4.48 mol Cl2. A reaction that produces 0.35 kg of BCl3 will use 4.48 mol of Cl2.
(a) The <em>balanced chemical equation </em>is
2B + 3Cl2 → 2BCl3
(b) Convert kilograms of BCl3 to moles of BCl3
MM: B = 10.81; Cl = 35.45; BCl3 = 117.16
Moles of BCl3 = 350 g BCl3 x (1 mol BCl3/117.16 g BCl3) = 2.987 mol BCl3
(c) Use the <em>molar ratio</em> of Cl2:BCl3 to calculate the moles of Cl2.
Moles of Cl2 = 2.987 mol BCl3 x (3 mol Cl2/2 mol BCl3) = 4.48 mol Cl2
Answer:
In general, atomic radius decreases across a period and increases down a group. ... Down a group, the number of energy levels (n) increases, so there is a greater distance between the nucleus and the outermost orbital. This results in a larger atomic radius.
Answer: the minimum amount of energy required to break bonds and start a chemical reaction
Explanation: got a 100% on the quick check
Sulfur is a group six element in period 3 with atomic number 16 and an electronic configuration of 2:8:6. Therefore, to attain a stable configuration it requires to gain two electrons forming an ion with a charge of -2. The negative charge is due to the gaining of electrons.
<span>70.4 mg CO2 x 1.0 g /1000 mg x 1 mole CO2/ 44 gCO2 x 1 mole C/1 mole CO2 = 0.0016 moles C
14.4 mg H2O x 1.0 g/1000 mg x 1 mole H2O/18 g H2O x 2 moles H/ 1 mole H2O = 0.0016 moles O
molar mass of C=12 g/mole
molar mass of H=1 g/mole
0.0016 moles C x 12 g C/ 1 mole C = 0.0192 g C or 19.2 mg C
0.00156 moles H x 1 g H/1 mole H = 0.00156 g H or 1.56 mg H
mg O= 30.4 mg vanillin - 19.2 mg C – 1.56 mg H = 9.64 mg O
molar mass of O=16 g/mole
9.64 mg O x 1 g/1000 mg x 1 mole O/16.0 g = 0.000602
C.0016 H.0016 O.000602; divide all the moles by the smallest value of0.000602
C2.66H2.66O1 is the empirical formula;
to obtain whole numbers multiply by 3
3[C2.66H2.66O1] = C8H8O3
above formula weight: 8(C) + 8(H) + 3(O) = 8(12) + 8(1) + 3(16) = 152 amu
The empirical formula weight and the molecular formula weight are the same .
Molecular formula is C8H8O3.</span>
