Answer:
In the third tube, the concentration is 0.16 ug/mL
Explanation:
In the first step, the solution is diluted by 5. Then, the concentration will be
20 ug/mL / 5 = 4 ug/mL
Then, in the second step this 4 ug / ml solution is diluted by a factor of five again:
4 ug /ml / 5 = 0.8 ug/mL
This solution is then diluted again by 5 and the concentration in the third tube will be then:
0.8 ug/mL / 5 = <u>0.16 ug/mL </u>
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Another way to calculate this is to divide the original concentration by the dilution factor ( 5 in this case) elevated to the number of dilutions. In this case:
Concentration in the third tube = 20 ug/mL / 5³ = 0.16 ug/mL
Sorry I’m only answering so I could upload
If you would draw the Lewis structures of these atoms, you would see that A has 2 electron pairs and 2 lone electrons (that can bond). For B you’d see that you only have 1 electron that can form a bond. This means that 1 atom of A (2 lone electrons) can bond with 2 atoms of B. To know the kind of bond you have to know wether or not there will be a ‘donation’ of an electron from one atom to another. This happens when the number of electrons on one atoms is equal to the number of electrons another atom needs to reach the noble gas structure. As you can see, this is not the case here. This means that you get an AB2 structure with covalent character.
2 C₅H₁₀ (l) + 15 O₂ (g) → 10 CO₂ (g) + 10 H₂O (g)
Explanation:
Balanced chemical equation for combustion of pentane C₅H₁₀:
C₅H₁₀ (l) + (15/2) O₂ (g) → 5 CO₂ (g) + 5 H₂O (g)
to get integer numbers for the stoechiometric coefficients we multiply with 2:
2 C₅H₁₀ (l) + 15 O₂ (g) → 10 CO₂ (g) + 10 H₂O (g)
where:
(l) - liquid
(g) - gas
Learn more about:
combustion of hydrocarbons
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