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Vladimir [108]
1 year ago
14

If 3.00 mL of 0.0250 M CuSO4 is diluted to 25.0 mL with pure water, what is the molarity of copper(II) sulfate in the diluted so

lution
Chemistry
1 answer:
g100num [7]1 year ago
8 0

Answer:

0.00268 M

Explanation:

To find the new molarity, you need to (1) find the moles of CuSO₄ (via the molarity equation using the beginning molarity and volume) and then (2) find the new molarity (using the moles and combined volume). Your final answer should have 3 sig figs to match the given values.

<u>Step 1:</u>

3.00 mL / 1,000 = 0.00300 L

Molarity = moles / volume (L)

0.0250 M = moles / 0.00300 L

(0.0250 M) x (0.00300 L) = moles

7.50 x 10⁻⁵ = moles

<u>Step 2:</u>

25.0 mL / 1,000 = 0.0250 L

0.0250 L + 0.00300 L = 0.0280 L

Molarity = moles / volume (L)

Molarity = (7.50 x 10⁻⁵ moles) / (0.0280 L)

Molarity = 0.00268 M

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How many mol of sodium corresponds to 1.0 x 10^15 atoms of sodium?
andrew-mc [135]
The Avogadro number represents the number of units in one mole of a chemical substance.
So to find the mole number of a chemical element, you divide its atom number of the Avogadro number which Na = 6.02*10^23 approx.
So n=N/Na (n=mole number, N=number of atoms, Na=Avogadro number)
n=1.0*10^15/6.02*10^23
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4Fe (s) + 3 O2 (g) → 2 Fe2O3 (s)
Novay_Z [31]

Explanation:

The number acquired by an element after the lose or gain of an electron is called oxidation number.

For example, 4Fe(s) + 3O_{2}(g) \rightarrow 2Fe_{2}O_{3}(s)

Here, oxidation number of Fe(s) is 0 and Fe in Fe_{2}O_{3} is +3.

Oxidation number of O in O_{2}(g) is 0 as it is present in its elemental state.

The oxidation number of O in Fe_{2}O_{3} is calculated as follows.

2(3) + 3x = 0\\6 + 3x = 0\\x = \frac{-6}{3}\\= -2

Hence,  oxidation number of O in Fe_{2}O_{3} is -2.

  • The loss of electrons by an element or substance is called oxidation. Here, electrons are being lost by Fe(s) as an increase in oxidation state is occurring. So, Fe(s) is oxidized.
  • The gain of electrons by an element or substance is called reduction. Here, electrons are being added to O_{2} as a decrease in its oxidation state is occurring. So, O_{2}  is reduced.
  • An element or compound which is being reduced is called oxidizing agent. Here, O_{2}  is the oxidizing agent.
  • An element or compound which is being oxidized is called reducing agent. Here, Fe(s) is the reducing agent.
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