Question #42257
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"What balanced equation represents a redox reaction?"
Let's take #r=0.3dot6#.
To cancel out the recurring part of the decimal we must multiply by 10, so #10r=3.dot6#.
#9r=10r-r=3.dot6-0.3dot6=3.3#
To make it a whole number we multiply by 10 again, #90r=33#.
Now just divide both sides by 90 to get r, #r=33/90=11/30#
See a solution process below:
First, we can let:
#a = 0.3bar6#
We can then multiply each side of the equation by #10# to give:
#10 * a = 10 * 0.3bar6#
#10a = 3.bar6#
We can next subtract the left sides of the equation and right sides of the equation to give:
#10a - a = 3.bar6 - 0.3bar6#
#10a - 1a = 3.bar6 - 0.3bar6#
#(10 - 1)a = 3.bar6 - 0.3bar6#
#9a = 3.bar6 - 0.3bar6#
We can then rewrite the right side of the equation as:
#9a = (3.6 + 0.0bar6) - (0.3 + 0.0bar6)#
#9a = 3.6 + 0.0bar6 - 0.3 - 0.0bar6#
#9a = 3.6 - 0.3 + 0.0bar6 - 0.0bar6#
#9a = (3.6 - 0.3) + (0.0bar6 - 0.0bar6)#
#9a = 3.3 + 0#
#9a = 3.3#
Next, divide each side of the equation by #color(red)(9)# to solve for #a# while keeping the equation balanced:
#(9a)/color(red)(9) = 3.3/color(red)(9)#
#(color(red)(cancel(color(black)(9)))a)/cancel(color(red)(9)) = 3.3/9#
#a = 3.3/9#
We can now convert the fraction to the simplest form by multiplying the fraction by the appropriate form of #1# and then cancelling common terms in the numerator and denominator:
#a = (10/10 xx 3.3/9)#
#a = (10 xx 3.3)/(10 xx 9)#
#a = 33/90#
#a = (11 xx 3)/(30 xx 3)#
#a = (11 xx color(red)(cancel(color(black)(3))))/(30 xx color(red)(cancel(color(black)(3))))#
#a = 11/30#
Therefore:
#0.3bar6 = 11/30#
