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kiruha [24]
3 years ago
11

Which of these are true for the reaction above?

Chemistry
1 answer:
agasfer [191]3 years ago
6 0

Answer:

a, d, f

Explanation:

ΔHrxn = ΔH(CCl4) -ΔH(CH4) = - 106.7 -(-74.8) = - 31.9 kJ/mol

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Troyanec [42]

Answer:

c

Explanation:

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2 years ago
It is possible to prepare ethanol by nucleophilic substitution. For an electrophile, you could choose EtCl, EtBr, or EtI; for a
tensa zangetsu [6.8K]

Answer:

The reactant/reagent that would be most atom economical is EtI (Ethy Iodide) and KOH (potassium oxide) as base

This is because the iodo group are weak base hence they have a good leaving character (i.e they are unstable on their own ) which would increase the rate of reaction and the strong base KOH give the most atom economical  

Explanation:

4 0
3 years ago
What is the molarity of a KF(aq) solution containing 3.0 mol of KF in 2.00L of solution?
sergeinik [125]

Answer : The molarity of KF in the solution is 1.5 M.

Explanation : Given,

Moles of KF = 3.0 mol

Volume of solution = 2.00 L

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

\text{Molarity}=\frac{\text{Moles of }KF}{\text{Volume of solution (in L)}}

Now put all the given values in this formula, we get:

\text{Molarity}=\frac{3.0mol}{2.00L}=1.5mole/L=1.5M

Therefore, the molarity of KF in the solution is 1.5 M.

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3 years ago
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2 years ago
Read 2 more answers
How many atoms of mercury are present in 3.4 cubic centimeters of liquid mercury? the density of mercury is 13.55 g/cc?
maks197457 [2]

Answer:

             1.37 × 10²³ Atoms of Mercury  

Solution:

Step 1: Calculate Mass of Mercury using following formula,

                               Density  =  Mass ÷ Volume

Solving for Mass,

                               Mass  =  Density × Volume

Putting values,

                               Mass  =  13.55 g.cm⁻³ × 3.4 cm³                ∴ 1 cm³ = 1 cc

                               Mass  =  46.07 g

Step 2: Calculating number of Moles using following formula;

                               Moles  =  Mass ÷ M.mass

Putting values,

                               Moles  =  46.07 g ÷ 200.59 g.mol⁻¹

                               Moles  =  0.229 mol

Step 3: Calculating Number of Atoms using following formula;

                               Number of atoms  =  Moles × 6.022 ×10²³

Putting value of moles,

                               Number of Atoms  =  0.229 mol × 6.022 × 10²³

                              Number of Atoms  =  1.37 × 10²³ Atoms of Hg

8 0
2 years ago
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