Unfortunately the data provided doesn't include the DENSITY of the ammonium chloride solution and molarity is defined as moles per volume. So without the density, the calculation of the molarity is impossible. But fortunately, there are tables available that do provide the required density and for a 20% solution by weight, the density of the solution is 1.057 g/ml.
So 1 liter of solution will mass 1057 grams and the mass of ammonium chloride will be 0.2 * 1057 g = 211.4 g. The number of moles will then be 211.4 g / 53.5 g/mol = 3.951401869 mol. Rounding to 3 significant digits gives a molarity of 3.95.
Now assuming that your teacher wants you to assume that the solution masses 1.00 g/ml, then the mass of ammonium chloride will only be 200g, and that is only (200/53.5) = 3.74 moles.
So in conclusion, the expected answer is 3.74 M, although the correct answer using missing information is 3.95 M.
The answer is 3 It is a body of knowledge gained using inquiry and experimentation. Hope this helped!
Answer:
No one is correct. The correct expression is:
Keq = [H₂]² . [O₂]² / [H₂O]²
Explanation:
To build the Keq expression in a chemical equilibrium you must consider the molar concentrations of reactants / products, and they must be elevated to the stoichiometric coefficient.
The balance reaction is:
<u>2</u> H₂O (g) ⇄ <u>2</u> H₂ (g) + O₂ (g)
Keq = [H₂]² . [O₂] / [H₂O]²
In opposite side: <u>2</u> H₂ (g) + O₂ (g) ⇄ <u>2</u> H₂O (g)
Keq = [H₂O]² / [H₂]² . [O₂]
Explanation:
A. Hydrogen bonding is present in CS2 but not in CO2.
B. CS2 has greater dipole moment than CO2 and thus the dipole-dipole forces in CS2 are stronger.
C. CS2 partly dissociates to form ions and CO2 does not. Therefore, ion-dipole interactions are present in CS2 but not in CO2.
D. The dispersion forces are greater in CS2 than in CO2.
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Answer:
No, it is not enthalpy favored since the chemical system gains energy.
Explanation:
The dissolution of ammonium nitrate in water is an endothermic process.
Endothermic process requires the system to gain energy to can dissolve the particles in water.
So, the reaction is not enthalpy favored.