B KOH
I would say this is the base for the compound substance
Answer:
- They are highly reactive metals
- They have low electro negativity
- They have low ionization energy
- They don't exist alone in nature
- They have low densities
Explanation:
Alkali metals are the elements in group 1 of the periodic table. They include Sodium, Lithium, Potassium e.t.c.
Due to the fact they have one atom in their outermost shell, they are very unstable because they easily react with other elements and are therefore don't exist alone in nature but combined with other elements for this same reason.
Since alkali metals don't easily attract other elements due to it's lone pair in the outer most shell, it can be said to have low electro negativity.
Also, they don't need energy to discharge their electrons since they are highly reactive due to their lone pair in the outermost shell and so we say they have low ionization energy.
Due to this reason, they also have low densities.
Answer:
oh I'm here thanks for your points
Answer:
65.08 g.
Explanation:
- For the reaction, the balanced equation is:
<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>
2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.
- Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:
<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>
<u><em>Using cross multiplication:</em></u>
2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.
0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.
∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.
<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol
) = <em>65.08 g.</em>
Additive color mixing involves multiple sources of light with different colors in each source. Subtractive color mixing involves a single source of light with different colors absorbing various wavelengths of the color spectrum. Secondary colors of one system serve as the primary colors for the other.
