Answer:
<h2>- It could be stretched into a thin wire.</h2>
Explanation:
As per the question, the most rational claim that the student can make about the aluminum metal is that 'it could be stretched into a thin wire' without breaking which shows its ductility. It is one of the most significant characteristics of a metal. Metals can conduct electricity in any state and not only when melted. Thus, option A is wrong. Options C and D are incorrect as metals neither have the same shape always nor do they break on hitting with a hammer. Therefore, <u>option E</u> is the correct answer.
Answer: assume pathogens are present and treat the samples accordingly
Explanation:
When investigators are unable to conclusively ascertain the source of a biological sample found at a crime scene, the correct thing to do is to treat it as if pathogens are present in it and handle it according to set rules on how to handle pathogens.
This is done to ensure that if a pathogen is indeed present, it would not cause a health emergency by infecting those who come in contact with the samples at the scene.
Answer:
Negative sign says that release of heat.
Explanation:
The expression for the calculation of the heat released or absorbed of a process is shown below as:-
Where,
is the heat released or absorbed
m is the mass
C is the specific heat capacity
is the temperature change
Thus, given that:-
Mass = 25.2 g
Specific heat = 0.444 J/g°C
So,
Negative sign says that release of heat.
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.
