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MariettaO [177]
2 years ago
12

Calculate the volume in liters of a 0.00231M copper(II) fluoride solution that contains 175.g of copper(II) fluoride CuF2. Be su

re your answer has the correct number of significant digits.
Chemistry
1 answer:
Free_Kalibri [48]2 years ago
7 0

Answer:

Volume = 746 L

Explanation:

Given that:- Mass of copper(II) fluoride = 175 g

Molar mass of copper(II) fluoride = 101.543 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{175\ g}{101.543\ g/mol}

Moles_{copper(II)\ fluoride}= 1.7234\ mol

Also,

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

So,,

Volume =\frac{Moles\ of\ solute}{Molarity}

Given, Molarity = 0.00231 M

So,

Volume =\frac{1.7234}{0.00231}\ L

<u>Volume = 746 L</u>

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Draw the alkene formed when 1-heptyne is treated with hbr in the presence of peroxide.
Nimfa-mama [501]
<h2>Heptene formed is -</h2><h2>CH_3-CH_2-CH_2-CH_2-CH_2-CH=CHBr</h2>

Explanation:

The two possibilities when the peroxide is not present

  • CH_{3}-CH_{2}-CH_{2}-CH_{2}-CH_{2}-C≡CH +HBr → CH_3-CH_2-CH_2-CH_2-CH_2-CBr=CH_{2}

  • CH_3-CH_2-CH_2-CH_2-CH_2-CBr=CH_2 + HBr →CH_3-CH_2-CH_2-CH_2-CH_2-CBr_2-CH_3

In presence peroxide,

CH_3-CH_2-CH_2-CH_2-CH_2-C≡CH+ HBr →CH_3-CH_2-CH_2-CH_2-CH_2-CH=CHBr

  • When peroxides are present in the reaction mixture, hydrogen bromide adds to the triple bond of heptane with regioselectivity.
  • This reaction is opposite to that of Markovnikov's rule which says that when asymmetrical alkene reacts with a protic acid HX, then the hydrogen of an acid is attached to the carbon with more in number of hydrogen substituents, and the halide (X) group is attached to the carbon with more in number of substituents of alkyl.
  • One mole of HBr adds to one mole of 1-heptane.
  • The structure of heptene formed is -

CH_3-CH_2-CH_2-CH_2-CH_2-CH=CHBr

5 0
3 years ago
g Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.71 g of ethane i
Kamila [148]

Answer:

6.05g

Explanation:

The reaction is given as;

Ethane + oxygen --> Carbon dioxide + water

2C2H6 + 7O2 --> 4CO2 + 6H2O

From the reaction above;

2 mol of ethane reacts with 7 mol of oxygen.

To proceed, we have to obtain the limiting reagent,

2,71g of ethane;

Number of moles = Mass / molar mass = 2.71 / 30 = 0.0903 mol

3.8g of oxygen;

Number of moles = Mass / molar mass = 3.8 / 16 = 0.2375 mol

If 0.0903 moles of ethane was used, it would require;

2 = 7

0.0903 = x

x = 0.31605 mol of oxygen needed

This means that oxygen is our limiting reagent.

From the reaction,

7 mol of oxygen yields 4 mol of carbon dioxide

0.2375 yields x?

7 = 4

0.2375 = x

x = 0.1357

Mass = Number of moles * Molar mass = 0.1357 * 44 = 6.05g

8 0
3 years ago
What is the molarity of a solution composed of 5.85 g of potassium iodide, KI, dissolved
Troyanec [42]

Answer:

0.282 M

General Formulas and Concepts:

<u>Chemistry - Solutions</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
  • Molarity = moles of solute / liters of solution

Explanation:

<u>Step 1: Define</u>

5.85 g KI

0.125 L

<u>Step 2: Identify Conversions</u>

Molar Mass of K - 39.10 g/mol

Molar Mass of I - 126.90 g/mol

Molar Mass of KI - 39.10 + 126.90 = 166 g/mol

<u>Step 3: Convert</u>

<u />5.85 \ g \ KI(\frac{1 \ mol \ KI}{166 \ g \ KI} ) = 0.035241 mol KI

<u>Step 4: Find Molarity</u>

M = 0.035241 mol KI / 0.125 L

M = 0.281928

<u>Step 5: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

0.281928 M ≈ 0.282 M

7 0
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leonid [27]
I’m pretty sure the answer is that there are equal number of protons and electrons
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