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Leokris [45]
2 years ago
11

1) For the following molecules: - Draw the Lewis structures/dots - Determine the polarity (polar or non-polar). - List all inter

molecular forces. a) CF4 b) CH2Br2 c) H2CO3 d) N2 e) PCl3 f) XeF2 g) NH4
Chemistry
1 answer:
4vir4ik [10]2 years ago
8 0

Answer:

ed

Explanation:

ed

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Balance this equation. If a coefficient of "1" is required, choose "blank" for that box. ____CO +____ O2 → ____CO2
Black_prince [1.1K]
<h3><u>Answer;</u></h3>

2, Blank, 2 ;

<h3><u>Explanation;</u></h3>

The balanced chemical equation would be;

2 CO + O2 → 2 CO2

Balancing a chemical equation ensures that the number of atoms of each element are equal on both the reactants side and the products side. This ensures that the law of conservation of mass is obeyed in chemical reactions.

4 0
3 years ago
Read 2 more answers
Wolff-kishner reduction (hydrazine, koh, ethylene glycol, 130°c) of the compound shown gave compound
telo118 [61]
The scheme is shown below, the steps involved are as follow,

Step one: Reduction:
               The carbonyl group of given compound on reduction using Wolf Kishner reagent converts the carbonyl group into -CH₂- group.

Step two: Epoxidation:
              The double bond present in starting compound when treated with m-CPBA (<span>meta-Chloroperoxybenzoic acid) gives corrsponding epoxide.

Step three: Reduction:
                The epoxide is reduced to alcohol on treatment with Lithium Aluminium Hydride (LiAlH</span>₄)<span> followed by hydrolysis.

Step four: Oxidation:
               The hydroxyl group (alcohol) is oxidized to carbonyl (ketonic group) using oxidizing agent Chromic acid (H</span>₂CrO₄).

8 0
3 years ago
How do you convert 134kj to Calories?
Sonja [21]
1 kilo joule = 0.239006 calories
134 kilo joule = 134 x 0.239006
                      = 32.026804 calories
5 0
3 years ago
Mixtures of benzene and cyclohexane exhibit ideal behavior. A solution was created containing 1.5 moles of liquid benzene and 2.
goblinko [34]

Answer:

Vapour pressure of cyclohexane at 50°C is 490torr

Vapour pressure of benzene at 50°C is 90torr

Explanation:

Using Raoult's law, pressure of a solution is defined by the sum of the product sbetween mole fraction of both solvents and pressure of pure solvents.

P_{solution} = X_{A}P^0_{A}+X_{B}P^0_{B}

In the first solution:

X_{cyclohexane}=\frac{2.5mol}{2.5mol+1.5mol} =0.625

X_{benzene}=\frac{1.5mol}{2.5mol+1.5mol} =0.375

340torr = 0.625P^0_{A}+0.375P^0_{B} <em>(1)</em>

For the second equation:

X_{cyclohexane}=\frac{3.5mol}{3.5mol+1.5mol} =0.700

X_{benzene}=\frac{1.5mol}{3.5mol+1.5mol} =0.300

370torr = 0.700P^0_{A}+0.300P^0_{B}<em>(2)</em>

Replacing (2) in (1):

340torr = 0.625P^0_{A}+0.375(1233.3-2.333P^0_{A})

340torr = 0.625P^0_{A}+462.5-0.875P^0_{A}

-122.5torr = -0.250P°A

P^0_{A} = 490 torr

<em>Vapour pressure of cyclohexane at 50°C is 490torr</em>

And for benzene:

370torr = 0.700*490torr+0.300P^0_{B}

P^0_{B}=90torr

<em>Vapour pressure of benzene at 50°C is 90torr</em>

3 0
2 years ago
A solution is made by dissolving 10.20 grams of glucose (C6H12O6) in 355 grams of water. What is the freezing point depression o
Ludmilka [50]

Answer:

0.297 °C

Step-by-step explanation:

The formula for the <em>freezing point depression </em>ΔT_f is

ΔT_f = iK_f·b

i is the van’t Hoff factor: the number of moles of particles you get from a solute.

For glucose,

       glucose(s) ⟶ glucose(aq)

1 mole glucose ⟶ 1 mol particles     i = 1

Data:

Mass of glucose = 10.20 g

  Mass of water = 355 g

                 ΔT_f = 1.86 °C·kg·mol⁻¹

Calculations:

(a) <em>Moles of glucose </em>

n = 10.20 g × (1 mol/180.16 g)

  = 0.056 62 mol

(b) <em>Kilograms of water </em>

m = 355 g × (1 kg/1000 g)

   = 0.355 kg

(c) <em>Molal concentration </em>

b = moles of solute/kilograms of solvent

  = 0.056 62 mol/0.355 kg

  = 0.1595 mol·kg⁻¹

(d) <em>Freezing point depression </em>

ΔT_f = 1 × 1.86 × 0.1595

        = 0.297 °C

3 0
3 years ago
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