Al S
1) <u> 35.94 </u> =1.33111 <u> 64.06 </u> = 2.001875
27 32
2) = <u> 1.33111 </u> = <u> 2.001875 </u>
1.33111 1.33111
3) = ( 1 ) × 2 = ( 1.5 ) ×2
4) = 2 = 3
Empirical Formula = Al2S3
1) Divide the percentage given in question by the Relative Atomic Mass (RAM)
of the given elements.
2) When you find the answers of the first part of question, divide these once again but this time, by the lowest number you found in part 1.
3) and 4) Write down the values. If you get a decimal which is in between 0.3-0.7 (including the 0.3 and 0.7), you cannot make it a whole number by rounding of. Therefore, multiply the decimal with a whole number until you get a whole number as your answer. In this question, when you multiply 1.5 by 2, the answer is 3 which is a whole number. Multiply the other whole number by the same number as that you multiplied for 1.5. And use these numbers in part 4 to make the empirical formula which is Aluminium Sulfide (Al2S3)
Hope this helps you :)))
Please give this a brainliest
Mobile phase is the liquid or organic solvent present in the developing tank or beaker by Ayesha zulfiqar
(Hoped that helped)
Answer:
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Answer:
The pressure of CH3OH and HCl will decrease.
The final partial pressure of HCl is 0.350038 atm
Explanation:
Step 1: Data given
Kp = 4.7 x 10^3 at 400K
Pressure of CH3OH = 0.250 atm
Pressure of HCl = 0.600 atm
Volume = 10.00 L
Step 2: The balanced equation
CH3OH(g) + HCl(g) <=> CH3Cl(g) + H2O(g)
Step 3: The initial pressure
p(CH3OH) = 0.250atm
p(HCl) = 0.600 atm
p(CH3Cl)= 0 atm
p(H2O) = 0 atm
Step 3: Calculate the pressure at the equilibrium
p(CH3OH) = 0.250 - X atm
p(HCl) = 0.600 - X atm
p(CH3Cl)= X atm
p(H2O) = X atm
Step 4: Calculate Kp
Kp = (pHO * pCH3Cl) / (pCH3* pHCl)
4.7 * 10³ = X² /(0.250-X)(0.600-X)
X = 0.249962
p(CH3OH) = 0.250 - 0.249962 = 0.000038 atm
p(HCl) = 0.600 - 0.249962 = 0.350038 atm
p(CH3Cl)= 0.249962 atm
p(H2O) = 0.249962 atm
Kp = (0.249962 * 0.249962) / (0.000038 * 0.350038)
Kp = 4.7 *10³
The pressure of CH3OH and HCl will decrease.
The final partial pressure of HCl is 0.350038 atm
Types of Bonds can be predicted by calculating the
difference in electronegativity.
If, Electronegativity difference is,
Less
than 0.4 then it is Non Polar Covalent Bond
Between 0.4 and 1.7 then it is Polar Covalent Bond
Greater than 1.7 then it is Ionic
For Br₂;
E.N of Bromine = 2.96
E.N of Bromine = 2.96
________
E.N Difference
0.00 (Non Polar Covalent Bond)
For MgS;
E.N of Sulfur = 2.58
E.N of Magnesium = 1.31
________
E.N Difference 1.27 (Ionic Bond)
For SO₂;
E.N of Oxygen = 3.44
E.N of Sulfur = 2.58
________
E.N Difference 0.86 (Polar Covalent Bond)
For KF;
E.N of Fluorine = 3.98
E.N of Potassium = 0.82
________
E.N Difference 3.16 (Ionic Bond)
Result: The Bonds in Br₂ and SO₂ are Covalent in Nature.
