<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.
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).
1 kilo joule = 0.239006 calories
134 kilo joule = 134 x 0.239006
= 32.026804 calories
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.

In the first solution:


<em>(1)</em>
For the second equation:


<em>(2)</em>
Replacing (2) in (1):


-122.5torr = -0.250P°A

<em>Vapour pressure of cyclohexane at 50°C is 490torr</em>
And for benzene:


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