Gases take the shape of their container. When you have a large container, the spaces between molecules (particles) can be further apart than if they were close together. In small containers, the particles are forced to be closer together, or compressed.
Think of it like a pep rally in a gym v.s. a classroom. In the gym, everyone has a bit of wiggle room. With the same number of people in a classroom, everyone would need to be packed in there. This can also explain why a smaller pot over boils from steam before a larger one does, even if the amount of water is the same.
Answer:
THE NEW PRESSURE OF THE HELIUM GAS IS 124kPa AFTER THE VOLUME WAS INCREASED FROM 2.48 L TO 2.98 L
Explanation:
Using Boyle's law which states that at constant temperature, the pressure of a given gas is inversely proportional to the volume occupied by the gas.
Mathematically,
P1 V1 = P2 V2
P1 = 150 kPa = 150 * 10^3 Pa
V1 = 2.48 L
V2 = 2.98 L
P2 = ?
Rearranging the formula making P2 the subject of the equation, we obtain;
P2 = P1 V1 / V2
P2 = 150 * 10^3 * 2.48 / 2.98
P2 = 372 * 10 ^3 / 2.98
P2 = 124.83 * 10^3 Pa or 124.8kPa
In other words, the new pressure of the helium gas after its volume was increased from 2.48 L to 2.98 L is 124.8kPa.
Answer:
Explanation:
soluble and insoluble salts are prepared by processes like neutralization reactions, simple displacement reactions and double displacement reactions.
The correct option is C.
A Lewis dot diagram is a representation of the valence electron of an atom, which uses dot around the symbol of the atom. Chlorine has seven electrons in its outermost shell, these seven electrons are arranged in form of dot around the atom of chlorine. If you count the number of dot given in option C, you will notice that they are seven.
Answer:
The molar concentration of Cu²⁺ in the initial solution is 6.964x10⁻⁴ M.
Explanation:
The first step to solving this problem is calculating the number of moles of Cu(NO₃)₂ added to the solution:

n = 1.375x10⁻⁵ mol
The second step is relating the number of moles to the signal. We know the the n calculated before is equivalent to a signal increase of 19.9 units (45.1-25.2):
1.375x10⁻⁵ mol _________ 19.9 units
x _________ 25.2 units
x = 1.741x10⁻⁵mol
Finally, we can calculate the Cu²⁺ concentration :
C = 1.741x10⁻⁵mol / 0.025 L
C = 6.964x10⁻⁴ M