28.4 grams of P₂O₅ will be obtained from the interaction of 11.2 liters of oxygen with phosphorus.
Let's consider the following balanced equation.
4 P + 5 O₂ ⇒ 2 P₂O₅
Since the conditions are not specified, we will assume that we are working at standard temperature and pressure. At STP, 1 mole of an ideal gas occupies 22.4 L. The volume of 11.2 L of oxygen at STP, assuming ideal behavior, is:
![11.2 L \times \frac{1mol}{22.4L} = 0.500 mol](https://tex.z-dn.net/?f=11.2%20L%20%5Ctimes%20%5Cfrac%7B1mol%7D%7B22.4L%7D%20%3D%200.500%20mol)
The molar ratio of O₂ to P₂O₅ is 5:2. The moles of P₂O₅ obtained from 0.500 moles of O₂ are:
![0.500 mol O_2 \times \frac{2molP_2O_5}{5molO_2} =0.200 mol P_2O_5](https://tex.z-dn.net/?f=0.500%20mol%20O_2%20%5Ctimes%20%5Cfrac%7B2molP_2O_5%7D%7B5molO_2%7D%20%3D0.200%20mol%20P_2O_5)
The molar mass of P₂O₅ is 141.94 g/mol. The mass corresponding to 0.200 moles of P₂O₅ is:
![0.200 mol \times \frac{141.94 g}{mol} = 28.4 g](https://tex.z-dn.net/?f=0.200%20mol%20%5Ctimes%20%5Cfrac%7B141.94%20g%7D%7Bmol%7D%20%3D%2028.4%20g)
28.4 grams of P₂O₅ will be obtained from the interaction of 11.2 liters of oxygen with phosphorus.
You can learn more about stoichiometry here: brainly.com/question/22288091
Answer:
The value of the activation energy is 240.96 kJ/mol.
Explanation:
According to the Arrhenius equation,
![k=A\times e^{\frac{-Ea}{RT}}](https://tex.z-dn.net/?f=k%3DA%5Ctimes%20e%5E%7B%5Cfrac%7B-Ea%7D%7BRT%7D%7D)
![\ln k=-\frac{E_a}{RT}+\ln A](https://tex.z-dn.net/?f=%5Cln%20k%3D-%5Cfrac%7BE_a%7D%7BRT%7D%2B%5Cln%20A)
![\log k=-\frac{E_a}{2.303RT}+\log A](https://tex.z-dn.net/?f=%5Clog%20k%3D-%5Cfrac%7BE_a%7D%7B2.303RT%7D%2B%5Clog%20A)
The graph between log(k) and (1/T) will give straight line with negative slope along with the intercept corresponding to the value of A.
Slope f the line =
The slope of the line :
![=\frac{\log (k_2)-\log(k_1)}{\frac{1}{T_2}-\frac{1}{T_1}}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%5Clog%20%28k_2%29-%5Clog%28k_1%29%7D%7B%5Cfrac%7B1%7D%7BT_2%7D-%5Cfrac%7B1%7D%7BT_1%7D%7D)
![T_1=600^oC=600+273K=873 K](https://tex.z-dn.net/?f=T_1%3D600%5EoC%3D600%2B273K%3D873%20K)
![k_1=1.87\times 10^{-3} (Ms)^{-1}](https://tex.z-dn.net/?f=k_1%3D1.87%5Ctimes%2010%5E%7B-3%7D%20%28Ms%29%5E%7B-1%7D)
![T_2=650^oC=650+273K=923K](https://tex.z-dn.net/?f=T_2%3D650%5EoC%3D650%2B273K%3D923K)
![k_2=0.0113 (Ms)^{-1}](https://tex.z-dn.net/?f=k_2%3D0.0113%20%28Ms%29%5E%7B-1%7D)
![-\frac{E_a}{2.303R}=\frac{\log (k_2)-\log(k_1)}{\frac{1}{T_2}-\frac{1}{T_1}}](https://tex.z-dn.net/?f=-%5Cfrac%7BE_a%7D%7B2.303R%7D%3D%5Cfrac%7B%5Clog%20%28k_2%29-%5Clog%28k_1%29%7D%7B%5Cfrac%7B1%7D%7BT_2%7D-%5Cfrac%7B1%7D%7BT_1%7D%7D)
![-\frac{E_a}{2.303\times 8.314J/K mol}=\frac{\log (0.0113 (Ms)^{-1})-\log(1.87\times 10^{-3} (Ms)^{-1})}{\frac{1}{923 K}-\frac{1}{873 K}}](https://tex.z-dn.net/?f=-%5Cfrac%7BE_a%7D%7B2.303%5Ctimes%208.314J%2FK%20mol%7D%3D%5Cfrac%7B%5Clog%20%280.0113%20%28Ms%29%5E%7B-1%7D%29-%5Clog%281.87%5Ctimes%2010%5E%7B-3%7D%20%28Ms%29%5E%7B-1%7D%29%7D%7B%5Cfrac%7B1%7D%7B923%20K%7D-%5Cfrac%7B1%7D%7B873%20K%7D%7D)
![-E_a=-240,959.466J/mol](https://tex.z-dn.net/?f=-E_a%3D-240%2C959.466J%2Fmol)
![E_a=240,959.466 J/mol=240.96 kJ/mol](https://tex.z-dn.net/?f=E_a%3D240%2C959.466%20J%2Fmol%3D240.96%20kJ%2Fmol)
The value of the activation energy is 240.96 kJ/mol.
Answer:
3.39g of MgBr
Explanation:
Reaction between magnesium and bromine
Mg + Br₂ → MgBr
Mass of MgBr = ?
1g of Mg + 5.0g of Br₂
6g of Mg and Br₂
Molar mass of Mg = 24g/mol
Molar mass of Br = 80g/mol
Number of moles = mass / molar mass
Mass = number of moles * molarmass
Mass of Mg = 1 * 24 = 24g
Mass of Br₂ = 1 * (2*80) = 160g (diatomic molecule)
Molarmass of MgBr = (24 + 80) = 104g/mol
Mass of MgBr = 1 * 104 = 104g/mol
24g of Mg + 160g of Br₂ = 104g of MgBr
184g of (mg and Br) = 104g of mgBr
6g of (mg and Br) = y g of MgBr
y = (6 * 104) / 184
y = 3.39g of MgBr
Answer:
Metals on the left of the Periodic Table.
Non-Metals on the top-right, plus Hydrogen.