Answer:
Lengths. of. Naphthalene. Figure 3.20 shows that there are two equivalent ... all the carbon–carbon bonds of benzene are identical and are intermediate in length ... A typical carbon–carbon single bond has a length of 1.54 Å, and a double ... of how resonance can be used to explain or predict experimental observations.Explanation:
Answer:
What are the advantages of titration?
Titrimetric analysis commonly referred to as volumetric analysis offers distinct advantages over cumbersome gravimetric methods:
Speed of analysis.
Instantaneous completion of reactions.
Greater accuracy due to minimization of material loss involved in decanting, filtration, precipitation or similar operations.
Explanation:
Disadvantages
It is a destructive method often using up relatively large quantities of the substance being analysed.
It requires reactions to occur in a liquid phase, often the chemistry of interest will make this inappropriate.
It can produce significant amounts of chemical waste which has to be disposed of.
It has limited accuracy.
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Answer:
8. the answer is B.
9. the answer is A.
Explanation:
Hello!
8. In this case, by bearing to mind that the limiting reactant is always completely consumed and the excess one remain as a leftover at the end of the reaction, we can also infer that as all the limiting reactant is consumed, it must determine the maximum amount of product as the excess reactant will hypothetically produce a greater mass than expected; thus, the answer to this question is B.
9. In this case, since the mole ratio of oxygen to water is 1:2, the following proportional factor is used to calculate the produced mass of water:
Thus, the answer is this case is A.
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Answer:
Below!
Explanation:
A planet with oceans, continents, and plate tectonics maximizes opportunities for speciation and natural selection, whereas a similar planet without plate tectonics provides fewer such opportunities. Plate tectonics exerts environmental pressures that drive evolution without being capable of extinguishing all life.
Given:
1.50 L
62.5 grams
and the MM of MgO: 40.31 g/mol
Molarity: mol/L
First, find mol.
62.5 g x 1mole ÷ 40.31 g = 1.55 mol
then divide mol and the given liters
1.55mol ÷ 1.50 L= 1.03 M
