Explanation:
When we add chlorine to a substance or compound then this process is known as chlorination.
For example, a process of chlorination is as follows.
Initiation :
where, Cl* is a free radical.
Propagation:


Termination:



Thus, we can conclude that out of the given options
is not formed through a termination reaction in the chlorination of methane.
The chemical reaction would be written as
2 AsF3<span> + 3 CCl4 = 2 AsCl3 + 3 CCl2F2
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We use the given amounts of the reactants to first find the limiting reactant. Then use the amount of the limiting reactant to proceed to further calculations.
150 g AsF3 ( 1 mol / 131.92 g) = 1.14 mol AsF3
180 g CCl4 (1 mol / 153.82 g) = 1.17 mol CCl4
Therefore, the limiting reactant would be CCl4 since it would be consumed completely. The theoretical yield would be:
1.17 mol CCl4 ( 3 mol CCl2F2 / 3 mol CCl4 ) = 1.17 mol CCl2F2
Answer:
Percent loss of water = 25%
Explanation:
Given data:
Mass of hydrated salt = 15.6 g
Mass of anhydrous salt = 11.7 g
Percentage of water lost = ?
Solution:
First of all we will calculate the mass of water in hydrated salt.
Mass of water = Mass of hydrated salt - Mass of anhydrous salt
Mass of water = 15.6 g - 11.7 g
Mass of water = 3.9 g
Now we will calculate the percentage.
Percent loss of water = mass of water / total mass × 100
Percent loss of water = 3.9 g/ 15.6 g × 100
Percent loss of water = 25%
Answer:
2.4 ×10^24 molecules of the herbicide.
Explanation:
We must first obtain the molar mass of the compound as follows;
C3H8NO5P= [3(12) + 8(1) + 14 +5(16) +31] = [36 + 8 + 14 + 80 + 31]= 169 gmol-1
We know that one mole of a compound contains the Avogadro's number of molecules.
Hence;
169 g of the herbicide contains 6.02×10^23 molecules
Therefore 669.1 g of the herbicide contains 669.1 × 6.02×10^23/ 169 = 2.4 ×10^24 molecules of the herbicide.
Equilibrium equation is
<span>Ag2CO3(s) <---> 2 Ag+(aq) + CO32-(aq) </span>
<span>From the reaction equation above, the formula for Ksp: </span>
<span>Ksp = [Ag+]^2 [CO32-] = 8.1 x 10^-12 </span>
<span>You know [CO32-], so you can solve for [Ag+] as: </span>
<span>(8.1 x 10^-12) = [Ag+]^2 (0.025) </span>
<span>[Ag+]^2 = 3.24 x 10^-10 </span>
<span>[Ag+] = 1.8 x 10^-5 M </span>