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Annabel K.
Tutor for three years
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Chemistry
TutorMe
Question:

CO2(g)+H2(g)-->2CO(g)+H2O(g) How would you balance this equation?

Annabel K.
Answer:

This is a redox reaction so balancing this requires a little more thought than just looking at the numbers on either side. Step 1. Pull out the first half reaction. Arbitrarily pick out any one you'd like. Make sure that the products and reactants at least somewhat resemble eachother. For example, I would put CO2 with 2CO not with H2O. So the first half reaction is: CO2(g)-->2CO(g) first balance non hydrogen and non oxygen atoms 2CO2(g)-->2CO(g) second balance oxygen by adding water to the side that needs more oxygen. 2CO2(g)-->2CO(g)+2H2O(l) third balance the hydrogens by adding H+ to the side that needs hydrogen 4H+(aq)+2CO2(g)-->2CO(g)+2H2O(l) fourth balance the charge by adding electrons to either balance out the charge or cancel it out. Since we have 4+ charge from the H+, we must add 4 electrons to the left side. Note that electrons do not have a phase. 4H+(aq)+4e-+2CO2(g)-->2CO(g)+2H2O(l) This is a reduction half reaction because the electrons are on the left side. Step 2. Repeat with the other half reaction H2(g)-->H2O(g) first balance non hydrogen non oxygen atoms (cant. there are no other atoms except hydrogen and oxygen) second balance oxygen by adding water to the side that needs more oxygen. H2O(l)+H2(g)-->H2O(g) third balance the hdyrogens by adding H+ to the side that needs more hydrogen H2O(l)+H2(g)-->H2O(g)+2H+(aq) fourth balance the charge by adding electrons to either balance out the charge or cancel it out. Since we have 2+ charge from the H+, we must add 2 electrons to the right side. Note that electrons do not have a phase. H2O(l)+H2(g)-->H2O(g)+2H+(aq)+2e- This is an oxidation half reaction because the electrons are on the right side. Step 3. Cancel out electrons by making both half reactions have same number of electrons. Oxidation: 4H+(aq)+4e-+2CO2(g)-->2CO(g)+2H2O(l) Reduction: H2O(l)+H2(g)-->H2O(g)+2H+(aq)+2e- we need to multiply every term in the reduction reaction by 2 so that there are 4 electrons on the right side so that it may cancel out with the 4 electrons on the left side from the oxidation reaction. 2H2O(l)+2H2(g)-->2H2O(g)+4H+(aq)+4e- Great. Now we can add the reactants of both reactions together and the products of both reactions together to make one big equation: 4H+(aq)+4e-+2CO2(g)+2H2O(l)+2H2(g) -->2H2O(g)+4H+(aq)+4e-+2CO(g)+2H2O(l) There are 4H+ on both sides so cancel them out 4e-+2CO2(g)+2H2O(l)+2H2(g) -->2H2O(g)+4e-+2CO(g)+2H2O(l) There are 4e- on both sides so cancel them out 2CO2(g)+2H2O(l)+2H2(g) -->2H2O(g)+2CO(g)+2H2O(l) There are 2H2O(l) on both sides so cancel them out to get the final balanced equation: 2CO2(g)+2H2(g) -->2H2O(g)+2CO(g)

Biology
TutorMe
Question:

How does a transduction work?

Annabel K.
Answer:

A transduction is an experimental process that involves changing the genome of a particular cell. Typically we do transductions to change the DNA of simple organisms such as E. coli because E. coli has plasmid (circular) DNA and can be easily manipulated. We "transduce" new DNA into the existing genome through a viral vector called a phage. So a virus has foreign DNA that it transduces into the E. coli's existing genome. Scientists take a few microliters of their E. coli of interest, and mix it with a few microliters of the virus of interest and let it grow overnight. Then, the mixture is spread onto an agar plate with all the required growth materials (this varies depending on the type of transduction). There is also a selectable marker that kills cells that have not acquired the new DNA. Eventually, the E. coli will reproduce and form colonies, and only the ones that grow on the given conditions have acquired the new viral DNA. The colonies are visible to the naked eye and so these new mutants are now ready to be used in further experiments.

Algebra
TutorMe
Question:

How would you solve for x in this equation? 3x+8=9x+84

Annabel K.
Answer:

First, you should isolate for x. Meaning you should put all the terms that have "x" on the same side, and only x on that side. Let's arbitrarily pick x to be on the left side, and the integers to be on the right side. 3x is already on the left, but we have to move 9x over to the left side. We do that by subtracting 9x from both sides: 3x-9x+8=9x-9x+84 as you can see on the right side, 9x-9x=0 so now on the right side we have: 3x-9x+8=84 Great! So now let's simplify the left side x variables. 3-9 is -6, so 3x-9x=-6x -6x+8=84 So now we have all the x terms together, but they're not "isolated" until the 8 on the left is on the right side. So we must subtract 8 from both sides -6x+8-8=84-8 -6x=76 And now we have to divide both sides by negative 6 x=-12.667 or -12 and 2/3

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