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2 years ago

A base

Chemistry

1 answer:

hichkok12 [17]2 years ago

5 0

B, turns red litmus paper to blue

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Linolenic acid (C18H30O2 - M.W. = 278.42 g/mol) reacts with hydrogen gas according to the equation: C18H30O2 + 3H2 (g) → C18H36O

Ludmilka [50]

Answer:

2.53 L is the volume of H₂ needed

Explanation:

The reaction is: C₁₈H₃₀O₂ + 3H₂ → C₁₈H₃₆O₂

By the way we can say, that 1 mol of linolenic acid reacts with 3 moles of oxygen in order to produce, 1 mol of stearic acid.

By stoichiometry, ratio is 1:3

Let's convert the mass of the linolenic acid to moles:

10.5 g . 1 mol / 278.42 g = 0.0377 moles

We apply a rule of three:

1 mol of linolenic acid needs 3 moles of H₂ to react

Then, 0.0377 moles will react with (0.0377 . 3 )/1 = 0.113 moles of hydrogen

We apply the Ideal Gases Law to find out the volume (condition of measure are STP) → P . V = n . R . T → V = ( n . R .T ) / P

V = (0.113 mol . 0.082 L.atm/mol.K . 273.15K) 1 atm = 2.53 L

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2 years ago

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sergij07 [2.7K]

Sound, heat, light etc

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A group of students working in a chemistry lab are planning a procedure to

NARA [144]

use a strong base to neutralize, like NaOH, KOH, etc

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At what temperature does water evaporate?

Alexus [3.1K]

It evaporates at 100°

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Read 2 more answers

The molar heat capacity of ethane is represented in the temperature range 298 K to 400 K by the empirical expression Cp,m in J K

BabaBlast [244]

Answer:

-88.66 kJ/mol

Explanation:

The expressions of heat capacity (Cp,m) for C(s) and for H₂(g) are:

C(s): Cp,m/(J K-1 mol-1) = 16.86 + (4.77T/10³) - (8.54x10⁵/T²)

H₂(g): Cp,m/(J K-1 mol-1) = 27.28 + (3.26T/10³) + (0.50x10⁵/T²)

Cp = A + BT + CT⁻²

For the Kirchoff's Law:

ΔHf = ΔH°f + $\int\limits^{T2}_{T1} {DCp(T)} \, dT$

Where ΔH°f is the enthalpy at 298 K, T1 is 298 K, T2 is the temperature given (373 K), and DCp is the variation of Cp (products less reactants). ΔH°f for ethene is -84.68 kJ/mol and the reaction is:

2C(s) + 3H₂(g) → C₂H₆

So, DCp:

dA = A(C₂H₆) - [2xA(C) + 3xA(H₂)] = 14.73 - [2x16.86 + 3x27.28] = -100.83

dB = B(C₂H₆) - [2xB(C) + 3xB(H₂)] = 0.1272 - [2x4.77x10⁻³ + 3x3.26x10⁻³] = 0.10788

dC = C(C₂H₆) - [2xC(C) + 3xC(H₂)] = 0 - (2x(-8.54x10⁵) + 3x0.50x10⁵) = 15.58x10⁵

dCp = -100.83 + 0.10788T + 15.58x10⁵T⁻²

$\int\limits^{373}_{298} {-100.83 + 0.10788T + 15.58x10^5T^{-2}} \, dT$ = -3796.48 J/mol = -3.80 kJ/mol (solved by a graphic calculator)

ΔHf = -84.68 - 3.80

ΔHf = -88.66 kJ/mol

7 0

2 years ago