This uses something called the combined gas law. The combined gas law is as follows: (P1*V1/T1) = (P2*V2/T2)
According to question 2, you are given the following values initially:
P1 = 680 mm Hg * (1 atm/760 mm Hg) = 0.895 atm
V1 = 20.0 L
T1 = 293 K
STP or standard temperature and pressure implies that the other values we know are:
P2 = 1 atm
T2 = 273 K
Our unknown is V2
If we plug in our known values into the combined gas law:
(P1*V1/T1) = (P2*V2/T2)
(0.895 atm * 20.0 L)/293K = (1 atm * X liters)/273 K
0.0611 L*atm/K = (1 atm * X liters)/273 K
16.7 L = X liters
Therefore, the volume occupied at STP is 16.7 liters
This makes sense because the gas would occupy a smaller volume at a lower temperature, since the gas would have a lower average kinetic energy.
Cornea - lens - retina
Explanation:
The pathway of light into the eyes goes from the cornea through the lens where it gets focus on the retina and optic nerves picks up the signal where the brain interprets the image. The eye is the organ of sight.
- When light rays from objects hits the human eye, it goes through the cornea first. In the cornea the image begins to form and using the pupil located in it the amount of light that enters is regulated.
- From the cornea, light goes into the lens. The lens allows for the rays to focus on the retina.
- The retina receives the rays and the image forms .
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The balanced chemical reaction is written as:
2K + F2 = 2KF
We are given the amount of potassium to be used in the reaction. THis will be the starting point. We do as follows:
23.5 g K ( 1 mol / 39.10 g) (1 mol F2 / 2 mol K ) ( 22.4 L / 1 mol ) = 6.73 L F2 gas needed
Answer:
Anode (oxidation): Cr(s) ⇒ Cr³⁺(aq) + 3 e⁻
Cathode (reduction): Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
Explanation:
Let's consider the notation of a galvanic cell.
Cr(s) | Cr³⁺(aq) || Ag⁺(aq) | Ag(s)
On the left, it is represented the anode (oxidation) and on the right, it is represented the cathode (reduction).
The half-reactions are:
Anode (oxidation): Cr(s) ⇒ Cr³⁺(aq) + 3 e⁻
Cathode (reduction): Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
To have the global reaction, we have to multiply the reduction by 3 (so the number of electrons gained and lost are the same) and add both half-reactions.
Global reaction: Cr(s) + 3 Ag⁺(aq) ⇒ Cr³⁺(aq) + 3 Ag(s)
Answer:
Explanation:
N2(g)+O2(g)⇌2NO(g),
N2(g)+2H2(g)⇌N2H4(g),
2H2O(g)⇌2H2(g)+O2(g),
If we add above reaction we will get:
2N2(g)+2H2O(g)⇌2NO(g)+N2H4(g) Eq (1)
Equilibrium constant for Eq (1) is
Divide Eq (1) by 2, it will become:
N2(g)+H2O(g)⇌NO(g)+1/2N2H4(g) Eq (2)
Equilibrium constant for Eq (2) is