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WHAT IS 81 BR?

Updated: 9/18/2023
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Q: WHAT IS 81 BR?
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Related questions

Is the isotope 79-Br or 81-Br more common and why?

79-Br is more common because its abudance is 50.69%. 81-Br's abudance is 49.31% making it less common. Abudance is the rate at which the isotope is found in nature. In this case out of every 100 Bromine atoms you found 51 would be 79-Br and 41 would be 81-Br.


What is the most common isotope Br?

It may be Bromine-79 (79Br) because it is one of the 2 stable isotopes. The other one is Bromine-81 (81Br).


What is the natural abundance of Br-81?

the abundance is 50%.


How many peaks will be present in a mass spectrum for?

There will be four peaks in the mass spectrum.


What is Br number of protons?

Br-80 has 35 protons.


What are the protons and the neutrons for the element Br?

Bromine-79 has 44 neutrons and bromine-81 has 46 neutrons. All the isotopes of bromine has 35 protons.


If the mass of Br-81 is 80.9163 amu what is the mass of Br 79 I know the answer is 78.9183 but how does one find the answer?

Problems like these are somewhat tricky. Yes, the mass of the Br-79 isotope is 78.9183. Bromine has only two naturally occurring isotopes Br-79 and Br-81. Br-81 is 49.31% abundant. Knowing this we can determine Br-79's percent abundance. We determine Br-79's percentage abundance by taking 100.00% and subtracting 49.31%. We can do this safely because all % are out of 100 and we know that there are only two Bromine isotopes. When the math problem has been worked out, we conclude that Br-79 is 50.69% abundant. 100.00% - 49.31% = 50.69% If we start to do weighted averages (the way these problems are solved), we end up with this: Isotope - Weight amu - Percentage Abundance Br-79 - X amu - 50.69% Br-81 - 80.9163 amu - 49.31% Amu means Atomic Mass Unit. We need to solve for x Now, when we begin our weighted average problems we must convert our percentages to decimal form. To do this we divide each percent by 100. (Simply move the decimal place two points to the left.) 50.69/100 = .0569 and 49.31/100 = .4931 With this information, we can set up an equation. Letting X = the Amu weight of Br-79. Also, recall that our average atomic mass for Br = 79.904 (from periodic table). This equation looks like this, (Mass in Amu of Br-81) * (percentage abundance in decimal of Br-81) + (X mass in amu of Br-79) * (percentage abundance in decimal of Br-79) = 79.904. Or simply (80.9163)(.4931) + (X)(.5069) = 79.904. To solve this we need to get x by itself. Step 1 Multiply 80.9163 by .4931 Our equation now looks like: (39.89982753) + (X)(.5069) = 79.904 Step 2 Subtract 39.89982753 from each side. Remember what we do to one side must be done to the other. 39.89982753 + (X)(.5069) = 79.904 -39.89982753 - 39.89982753 Now our equation is: (X)(.5069) = 40.00417247 Step 3 Divide each side by .5069 to get x by itself. The equation then looks like this (X)(.5069) = 40.00417247 (.5069) = (.5069) Then X = 78.91925916 Depending on the calculator used, and where a person rounds the answers may vary slightly. This problem if it came from where I think it came needs to be rounded to 4 sig figs. If done properly when rounded the answer is 78.91 amu.


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