TheXSquaredFactor

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 #5
avatar+2439 
+2

This method is somewhat different than the guest's because it does not need to consider the amount of Cl2 reacted. The general conversion strategy for this particular procedure goes as follows. The table displays the general layout, and all unknowns are marked with question marks:

 

\(\text{grams}\rightarrow\text{moles}\rightarrow\text{moles}\rightarrow\text{grams}\)

 

\(16.2g\text{H}_2\)\(1\text{molH}_2\)\(?\text{molHCl}\)\(?g\text{HCl}\)
 \(?g\text{H}_2\)\(?\text{molH}_2\)\(1\text{molHCl}\)

 

Let's travel through this table column by column. We start with the given information, 16.2gH2, and the eventual goal is to perform a series of conversions. The first column of the table is already finished. 

 

The second column asks the following question: How many grams of Hare in one mole of H2? In order to answer this question, we have to reference the indispensable periodic table of the elements. I generally use https://www.ptable.com/

as an electronic version of the table. 

 

The atomic mass of an element is also the molar mass represented in grams, so H has a molar mass of 1.008g. However, realize that we are finding the molar mass of H2. The subscript indicates that there are two hydrogen molecules, so double the original molar mass, 1.008g, to obtain the molar mass of H2\(1\text{molH}_2=1.008g*2=2.016g\).\(\) 

 

The third column is quite a simple step, actually. It compares the molar ratio of the two molecules in question, HCl and H2, in this case. Determining this information requires some basic knowledge of a balanced equation. In the given chemical reaction, it is possible to perceive it in the following sense: One molecule of H2 reacts and yields two molecules of HCl. The number of molecules contained in a mole equals Avagadro's constant, or \(1\text{mol}=6.02*10^{23}\text{ molecules}\). If you continue this logic, the original balanced equation indicates \(1\text{molH}_2=2\text{molHCl}\).

 

The procedure for the fourth column is identical to the procedure for the second column. How many grams of HCl equals one mole of HCl? Because HCl is a compound, the combined mass of the elements equals its molar mass. As aforementioned, H has a mass of 1.008g per mole. Cl, according to the trusty periodic table, has a mass of 35.45g per mole. Therefore, \(1\text{molHCl}=1.008g+35.45g=36.458g\)

 

After all this work, we have finally determined all the missing values in the original conversion that I suggested earlier. The table now looks complete. 

 

\(16.2g\text{H}_2\)\(1\text{molH}_2\)\(2\text{molHCl}\)\(36.458g\text{HCl}\)
 \(2.016g\text{H}_2\)\(1\text{molH}_2\)\(1\text{molHCl}\)

 

This table is really a fancy representation of three ratios. 

 

\(16.2g\text{H}_2* \frac{1\text{molH}_2}{2.016g\text{H}_2}* \frac{2\text{molHCl}}{1\text{molH}_2}* \frac{36.458g\text{HCl}}{1\text{molHCl}}\)First and foremost, let's cancel out all the common units. Doing this shows that the only unit remaining is the desired unit. 
\(16.2* \frac{1}{2.016}* \frac{2}{1}* \frac{36.458g\text{HCl}}{1}\)Now it is a matter of simplifying. When I input this entire expression into the calculator, I get an answer close to what the guest got.
\(586.0g\text{HCl}\) 
  

 

I default to this method because it removes the need to approximate halfway through the calculation.

Feb 16, 2018
 #1
avatar+2439 
+2

In order to add these radicals, the radicals must be placed in simplest radical form. When a radical is in this form, the radicand (the number or expression contained inside the radical) must not have any perfect square factors, excluding 1. This process changes the way the radical looks, but it does not actually change the value.

 

Putting the radicals in simplest form requires one to identify a perfect square factor of the radicand. Let's try the first radical, \(2\sqrt{18}\). In this case, the radicand is 18. A perfect square factor of 18 is 9 since 9*2=18. Let's break up the radicand in this fashion. 

 

\(2\sqrt{18}\Rightarrow 2\sqrt{9*2}\)

 

It is probably obviousn that both expressions are equivalent in value. Because a multiplication operation is being performed inside the radical and the multiplicand and multiplier are both positive, it is possible to split up the radical into two separate parts.

 

\(2\sqrt{9*2}\Rightarrow 2\sqrt{9}*\sqrt{2}\)

 

This manipulation of the original term has now created two radicals: \(\sqrt{9}\text{ and }\sqrt{2}\). Notice that \(\sqrt{9}\) can be simplified into something rational. In other words, \(\sqrt{9}=3\). We can use this to simplify the radical even further.

 

\(2\textcolor{red}{\sqrt{9}}*\sqrt{2}=2*\textcolor{red}{3}*\sqrt{2}=6\sqrt{2}\) 

 

Notice that the square root of 2 remains. The radicand, 2, has no perfect square factors, other than 1. I explained in the beginning that this indicates that a particular radical is in simplest form. In fact, we can use this exact process to simplify \(\sqrt{32}\). I will use the same process, but I will skip the explanation portion.

 

\(\sqrt{32}=\sqrt{4*8}=\sqrt{4}\sqrt{8}=2\sqrt{8}\)

 

Actually, I need to explain something else for you! Ideally, you want to identify the LARGEST perfect square factor. In the preceding work I showed, I identified that 4 is a perfect square factor of the radicand, 32. This is true, but 4 is not the largest perfect square factor. If you keep searching, you will quickly realize that 16 is the LARGEST perfect square factor of 32. 

 

\(\sqrt{32}=\sqrt{16*2}=\sqrt{16}\sqrt{2}=4\sqrt{2}\)

 

Now,\(\sqrt{32}\) is in simplest radical form. Sometimes, though, identifying the LARGEST perfect square factor may be impractical--especially if you are dealing with enormous numbers. Let's go back to the original approach I showed:

 

\(\sqrt{32}=\sqrt{4*8}=\sqrt{4}\sqrt{8}=2\sqrt{8}\)

 

Now, the radicand is 8. It, too, has a perfect square factor, which is 4. This means you can simplify further. 

 

\(2\sqrt{8}=2*\sqrt{4*2}=2\sqrt{4}*\sqrt{2}=2*2\sqrt{2}=4\sqrt{2}\)

 

Look at that! We arrived at the same answer as we did before. Therefore, always be sure that the resulting radical does not have any perfect square factors. Let's get back to the original problem. 

 

\(2\sqrt{18}+\sqrt{32}\\ 6\sqrt{2}+4\sqrt{2}\)

 

The whole point of putting both radicals into simplest form is that now they have identical radicals. We can now them together. 

 

\(6\sqrt{2}+4\sqrt{2}=10\sqrt{2}\)

 

I know that this is a long-winded explanation for such a simple problem. If you are confused, just ask away! I'll be glad to answer.

Feb 15, 2018
 #1
avatar+2439 
+1

To solve for a variable, in this case, simpl means to isolate it. That is the goal of solving for a variable. 

 

1) 

\(C=K\left(\frac{Rr}{R-r}\right)\) Let's first multiply both sides by (R-r). This will eliminate the denominator, which will ease the process of solving for "K."
\(C(R-r)=KRr\) Divide by Rr to isolate K.
\(K=\frac{C(R-r)}{Rr}\\\) You could distribute, if you so desired, but it is not necessary. This just creates more work for yourself, anyway. This is the final answer. 
   

 

2) This one will be harder since two "R's" exist in the original equation. This generally signals for the use of grouping. 

 

\(C=K\left(\frac{Rr}{R-r}\right)\) Let's do the exact same thing as before; multiply by R-r.
\(C(R-r)=KRr\) Let's distribute the C into the binomial. This will allow us to have two terms with "R's"
\(CR-Cr=KRr\) Let's subtract KRr from both sides. Let's add Cr to both means meanwhile.
\(CR-KRr=Cr\) Using grouping, we can change this into one "R."
\(R(C-Kr)=Cr\) Now, divide by C-Kr to isolate "R."
\(R=\frac{Cr}{C-Kr}\) This is finished because we have isolated the R.
   

 

3) We will have to utilize the same technique in number 2 in order to isolate "f."

 

\(F=\frac{fg}{f+g-d}\) Let's multiply the denominator on both sides. This will get rid of any and all pesky denominators in this equation,
\(F(f+g-d)=fg\) There is a "g" in the trinomial, so we will have to expand that part. 
\(Ff+Fg-Fd=fg\) Move all terms with a factor of f on the left hand side. Any other terms are moved to the right hand side.
\(Ff-fg=-Fg+Fd\) Factor out an "f" from both terms from the left hand side. 
\(f(F-g)=-Fg+Fd\) Finally, divide by F-g to isolate the wanted variable completey.
\(f=\frac{-Fg+Fd}{F-g}\) No more simplification is possible here. Leave the fraction as is. 
   
Feb 13, 2018
 #5
avatar+2439 
+3

For the second question, it is possible to utilize something known as the Pythagorean Inequality Theorem. The theorem actually has two parts, but the relevant portion for this particular problem is that if the square of a triangle's longest side length is greater than the the sum of the squares of the shorter side lengths, then the triangle is obtuse. This wording is quite verbose, I know, so I decided to create a visual aide as a reference for you! 

 

 

Let's first try and identify what the longest side length is of this particular triangle is. We know that 16 and 21 are already defined lengths, but we also have the unknown side length, which is labeled as s. I will follow the visual aide above, and I will consider the longest side length to be of length "c." In the visual aide above, the shorter side lengths are labeled "a" and "b."

 

It is easy to exclude the side with length 16 as the longest side length since we know that there exists a side with length 21, which is longer. However, beyond this, there are two possibilities. The possibility is that:

 

  • The side with length s is the longest side.
  • The side with length 21 is the longest side.

 

Of course, we do not know which one is the longest side, so we will just have to assume both cases. Let's begin with assuming that "s" is the longest side, or our "c."

 

\(c^2>a^2+b^2\)"s" will be the longest side, so plug it into "c." Plug the other two side lengths in any order.
\(s^2>21^2+16^2\)Now, solve for "s" to see what the possibilities are.
\(s^2>441+256\) 
\(s^2>697\)Take the square root of both sides. This results in an absolute value inequality. 
\(|s|>\sqrt{697}\)Of course, the absolute value splits the equation into two parts. 
\(s>\sqrt{697}\)\(s<-\sqrt{697}\)

 

Of course, let's remember that "s" is an integer side length of a triangle, so "s" must be positive. We can reject the second inequality because those values include ones less tha zero.
\(s>\sqrt{697}\) 

 

The above algebra only solved the first case where we assumed that "s" is the longest side. Now, let's assume that the side with length 21 is the indeed the longest. 

 

\(21^2>16^2+s^2\)Let's solve for "s" here by simplifying first and foremost. 
\(441>256+s^2\)Subtract 256 from both sides.
\(185>s^2\\ s^2<185\)For me, I can interpret the inequality better when the variable is placed on the left hand side of the equation. Take the square root of both sides.
\(|s|<\sqrt{185}\)Now, solve the absolute value inequality. 
\(s<\sqrt{185}\)\(s>-\sqrt{185}\)

 

Unlike the previous inequality, we can combine this into one compound inequality.
-√185 < s < √185Of course, yet again, "s" is a side length, so it should be greater than zero.
0 < s < √185 

 

Of course, let's take clean this up. We know, by the given information, that we only care about integer solutions, so let's calculate the radicals.We now have set restrictions for "s."  \(\)

 

√697 ≈ 26.40

√185 ≈ 13.60

 

We can then make the restrictions the following      

 

We are not done yet, though! We have to take into account the side length restriction that Cpill figured out in part a! He found that 5 < s < 37

 

Therefore, integer sides with lengths 6 to 13 or sides with lengths 27 to 36 are possible. We can find the total number of integer solutions by figuring out how many integers are in this range. 

 

From [6,13], there are 8 integers

From [27,36], there are 10 integers

 

Altogether, there are 18 integer solutions. 

 

*Unfortunately, LaTeX appears to be funky when compound inequalities are introduced, so I have resorted to a different method to portray them.

Feb 13, 2018