1) How many miles are covered in a 15km race? (1 mile=5280, 12 in=1 ft, 1in=2.54cm)
I know that I have to convert but I do not know where to start.
2) Silver (107.87amu) has two naturally occuring isotopes. In a typical sample, 51.84%of silver exists as Ag-107 (106.9051 amu). Calculate the atmoic mass of the other isotope.
I was going to set it up like this
(.5184 * 106.9051) + (x * 107.87) but I don't think this will work.
Start with what you must convert. That is 15 km. Than multiply by the factors to convert to another unit. For example, you don't have the conversion in your post from km to meters or from meters to cm.
There are 1000 m in a km and 100 cm in a meter. Here is how you do them.
15 km x (1000 m/km) x (100 cm/m) x .....
Note that the conversion factor is placed so that the km cancels with km (which converts to meters), then the next factor converts m (m cancels) to cm. Now add the factor for cm to inches, then inches to feet, then feet to miles.
You're on the right tract for the Ag isotope problem.
If 51.84% is Ag 107, then 100-51.84 = 48.16% must be the other isotope. So
(0.5184*106.9051) + (0.4816*x) = 107.87
For #2 I got 55.41 from (.5184 * 106.9051) so
55.41 + 0.4816x = 107.87
-55.41 -55.41
then divide by 0.4816
The answer is 108.91amu but I did not get this.
1) To convert kilometers to miles, you can use the conversion factor 1 mile = 1.60934 kilometers. Since you have a 15-kilometer race, you can multiply it by the conversion factor to find the equivalent distance in miles:
15 km * (1 mile / 1.60934 km) = 9.321566 miles
So, the 15 km race is approximately equal to 9.32 miles.
2) Let's set up the equation to find the atomic mass of the other naturally occurring isotope of silver. Let's assume the atomic mass of the other isotope is represented as 'x'.
The equation can be set up as follows:
(51.84% * 106.9051 amu) + (remaining % * x amu) = 107.87 amu
To solve for 'x', we can subtract the product of the known isotope mass and abundance from the total atomic mass and then divide by the remaining percentage:
x = (107.87 amu - (51.84% * 106.9051 amu)) / (1 - 51.84%)
x = (107.87 amu - 55.4377104 amu) / (48.16%)
x = 52.4322896 amu / 0.4816
x ≈ 108.799 amu
So, the atomic mass of the other naturally occurring isotope of silver is approximately 108.799 amu.
1) To convert kilometers to miles, we need to use the conversion factor of 1 mile = 0.621371 kilometers. So, to find out how many miles are covered in a 15 km race, you can use the following equation:
Miles = Kilometers * Conversion factor
Miles = 15 km * 0.621371 miles/km
Miles = 9.320565 miles
Therefore, the answer is approximately 9.32 miles.
2) To calculate the atomic mass of the other isotope of silver, we can set up the equation using the given information. Let's assume the atomic mass of the other isotope is x.
Total atomic mass = (Percentage of isotope 1 * atomic mass of isotope 1) + (Percentage of isotope 2 * atomic mass of isotope 2)
107.87amu = (0.5184 * 106.9051) + (0.4816 * x)
107.87 = 55.4373984 + 0.4816x
52.4326016 = 0.4816x
x = 52.4326016/0.4816
x ≈ 108.70 amu
Therefore, the atomic mass of the other isotope is approximately 108.70 amu.