Answer:
Mass = 3.6 g
Explanation:
Given data:
Number of atoms of scandium = 4.77×10²² atoms
Mass of arsenic = ?
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022× 10²³ is called Avogadro number.
one mole = 6.02×10²³ atoms
one mole × 4.77×10²² atoms / 6.02×10²³ atoms
0.08 mol
Mass of scandium
Mass = number of moles × molar mass
Mass = 0.08 mol × 45 g/mol
Mass = 3.6 g
Answer:
300,000km is the speed of light
Answer:
The correct appropriate will be Option 1 (Acid anhydrides are less stable than esters so the equilibrium favors the ester product.)
Explanation:
- Acid anhydride, instead of just a carboxyl group, is typically favored for esterification. The predominant theory would be that Anhydride acid is somewhat more volatile than acid. This is favored equilibrium changes more toward the right of the whole ester structure.
- Extremely responsive than carboxylic acid become acid anhydride as well as acyl chloride. Thus, for esterification, individuals were most favored.
The other options offered are not relevant to something like the scenario presented. So, the solution here is just the right one.
Answer:
Al + 3AgCl → AlCl₃ + 3Ag
Explanation:
The given equation is:
Al + AgCl →
We are to find the product and hence balance the equation. This problem is a simple single replacement reaction.
By virtue of this, Aluminum will displace Ag from the solution:
Al + AgCl → AlCl₃ + Ag
We then balance the equation:
Al + 3AgCl → AlCl₃ + 3Ag
Answer:
electron sea model for metals suggest that valence electrons drift freely around the metal cations.
Explanation:
Explanation: In electron sea model, the valence electrons in metals are delocalized instead of orbiting around the nucleus. ... These electrons are free to move within the metal atoms. Thus, we can conclude that the electron sea model for metals suggest that valence electrons drift freely around the metal cations.
