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3 years ago

How many atoms are in 1.50 moles of Hg?

Chemistry

2 answers:

Oxana [17]3 years ago

3 0

Answer:

Atoms=9.033*10^23 atoms

Explanation:

Atoms=no.of miles*Avogadro's no.

Atoms=1.5*6.022*10^23

Atoms=9.033*10^23 atoms

vesna_86 [32]3 years ago

3 0

Answer:

There are 9.0345*10²³ in 1.5 moles of Hg.

Explanation:

The Avogadro Number or Avogadro Constant is called the number of particles that constitute a substance (usually atoms or molecules) and that can be found in the amount of one mole of that substance. Its value is 6.023 * 10²³ particles per mole. Avogadro's number applies to any substance.

Then a rule of three can be applied to calculate the number of atoms in 1.5 moles of Hg. This rule applies as follows: if 1 mole of Hg contains 6,023 * 10²³ atoms, 1.5 moles how many atoms of the element will they contain?

$Number of atoms=\frac{1.5 moles* 6.023*10^{23}atoms }{1 mole}$

Number of atoms=9.0345*10²³

There are 9.0345*10²³ in 1.5 moles of Hg.

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Answer:

$13,500

Explanation:

The differential analysis of the proposal to replace the commercial oven is shown below:-

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3 years ago

1. The volume of a gas is 10.0 liters and its temperature is 1500 K. If the temperature is

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Answer:

5 L

Explanation:

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For your problem, it would look like this...

$\frac{10L}{1500k} = \frac{V_{F} }{750k}$

You would cross multiply and your answer would be 5L.

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3 years ago

Serhud [2]

Explanation:

Given that,

The frequency of electromagnetic spectrum is $2.73\times 10^{16}\ Hz$

(A) Let the wavelength of this radiation is $\lambda$ . We know that,

$c=f\lambda\\\\\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{2.73\times 10^{16}}\\\\\lambda=1.09\times 10^{-8}\ m$

So, the wavelength of this radiation is $1.09\times 10^{-8}\ m$ .

(B) Let E is the energy associated with this radiation. Energy of an electromagnetic radiation is given by :

$E=hf$

h is Planck's constant

$E=6.63\times 10^{-34}\times 2.73\times 10^{16}\\\\E=1.8\times 10^{-17}\ J$

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It means,

$1.8\times 10^{-17}\ J=\dfrac{1}{4184}\times 1.8\times 10^{-17}\\\\=4.3\times 10^{-21}\ \text{kcal}$

Hence, this is the required solution.

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3 years ago

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3 years ago

If an unknown sample contains 39.04% sulfuric acid by mass, then a 0.9368 g of that sample would require _____ mL of 0.2389 M Na

bonufazy [111]

Answer:

A) 31.22

Explanation:

The reaction of sulfuric acid with NaOH is:

H₂SO₄ + 2 NaOH → Na₂SO₄ + 2H₂O

To solve this problem we need to determine the moles of acid that will react, and, using the chemical equation we can determine the moles of NaOH and the volume that a 0.2389M NaOH solution would require to neutralize it.

Moles H₂SO₄ (Molar mass: 98.08g/mol):

0.9368g * 39.04% = 0.3657g H₂SO₄ * (1mol / 98.08g) =

3.7289x10⁻³moles H₂SO₄

And moles of NaOH that you require to neutralize the acid are:

3.7289x10⁻³moles H₂SO₄ * (2 moles NaOH / 1 mole H₂SO₄) =

7.4578x10⁻³ moles NaOH

Using a 0.2389M NaOH solution:

7.4578x10⁻³ moles NaOH * (1L / 0.2389mol) = 0.03122L = 31.22mL

Right answer is:

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3 years ago