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leonid [27]
3 years ago
11

One mole of a chemical element contains approximately 6.02 × 1023 atoms. How many atoms are present in 4.6 × 104 moles of oxygen

? Represent the answer in scientific notation. Round so the first factor goes to the hundredths place.
A.
2.77 × 10^27 atoms
B.
2.77 × 10^28 atoms
C.
27.7 × 10^27 atoms
D.
27.7 × 10^28 atoms

Mathematics
2 answers:
Paraphin [41]3 years ago
8 0

Answer:

B. 2.77 · 10^28

Step-by-step explanation:

The total number of atoms is ...

(6.02·10^23 atoms/mol)·(4.6·10^4 mol) = (6.02·4.6)·10^(23+4) atoms

≈ 27.7·10^27 atoms

≈ 2.77·10^28 atoms . . . . . . put in scientific notation form

_____

Scientific notation has one non-zero digit to the left of the decimal point and the exponent adjusted accordingly.

__

Your calculator can help you figure this out. If necessary, put the display into scientific notation mode. (For numbers this size, it is usually not necessary.)

frutty [35]3 years ago
4 0

Answer: B.  2.77\times10^{28\text{ atoms}}

Step-by-step explanation:

Given : One mole  of a chemical element contains approximately 6.02\times10^{23}\text{ atoms}

Then , the number of atoms present in 4.6\times10^{4}\text{ moles} is given by :-

6.02\times10^{23}\times4.6\times10^{4}\\\\=6.02\times4.6\times10^{23}\times10^{4}

Using exponents property:

a^m\times a^n=a^{m+n}

=27.692\times10^{23+4}\\\\\approx2.77\times10^{23+4+1}=2.77\times10^{28\text{ atoms}}

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To solve this we are going to use the future value of annuity due formula: FV=(1+ \frac{r}{n} )*P[ \frac{(1+ \frac{r}{n})^{kt}-1 }{ \frac{r}{n} } ]
where
FV is the future value
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n is the number of times the interest is compounded per year
k is the number of deposits per year

We know for our problem that P=420 and t=15. To convert the interest rate to decimal form, we are going to divide the rate by 100%: r= \frac{10}{100} =0.1. Since Ruben makes the deposits every 6 months, k=2. The interest is compounded semiannually, so 2 times per year; therefore, k=2.
Lets replace the values in our formula:

FV=(1+ \frac{r}{n} )*P[ \frac{(1+ \frac{r}{n})^{kt}-1 }{ \frac{r}{n} } ]
FV=(1+ \frac{0.1}{2} )*420[ \frac{(1+ \frac{0.1}{2})^{(2)(15)}-1 }{ \frac{01}{2} } ]
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