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geniusboy [140]
3 years ago
9

Suppose 40.8g of copper(II) acetate is dissolved in 200.mL of a 0.70 M aqueous solution of sodium chromate.

Chemistry
1 answer:
Contact [7]3 years ago
4 0

Answer:

0.42 M

Explanation:

The reaction that takes place is:

  • Cu(CH₃COO)₂ + Na₂CrO₄ → Cu(CrO₄) + 2Na(CH₃COO)

First we <u>calculate the moles of Na₂CrO₄</u>, using the <em>given volume and concentration</em>:

(200 mL = 0.200L)

  • 0.70 M * 0.200 L = 0.14 moles Na₂CrO₄

Now we <u>calculate the moles of Cu(CH₃COO)₂</u>, using its <em>molar mass</em>:

  • 40.8 g ÷ 181.63 g/mol = 0.224 mol Cu(CH₃COO)₂

Because the molar ratio of Cu(CH₃COO)₂ and Na₂CrO₄ is 1:1, we can directly <u>substract the reacting moles of Na₂CrO₄ from the added moles of Cu(CH₃COO)₂</u>:

  • 0.224 mol - 0.14 mol = 0.085 mol

Finally we <u>calculate the resulting molarity</u> of Cu⁺², from the <em>excess </em>cations remaining:

  • 0.085 mol / 0.200 L = 0.42 M

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In a process called photosynthesis, plants use carbon dioxide and water, along with energy from sunlight, to produce sugar and o
Alex Ar [27]

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<h2>What is photosynthesis?</h2>

The process by which plants convert carbon dioxide, water, and sunlight into oxygen and sugar-based energy is known as photosynthesis.

<h3>The photosynthesis equation is as follows:</h3>

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The following three significant activities are involved in photosynthesis:

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3. Carbon dioxide is converted to sugars.

         

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4 0
1 year ago
what is the percent yield of titanium (II) oxide if 20.0 grams of titanium (II) sulfide is reacted with water? The actual yield
earnstyle [38]

Answer : The percent yield of titanium (II) oxide is, 142.5 % and the impurities could have caused the percent yield to be so high.

Explanation : Given,

Mass of titanium(II) sulfide = 20.0 g

Molar mass of titanium(II) sulfide = 79.9 g/mole

Molar mass of titanium(II) oxide = 63.9 g/mole

First we have to calculate the moles of titanium(II) sulfide.

\text{ Moles of titanium(II) sulfide}=\frac{\text{ Mass of titanium(II) sulfide}}{\text{ Molar mass of titanium(II) sulfide}}=\frac{20.0g}{79.9g/mole}=0.2503moles

Now we have to calculate the moles of titanium(II) oxide.

The balanced chemical reaction is,

TiS+H_2O\rightarrow TiO+H_2S

From the reaction, we conclude that

As, 1 mole of titanium(II) sulfide react to give 1 mole of titanium(II) oxide

So, 0.2503 mole of titanium(II) sulfide react to give 0.2503 mole of titanium(II) oxide

Now we have to calculate the mass of titanium(II) oxide.

\text{ Mass of titanium(II) oxide}=\text{ Moles of titanium(II) oxide}\times \text{ Molar mass of titanium(II) oxide}

\text{ Mass of titanium(II) oxide}=(0.2503moles)\times (63.9g/mole)=15.99g

To calculate the percentage yield of titanium (II) oxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of titanium (II) oxide = 22.8 g

Theoretical yield of titanium (II) oxide = 15.99 g

Putting values in above equation, we get:

\%\text{ yield of titanium (II) oxide}=\frac{22.8g}{15.99g}\times 100\\\\\% \text{yield of titanium (II) oxide}=142.5\%

Hence, the percent yield of titanium (II) oxide is, 142.5 %

If the percent yields is greater than 100% that means the product of the reaction contains impurities which cause its mass to be greater than it actually.

5 0
3 years ago
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<u>Answer:</u> The volume of stock solution of calcium chloride required is 10 mL.

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To calculate the amount of solute needed, the formula used is:

C_1V_1=C_2V_2             ....(1)

where,

C_1\text{ and }V_1 are the concentration and volume of stock solution of calcium chloride

C_2\text{ and }V_2 are the concentration and volume of diluted solution of calcium chloride

Given values:

C_1=15.0M\\C_2=0.300M\\V_2=500mL

Plugging values in equation 1:

15.0\times V_2=0.300\times 500\\\\V_2=10mL

Hence, the volume of stock solution of calcium chloride required is 10 mL.

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harkovskaia [24]

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