b,f,h are already balanced
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:
K = 0.5
Explanation:
Based on the reaction:
PCl₃ + Cl₂ ⇄ PCl₅
The equilibrium constant, K, is defined as:
K = P PCl₅ / P PCl₃ * P Cl₂
<em>Where P represent the pressure at the equilibrium for each one of the gases involved in the equilibrium.</em>
<em />
As:
P PCl₅ = 1.0atm
P PCl₃ = 1.0atm
P Cl₂ = 2.0atm
K = 1.0atm / 1.0atm * 2.0atm
<h3>K = 0.5</h3>
Answer:
Ammonium bromide can be prepared by the direct action of hydrogen bromide on ammonia. It can also be prepared by the reaction of ammonia with iron(II) bromide or iron(III) bromide, which may be obtained by passing aqueous bromine solution over iron filings.
Explanation:
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The formation of chemical bonds occurs due to the attractive forces between oppositely charged ions (ionic bonds) or by sharing of electrons (covalent bonds).
An atom having tendency of attracting a shared pair of electrons towards itself and this chemical property is said to Electronegativity .
Thus, the attractive forces which draws in surrounding electrons for chemical bonds is electronegativity.
