To determine the mass, you need to know the molecular weight of the c8h10n4o2 . The molecular weight of <span>c8h10n4o2 would be: 8*12 + 10*1 + 4*14 + 2*16= 194g/mol.
To convert the number of molecules into moles, you need to divide it with 6.02 * 10^23. The calculation of the mass of </span>c8h10n4o2 would be:
(7.20×10^20 molecules) /(6.02 * 10^23 molecule/mol) * 194g/mol= 232 * 10^-3 grams= 0.232 grams
Can u explain it more plz.
Given the following equation; Cu + 2AgNO3 = Cu(NO3)2 + 2Ag, 48.97 grams of Cu are needed to react with 262g of AgNO3.
<h3>How to calculate mass of substances?</h3>
The mass of a substance can be calculated using the following steps:
Cu + 2AgNO3 = Cu(NO3)2 + 2Ag
1 mole of Cu react with 2 moles of AgNO3
- Molar mass of AgNO3 = 169.87 g/mol
- Molar mass of Cu = 63.5g/mol
moles of AgNO3 = 262g/169.87g/mol = 1.54mol
1.54 moles of AgNO3 will react with 0.77 moles of Cu.
mass of Cu = 0.77 × 63.5 = 48.97g
Therefore, given the following equation; Cu + 2AgNO3 = Cu(NO3)2 + 2Ag, 48.97 grams of Cu are needed to react with 262g of AgNO3.
Learn more about mass at: brainly.com/question/6876669
Answer:
productivity and water depth
Explanation:
The productivity and the depth of water are both equally important as it directly affects the accumulation of biogenic sediments such as the siliceous ooze and calcareous ooze. In the equator and the coastal upwelling areas, and at the site of divergence of oceans, there occurs a high rate and amount of productivity, and these are considered to be the primary productivity.
The siliceous oozes are a good indicator of extensively high productivity in comparison to the carbonate oozes. The main reason behind this is that the silica can be easily dissolved in the surface water. On the other hand, the carbonates dissolve at a relatively lower ocean water depth, so there requires a high amount of surface productivity in order to allow these siliceous oozes to reach the ocean bottom.
Thus, the water depth and productivity, both are considered as the limiting factor in determining the accumulation of biogenic oozes.
I believe the answer is 65.254%