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aniked [119]
3 years ago
9

A chemist must prepare 275. mL of 1967, ?? aqueous copper(II) fluoride (Cur) working solution. He'll do this by pouring out some

2.63 T-aqueous copper(II) fluoride stock solution into a graduated cylinder and diluting it with distilled water. Calculate the volume in mL of the copper(II) fluoride stock solution that the chemist should pour out. Be sure your answer has the correct number of significant digits mL

Chemistry
1 answer:
blagie [28]3 years ago
3 0

Answer:

206 mL

Explanation:

In the annexed picture you can see your same question, just in another format.

First we <u>calculate the total moles of CuF₂ that are required in the working solution</u>:

1967 μM ⇒ 1967 / 10⁶ = 1.967 x10⁻³M

1.967 x10⁻³M * 0.275 L = 5.409x10⁻⁴ mol

Now we divide those moles by the concentration of the stock solution, to <u>calculate the volume</u>:

5.409x10⁻⁴ mol ⇒ 5.409x10⁻⁴ * 1000 = 0.5409 mmol

0.5409 mmol ÷ (2.63 mmol/L) = 0.206 L

0.206 L ⇒ 0.206 * 1000 = 206 mL

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On four, in real life, she could put it in a strainer and put a bowl below the strainer. The liquid will fall through and into the bowl, but the solids will stay in the strainer. That may not be correct in terms of your reading because I don't know what the reading is.
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4 years ago
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Diagram of mass spectroscopy​
ra1l [238]

Explanation:

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3 years ago
An acetaminophen suspension for infants contains 80 mg/0.80 mL. The recommended dose is 15 mg/kg body weight. How many mL of thi
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4 years ago
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Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide.4HCl(aq)+MnO2(s)⟶Mn
mario62 [17]

Answer:

HCl is limiting reactant

Theoretical yield: 23.8g Cl₂

Actual yield: 17.6g C₂

Explanation:

Based on the reaction:

4HCl(aq)+MnO2(s)⟶MnCl2(aq)+2H2O(l)+Cl2(g)

<em>4 moles of HCl reacts per mole of MnO₂ to produce 1 mole of MnCl₂ and Cl₂ and 2 moles of water.</em>

To find the limiting reactant you must know the moles of each reactant and knowing that 4 moles of HCl reacts per mole of MnO₂ you can sikve this problem, thus:

Moles HCl (Molar mass: 36.46g/mol): 48.9g ₓ (1mol / 36.46g/mol) =

<em>1.341 moles HCl</em>

Moles MnO₂ (Molar mass: 86.937g/mol): 36.9g ₓ (1mol / 86.937g) =

<em>0.424 moles MnO₂</em>

<em />

For a complete reaction of 0.424 moles of MnO₂ you require:

0.424moles MnO₂ ₓ (4 moles HCl / 1 mole MnO₂) = 1.696 moles of HCl.

As you have just 1.341 moles of HCl. HCl is limiting reactant.

Theoretical yield means, in the reaction, that 4 moles of HCl will produce 1 mole of Cl₂. As moles of HCl are 1.341:

1.341 moles HCl ₓ (1 mole Cl₂ / 4 moles HCl) = 0.33525 moles Cl₂

In grams (Molar mass Cl₂: 70.9g/mol):

Theoretical yield: 0.33525 moles Cl₂ ₓ (70.9g / mol) = 23.8g Cl₂

As yield of reaction is 74.7%, the real mass of Cl₂ you obtain (Actual yield) is:

23.8g Cl₂ ₓ 74% = 17.6g C₂

6 0
3 years ago
Help please-
anzhelika [568]

Answer:

Oxygen and Carbon are in the non-metal section. Both are in 2nd period while Carbon is in group 4A and Oxygen is in group 6A.

Oxygen has more valence electrons than carbon. I wish I could give you a more accurate explanation why on paper, but all you need to do is count the electrons on the largest shell.

Here's the orbital notation of Carbon: 1s^2 2s^2 2px^1 py^1

Carbon's largest shell is 2. Count the electrons on shell 2, and you get 4 valence electrons.

Orbital notation of Oxygen: 1s^2 2s^2 2px^2 py^1 pz^1

Oxygen's largest shell is 2. So, like with carbon, count the electrons on the largest shell. You get 6.

There is a way easier way of interpreting this where you do not even have to write the orbital notation. I wish I can explain, but just know the periods and the group numbers.

7 0
3 years ago
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