Answer:
corbonization for dictionnal
Explanation:
correct
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
The answer is 34.1 mL.
Solution:
Assuming ideal behavior of gases, we can use the universal gas law equation
P1V1/T1 = P2V2/T2
The terms with subscripts of one represent the given initial values while for terms with subscripts of two represent the standard states which is the final condition.
At STP, P2 is 760.0torr and T2 is 0°C or 273.15K. Substituting the values to the ideal gas expression, we can now calculate for the volume V2 of the gas at STP:
(800.0torr * 34.2mL) / 288.15K = (760.0torr * V2) / 273.15K
V2 = (800.0torr * 34.2mL * 273.15K) / (288.15K * 760.0torr)
V2 = 34.1 mL
Answer:
Element Symbol Mass Percent
Chlorine Cl 65.571%
Iron Fe 34.429%
Explanation:
Ойлголоо, уучлаарай Ойлголоо, уучлаарай /; coo
