Answer:
Circular motion
Explanation:
Consider a car race, where you need to finish one lap in a circuit. Her the start line and the finish line are the same. So , where you start, is where you will be reaching after the race is done. So, the displacement here is <em>Zero</em> , since displacement is the distance from initial point to final point. But, as one can see, the distance travelled is not zero.
Say the race track is a circle of radius r meters. So the lap distance would be the circumference of the circle, which is 2πr m. So the distance travelled is 2πr m and the displacement is zero.
Hope this satisfies you :)
Answer:
The actual yield of iron is 0.208 moles.
Explanation:
Given data:
Moles of Fe₂O₃ = 0.301 mol
Mass of iron = 11.6 g
Actual yield of iron in moles = ?
Solution:
Number of moles = mass / molar mass
Number of moles = 11.6 g/ 55.85 g/mol
Number of moles = 0.208 mol
It is the actual yield of iron in moles.
Theoretical yield can be calculated from given data.
One mole of Fe₂O₃ produce two mole two mole of iron.
0.301 moles will produce:
0.301× 2 = 0.602 mol
Theoretical yield in moles will be 0.602 moles.
Answer:
Water is called the "universal solvent" because it is capable of dissolving more substances than any other liquid.
Explanation:
Answer:
addition polymerization
Explanation:
In addition polymerization, the monomers are simply joined to each other to form a polymer having the same empirical formula as the monomer but of higher relative molecular mass. The monomers in addition polymerization are usually simple unsaturated molecules such as alkenes.
We can deduce the reaction to be an addition polymerization because of the the attachment of n to both the unsaturated monomer and the saturated polymer without the loss of any small molecule. If it was a condensation polymerization, there would have been an accompanying loss of a small molecule such as water.
Explanation:
Set up the ion formation equations, with ionization energy values for each electron in the valence layer, of the atoms of the chemical elements below: a) Na z=11 b) Ca z= 20 c) Sr z 38 d) Li z= 3 e) Cs z= 55 f) Be z= 43
