Answer:
even though I am missing the question I believe that the correct answer to the question is; <em>his cats will be brown (Dominant)</em>
Explanation:
this statement is true to the fact that on of his cats has the homozygous dominant gene BB. For their to be a chance of a white cat as an offspring both cats would either have to be Hetrozygous (Bb) or Homozygous recesssive (bb)
Answer:
When naming molecular compounds prefixes are used to dictate the number of a given element present in the compound. ” Mono-” indicates one, “Di-” indicates two, “Tri-” is three, “Tetra-” is four, “Penta-” is five, and “Hexa-” is six, “Hepta-” is seven, “Octo-” is eight, “Nona-” is nine, and “Deca-” is ten.
Answer is: the percent composition of Hg in the compound is 71.5%.
Balanced chemical reaction: Hg + Br₂ → HgBr₂.
m(Hg) = 60.2 g; mass of the mercury.
m(Br₂) = 24.0; mass of the bromine.
m(HgBr₂) = m(Hg) + m(Br₂).
m(HgBr₂) = 60.2 g + 24 g.
m(HgBr₂) = 84.2 g; mass of the compound.
ω(Hg) = m(Hg) ÷ m(HgBr₂) · 100%.
ω(Hg) = 60.2 g ÷ 84.2 g · 100%.
ω(Hg) = 71.5%.
<span>The best answer is B. ICl experiences induced dipole-induced dipole interactions. Both iodine and chlorine belongs to the same group of the periodic table. Electronegativity decreases as you go down a group therefore Cl will have a greater attraction with the bond it forms with another atom. Dipole-dipole interactions form between I and Cl. For the Br2 molecule, no dipole occurs because they are two identical atoms. Therefore we will be expecting ICl will have a higher boiling point due to higher binding energy it forms.</span>
Answer:
pH = 6.999
The solution is acidic.
Explanation:
HBr is a strong acid, a very strong one.
In water, this acid is totally dissociated.
HBr + H₂O → H₃O⁺ + Br⁻
We can think pH, as - log 7.75×10⁻¹² but this is 11.1
acid pH can't never be higher than 7.
We apply the charge balance:
[H⁺] = [Br⁻] + [OH⁻]
All the protons come from the bromide and the OH⁻ that come from water.
We can also think [OH⁻] = Kw / [H⁺] so:
[H⁺] = [Br⁻] + Kw / [H⁺]
Now, our unknown is [H⁺]
[H⁺] = 7.75×10⁻¹² + 1×10⁻¹⁴ / [H⁺]
[H⁺] = (7.75×10⁻¹² [H⁺] + 1×10⁻¹⁴) / [H⁺]
This is quadratic equation: [H⁺]² - 7.75×10⁻¹² [H⁺] - 1×10⁻¹⁴
a = 1 ; b = - 7.75×10⁻¹² ; c = -1×10⁻¹⁴
(-b +- √(b² - 4ac) / (2a)
[H⁺] = 1.000038751×10⁻⁷
- log [H⁺] = pH → 6.999
A very strong acid as HBr, in this case, it is so diluted that its pH is almost neutral.
