This would be C. A car rusting
A. As pressure on the gas increases, the volume and temperature will both decrease
Answer:
- <u>Cadmium has larger atomic radius than sulfur.</u>
Explanation:
Down a period, atomic radii decrease from left to right due to the increase in the number of protons and electrons across a period: when a proton is added the pull of the electrons towards the nucleus is larger, so the size of the atom decreases.
Hence, you can compare the elements that belong to a same period and predict that the atom with lower atomic number (number of protons) will haver larger atomic radius. With that:
- Oxygen and fluorine are in the period 3, being oxygen to the left of fluorine, so oxygen is larger than fluorine.
- Sulfur and chlorine are in the period 4, being sulfur to the left of chlorine, so sulfur is larger than chlorine.
Now see whan happens down a group. Atomic radius increases from top to bottom within a group due to electron shielding. That permits you to compare the size of the elements in a group:
- Fluorine and chlorine are in the same group (17), with chlorine directly below fluorine, so the atomic radius of chlorine is larger than the atomic radius of fluorine.
- Sulfur and oxygen are in the same group (16), with sulfur directlly below oxygen, so sulfur the atomic radius of sulfur is larger than the atocmi radius of oxygen.
So far, you can rank the atomic radius of sulfur, chlorine, fluorine, and oxygen, in increasing order as:
- O < F < Cl < S, concluding that O, F, and Cl have smaller atomic radius than S.
Cadmiun, Cd, is to the left and below sulfur, so both electron shielding (down a group) and increase of the number of protons (down a period) lead to predict the cadmium has a larger atomic radius than sulfur.
Answer:
0.289J of heat are added
Explanation:
We can relate the change in heat of a substance with its increasing in temperature using the equation:
q = m*ΔT*S
<em>Where Q is change in heat</em>
<em>m is mass of substance (In this case, 0.0948g of water)</em>
<em>ΔT = 0.728°C</em>
<em>S is specific heat (For water, 4.184J/g°C)</em>
Replacing:
q = 0.0948g*0.728°C*4.184J/g°C
q = 0.289J of heat are added
Oxidation
It is oxidation because two O's 1 x 1 d 2 i's 1 n and 1 a
