Answer:
The mass of 10 cm³of a 0.4 g/dm³ solution of sodium carbonate is 0.004 grams
Explanation:
The question is with regards to density calculations
The density of the given sodium carbonate solution, ρ = 0.4 g/dm³
The volume of the given solution of sodium carbonate, V = 10 cm³ = 0.01 dm³
![Density \ of \ an \ object, \rho = \dfrac{The \ mass \ of \ the \ object, \ m }{\ The \ volume \ of \ the \ object, \ V }](https://tex.z-dn.net/?f=Density%20%5C%20of%20%5C%20an%20%5C%20object%2C%20%5Crho%20%20%3D%20%5Cdfrac%7BThe%20%5C%20mass%20%5C%20of%20%5C%20the%20%5C%20object%2C%20%5C%20m%20%7D%7B%5C%20The%20%5C%20volume%20%5C%20of%20%5C%20the%20%5C%20object%2C%20%5C%20V%20%7D)
![\rho = \dfrac{m}{V}](https://tex.z-dn.net/?f=%5Crho%20%3D%20%5Cdfrac%7Bm%7D%7BV%7D)
Therefore, we have;
![The \ Density \ of \ the \ sodium \ carbonate, \ \rho = 0.4 \ g/dm^3 = \dfrac{m }{ 0.01 \ dm^3 }](https://tex.z-dn.net/?f=The%20%5C%20Density%20%5C%20of%20%5C%20the%20%5C%20sodium%20%5C%20carbonate%2C%20%5C%20%5Crho%20%20%3D%200.4%20%5C%20g%2Fdm%5E3%20%3D%20%20%5Cdfrac%7Bm%20%7D%7B%200.01%20%5C%20dm%5E3%20%7D)
The mass, "m", of the sodium carbonate in = ρ×V = 0.4 g/dm³ × 0.01 dm³ = 0.004 g
The mass of 10 cm³ (10 cm³ = 0.01 dm³) of a 0.4 g/dm³ solution of sodium carbonate, m = 0.004 g.
Li2S + 2 HNO3 --> 2 LiNO3 + H2S
Li2 S + H2 N2 O2 --> Li2 N2 O5 + H2 S
Li S + H2 N2 O5 -> Li N2 O5 + H2 S
Li2 S2 + H4 N4 O10 --> Li2 N4 O10 + H4 S2
Li^2 S^2 + H^4 N^4 O^10 --> Li^2 N^4 O^10 + H^4 S^2
Answer:
B
Explanation:
All organic material has carbon.
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