Hey there, The answer is A. Free electrons.
Increase the Surface Area of the Reactants. Increasing the surface area of the reactants increases the rate of the reaction. More surface area means more collisions of the reactant molecules and an increased rate of the reaction. This occurs when reactants are made to react in powdered form
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.
To find the number of neutrons, subtract the number of protons from the mass number. number of neutrons=40−19=21.
Answer:
The initial rate of the reaction between substances P and Q was measured in a series of
experiments and the following rate equation was deduced.
Complete the table of data below for the reaction between P and Q
Explanation:
Given rate of the reaction is:
![rate= k[P]^{2} [Q]\\=>[Q]=\frac{rate}{k.[P]^{2} } \\and \\\\\\\ [P]=\sqrt{\frac{rate}{k.[Q]} }](https://tex.z-dn.net/?f=rate%3D%20k%5BP%5D%5E%7B2%7D%20%5BQ%5D%5C%5C%3D%3E%5BQ%5D%3D%5Cfrac%7Brate%7D%7Bk.%5BP%5D%5E%7B2%7D%20%7D%20%5C%5Cand%20%5C%5C%5C%5C%5C%5C%5C%20%5BP%5D%3D%5Csqrt%7B%5Cfrac%7Brate%7D%7Bk.%5BQ%5D%7D%20%7D)
Substitute the given values in this formulae to get the [P], [Q] and rate values.
From the first row,
the value of k can be calulated:
![k=\frac{rate}{[P]^{2}[Q] } \\ =\frac{4.8*10^-3}{(0.2)^{2} 2. (0.30)} \\ =0.4](https://tex.z-dn.net/?f=k%3D%5Cfrac%7Brate%7D%7B%5BP%5D%5E%7B2%7D%5BQ%5D%20%7D%20%5C%5C%20%20%3D%5Cfrac%7B4.8%2A10%5E-3%7D%7B%280.2%29%5E%7B2%7D%202.%20%280.30%29%7D%20%5C%5C%20%3D0.4)
Second row:
2. Rate value:

3.Third row:
![[Q]=\frac{rate}{k.[P]^{2} } \\ =9.6*10^-3 / (0.4 *(0.40)^{2} \\ =0.15mol.dm^{-3}](https://tex.z-dn.net/?f=%5BQ%5D%3D%5Cfrac%7Brate%7D%7Bk.%5BP%5D%5E%7B2%7D%20%7D%20%5C%5C%20%20%20%20%20%3D9.6%2A10%5E-3%20%2F%20%280.4%20%2A%280.40%29%5E%7B2%7D%20%5C%5C%20%20%20%20%3D0.15mol.dm%5E%7B-3%7D)
4. Fourth row:
![[P]=\sqrt{\frac{rate}{k.[Q]} }\\=>[P]=\sqrt{\frac{19.2*10^-3}{0.60*0.4} } \\=>[P]=0.283mol.dm^{-3}](https://tex.z-dn.net/?f=%5BP%5D%3D%5Csqrt%7B%5Cfrac%7Brate%7D%7Bk.%5BQ%5D%7D%20%7D%5C%5C%3D%3E%5BP%5D%3D%5Csqrt%7B%5Cfrac%7B19.2%2A10%5E-3%7D%7B0.60%2A0.4%7D%20%7D%20%5C%5C%3D%3E%5BP%5D%3D0.283mol.dm%5E%7B-3%7D)