Answer:
0.295 g Co
General Formulas and Concepts:
<u>Chemistry - Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
Explanation:
<u>Step 1: Define</u>
3.01 × 10²¹ atoms Co
<u>Step 2: Identify Conversions</u>
Avogadro's Number
Molar Mass of Co - 58.93 g/mol
<u>Step 3: Convert</u>
<u />
= 0.294552 g Co
<u>Step 4: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
0.294552 g Co ≈ 0.295 g Co
Divide both sides to get 104 and that’s your answer
Answer: When using 645 L /s of O2 in a temperature and pressure of 195°C, 0.88 atm respectively, we will get 0.355Kg /s NO
Explanation:
- First we review the equation that represents the oxidation process of the NH3 to NO.
4NH3(g) + 5O2(g) ⟶4 NO(g) +6 H2O(l)
- Second we gather the information what we are going to use in our calculations.
O2 Volume Rate = 645 L /s
Pressure = 0.88 atm
Temperature = 195°C + 273 = 468K
NO molecular weight = 30.01 g/mol
- Third, in order to calculate the amount of NO moles produced by 645L/ s of O2, we must find out, how many moles (n) are 645L O2 by using the general gas equation PV =n RT
Let´s keep in mind that using this equation our constant R is 0.08205Lxatm/Kxmol
PV =n RT
n= PV / RT
n= [ 0.88atm x 645L/s] / [ (0.08205 Lxatm/Kxmol) x 468K]
n= 14.781 moles /s of O2
-
Fourth, now by knowing the amount of moles of O2, we can use the equation to calculate how many moles of NO will be produced and then with the molecular weight, we will finally know the total mass per second .
14.781 moles /s of O2 x 4moles NO / 5 moles O2 x 30.01g NO / 1 mol NO x 1Kg NO /1000g NO = 0.355Kg /s NO
In order to find out the number of electrons present in s orbital of Phosphorous we have to write the electronic configuration of phosphorous.
As the Atomic number of Phosphorous is 15 so it will have 15 electrons in neutral state. So,
P = 15 = 1s², 2s², 2p⁶, 3s², 3p³
Now analyzing electronic configuration, we found that it contains 3 s orbitals, i.e. 1s, 2s and 3s, and each s orbital is completely filled with two electrons each.
Result:
Therefore, a total of six electrons are present in all s orbitals of neutral Phosphorous Atom.
