Answer:
1.13 moles Au
Explanation:
Moles Au = 6.80x10²³atoms / 6.023x10²³atoms/mole = 1.13 moles Au
Answer:
magnitude means absolute value, so the one that is greastest, like |-7| and |4| even id |-7| is a negative number, but it is still the one farthest away from 0, so |-7| is greater than |4|.
That is the way to find the greatest magnitude, but because I don't know your numbers so I can not answer your question, but this is the way to solve for it.
HOPE THIS HELPS!!!!!!!!!( IF IT DOES <u><em>PLEASE MARK ME AS BRAINLIEST )</em></u>
Answer:
- 0.99 °C ≅ - 1.0 °C.
Explanation:
- We can solve this problem using the relation:
<em>ΔTf = (Kf)(m),</em>
where, ΔTf is the depression in the freezing point.
Kf is the molal freezing point depression constant of water = -1.86 °C/m,
m is the molality of the solution (m = moles of solute / kg of solvent = (23.5 g / 180.156 g/mol)/(0.245 kg) = 0.53 m.
<em>∴ ΔTf = (Kf)(m)</em> = (-1.86 °C/m)(0.53 m) =<em> - 0.99 °C ≅ - 1.0 °C.</em>
Answer: option <span>A) increases from bottom to top within the group.
Explanation:
</span>It is a known trend that the metallic character of the elements increase from let to right and from top to bottom.
The greater the metallic character the greater the reactivity of the metal.
So, the elements of the columns 1 and 2 are the most reactive metals and among them the elements at the bottom are yet more reactive.
<span>The higher reactivity of the metals that are lower in the periodic table is attributed to the greater total number of electrons.
The greater the total number of electrons the more reactive the metals
as their outermost electrons (the valence electrons which are those that react) are located further from the nucleus and therefore they are held less
strongly, which makes them react more easily.</span>
Answer:
Electrons will flow from left to right through the wire.
Pb^2+ ions will be reduccd to Pb metal.
The concentration of Sn2+ ions in the left compartment will increase.
Explanation:
Looking at the relative electrode potentials of the two metals
Sn= -0.14
Pb=-0.13
Tin is expected to function as the anode (left hand half cell) and lead as the anode (right hand half cell) tin oxidizes to sn^2+ hence its concentration increases on the left compartment while lead is reduced to ordinary lead metal on the right hand half cell . since oxidation occurs on the left hand side, electrons flow from left to right.
