The question requires us to explain the differences in radii of neutral atoms, cations and anions.
To answer this question, we need to keep in mind that a neutral atom presents the same number of protons (positive particles) and electrons (negative particles). Another important information is that the protons are located in the nucleus of the atom, while the electrons are around the nucleus. Also, there is an electrostatic force between protons and electrons, which means that they the protons tend to attract the electrons to the nucleus.
While a neutral atom presents the same number of protons and electrons, a cation is an ion with positive charge, which means it has lost one or more electrons. In a cation, the balance between protons and electrons doesn't exist anymore: now, there is more positive than negative charge (more protons than electrons), and the overall attractive force that the protons have for the electrons is increased. As a result, the electrons stay closer to the nucleus and the radius of a cation is smaller than the neutral atom from which it was derived.
On the other side, anions present negative charge, which means they have received electrons. Similarly to cations, the balance between protons and electrons doesn't exist anymore, but in this case, there are more electrons than protons. In an anion, the overall attractive force that the protons have for the electrons is decreased. As a result, the electrons are "more free" to move and, as they are not so attracted to the nucleus, they tend to stay farther from the positive nucleus compared to the neutral atom - because of this, the radius of an anion is larger than the neutral atom from which it was derived.
Answer:
a) Volume of vial= 9.626cm3
b) Mass of vial with water = 62.92 g
Explanation:
a) Mass of empty vial = 55.32 g
Mass of Vial + Hg = 185.56 g
Therefore,
Density of Hg = 13.53 g/cm3
b) Volume of water = volume of vial = 9.626 cm3
Density of water = 0.997 g/cm3
Answer:
The endpoint volume is 50.52 ± 0.14 mL
Explanation:
In a titration always is necessary to subtract the blank volume to the titrant volume to obtain the real volume of the titrant. Thus in this case, the total endpoint volume is the sum of the initial volume delivered and the second volume delivered, minus the blank volume:
V = (49.16±0.06 mL) + (1.69±0.04 mL) - (0.33±0.04 mL)
V = (49.16 + 1.69 - 0.33) ± (0.06+0.04+0.04) mL
V = 50.52 ± 0.14 mL
It is necessary to consider the sum of the errors too.
Answer:
The volume of the stock solution needed is 1L
Explanation:
Step 1:
Data obtained from the question. This include the following:
Concentration of stock solution (C1) = 6M
Volume of stock solution needed (V1) =?
Concentration of diluted solution (C2) = 1M
Volume of diluted solution (V2) = 6L
Step 2:
Determination of the volume of the stock solution needed.
With the dilution formula C1V1 = C2V2, the volume of the stock solution needed can be obtained as follow:
C1V1 = C2V2
6 x V1 = 1 x 6
Divide both side by 6
V1 = 6/6
V1 = 1L
Therefore, the volume of the stock solution needed is 1L
The delta H of -484 kJ is the heat given off when 2 moles of H2 react with 1 mole of O2 to make 2 moles of H2O. You don't have anywhere near that much reactants, only 1/4 as much
<span>actual delta H = 0.34 moles H2 x (-484 kJ / 2 moles H2) = 823 kJ </span>
<span>delta E = delta H - PdeltaV = 823 kJ - 0.41 kJ = 822 kJ</span>
