Answer:
You can increase the validity of an experiment by controlling more variables, improving measurement technique, increasing randomization to reduce sample bias, blinding the experiment, and adding control or placebo groups.
Answer:
A; deciduous
Explanation:
Trees that lose all of their leaves for part of the year are known as deciduous trees. Those that don't are called evergreen trees. Common deciduous trees in the Northern Hemisphere include several species of ash, aspen, beech, birch, cherry, elm, hickory, hornbeam, maple, oak, poplar and willow
Answer:
tetrahedral
As carbon has no 4 valence electrons that are occupied by 4 chlorine atoms, we get 0 lone pair in carbon and sp3 hybridisation which leads to tetrahedral shape.
*Note: there seems to be a "typo" in the given problem. The "1000" must probably be "100" only since summing up the total number of atoms given for the different isotopes would not equal 1000 but 100 instead. Using 1000 as the basis for the mass fractions would give an answer which is 10 times less than should be.
To determine the average atomic mass of halfnium, the mass fractions of the isotopes multiplied by their respected atomic masses must all be added.
1. Determining the mass fractions (m1, m2...m5) with 100 atoms total as the basis:
m1 = 5/100 = 0.05
m2 = 19/100 = 0.19
m3 = 27/100 = 0.27
m4 = 14/100 = 0.14
m5 = 35/100 = 0.35
2. Multiplying the mass fractions with the atomic masses of the respective isotopes.
Average atomic mass of Halfnium
= (m1*176) + (m2*177) + (m3*178) + (m4*179) + (m5*180)
= (0.05*176) + (0.19*177) + (0.27*178) + (0.14*179) + (0.35*180)
= 178.55 amu
Thus, the average atomic mass of Halfnium based on the given data for its isotopes is 178.55 amu.
Answer:
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