Fe(ii) can be precipitated from a slightly basic aqueous solution by bubbling oxygen through the solution, which converts fe(ii) to insoluble fe(iii): $$4fe(oh)+(aq)+4oh−(aq)+o2​(g)+2h2​o(l) 4fe(oh)3​(s) how many grams of o2 are consumed to precipitate all of the iron in 50.0 ml of 0.0850 m fe(ii)?

Answers

Answer 1
Answer:

0.0340 g O2

Step 1. Write the balanced chemical equation

4Fe(OH)^(+) + 4OH^(-) + O2 + 2H2O → 4Fe(OH)3

Step 2. Calculate the moles of Fe^(2+)

Moles of Fe^(2+) = 50.0 mL Fe^(2+) × [0.0850 mmol Fe^(2+)/1 mL Fe^(2+)]

= 4.250 mmol Fe^(2+)

Step 3. Calculate the moles of O2

Moles of O2 = 4.250 mmol Fe^(2+) × [1 mmol O2/4 mmol Fe^(2+)]

= 1.062 mmol O2

Step 4. Calculate the mass of O2

Mass of O2 = 1.062 mmol O2 × (32.00 mg O2/1 mmol O2) = 34.0 mg O2

= 0.0340 g O2

Answer 2
Answer:

0.0342 grams of O2 are consumed to precipitate all of the iron in 50.0 mL of 0.0850 M Fe(II) solution.

To solve this problem, we need to first calculate the number of moles of Fe(II) in 50.0 mL of 0.0850 M Fe(II) solution.

Moles of Fe(II) = (0.0850 mol/L) * (50.0 mL) = 0.00425 mol

According to the balanced chemical equation, 4 moles of Fe(II) react with 1 mole of O2. Therefore, the number of moles of O2 required to precipitate all of the iron in 50.0 mL of 0.0850 M Fe(II) solution is:

Moles of O2 = (0.00425 mol Fe(II)) * (1 mol O2 / 4 mol Fe(II)) = 0.00106 mol O2

Now we can convert the moles of O2 to grams using the molar mass of O2 (32.00 g/mol):

Grams of O2 = (0.00106 mol O2) * (32.00 g/mol) = 0.0342 g O2

Therefore, 0.0342 grams of O2 are consumed to precipitate all of the iron in 50.0 mL of 0.0850 M Fe(II) solution.

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Answers

Answer:

Explanation:

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Write the numbers in scientific notation.291.7 = 2.917
X10%
where x =
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X10%
where x =​

Answers

The numbers in scientific notation are 291.7 = 2.917 × 10₂ and 0.0960 = 9.60 × 10⁻².

Scientific notation is commonly used in scientific and mathematical calculations, as well as when dealing with very large or very small numbers. It allows for a more compact and manageable representation of these numbers.

To write the numbers in scientific notation, we need to express them in the form of "a × 10^b," where "a" is a number between 1 and 10, and "b" is an integer.

For the number 291.7, write it in scientific notation as:

291.7 = 2.917 × 10²

For the number 0.0960, write it in scientific notation as:

0.0960 = 9.60 × 10⁻²

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2.917 x 10^2 0.960x 10^1

Which is an example of plasmas in nature? solar cells plasma balls auroras clouds

Answers

The best example of natural plasma is the ionosphere of the atmosphere including solar corona and lightning including auroras cloud.

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the heated or hot  matter that is so hot that the electrons are thrown away from the atoms and forms the ionized gas. Comprises over 99% of the visible universe.

Lightning strikes create plasma by doing a stricking of electricity. Mostly the Sun, and some stars, are in a plasma state. Certain regions of Earth's atmosphere contain some plasma created primarily by ultraviolet radiation from the Sun.

One reason plasma is not so common is because this needs high temperatures required to keep gas in the plasma state. At average temperatures on Earth, there just isn't enough energy for atoms to remain ionized.

However, at thousands to millions of degrees Kelvin, these energies are available, and plasmas dominate

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Answer:

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Explanation:

A rock contains 0.37 mg of Pb-206 and 0.95 mg of U-238. Approximately how many U-238 atoms were in the rock when it was formed billions of years ago? (The half-life for 238U  206Pb is 4.5  109 yr.)

Answers

Half-life of a radioactive element

The Half-Life of a radioactive element osbtime taken for half the nucleus of the atom of the element to decay

Calculating the original amount of U-238

  • The number of moles present in each element is first determined:
  • Number of moles = mass/molar mass

For Pb-206:

mass = 0.37 mg = 0.37 * 10⁻³ g

molar mass = 206 g/mol

number of moles = 0.37 * 10⁻³ g/206 g/mol

number of moles of Pb-206 = 1.79 * 10⁻⁶ moles

For U-238:

mass = 0.95 mg = 0.95 * 10⁻³ g

molar mass = 238 g/mol

number of moles = 0.95 * 10⁻³ g/238 g/mol

number of moles = 3.99 * 10⁻⁶ moles

Assuming that all the Pb-206 were formed from U-238

  • Initial moles of U-238 = Moles of present U-238 + molesw of present Pb-208

Initial moles of U-238 = 3.99 * 10⁻⁶ moles + 1.79 * 10⁻⁶ moles

Initial moles of U-238 = 5.78 * 10⁻⁶ moles

One mole of U-238 contains = 6.02 * 10²³ atoms

5.78 * 10⁻⁶ moles of U-238 will contain 6.02 * 10²³ * 5.78 * 10⁻⁶ atoms

Number of atoms of U-238 initially present = 3.48 * 10¹⁸ atoms

Therefore, the number of atoms of U-238 initially present is 3.48 * 10¹⁸ atoms

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How many molecules are in 2NaO? please someone help i dont know this :')​

Answers

Answer:

One molecule

Explanation

But there is three different atoms forming this one NaOH. The three atoms are Na, O and H, that is one sodium, one oxygen, and one hydrogen.