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3 years ago

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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

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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Match the parabolas represented by the equations with their foci.

Brrunno [24]

Answer:

Step-by-step explanation:

Before we begin this, there are a few things that need to be said and a few formulas you need to know. First is that we need to use the work form of a parabola, which is

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All of the parabolas listed in blue highlight open either up or down, and this work form represents those 2 options. The only thing we need to know is that if there is a negative sign in front of the a, the parabola opens upside down like a mountain instead of up like a cup.

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$y=-x^2+4x+8$ and, completing the square one step at a time:

$-x^2+4x=-8$ and

$-(x^2-4x+4)=-8-4$ and

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$-(x-2)^2+12=y$

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$(2, 12-\frac{1}{4})$ which simplifies to (2, 11.75) which is choice 2 in your options.

Now for the second one (yes, this takes forever...)

$y=2x^2+16x+18$ and completing the square one step at a time:

$2x^2+16x=-18$ and

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$2(x+4)^2-14=y$

From this we can see that the vertex is h = -4 and k = -14. Now to find p from a.

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Now for the third one...

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$-2x^2+5x=-14$ and

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$-2(x-\frac{5}{4})^2=-\frac{137}{8}$ and

$-2(x-\frac{5}{4})^2+\frac{137}{8}=y$

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$p=\frac{1}{4(2)}=\frac{1}{8}$

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$(1.25, 17.125-\frac{1}{8})$ which simplifies down to (1.25, 17) which is choice 4 in your options.

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$-x^2+17x=-7$ and

$-(x^2-17x)=-7$ and

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$-(x-8.5)^2+79.25=y$

From that we see that the vertex is h = 8.5 and k = 79.25. Now to find p.

$|a|=1$ , ∴

$p=\frac{1}{4(1)}=\frac{1}{4}$

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$(8.5, 79.25-\frac{1}{4})$ which simplifies down to (8.5, 79) and I don't see a choice from your available options there.

On to the fifth one...

$y=2x^2+11x+5$ and again step by step:

$2x^2+11x=-5$ and

$2(x^2+\frac{11}{2}x+\frac{121}{16})=-5+\frac{242}{16}$ and

$2(x+\frac{11}{4})^2=\frac{81}{8}$ and

$2(x+\frac{11}{4})^2-\frac{81}{8}=y$

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$|a|=2$ , ∴

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Good luck with your conic sections!!!

7 0

3 years ago

PLEASE CHECK AND SEE IF I'M CORRECT!!!<br> CORRECT ANSWERS ONLY PLEASE!!!

jeka57 [31]

I'm thinking this is the anwer if im right

8 0

3 years ago

Read 2 more answers