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Sedaia [141]
2 years ago
14

Suppose 'A' is a liquid aromatic compound with molecular weight 78 and burns with sooty flame. a.Give the name of the compound '

A' b.write the molecular structure of 'A' C. What is the product when 'A' is treated with ? i.conc.HNO3 with conc.H2SO4 as catalyst ii.Halogen (cl2)in presence of sunlight and mention the use of the product obtained​
Chemistry
1 answer:
Oliga [24]2 years ago
3 0

Aromatic compounds are compounds that contain carbon-carbon multiple bonds.

The question did not mention that a heteroatom is present in the compound so we can assume that there is none of such. In that case, the compound contains only hydrogen and carbon.

So,

(CH)n = 78

where n is the number of each atom present.

(12 +1)n = 78

n = 78/13

n = 6

The molecular formula of the compound is C6H6

When C6H6 is treated with .conc.HNO3/conc.H2SO4 the compound shown in image 1 is formed. The reaction occurs at the C-C multiple bond.

When C6H6 is reacted with chlorine in the presence of sunlight, hexachlorobenzene (shown in image 2 attached) is formed.

brainly.com/question/24305135

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1. The heat of fusion for the ice-water phase transition is 335 kJ/kg at 0°C and 1 bar. The density of water is 1000 kg/m3 at th
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Answer:

Expression for the change of melting temperature with pressure..> T₂ = T₁exp(-(P₂-P₁)/(3.61x10⁹ Pa), Freezing Point = 0°C

Explanation:

Derivation from state postulate

Using the state postulate, take the specific entropy,  , for a homogeneous substance to be a function of specific volume  and temperature  .

ds = (partial s/partial v)(t) dv + (partial s/partial T)(v) dT

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ds = (partial s/partial v)(T)  dv

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ds = (partial P/partial T)(v) dv

s(β) – s(aplαha) = dP/dT (v(β) – v(α))

dP/dT = s(β) – s(α)/v(β) – v(α) = Δs/Δv

Here Δs and Δv are respectively the change in specific entropy and specific volume from the initial phase α to the final phase β.

For a closed system undergoing an internally reversible process, the first law is

du = δq – δw = Tds - Pdv

Using the definition of specific enthalpy, h and the fact that the temperature and pressure are constant, we have

du + Pdv = dh Tds,

ds = dh/T,

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After substitution of this result into the derivative of the pressure, one finds

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<u>This last equation is the Clapeyron equation.</u>

a)

(dP/dT) = dH/TdV => dP/dlnT = dH/dV

=> dP/dlnT = dH/dV = [H(liquid) - H(solid)]/[V(liquid) - V(solid)]

= [335,000 J/kg]/[1000⁻¹ - 915⁻¹ m³/kg]

= -3.61x10⁹ J/m³ = -3.61x10⁹ Pa

=> P₂ = P₁ - 3.61x10⁹ ln(T₂/T₁) Pa

or

T₂ = T₁exp(-(P₂-P₁)/(3.61x10⁹ Pa)

b) if the pressure in Denver is 84.6 kPa:

T₂(freezing) = 273.15exp[-(84,600-100,000)/(3.61x10⁹)]

≅ 273.15 = 0°C T₁(freezing) essentially no change

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3 years ago
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