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xxTIMURxx [149]
3 years ago
14

How many moles of Fe contains 3.41 x 1023 Fe atoms?

Chemistry
2 answers:
Liula [17]3 years ago
8 0

Answer:

Explanation:

28个

ValentinkaMS [17]3 years ago
5 0

Answer:

\boxed {\boxed {\sf  0.566 \ mol \ Fe}}

Explanation:

We are asked to convert a number of atoms to moles.  

We can convert atoms to moles using Avogadro's Number, which is 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this problem, the particles are atoms of iron (Fe). There are <u>6.022 ×10²³ atoms of iron in 1 mole of iron</u>.  

We use dimensional analysis to convert atoms to moles. This involves setting up ratios. Use Avogadro's Number and the underlined information to make a ratio.

\frac {6.022 \times 10^{23}\ atoms \ Fe}{1 \ mol \ Fe}

We are converting 3.41 × 10²³ atoms of iron to moles, so we multiply by this value.

3.41 \times 10^{23} \ atoms \ Fe *\frac {6.022 \times 10^{23}\ atoms \ Fe}{1 \ mol \ Fe}

Flip the ratio. It stays equivalent, but it allows the units of atoms of iron to cancel.

3.41 \times 10^{23} \ atoms \ Fe *\frac{1 \ mol \ Fe} {6.022 \times 10^{23}\ atoms \ Fe}

3.41 \times 10^{23}*\frac{1 \ mol \ Fe} {6.022 \times 10^{23}}

\frac{3.41 \times 10^{23}} {6.022 \times 10^{23}} \ mol \ Fe

0.5662570575\ mol \ Fe

The original measure ment of iron atoms ( 3.41 × 10²³ ) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandths place. The 2 in the ten-thousandths place ( 0.566<u>2</u>570575) tells us to leave the 6 in the thousandths place.

0.566 \ mol \ Fe

3.41 × 10²³ atoms of iron is equal to approximately <u>0.566 moles of iron.</u>

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PilotLPTM [1.2K]
KOH is a strong base and HBr is a strong acid and completely dissociates.
The balanced equation for the reaction is;
KOH + HBr ---> KBr + H₂O
Stoichiometry of acid to base is 1:1
The number of KOH moles reacted - 0.50 M / 1000 mL/L x 48.0 mL = 0.024 mol
number of HBr moles reacted - 0.25 M/ 1000 mL/L x 96.0 mL  = 0.024 mol
the number of H⁺ ions are equal to number of OH⁻ ions. 
Then the solution is neutral.
pH of neutral solutions at 25 °C is 7. 
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8 0
3 years ago
Calculate the molar solubility of fe(oh)2 in pure water. (the value of ksp for fe(oh)2 is 4.87×10−17.)
rosijanka [135]
When we have this balanced equation for a reaction:
Fe(OH)2(s) ↔ Fe+2  + 2OH-

when Fe(OH)2 give 1 mole of Fe+2 & 2 mol of OH-
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 when Ksp = [Fe+2][OH-]^2
and have Ksp = 4.87x10^-17 
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3 years ago
1. How many milliliters of 10.0 M HNO 3 are needed to prepare 0.350 L of 0.400 M solution?
tatyana61 [14]
<h3>Answer:</h3>

14 milliliters

<h3>Explanation:</h3>

We are given;

  • 10.0 M HNO₃

Prepared solution;

  • Volume of solution as 0.350 L
  • Molarity as 0.40 M

We are required to determine the initial volume of HNO₃

  • We are going to use the dilution formula;
  • The dilution formula is;

M₁V₁ = M₂V₂

Rearranging the formula;

V₁ = M₂V₂ ÷ M₁

    =(0.40 M × 0.350 L) ÷ 10.0 M

   = 0.014 L

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Volume = 14 mL

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5 0
2 years ago
Pls help
Irina-Kira [14]

Answer:

CrO₂ --------------------> Cr⁴⁺ and O²⁻

VCO₃ -------------------> V²⁺ and CO₃²⁻

Cr₂(SO₄)₃ -------------> Cr³⁺ and SO₄²⁻

(NH₄)₂S ----------------> NH₄⁺ and S²⁻

Explanation:

Within ionic compounds, the cation is listed first, followed by the anion. Some of the ions are polyatomic, meaning they are covalently bonded to other elements. Polyatomic ions always have a specific charge.

All of these ionic compounds have an overall charge of 0. As such, the charges of the cations and anions must balance out. In order to do so, there are some compounds which have more than one atom of each ion.

2.) CrO₂

------> Oxygen (O) always forms the anion, O²⁻.

------> Therefore, if there are 2 oxygen anions, the chromium (Cr) must have the cationic form of Cr⁴⁺.

------> +4 + (-2) + (-2) = 0

3.) VCO₃

------> Carbonate (CO₃), a polyatomic ion, always has the state CO₃²⁻.

------> If there is only one atom of each ion, the charges must perfectly balance, making vanadium (V) be the cation V²⁺.

------> +2 + (-2) = 0

4.) Cr₂(SO₄)₃

------> Sulfate (SO₄), a polyatomic ion, always has the state SO₄²⁻.

-------> The only way the charges could balance out is if the chromium (Cr) is in the cationic form Cr³⁺.

------> +3 + 3 + (-2) + (-2) + (-2) = 0

5.) (NH₄)₂S

------> Ammonium (NH₄), a polyatomic ion, always has the state NH₄⁺.

------> Sulfur (S) always forms the anion S²⁻.

------> +1 + 1 + (-2) = 0

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