When molten rock undergoes crystallization above ground, what type of rock results?
2 answers:
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Answer:
( About ) 0.03232 M
Explanation:
Based on the units for this reaction it should be a second order reaction, and hence you would apply the integrated rate law equation "1 / [X] = kt + 1 / []"
This formula would be true for the following information -
{ = the initial concentration of X, k = rate constant, [ X ] = the concentration after a certain time ( which is what you need to determine ), and t = time in minutes }
________
Therefore, all we have left to do is plug in the known values. The initial concentration of X is 0.467 at a time of 0 minutes, as you can tell from the given data. This is not relevant to the time needed in the formula, as we need to calculate the concentration of X after 18 minutes ( time = 18 minutes ). And of course k, the rate constant = 1.6
1 / [X] = ( 1.6 )( 18 minutes ) + 1 / ( 0.467 ) - Now let's solve for X
1 / [X] = 28.8 + 1 / ( 0.467 ),
1 / [X] = 28.8 + 2.1413...,
1 / [X] = 31,
[X] = 1 / 31 = ( About ) 0.03232 M
Now for this last bit here you probably are wondering why 1 / 31 is not 0.03232, rather 0.032258... Well, I did approximate one of the numbers along the way ( 2.1413... ) and took the precise value into account on my own and solved a bit more accurately. So that is your solution! The concentration of X after 18 minutes is about 0.03232 M
Find the oxidation state for each element.
The oxidation state of H is +1 in most compounds.
- The S atom in H₂S is bonded to two H atoms. The oxidation state of S in H₂S will be -(2 × (+1)) = -2.
- Two O atom in H₂O₂ are bonded to two H atoms. On average, the oxidation state for each O atom in H₂O₂ will be - 1/2 × (2 × (+1)) = -1.
- S has an oxidation state of 0 when it is not bonded to any other element;
- The O atom in H₂O is bonded to two H atoms. The oxidation state of O in H₂O will be -(2 × (+1)) = -2.
Changes in oxidation states:
- S: from -2 to 0.
- O: from -1 to -2.
Each S atom will reduce two O atoms. There are two O atoms in each H₂O₂ molecule. There's only one S atom in each H₂S molecule. As a result, pairing one H₂O₂ with one H₂S will balance the change in oxidation state.
Each H₂O₂ contains two O atoms. The two O atoms will produce two H₂O molecules. Each H₂S molecule will lead to one S molecule. Thus the coefficient for H₂O will be two and the coefficient for the rest three species will be one.