There are 3.98 × 10^23 atoms of oxygen in the sample.
Given that;
1 mole of Mo(NO3)6 contains 6.02 × 10^23 atoms of Nitrogen
x moles of Mo(NO3)6 contains 2.22 x 10^22 atoms of nitrogen
x = 1 mole × 2.22 x 10^22 atoms/6.02 × 10^23 atoms
x = 0.0368 moles
The number of oxygen atoms in the sample is given by; 0.0368 × 6.02 × 10^23 × 18
Therefore, there are 3.98 × 10^23 atoms of oxygen in the sample.
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Answer:
Its basically a series of events that take place in a cell. A cell spends most of its time in what is called inter phase, and during this time it grows, replicates its chromosomes, and prepares for cell division. Then after that The cell then leaves interphase, undergoes mitosis, and completes its division.
Explanation:
-Hailey: )
Answer:
Explanation:
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In this case, since the density is computed by dividing the mass of the substance by its occupied volume (d=m/V), we first need to realize that 0.8206 g/mL is the same to 0.8206 kg/L, which means we first need to compute the volume in L:
Then, solving for the mass in d=m/V, we get m=d*V and therefore the mass of gasoline in that full tank turns out:
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Answer:
See explanation
Explanation:
The first step in this reaction is a unimolecular reaction. It involves the formation of the carbocation. This is so because tertiary alkyl halides only undergo substitution by SN1 mechanism due to sterric crowding.
The second step in the reaction is bi molecular. In this step, the carbocation now combines with the OH^- to yield the alcohol.
Net equation of the reaction is;
(CH3)3CBr + OH^- -------> (CH3)3COH + Br^-
The intermediate here is the carbocation, (CH3)3C^+
total pressure = ( 705 torr + 231.525torr) = 936.525torr
