7. oxygen is composed of three isotopes: one has a a mass of 16.999 amu. this isotope
makes up 0.037% of oxygen. of the other two, one has a mass of 15.995 amu, and the
other has a mass of 17.999 amu. calculate the abundance of the other two isotopes, using
the average atomic mass of 15.9994 amu.

Answer : The abundance in % for other two isotopes of the oxygen is 99.762% and 0.201%

Explanation :

The average atomic mass of an element is the total of isotopic mass times the percent abundance for all the isotopes.

Oxygen has 3 isotope, ¹⁶O, ¹⁷O and ¹⁸O.

The average atomic mass of oxygen can be calculated as follows.

Average Atomic Mass = (Isotopic mass of ¹⁶O x % abundance of ¹⁶O) + (isotopic mass of ¹⁷O x % abundance of ¹⁷O) + (Isotopic mass of ¹⁸O x % abundance of ¹⁸O)

The mass of ¹⁷O = 16.999 amu and its % abundance is 0.037% which is 0.00037

The mass of ¹⁶O = 15.995 amu. Let us assume its % abundance as a

The mass of ¹⁸O = 17.999 amu

The total % abundance for all the isotopes is 100% or 1 .

Therefore % abundance for ¹⁸O would be 1 - ( a + 0.00037) = 0.99963 - a

Let us plug in the above values in the average atomic mass formula .

Average atomic mass = ( 15.995)(a) + (16.999)(0.00037) + (17.999)(0.99963 - a)

The average atomic mass of element O is 15.9994

Therefore we have,

15.9994 = 15.995a + 0.0062896 + 17.99234 - 17.999a

15.9994 = -2.004a + 17.99863

-1.99923 = -2.004a

a = 0.99762

a = 99.762%

The percent abundance of ¹⁶O is 99.762%

The percent abundance of ¹⁸O is 0.99963 - 0.99762 = 0.00201 or 0.201%

Hope this helps

Please give me Brainliest

no not all of them.

step-by-step explanation:

some of them have a square base.

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

=====================================

part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

=====================================

part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).