Here we have to get the 
of the reaction at 520 K temperature.
The of the reaction is 1.705 atm
We know the relation between and 
is 
, where = The equilibrium constant of the reaction in terms of partial pressure, = The equilibrium constant of the reaction in terms of concentration and N = number of moles of gaseous products - Number of moles of gaseous reactants.
Now in this reaction, PCl₃ + Cl₂ ⇄ PCl₅
Thus number of moles of gaseous product is 1, and number of moles of gaseous reactants are 2. Thus N = |1 - 2| = 1 mole
The given value of is 4.0×10⁻²
The molar gas constant, R = 0.082 L. Atm. mol⁻¹. K⁻¹ and temperature, T = 520 K.
On plugging the values in the equation we get,
Or, = 1.705 atm
Thus, the of the reaction is 1.705 atm
Answer:1 mol of Mg(NO3)2 contains 6.022*10^23 molecules
3 mol Mg(NO3)2 contains 3*6.022*10^23 = 1.81*10^24 molecules
Each Mg(NO3)2 molecule contains 2 N atoms
Number of N atoms = 2*1.81*10^24 = 3.62*10^24 N atoms.
Are there options for answers
Answer:
Fluorine
General Formulas and Concepts:
<u>Chemistry</u>
- Reading a Periodic Table
- Periodic Trends
- Electronegativity - the tendency for an element to attract an electron to itself
- Z-effective and Coulomb's Law, Forces of Attraction
Explanation:
The Periodic Trend for Electronegativity is up and to the right of the Periodic Table.
Fluorine is Element 9 and has 9 protons. Radium is Element 88 and has 88 protons. Therefore, Radium has a bigger Zeff than Flourine.
However, since Radium is in Period 7 while Fluorine is in Period 2, Radium has more core e⁻ than Fluorine does. This will create a much larger shielding effect, causing Radium's outermost e⁻ to have less FOA between them. Fluorine, since it has less core e⁻, the FOA between the nucleus and outershell e⁻ will be much stronger.
Therefore, Fluorine would attract an electron more than Radium, thus bringing us to the conclusion that Fluorine has a higher electronegativity.
Answer:
2.038 seconds.
Explanation:
So, in the question above we are given the following parameters in order to solve this question. We are given a rate constant of 0.500 s^-, initial concentration= 0.860 M and final concentration= 0.310 M,the time,t =??.
Assuming that the equation for the first order of reaction is given below,that is;
A ---------------------------------> products.
Recall the formula below;
B= B° e^-kt.
Therefore, e^-kt = B/B°.
-kt = ln B/B°.
kt= ln B°/B.
Where B° and B are the amount of the initial concentration and the amount of the concentration remaining, k is the rate constant and t = time taken for the concentration to decrease.
So, we have; time taken,t = ln( 0.860/.310)/0.500.
==> ln 2.77/0.500.
==> time taken,t =2.038 seconds.
