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.
Okay... what are the following
Answer:
The total change in enthalpy for the reaction is - 81533.6 J/mol
Explanation:
Given the data in the question;
Reaction;
HCl + NaOH → NaCl + H₂O
Where initial temperature is 21.2 °C and final temperature is 28.0 °C. Ccal is 1234.28 J
Moles of NaOH = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol
Moles of HCl = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol
so, 0.0500 moles of H₂O produced
Volume of solution = 50.mL + 50.mL = 100.0 mL
Mass of solution m = volume × density = 100.0mL × 1.0 g/mL = 100 g
now ,
Heat energy of Solution q= (mass × specific heat capacity × temp Δ) + Cal
we know that; The specific heat of water(H₂O) is 4.18 J/g°C.
so we substitute
q_soln = (100g × 4.18 × ( 28.0 °C - 21.2 °C) ) + 1234.28
q_soln = 2842.4 + 1234.28
q_soln = 4076.68 J
Enthalpy change for the neutralization is ΔH
ΔH = -q_soln / mole of water produced
so we substitute
ΔH = -( 4076.68 J ) / 0.0500 mol
ΔH = - 81533.6 J/mol
Therefore, the total change in enthalpy for the reaction is - 81533.6 J/mol
Main function of a restriction enzyme is to cleve (Cut) DNA bases.
Hope this helps!
Answer:
15.98 L
Explanation:
First, you need to find T1, T2, V1 and V2.
T1 = 25 C = 298.15 K (25C + 273.15K)
T2 = 100 C = 373.15 K (100C + 273.15K)
V1 = 20. L
V2 = ? (we are trying to find)
Next, rearrange to fit the formula
V2 = V1 x T1 / T2
Next, fill in with our numbers
V2 = 20. L x 298.15 K / 373.15 K
Do the math and you should get...
15.98 L
- If you need more help or futher explanation please let me know. I would be glad to help!
