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navik [9.2K]
4 years ago
9

Determine the number of atoms in 98.3 g mercury, Hg

Mathematics
2 answers:
Arisa [49]4 years ago
8 0
The answer is 2.95 × 10²³ atoms

Atomic mass is 200.59 g.
So, 1 mole has 200.59 g. Let's calculate how many moles have 98.3 g:
1M : 200.59g = x : 98.3g
x = 98.3 g * 1 M : 200.59 g = 0.49 M

To calculate this, we will use Avogadro's number which is the number of units (atoms, molecules) in 1 mole of substance:

6.023 × 10²³ atoms per 1 mole
<span>How many atoms are in 0.49 mole:
</span>6.023 × 10²³ atoms : 1M = x : 0.49M
x = 6.023 × 10²³ atoms : 1M * 0.49M = 2.95 × 10²³ atoms
Serjik [45]4 years ago
7 0
<span>1.22875 x 6.02 x 10^23=7.397075 x 10^23</span>
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Evaluate 5^2 + 10 ÷ 2.
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Answer:

Step-by-step explanation:

5^2 + 10 ÷ 2

25 + 10 ÷ 2

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30

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3 years ago
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3 years ago
Justin drinks 1 litter of water.He drank 2,000 litters at his game.How many during his game and his practice?Explain.
djverab [1.8K]

The question is incomplete. The complete question is :

Justin drinks 1 litter of water during the soccer practice. He drank 2,000 milliliters of water at his game. How many liters of water did he drink during his game and his practice? Explain.

Solution :

It is given that :

During practice, Justine drank = 1 liter of water

During game, Justine drank = 2000 milliliters of water

We know that,

1 liter = 1000 mL

Therefore, during the game, Justine drank :

1000 mL = 1 liter

∴ 2000 mL = 2 liter

So Justine drank 2 liters of water during his soccer game and 1 liter of water during his practice.

7 0
3 years ago
Plz help me, brainliest to be given
Leokris [45]

Answer:

4%

Step-by-step explanation:

R=INTEREST X 100% / TIME X PRINCIPAL

r=40 x 100=4000

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5 0
3 years ago
Solve the equation below.<br> 5x^2 − 7x + 8 = 2
zavuch27 [327]

Answer:

\frac{7- \sqrt{71}i }{10}

\frac{7+ \sqrt{71}i }{10}

Step-by-step explanation:

5x^{2} - 7x + 6= 0

By quadratic formula,

Given in this question a = 5,

                                   b = -7

                                    c = 6

x can be

\frac{-b + \sqrt{b^{2} - 4ac } }{2a}

\frac{-b - \sqrt{b^{2} - 4ac } }{2a}

Putting these values in the formula we get,

x1 = \frac{7+ \sqrt{49 - 120} }{10}

x2 = \frac{7- \sqrt{49 - 120} }{10}

6 0
3 years ago
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