Answer:
CH₅N
Explanation:
In the combustion, all of the C in the compound was used to produce CO₂ in a 1:1 ratio. Thus, the moles of CO₂ (MW 44.01 g/mol) produced equals the moles of C in the compound:
(44.0 g)(mol/44.01g) = 0.99977 mol CO₂ = 0.99977... mol C
Similarly, all of the H in the compound was used to produce H₂O in a ratio of 2H:1H₂O. The moles of H₂O (MW 18.02 g/mol) produced was:
(45.0 g)(mol/18.02g) = 2.497...mol H₂O
Moles of H is found using the molar ratio of 2H:1H₂O:
(2.497...mol H₂O)(2H/1H₂O) = 4.994...mol H
The ratio of H to C in the compound is:
(4.994...mol H)/(0.99977... mol C) = 5 H:C
Some NO₂ was produced from the N in the compound. Assuming a 1:1 ratio of C:N, the simplest empirical formula is: CH₅N.
You forgot to add pictures ..
Mass percentage of a solution is the amount of solute present in 100 g of the solution.
Given data:
Mass of solute H2SO4 = 571.3 g
Volume of the solution = 1 lit = 1000 ml
Density of solution = 1.329 g/cm3 = 1.329 g/ml
Calculations:
Mass of the given volume of solution = 1.329 g * 1000 ml/1 ml = 1329 g
Therefore we have:
571.3 g of H2SO4 in 1329 g of the solution
Hence, the amount of H2SO4 in 100 g of solution= 571.3 *100/1329 = 42.987
Mass percentage of H2SO4 (%w/w) is 42.99 %
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.
Remember that a cation will be smaller than its neutral atom, and an anion will be larger than its neutral atom. This would automatically eliminate answer choices A and D.
Also keep in mind that atomic radii decreases from left to right as you move along a periodic table. It also decreases from bottom up.
Atomic radii increases as you move from right to left and as you go from up to down.
As bromine is higher up in the periodic table than Iodine, it would have a smaller radius. Iodine would have a larger radius.
The correct answer is B. Br
