Answer:
It would have a charge of +2.
Explanation:
A proton has a charge of +1, an elctron has a charge of -1, and a neutron has a charge of 0. Ignore the neurons since they have a charge of 0. Since there are 20 protons and 18 electrons (20-18), the overall charge of the atom is +2.
His mass will remain constant; his weight will decrease
This uses the concept of freezing point depression. When faced with this issue, we use the following equation:
ΔT = i·Kf·m
which translates in english to:
Change in freezing point = vant hoff factor * molal freezing point depression constant * molality of solution
Because the freezing point depression is a colligative property, it does not depend on the identity of the molecules, just the number of them.
Now, we know that molality will be constant, and Kf will be constant, so our only unknown is "i", or the van't hoff factor.
The van't hoff factor is the number of atoms that dissociate from each individual molecule. The higher the van't hoff factor, the more depressed the freezing point will be.
NaCl will dissociate into Na+ and Cl-, so it has i = 2
CaCl2 will dissociate into Ca2+ and 2 Cl-, so it has i = 3
AlBr3 will dissociate into Al3+ and 3 Br-, so it has i = 4
Therefore, AlBr3 will lower the freezing point of water the most.
Answer:
I believe it is D
Explanation:
since Rutherford's explanation, when he made it in 1911, was that scattering was caused by a hard, dense court to centre of the Adam, which is the nucleus and he used Alpha particles to observe the scattered backwards from a gold foil
Answer:
Three hydrogen atoms to form PH₃.
Explanation:
Hello!
In this case, since the elements belonging to the nitrogen family (N, P, As, Sb and Bi) show five valence electrons, because there are five electrons at their outer shell, it is clear that if phosphorous bonds with hydrogen, it is going to require the same amount of oxygen atoms (3) because elements having five valence electrons need 3 bonds in order to attain the octet (5+3=8).
Therefore the compound would be:
Which is phosphine.
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