Answer:
The new temperature of the nitrogen gas is 516.8 K or 243.8 C.
Explanation:
Gay-Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of collisions with the walls increases, that is, the pressure increases. That is, the pressure of the gas is directly proportional to its temperature.
Gay-Lussac's law can be expressed mathematically as follows:
Where P = pressure, T = temperature, K = Constant
You want to study two different states, an initial state and a final state. You have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment. By varying the temperature to a new value T2, then the pressure will change to P2, and the following will be fulfilled:

In this case:
- P1= 2 atm
- T1= 50 C= 323 K (being 0 C= 273 K)
- P2= 3.2 atm
- T2= ?
Replacing:

Solving:


T2= 516.8 K= 243.8 C
<u><em>The new temperature of the nitrogen gas is 516.8 K or 243.8 C.</em></u>
As has an atomic number of 33, so it has 33 protons.
It has a charge of 3-, so there are three more electrons than protons. Thus, there are 36 electrons.
It has a mass of 75, which is the sum of neutrons and protons.
33+n=75 ---> n = 75 - 33 = 42 neutrons
The answer is e) 33 protons, 42 neutrons and 36 electrons.
Answer:
0.8 mL of protein solution, 9.2 mL of water
Explanation:
The dilution equation can be used to relate the concentration C₁ and volume V₁ of the stock/undiluted solution to the concentration C₂ and volume V₂ of the diluted solution:
C₁V₁ = C₂V₂
We would like to calculate the value for V₁, the volume of the inital solution that we need to dilute to make the required solution.
V₁ = (C₂V₂) / C₁ = (2mg/mL x 10mL) / (25 mg/mL) = 0.8 mL
Thus, a volume of 0.8 mL of protein solution should be diluted with enough water to bring the total volume to 10 mL. The amount of water needed is:
(10 mL - 0.8 mL) = 9.2 mL
The grams of aluminium extracted from 5000g of alumina is 2647 grams
<h3>Chemical formula of alumina:</h3>
Let's calculate the molecular mass of Al₂O₃
Al₂O₃ = 27 × 2 + 16 × 3 = 54 + 48 = 102 g/mol
Therefore,
102 g of Al₂O₃ = 54 g of aluminium
5000g of Al₂O₃ = ?
mass of aluminium produced = 5000 × 54 / 102
mass of aluminium produced = 270000 / 102
mass of aluminium produced = 2647.05882353
mass of aluminium produced = 2647 grams
learn more on mass here: brainly.com/question/14627327