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# Tutor profile: Raven B.

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Raven B.
Science & Math Tutor
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## Questions

### Subject:Chemistry

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Question:

Balance the following chemical equation: AlBr3 + Cl2 -> AlCl3 + Br2

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Raven B.

The first thing I like to do when balancing an equation is to list the elements involved on either side of the reaction with the number of atoms already involved like so: AlBr3 + Cl2 -> AlCl3 + Br2 Left: __Al(1) __Br(3) __Cl(2) Right: __Al(1) __Cl(3) __Br(2) Next I look at the elements that are unbalanced and determine the lowest common multiple. For Bromine, you have 3 atoms on the left and 2 on the right, the common multiple is 6. Now you need to determine what you would need to multiply each bromine molecule by to get 6 and enter this number on your elements list. Make sure to change the number of atoms for any elements attached to bromine. In this problem, bromine and aluminum are part of the same molecule on the left of the equation. Thus increasing bromine on the left will increase aluminum. 2 AlBr3 + Cl2 -> AlCl3 + 3 Br2 Left: _2_Al(1) _6_Br(3) __Cl(2) Right: __Al(1) __Cl(3) _6_Br(2) Now bromine is balanced but you have unbalanced aluminum. Repeat the above process to find the common multiple for aluminum atoms. You should get 2. You already have 2 on the left, so you only need to increase aluminum on the right. Don't forget to adjust any elements that are attached to aluminum- in this case chlorine. 2 AlBr3 + Cl2 -> 2 AlCl3 + 3 Br2 Left: _2_Al(1) _6_Br(3) __Cl(2) Right: _2_Al(1) _6_Cl(3) _6_Br(2) The last element to balance is chlorine. Repeat the above process and you will find that the common multiple for chlorine is 6. You already have 6 chlorine molecules on the right, so you only need to adjust the left. 2 AlBr3 + 3 Cl2 -> 2 AlCl3 + 3 Br2 Left: _2_Al(1) _6_Br(3) _3_Cl(2) Right: _2_Al(1) _6_Cl(3) _6_Br(2) Double check that you have the same number of atoms of each element on either side of the equation- this equation is now balanced.

### Subject:Biology

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Question:

A population of flowers have 2 unlinked genes which determine flower color and number of petals. The dominant C allele results in blue flowers while c results in white. The dominant P gene results in 6 petals while p results in 5. If you cross a CcPp plant with a Ccpp plant, what is the probability that the offspring will have 5 blue petals?

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Raven B.

The simplest way to evaluate the results of a genetic cross such as this are by creating a Punnet square. Since you are crossing two traits you will need a 4x4 grid. On the left side let's list the allele combinations for the first parent. Since the genes are unlinked, there is an equal probability of each allele combination. Those 4 possibilities are CP, Cp, cP, and cp. For the second parent, we can determine that there are only 2 possibilities: Cp and cp. However, there are 2 p alleles and either allele can pair with C or c. So really we have 4 possibilities: Cp, Cp, cp, and cp. Now you can cross the parental alleles and fill in the grid with possible genetic combinations for the offspring. Once you have done this you should have 6 genetic combinations. List the possibilities along with how many squares they appear in: 2 CCPp 4 CcPp 2 CCpp 4 Ccpp 2 ccPp 2 ccpp The question wants to know what the probability is of a blue 5-petal flower so we need to determine what phenotype will result from each genotype: 2 CCPp = blue, 6 petals 4 CcPp = blue, 6 petals 2 CCpp = blue, 5 petals 4 Ccpp = blue, 5 petals 2 ccPp = white, 6 petals 2 ccpp = white, 5 petals You can see that 2 of the genotypes will give you the blue, 5 petal phenotype. This means that 6 of the 16 possible genotypes will give you the flow type you want. Now simply determine what percentage of genotypes will give you your phenotype: 6/16 x 100% = 37.5% Thus by crossing these 2 parental plants, you have a 37.5% probability of the offspring having 5 blue petals.

### Subject:Algebra

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Question:

Find the value(s) of X by factoring the following expression: 6x^3+9x+15x^2=3x

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Raven B.

The first step when factoring is to have all numerical values on one side of the equation so that everything equals 0: 6x^3+9x+15x^2-3x=0 Now you can combine similar terms: 6x^3+6x+15x^2=0 If we notice that there is a common term on the left side of the equation we can simplify: 3x(2x^2+2+5x)=0 We can rearrange the expression within the parentheses to resemble a quadratic expression (ax^2+bx+c=0): 3x(2x^2+5x+2)=0 Now we can ignore the 3x (for now) and factor just the quadratic expression. First multiply the 'a' and 'c' terms and list the factors of the product: a=2 c=2 2*2=4 factors of 4: 1,2,4 Now determine what 2 factors of 4 add up to 'b' which in this case is 5: factors of 4: 1,2,4 1+4=5 Now you will use the factors you have picked to split the value of 'b' with the greater term on the left: 3x(2x^2+4x+1x+2)=0 Still ignoring the '3x' we can factor the terms within the parentheses by factoring together the first two terms and the last two terms: 3x(2x(x+2)+1(x+2))=0 Now that you have factored the quadratic expression, you will notice that your new terms each have (x+2) in common. You can pull this common term out of the expression: 3x(x+2)(2x+1)=0 This expression is now completely simplified. To find the values of x, you set each separate term equal to 0 and solve for x: 3x=0 gives x=0 (because 0 divided by any number is still 0) x+2=0 gives x= -2 2x+1=0 becomes 2x= -1 which gives x= -1/2 So your x values are 0, -2, and -1/2 To confirm that these values are correct, return to your original expression and substitute in these values one at a time. When you solve each side of the expression, you will get equivalent values as demonstrated below: 6(-2)^3+9(-2)+15(-2)^2=3(-2) 6(-8)+(-18)+15(4)= -6 -48 - 18 + 60 = -6 -6 = -6

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