Answer: 11.5 moles of carbon
Explanation:
Based on Avogadro's law:
1 mole of any substance has 6.02 x 10^23 atoms
So, 1 mole of carbon = 6.02 x 10^23 atoms
Z moles = 6.93 x 10^24 atoms
To get the value of Z, cross multiply:
(6.93 x 10^24 atoms x 1mole) = (6.02 x 10^23 atoms x Z moles)
6.93 x 10^24 = (6.02 x 10^23 x Z)
Z = (6.93 x 10^24) ➗ (6.02 x 10^23)
Z = 1.15 x 10
Z = 11.5 moles
Thus, there are 11.5 moles of carbon.
T is amount after time t
<span>Ao is initial amount </span>
<span>t is time </span>
<span>HL is half life </span>
<span>log (At) = log [ Ao x (1/2)^(t/HL) ] </span>
<span>log (At) = log Ao + log (1/2)^(t/HL) </span>
<span>log (At) = log Ao + (t/HL) x log (1/2) </span>
<span>( log At - log Ao) / log (1/2) = t / HL </span>
<span>log (At/Ao) / log (1/2) = t / HL </span>
<span>HL = t / [( log (At / Ao)) / log (1/2) ] </span>
<span>HL = 14.4 s / [ ( log (12.5 / 50) / log (1/2) ] </span>
<span>HL = 14.4 s / 2 = 7.2 seconds </span>
Neon I think. Go to the periodic table and see which one is the 11th
Answer: Chloroplasts
Reasoning: I just had my 830th lesson in school about these
Answer:
The correct answer is "D. The difference in their electronegativities is greater than 1.7"
Explanation:
An ionic bond is produced between metallic and non-metallic atoms, where electrons are completely transferred from one atom to another. During this process, one atom loses electrons and another one gains them, forming ions. Usually, the metal gives up its electrons forming a cation to the nonmetal element, which forms an anion.
So, this type of bond consists of the attraction of atoms with electric charges of different signs, that is, the bond is established when atoms of low electronegative elements (metals) are bonded with atoms of highly electronegative elements (nonmetals).
So you can say that atoms of electronegativity difference equal to or greater than 1.7 participate in this type of bond (with its exceptions).
<u><em>The correct answer is "D. The difference in their electronegativities is greater than 1.7"</em></u>