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algol13
4 years ago
10

the student records the concentration of the stock solution of H2O2 to be 0.950 M. The student proceeds to prepare the reaction

solution as stated in Step 9: They add 10.0 mL of 0.10 M potassium iodide and 15.0 mL of distilled water to their flask. They then add 5.0 mL of the stock H2O2 solution and mix the solution to initiate the reaction. Calculate the molarity of I- in the initial reaction solution.
Chemistry
1 answer:
lana [24]4 years ago
6 0

Answer:

0.0400M of KI

Explanation:

Molarity is an unit of concentration defined as the ratio between moles of solute and liters of solution.

When you add 10.0 mL of 0.10M KI and 15.0mL, total volume is:

25.0mL = <em>0.025L of solution</em>

<em />

And moles of KI are:

0.0100L × 0.10M = <em>0.00100 moles of KI</em>

<em />

Thus, molarity is:

0.00100 moles / 0.025L = <em>0.0400M of KI</em>

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Answer:

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Explanation:

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Answer:

2812.6 g of H₂SO₄

Explanation:

From the question given above, the following data were obtained:

Mole of H₂SO₄ = 28.7 moles

Mass of H₂SO₄ =?

Next, we shall determine the molar mass of H₂SO₄. This can be obtained as follow:

Molar mass of H₂SO₄ = (1×2) + 32 + (16×4)

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Finally, we shall determine the mass of H₂SO₄. This can be obtained as follow:

Mole of H₂SO₄ = 28.7 moles

Molar mass of H₂SO₄ =

Mass of H₂SO₄ =?

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Mass of H₂SO₄ = 28.7 × 98

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timofeeve [1]

Answer:

\boxed {\boxed {\sf 0.80 \ mol\ F}}

Explanation:

We are asked to find how many moles are in 4.8 × 10²³ fluorine atoms. We convert atoms to moles using Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are atoms of fluorine.

We will convert using dimensional analysis and set up a ratio using Avogadro's Number.

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

We are converting 4.8 × 10²³ fluorine atoms to moles, so we multiply the ratio by this number.

4.8 \times 10^{23} \ atoms \ F *\frac {6.022 \times 10^{23} \ atoms \ F}{ 1 \ mol \ F}

Flip the ratio so the units of atoms of fluorine cancel each other out.

4.8 \times 10^{23} \ atoms \ F *\frac { 1 \ mol \ F}{6.022 \times 10^{23} \ atoms \ F}

4.8 \times 10^{23}  *\frac { 1 \ mol \ F}{6.022 \times 10^{23} }

Condense into 1 fraction.

\frac { 4.8 \times 10^{23} }{6.022 \times 10^{23} } \ mol \ F

Divide.

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0.80 \ mol \ F

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