Answer : The oxidation state of Mg in Mg(s) is (0).
Explanation :
Oxidation number or oxidation state : It represent the number of electrons lost or gained by the atoms of an element in a compound.
Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.
Rules for Oxidation Numbers are :
The oxidation number of a free element is always zero.
The oxidation number of a monatomic ion equals the charge of the ion.
The oxidation number of Hydrogen (H) is +1, but it is -1 in when combined with less electronegative elements.
The oxidation number of oxygen (O) in compounds is usually -2.
The oxidation number of a Group 17 element in a binary compound is -1.
The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.
The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.
The given chemical reaction is:
In the given reaction, the oxidation state of Mg in Mg(s) is (0) because it is a free element and the oxidation state of Mg in is (+2).
Hence, the oxidation state of Mg in Mg(s) is (0).
Answer: When an atom becomes a positive ion it pulls it's electrons closer, decreasing it's radius moreover, when it becomes a negative ion, it pulls it's electrons closer and decreases the radius
I hope this helps :)
Chemical reactions that require oxygen are called aerobic, while those that do not require oxygen are called anaerobic.
Answer:
48 molecules of CO₂
Explanation:
I think you made a mistake in your question. The formula for propane is C₃H₈, not C₃H₃. But, I will give you the answer for both cases.
For C₃H₃:
First you have to balance the equation.
4 C₃H₃ + 15 O₂ ⇒ 12 CO₂ + 6 H₂O
Next, you need to use the mole ratios between C₃H₃ and CO₂ to find the amount of molecules of CO₂ you will produce with the given amount of C₃H₃.
(16 mol's C₃H₃) × (12 mol's CO₂/4 mol's C₃H₃) = 48 mol's CO₂
You will get 48 molecules of CO₂.
For C₃H₈:
Balance the equation.
C₃H₈ + 5 O₂ ⇒ 3 CO₂ + 4 H₂O
Use the mole ratios between C₃H₈ and CO₂.
(16 mol's C₃H₈) × (3 mol's CO₂/1 mol's C₃H₈) = 48 mol's CO₂
You will get 48 molecules of CO₂ for this equation as well.
Here,
Heat gain by the first sample of water + Heat lost by the second sample of water is equal to zero (0).
Now, Mass of water sample one = 108 g (given)
Mass of water sample two = 66.9 g (given)
Temperature for water sample one =
Let, temperature for water sample two =x
And, final temperature =
Now,
where, = final temperature
= initial temperature
Substitute all the given values in above formula:
[tex x =84.81^{0}C [/tex]