The new volume when pressure increases to 2,030 kPa is 0.8L
BOYLE'S LAW:
The new volume of a gas can be calculated using Boyle's law equation:
P1V1 = P2V2
Where;
- P1 = initial pressure (kPa)
- P2 = final pressure (kPa)
- V1 = initial volume (L)
- V2 = final volume (L)
According to this question, a 4.0 L balloon has a pressure of 406 kPa. When the pressure increases to 2,030 kPa, the volume is calculated as:
406 × 4 = 2030 × V2
1624 = 2030V2
V2 = 1624 ÷ 2030
V2 = 0.8L
Therefore, the new volume when pressure increases to 2,030 kPa is 0.8L.
Learn more about Boyle's law calculations at: brainly.com/question/1437490?referrer=searchResults
Density describes how compact or concentrated something is. For example, suppose you have two boxes, one large and one small. ... That means the small box has a higher density than the large box. Density also tells how concentrated or crowded something is.
Answer:
There are 11 protons in a sodium atom but only 3 in a lithium atom, so the nuclear charge is much greater. ... The only factor left is the extra distance between the outer electron and the nucleus in sodium's case. That lowers the ionization energy.
Answer:
1.63425 × 10^- 18 Joules.
Explanation:
We are able to solve this kind of problem, all thanks to Bohr's Model atom. With the model we can calculate the energy required to move the electron of the hydrogen atom from the 1s to the 2s orbital.
We will be using the formula in the equation (1) below;
Energy, E(n) = - Z^2 × R(H) × [1/n^2]. -------------------------------------------------(1).
Where R(H) is the Rydberg's constant having a value of 2.179 × 10^-18 Joules and Z is the atomic number= 1 for hydrogen.
Since the Electrons moved in the hydrogen atom from the 1s to the 2s orbital,then we have;
∆E= - R(H) × [1/nf^2 - 1/ni^2 ].
Where nf = 2 = final level= higher orbital, ni= initial level= lower orbital.
Therefore, ∆E= - 2.179 × 10^-18 Joules× [ 1/2^2 - 1/1^2].
= -2.179 × 10^-18 Joules × (0.25 - 1).
= - 2.179 × 10^-18 × (- 0.75).
= 1.63425 × 10^- 18 Joules.
