STRUCTURE OF BROMOUS ACID: H–O–Br=O
<span>In this structure, all the elements have a formal charge of
zero. The formal charge of each element is calculated below: </span><span>
H: 1 – 1/2(2) – 0 = 0
O: 6 – 1/2(4) – 4 = 0
Br: 7 – 1/2(6) – 4 = 0
<span>O: 6 – 1/2(4) – 4 = 0</span></span>
Answer:
e) pH is independent of concentration.
Explanation:
a) It is a mixture of a weak acid and its conjugate base. <em>TRUE. </em>A buffer is defined as a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid.
b) Resists pH changes because it reacts with added acid or base. <em>TRUE. </em>Thermodynamically, the reaction of added acid or base is faster with the buffer mixture than with H⁺ or OH⁻ ions of the solutions.
c) The maximum buffer capacity is at pH = pKa. <em>TRUE. </em>The buffer capacity is pka±1. For this, buffer capacity is maximum in pka.
d) pH is dependent on the solution ionic strength and temperature. <em>TRUE.</em> Ionic strength and temperature are factors that influence concentrations of ions in solutions as the H⁺ ion that is the responsible
e) pH is independent of concentration. <em>FALSE. </em>pH in a buffer depends completely of concentrations of the acid and its conjugate base or vice versa.
I hope it helps!
Answer:
3.011×1023 individual carbon dioxide molecules
<span>The functional group in organic chemistry is the set of atoms attached to the carbon skeleton that often determine the properties of an organic compound.
Following are major functional groups that are present in an organic compound. carboxylic acid, ester, ether, amine, amide, thiol, halide, alkene, alkyne, phenol, alcohol, aldehyde, ketone, etc.</span>
Answer:
see explanations
Explanation:
4NH₃(g) + 5O₂(g) => 4NO(g) + 6H₂O(g)
Ci(NH₃) = 3.5mole/4L = 0.875M
Cf(NH₃) = 1.6mole/4L = 0.400M
Rate-1 => Δ[NH₃]/Δt = |(0.400M - 0.875M)/3min| = 0.158M/s
Rate-2 => 6(Δ[NH₃]/Δt) = 4(Δ[H₂O]/Δt) => 6/4(0.158M/s) = 0.237M/s
Rate-3 => 5(Δ[NH₃]/Δt) = 4(Δ[O₂]/Δt) => 5/4(0.158M/s) = 0.237M/s
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NOTE: When setting up comparative rate expressions for a given reaction, set the rates expressions as equal then swap coefficient values. Then solve for rate of interest and substitute givens.
example: for NH₃ and H₂O
- set rates expressions equal => Δ[NH₃]/Δt = Δ[H₂O]/Δt
- then swap and insert coefficients from given rxn ...
- solve for rate of interest ...
4NH₃(g) + 5O₂(g) => 4NO(g) + 6H₂O(g)
=> 6(Δ[NH₃]/Δt) = 4(Δ[H₂O]/Δt)
=> Δ[H₂O]/Δt = 6/4(Δ[NH₃]/Δt) = 6/4(0.237M/s) = 0.237M/s
