Answer:
0.287 mole of PCl5.
Explanation:
We'll begin by calculating the number of mole in 51g of Cl2. This is illustrated below:
Molar mass of Cl2 = 2 x 35.5 = 71g/mol
Mass of Cl2 = 51g
Number of mole of Cl2 =..?
Mole = Mass /Molar Mass
Number of mole of Cl2 = 51/71 = 0.718 mole
Next, we shall write the balanced equation for the reaction. This is given below:
P4 + 10Cl2 → 4PCl5
Finally, we determine the number of mole of PCl5 produced from the reaction as follow:
From the balanced equation above,
10 moles of Cl2 reacted to produce 4 moles of PCl5.
Therefore, 0.718 mole of Cl2 will react to produce = (0.718 x 4)/10 = 0.287 mole of PCl5.
Therefore, 0.287 mole of PCl5 is produced from the reaction.
Answer:
Explanation:
A 30.0g sample of O2 at standard temperature and pressure (STP) would occupy what volume in liters
PV =nRT
at STP P= 1atm. T= 273 K
n is the number of moles. O2 has a molar mass of 32.
30 gm of O2 is 30/32= 0.94 =n
PV = nRT
at STP: P= 1 atm, T=273 K, R is always 0.082 for P in atm and T in K
SO
1 X V = 0.94 X 0.082 X 273
using high school freshman algebra,
V= 0.94 X 0.082 X 273 = 21L
using high school algebra I,
V=
<u>Answer:</u> The equilibrium constant for this reaction is 
<u>Explanation:</u>
The equation used to calculate standard Gibbs free change is of a reaction is:
![\Delta G^o_{rxn}=\sum [n\times \Delta G^o_{(product)}]-\sum [n\times \Delta G^o_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G%5Eo_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G%5Eo_%7B%28reactant%29%7D%5D)
For the given chemical reaction:
The equation for the standard Gibbs free change of the above reaction is:
![\Delta G^o_{rxn}=[(1\times \Delta G^o_{(Ni(CO)_4(g))})]-[(1\times \Delta G^o_{(Ni(s))})+(4\times \Delta G^o_{(CO(g))})]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20G%5Eo_%7B%28Ni%28CO%29_4%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20G%5Eo_%7B%28Ni%28s%29%29%7D%29%2B%284%5Ctimes%20%5CDelta%20G%5Eo_%7B%28CO%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta G^o_{rxn}=[(1\times (-587.4))]-[(1\times (0))+(4\times (-137.3))]\\\\\Delta G^o_{rxn}=-38.2kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-587.4%29%29%5D-%5B%281%5Ctimes%20%280%29%29%2B%284%5Ctimes%20%28-137.3%29%29%5D%5C%5C%5C%5C%5CDelta%20G%5Eo_%7Brxn%7D%3D-38.2kJ%2Fmol)
To calculate the equilibrium constant (at 58°C) for given value of Gibbs free energy, we use the relation:
where,
= Standard Gibbs free energy = -38.2 kJ/mol = -38200 J/mol (Conversion factor: 1 kJ = 1000 J )
R = Gas constant = 8.314 J/K mol
T = temperature = ![58^oC=[273+58]K=331K](https://tex.z-dn.net/?f=58%5EoC%3D%5B273%2B58%5DK%3D331K)
= equilibrium constant at 58°C = ?
Putting values in above equation, we get:
Hence, the equilibrium constant for this reaction is
Spectroscopy be used to distinguish between the following is the compound B has a peak at 3200 – 3500 cm⁻¹ in its IR spectrum.
<h3>What is spectroscopy?</h3>
Spectroscopy is the study of emission or absorption of light. It is used to study the structure of atoms and molecules.
The three types of spectroscopy are:
- atomic absorption spectroscopy (AAS)
- atomic emission spectroscopy (AES)
- atomic fluorescence spectroscopy (AFS)
Thus, the correct option is B, the compound B has a peak at 3200 – 3500 cm⁻¹ in its IR spectrum.
Learn more about spectroscopy
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