Answer:54.36 K (−218.79 °C; −361.82 °F)
Explanation:
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The concentration of the drug stock solution is 1.5*10^-9 M i.e. 1.5 * 10^-9 moles of the drug per Liter of the solution
Therefore, the number of moles present in 1 ml i.e. 1*10^-3 L of the solution would be = 1 *10^-3 L * 1.5 * 10^-9 moles/1 L = 1.5 * 10^-12 moles
1 mole of the drug will contain 6.023*10^23 drug molecules
Therefore, 1.5*10^-12 moles of the drug will correspond to :
1.5 * 10^-12 moles * 6.023*10^23 molecules/1 mole = 9.035 * 10^11 molecules
The number of cancer cells = 2.0 * 10^5
Hence the ratio = drug molecules/cancer cells
= 9.035 *10^11/2.0 *10^5
= 4.5 * 10^6
The answer is 79.9 g.
Copper takes 92.0% of aluminum bronze and it is a limiting factor. We have aluminum in excess, so we need to make a proportion.
If 73.5 g of copper are 92.0%, how many g of aluminum bronze will be 100%:
73.5 g : 92.0% = x : 100%.
x = 73.5 g : 100% * 92.0%
x = 79.9 g
Therefore, from 73.5 g of copper and 6.4 g of aluminum (since 79.9 g - 73.5 g = 6.4 g), maximum 79.9 g of aluminum bronze can be prepared.
Explanation:
It's (D), nuclear fission................
