Answer:
C) H2S
Explanation:
In chemistry, the dissolution of one substance in another is dependent on the magnitude of intermolecular interaction between the two substances. Hence, if two substances do not interact in one way or the other, then one can not dissolve the other.
Let us consider the fact that NH3 is a polar molecule and it is a general principle that like dissolves like. Hence, only H2S which is also a polar molecule can effectively interact with NH3 due to dipole-dipole interaction between the two molecules.
Also, ammonia reacts with hydrogen sulphide as follows;
2NH3 + H2S → (NH4)2S
Hence H2S is more likely to dissolve in NH3.
Um i think gold... i think?
The answer is 6.88.
Solution:
We can calculate for the percent composition of CaCl2 by mass by dividing the mass of the CaCl2 solute by the mass of the solution and then multiply by 100. The total mass of the resulting solution is the sum of the mass of CaCl2 solute and the mass of water solvent. Therefore, the percent composition of CaCl2 by mass is
% by mass = (mass of the solute / mass of the solution)*100
= mass of solute / (mass of the solute + mass of the solvent)*100
= (27.7 g CaCl2 / 27.7g + 375g) * 100
= 6.88
Proton plus neutron is the correct answer. Protons and neutrons have a mass of 1 and electrons have a mass of 0. So in order to find the mass of an atom you need to add the number of protons and the number of neutrons.
1) Balanced chemical equation:
2SO2 (g) + O2 (g) -> 2SO3 (l)
2) Molar ratios
2 mol SO2 : 1 mol O2 : 2 mol SO3
3) Convert 6.00 g O2 to moles
number of moles = mass in grams / molar mass
number of moles = 6.00 g / 32 g/mol = 0.1875 mol O2.
4) Use proportions with the molar ratios
=> 2 moles SO2 / 1 mol O2 = x / 0.1875 mol O2
=> x = 0.1875 mol O2 * 2 mol SO2 / 1 mol O2 = 0.375 mol SO2.
5) Convert 0.375 mol SO2 to grams
mass in grams = number of moles * molar mass
molar mass SO2 = 32 g/mol + 2*16 g/mol = 64 g/mol
=> mass SO2 = 0.375 mol * 64 g / mol = 24.0 g
Answer: 24.0 g of SO2 are needed to react completely with 6.00 g O2.
