Graph D.
Charles's law states that at the same pressure, the Volumen of a gas is proportional to its temperature (absolute temperature).
This is
V = K T, where K is the constant of proportionallity.
You can also write it in the form
V1 / V2 = T1/ Ts
When Temperature rise Volumen rise in a proportional way. If Temperature doubles Volumen doubles, if Tempearature is decreases to on third, Volume will decrease to one third.
<span>led to the discovery of many new sea floor features of smaller scales. Deep sea trenches and sea volcanoes.
If that helped please mark brainliest! <3
-Procklownyt</span>
This is an incomplete question, here is a complete question.
The Henry's law constant for oxygen dissolved in water is 4.34 × 10⁹ g/L.Pa at 25⁰C.If the partial pressure of oxygen in air is 0.2 atm, under atmospheric conditions, calculate the molar concentration of oxygen in air-saturated and oxygen saturated water.
Answer : The molar concentration of oxygen is, 
Explanation :
As we know that,
where,
= molar solubility of 
= ?
= partial pressure of = 0.2 atm = 1.97×10⁻⁶ Pa
= Henry's law constant = 4.34 × 10⁹ g/L.Pa
Now put all the given values in the above formula, we get:
Now we have to molar concentration of oxygen.
Molar concentration of oxygen = 
Therefore, the molar concentration of oxygen is,
Answer:
The volume of water to be added is 0.175 liters of water
Explanation:
The given concentration of the nitric acid = 55% (M/M)
The mass of the nitric acid solution = 100 gm
The concentration solution is to diluted to = 20% (M/M)
The 100 g 55%(M/M) nitric acid solution gives 55g nitric acid in 100 g of solution
Therefore, to have 20% (M/M) nitric acid solution with the 55 g nitric acid, we get
Let "x" represent the volume of the resulting solution, we have;
20% of x = 55 g of nitric acid
∴ 20/100 × x = 55 g
x = 55 g × 100/20 = 275 g
The mass of extra water to be added = The mass of the 20%(M/M) solution solution of nitric acid - The current mass of the 55%(M/M) solution of nitric acid
The mass of extra water to be added = 275 g - 100 g = 175 g
Volume = Mass/Density
The density of water ≈ 1 g/ml
∴ The volume of water to be added that gives 175 g of water = 175 g/(1 g/ml) = 175 ml. = 0.175 l
The volume of water to be added = 0.175 liters of water.
Reduction <span>always results in a lowering of the oxidation number. The reaction of the system above is written as:
</span><span>Cu2+(aq) + Fe(s) --> Cu(s) + Fe2+(aq)
</span>
From the reaction, we see that copper goes from the +2 to a neutral charge. Lowering of the oxidation number happens so this is the element that is being reduced.
