<h3><u>Answer</u>;</h3>
C3H4O
<h3><u>Explanation;</u></h3>
Empirical formula is the simplest formula of a compound;
Molar mass CO2 = 44.01
Mass of CO2 produced = 2.053 g
Mass of carbon in original sample = 12.01/44.01 × 2.053
= 0.5603g
Molar mass H2O = 18
Mass of H in original sample = 2/18 ×0.5601
= 0.0622 g
Thus; original sample contained 0.5603g C and 0.0622g H. The balance of the sample was O
Mass of O = 0.8715 - (0.5603 + 0.0622) = 0.249g
The mole ratio of C:H:O will be;
Moles C = 0.5603/12 = 0.0467
Moles H = 0.0622
Moles O = 0.249/16 = 0.01556
C:H:O = 0.0467:0.0622:0.01556
Divide through by 0.01556:
C:H:O = 3:4:1
Empirical formula is thus C3H4O
A neutral atom in its ground state contains 28 electrons. this element is considered a <span>transition </span>element, and has 8 electrons in orbitals with l = 2.
Electron configuration: ₂₈X 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s².
l is azimuthal quantum number, l = 2 describes d orbital. There is eight electrons in 3d orbital.
Mg(OH)2 + 2 HCl → MgCl2 + 2 H2O
(0.0500 L) x (0.500 mol/L HCl) x (1 mol Mg(OH)2 / 2 mol HCl) x (58.3197 g Mg(OH)2/mol) = 0.729 g = 729 mg Mg(OH)2
<span>(729 mg Mg(OH)2) x (5.00 mL / 400 mg) = 9.11 mL milk of magnesia</span>
<span>Answer:
2Al(s) + 3Cl2(g) ---> 2AlCl3(s)
Always work in moles
moles = mass of substane / molar mass
2Al (s) + 3Cl (g) --> 2AlCl3 (s)
moles Al = 23.0 g / 26.98 g/mol
= 0.852 moles Al</span>
im pretty sure the correct answer is A
