Answer:
the open bag allowed the gas from the reaction to escape
M(H2SO4)=n*M=4.75*98=465.5g
Answer:
0.186M
Explanation:
First, we need to obtain the moles of nitric acid that are given for each solution. Then, we need to divide these moles in total volume (120mL + 20mL = 140mL = 0.140L) to obtain molarity:
<em>Moles Nitric acid:</em>
0.0200L * (0.100mol / L) = 0.00200 moles
0.120L * (0.200mol / L)= 0.02400 moles
Total moles: 0.02400moles + 0.00200moles = 0.026 moles of nitric acid
Molarity: 0.026 moles / 0.140L
<h3>0.186M</h3>
Answer:
a. 3; b. 5; c. 10; d. 12
Explanation:
pH is defined as the negative log of the hydronium concentration:
pH = -log[H₃O⁺] (hydronium concentration)
For problems a. and b., HCl and HNO₃ are strong acids. This means that all of the HCl and HNO₃ would ionize, producing hydronium (H₃O⁺) and the conjugate bases Cl⁻ and NO₃⁻ respectively. Further, since all of the strong acid ionizes, 1 x 10⁻³ M H₃O⁺ would be produced for a., and 1.0 x 10⁻⁵ M H₃O⁺ for b. Plugging in your calculator -log[1 x 10⁻³] and -log[1.0 x 10⁻⁵] would equal 3 and 5, respectively.
For problems c. and d. we are given a strong base rather than acid. In this case, we can calculate the pOH:
pOH = -log[OH⁻] (hydroxide concentration)
Strong bases similarly ionize to completion, producing [OH⁻] in the process; 1 x 10⁻⁴ M OH⁻ will be produced for c., and 1.0 x 10⁻² M OH⁻ produced for d. Taking the negative log of the hydroxide concentrations would yield a pOH of 4 for c. and a pOH of 2 for d.
Finally, to find the pH of c. and d., we can take the pOH and subtract it from 14, giving us 10 for c. and 12 for d.
(Subtracting from 14 is assuming we are at 25°C; 14, the sum of pH and pOH, changes at different temperatures.)
Explanation:
Once solid ammonium nitrate interacts with water, the molecules of polar water intermingle with these ions and attract individual ions from the structure of the lattice, that actually will break down. E.g;-NH4NO3(s) — NH4+(aq)+ NO3-(aq) To split the ionic bonds that bind the lattice intact takes energy that is drained from the surroundings to cool the solution.
Some heat energy is produced once the ammonium and nitrate ions react with the water molecules (exothermic reaction), however this heat is far below that is needed by the H2O molecules to split the powerful ionic bonds in the solid ammonium nitrate.
Hence, we can say that the dissolution of ammonium nitrate in water is highly endothermic reaction.
