The correct answer is B,
<span>It is endothermic, with both positive enthalpy and entropy changes.</span>
The correct response is the second option.
6.1103x10^4. As this was the only answer that had the same number of significant figures as the starting value.
Answer:
[HF]₀ = 0.125M
Explanation:
NaOH + HF => NaF + H₂O
Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3. This is 0.089M NaF and 0.001M HF remaining.
=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.
HF ⇄ H⁺ + F⁻
C(eq) [HF] 10⁻³M 0.089M (<= soln after adding 20ml 0.200M NaOH)
Ka = [H⁺][F⁻]/[HF]₀ => [HF]₀ = [H⁺][F⁻]/Ka
[HF]₀ = (0.001)(0.089)/(7.1 x 10⁻⁴) M = 0.125M
Answer:
<h2>Hi there !</h2>
<h2>C. HCl</h2>
Explanation:
<h2>Reason :-</h2>
<h2>Salts are strong electrolytes, so they undergo complete dissociation.</h2><h3>Hope it helps u.....</h3><h3>Stay safe, stay healthy and blessed</h3><h3>Have a good day</h3><h3>Thank you ~</h3>
