Answer:
172 g Al
Step-by-step explanation:
We know we will need a balanced equation with masses and molar masses, so let’s <em>gather all the information</em> in one place.
M_r: 26.98 101.96
4Al + 3O₂ ⟶ 2Al₂O₃
m/g: 325
(a) Calculate the <em>moles of Al₂O₃
</em>
n = 325 g Al₂O₃ × 1 mol Al₂O₃ /39.10 g Al₂O₃
n = 3.188 mol Al₂O₃
(b) Calculate the <em>moles of Al
</em>
The molar ratio is (4 mol Al/2 mol Al₂O₃)
n = 3.188 mol Al₂O₃ × (4 mol Al/2 mol Al₂O₃)
n = 6.375 mol Al
(c) Calculate the <em>mass of Al</em>
m = 6.375 mol Al × (26.98 g Al/1 mol Al)
m = 172 g Al
Note: The answer can have only <em>three</em> significant figures because that is all you gave for the mass of Al₂O₃.
Because the alkali metals are the group 1 metals, they have only 1 valence electron that they want to lose, and the halogens are the group 17 nonmetals, they want to gain 1 valence electron to become stable.
Answer:
For each system listed in first column of the table below, decide (if possible) whether the change described in the second column will increase the entropy S of the system, decrease S , or leave S unchanged. If you don?t have enough information to decide, check the not enough information button in the last column
(2) a base because they accept H+ ions. NH3 is the conjugate base of NH4+.
Answer:
See below
Step-by-step explanation:
- Hydrogen either reacts with or is formed by reactions with many other elements, so chemists could use it directly to determine their relative masses.
- Hydrogen has the smallest atomic mass, so it was convenient to give H a relative atomic mass of 1 and assign those of other elements as multiples of this number.
The O = 16 scale became the standard in 1903 and carbon-12 was chosen in 1961.
