is find a set of equivalent figures for those units. In our reference here, we see that 251.996 calories is physically equal to 0.293071 watt hour. To convert from calories into watt-hours, we must form a “unity fraction” with these physically equal figures (a fraction composed of different figures and different units, the numerator and denominator being physically equal to one another), placing the desired unit in the numerator and the initial unit in the denominator, and then multiply our initial value of calories by that fraction.
Explanation:
Since both terms of the “unity fraction” are physically equal to one another, the fraction as a whole has a physical value of 1, and so does not change the true value of any figure when multiplied by it. When units are canceled, however, there will be a change in units. For example, 2500 calories multiplied by the unity fraction of (0.293071 w-hr / 251.996 cal) = 2.9075 watt-hours.
Aluminum oxide produced : = 79.152 g
<h3>Further explanation</h3>
Given
46.5g of Al
165.37g of MnO
Required
Aluminum oxide produced
Solution
Reaction
2 Al (s) + 3 MnO (s) → 3 Mn (s) + Al₂O₃ (s)
mol = mass : Ar
mol = 46.5 : 27
mol = 1.722
mol = 165.37 : 71
mol = 2.329
mol : coefficient ratio Al : MnO = 1.722/2 : 2.329/3 = 0.861 : 0.776
MnO as a limiting reactant(smaller ratio)
So mol Al₂O₃ based on MnO as a limiting reactant
From equation , mol Al₂O₃ :
= 1/3 x mol MnO
= 1/3 x 2.329
= 0.776
Mass Al₂O₃ (MW=102 g/mol) :
= 0.776 x 102
= 79.152 g
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