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.
I can help you but what is the question here?
Explanation:
There is variety of evidence that supports the claims that plate tectonics accounts for
(1) the distribution of fossils on different continents
,(2) the occurrence of earthquakes, and
(3) continental and ocean floor features including mountains, volcanoes, faults, and trenches.
MARK AS BRAINLIST IF IT IS USEFUL
Answer:
2HgO → 2Hg + O₂
The reaction is a decomposition reaction because from one reactant two products were obtained
Explanation:
The problem here involves balancing the chemical reaction given and also classifying the reaction.
The unbalanced expression is given as:
HgO → Hg + O₂
Assign the alphabets a,b and c as coefficients that will balance the equation
aHgO → bHg + cO₂
Conserving Hg: a = b
O: a = 2c
So, let c = 1, a = 2 b = 2
Therefore;
2HgO → 2Hg + O₂
The reaction is a decomposition reaction because from one reactant two products were obtained
The answer is it contains the electrodes. Without the salt scaffold, the arrangement in the anode compartment would turn out to be decidedly charged and the arrangement in the cathode compartment would turn out to be contrarily charged, on account of the charge lopsidedness, the terminal response would rapidly stop.
It keeps up the stream of electrons from the oxidation half-cell to a decrease half cell, this finishes the circuit.
