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:
2.63*10^23
Explanation:
1 mol rhodium = 102.91
44.8g/1 mol * 1 mol/ 102.91mol * 6.022*10^23/1 mol =
2.63*10^23
3! You have to ensure balance of all the different elements.
Answer:
- Absolute zero is - 459.67 °F
Explanation:
<u>1) Convert absolute zero to celsius:</u>
- 0 K = - 273.15°C ( this is per definition of the scale)
<u>2) Convert - 273.15°C to Fahrenheit:</u>
- T (°F) = T (°C) × 1.8 + 32 (this is the conversion equation=
- T (°F) = - 273.15 × 1.8 + 32 = - 459.67 °F ← answer
Answer:
D
Explanation:
When lead ions and sulfate ions bond, they form sediment so neither a nor b can be the answer.
The important thing is that two nitrate ions were originally bonded with one lead ion, while two potassium ions bonded with a sulfate ion.
Finally, since potassium and nitrate ions don't form sediment these two ions must remain. Therefore the answer is D
