<span>the noble gases are those who occupy the eighth group of the periodic table and are so called because having the complete valence shell do not need to form bonds with other atoms and are thus in atomic form ...
</span>
The noble gases (also called rare gases) are of the inert gases that constitute the eighteenth [1] group of the periodic table of the elements, ie, the right-most column. They consist of atoms with electron shells full. It includes the following elements: helium neon argon krypton xenon radon Ununoctium <span>Sometimes they (particularly helium) are located together with other gases (mostly nitrogen and methane) into endogenous sources; helium of endogenous origin comes from the decomposition of radioactive elements present in the subsurface that emit α particles (ie ions He2 +): These oxidized species present in the soil and become elio.I atoms of the noble gases are all monatomic gas, not easily liquefiable, present the atmosphere in different percentages; the most common is argon which is approximately the 0.932%.</span>
Answer:
1. No
2.a. Nothing will happen to figure 1 as both the sides have 30 N.
2.b. The force with 30 N will push 10 N because 10 N is less force than 30 N.
Answer is: a beaker contains <span>
heterogeneous mixture.
</span>
A heterogeneous mixture<span> have compounds that remain separate in the sample.</span>
Heterogeneous
mixture is not uniform in composition (in this mixture different sand and small pebbles), but proportions of its components (in this
mixture particles of different colors and size) vary throughout
the sample.
so, There are (1/12.0)moles with each mole having 6.022 x10^23 atoms. 1 mole of carbon-12 has 6.02 x 1023 iotas.
Answer:
The dilution factor of protein in tube # 4 is 125. Molar concentration is 0.0088 M protein
Explanation:
The dilution factor indicates how many times is more concentrated a main solution in relationship with a diluted solution. In this case, the main solution is in tube #1. For calculating the dilution factor and molar concentration in tube #4 we need the main solution concentration which comes from next equation:
Initial volume * initial concentration = final volume * final concentration
0.5 mL * 10M = 5mL * final concentration
1.1 M = final concentration = main solution concentration
Applying the same equation for remain tubes we have 0.22 M for tube #2, 0.044 M for tube # 4 and 0.0088 for tube # 4.
Dilution factor = Main solution concentration/tube 4 concentration
Dilution factor = 1.1/0.0088 = 125
I hope my answer helps you
