Answer:
<h3>The answer is 1.30288 × 10³ g</h3>
Explanation:
The mass of a substance when given the density and volume can be found by using the formula
<h3>mass = Density × volume</h3>
From the question
volume = 95.8mL
density = 13.6g/mL
We have
mass = 13.6 × 95.8 = 1302.88
We have the final answer as
<h3>1.30288 × 10³ g</h3>
Hope this helps you
HBr and HF are both monoprotic Arrhenius acids—that is, in aqueous solution, they dissociate and ionize to give hydrogen ions. A strong acid ionizes completely; a weak acid ionizes partially.
In this case, HBr, being a strong acid, would ionize completely in water to yield H+ and Br- ions. However, HF, being a weak acid, would ionize only to a limited extent: some of the HF molecules will ionize into H+ and F- ions, but most of the HF will remain undissociated.
pH is, by definition, a measurement of the concentration of hydrogen ions in solution (pH = -log[H+]). A higher concentration of hydrogen ions gives a lower pH, while a lower concentration of hydrogen ions gives a higher pH. At 25 °C, a pH of 7 indicates a neutral solution; a pH less than 7 indicates an acidic solution; and a pH greater than 7 indicates a basic solution.
If we have equal concentrations of HBr and HF, then the HBr solution will have a greater concentration of hydrogen ions in solution than the HF solution. Consequently, the pH of the HBr solution will be less than the pH of the HF solution.
Choice A is incorrect: Strong acids like HBr dissociate completely, not partially.
Choice B is incorrect: While the initial concentration of HBr and HF are the same, the H+ concentration in the HBr solution is greater. Since pH is a function of H+ concentration, the pH of the two solutions cannot be the same.
Choice C is correct: A greater H+ concentration gives a lower pH value. The HBr solution has the greater H+ concentration. Thus, the pH of the HBr solution would be less than that of the HF solution.
Choice D is incorrect for the reason why choice C is correct.
Answer:
60 moles of NaF
Explanation:
The balanced equation for the reaction is given below:
Al(NO3)3 + 3NaF —> 3NaNO3 + AlF3
From the balanced equation above,
3 moles of NaF reacted to produce 1 mole of AlF3.
Therefore, Xmol of NaF will react to produce 20 moles of AlF3 i.e
Xmol of NaF = 3 x 20
Xmol of NaF = 60 moles
Therefore, 60 moles of NaF are required to produce 20 moles of AlF3.