Based on its chemical formula, which of the following substances is an organic compound

When the fuel mixture contained in a 1.49 l tank, stored at 750 mmhg and 298 k, undergoes complete combustion, how much heat is

Ket [755]

Get to know first how many moles in the gas:n = pV/RT= (1.013*10^5*750/760) Pa *1.49*10^-3 m^3/(8.314 J/(molK)*298) n = 0.0601 moles.
The combustion energies are 889 kJ/mol (methane) and 2 220 kJ (propane) x = moles methane, y = moles propane
x*889 + y*2220 = 778 x + y = 0.0601----------- x = 0.267784 moles = 0.267784*100/0.0601 = 44.6 % y = 0.243216 moles = 0.243216*100/0.0601 = 55.4 %

4 0

3 years ago

Phosgene, COCl2, gained notoriety as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide w

jek_recluse [69]

The question is incomplete, here is the complete question:

Phosgene, $COCl_2$ , gained notoriety as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide with chlorine:

$CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)$

The value of $K_c$ for this reaction is 5.79 at 570 K. What are the equilibrium partial pressures of the three gases if a reaction vessel initially contains a mixture of the reactants in which $p_{CO}=p_{Cl_2}=0.265atm$ and $p_{COCl_2}=0.000atm$ ?

Answer: The equilibrium partial pressure of CO, $Cl_2\text{ and }COCl_2$ is 0.257 atm, 0.257 atm and 0.008 atm respectively.

Explanation:

The relation of $K_c\text{ and }K_p$ is given by:

$K_p=K_c(RT)^{\Delta n_g}$

$K_p$ = Equilibrium constant in terms of partial pressure

$K_c$ = Equilibrium constant in terms of concentration = 5.79

$\Delta n_g$ = Difference between gaseous moles on product side and reactant side = $n_{g,p}-n_{g,r}=1-2=-1$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

T = Temperature = 570 K

Putting values in above equation, we get:

$K_p=5.79\times (0.0821\times 570)^{-1}\\\\K_p=0.124$

We are given:

Initial partial pressure of CO = 0.265 atm

Initial partial pressure of chlorine gas = 0.265 atm

Initial partial pressure of phosgene = 0.00 atm

The given chemical equation follows:

$CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)$

Initial: 0.265 0.265

At eqllm: 0.265-x 0.265-x x

The expression of $K_p$ for above equation follows:

$K_p=\frac{p_{COCl_2}}{p_{CO}\times p_{Cl_2}}$

Putting values in above equation, we get:

$0.124=\frac{x}{(0.265-x)\times (0.265-x)}\\\\x=0.0082,8.59$

Neglecting the value of x = 8.59 because equilibrium partial pressure cannot be greater than initial pressure

So, the equilibrium partial pressure of CO = $(0.265-x)=(0.265-0.008)=0.257atm$

The equilibrium partial pressure of $Cl_2=(0.265-x)=(0.265-0.008)=0.257atm$

The equilibrium partial pressure of $COCl_2=x=0.008atm$

Hence, the equilibrium partial pressure of CO, $Cl_2\text{ and }COCl_2$ is 0.257 atm, 0.257 atm and 0.008 atm respectively.

6 0

3 years ago

A piston contains 5.00 L of gas at a pressure of 101 kPa. The piston is compressed until the gas occupies 2.00 L of space. What

galben [10]

Answer: 252.5 kPa

Explanation:

Given that:

initial volume of gas V1 = 5.00 L

initial pressure of gas P1 = 101 kPa.

new Volume V2 = 2.00 L

new pressure P2 = ?

Since, only pressure and volume are involved, apply the formula for Boyle's law

P1V1 = P2V2

101 kPa x 5.0L = P2 x 2.00L

505 = P2 x 2.00L

P2 = 505/2.00

P2 = 252.5 kPa

Thus, the new pressure of gas inside the piston is 252.5 kPa

4 0

3 years ago

When 78.6 g of urea CH4N2O are dissolved in 700. g of a certain mystery liquid X, the freezing point of the solution is 4.9 Â°C

iogann1982 [59]

Answer:

The van't Hoff factor of NaCl in liquid X is 1.69

Explanation:

Step 1: Data given

Mass of urea = 78.6 grams

Molar mass of urea = 60.06 g/mol

Mass of liquid X = 700 grams = 0.700 kg

he freezing point of the solution is 4.9°C lower than the freezing point of pure X

When 78.6 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 8.5°C lower than the freezing point of pure X.

Molar mass of NaCl = 58.44 g/mol

Step 2: Calculate moles

Moles urea = mass / molar mass

Moles urea = 78.6 grams / 60.06 g/mol

Moles urea = 1.31 moles

Moles NaCl = 78.9 grams / 58.44 g/mol

Moles NaCl = 1.35 moles

Step 3: Calculate molality

Molality = moles / mass of liquid

Molality urea = 1.31 moles / 0.700 kg

Molality = 1.87 molal

Molality NaCl = 1.35 moles / 0.700 kg

Molality NaCl = 1.92 molal

Step 4: Calculate the freezing point depression constant of X

ΔT = i*Kf*m

⇒with ΔT = the freezing point depression = 4.9 °C

⇒with i = the van't hoff factor of urea = 1

⇒with Kf =the freezing point depression consant of X = TO BE DETERMINED

⇒with m = the molality of urea solution = 1.87 molal

4.9 °C = 1 * Kf * 1.87 molal

Kf == 4.9 / 1.87

Kf = 2.62 °C/m

Step 5: Calculate the van't Hoff facotr of NaCl in X

ΔT = i*Kf*m

⇒with ΔT = the freezing point depression = 8.5 °C

⇒with i = the van't hoff factor of urea = TO BE DETERMINED

⇒with Kf =the freezing point depression consant of X = 2.62 °C/m

⇒with m = the molality of urea solution = 1.92 molal

8.5 °C = i * 2.62 °C/m * 1.92 m

i = 8.5 / (2.62 * 1.92)

i = 1.69

The van't Hoff factor of NaCl in liquid X is 1.69

5 0

3 years ago

Why is a spectrum for a given element unique for that element?

AfilCa [17]

Answer: B. Each element has a unique arrangement of electrons.

Explanation:

1) Since the number of electrons in neutral atoms is equal to the number of protons, and the number of protons are unique for each element, each element has a unique arrangement of electrons.

2) The spectrum of an element is the radiation emitted for the atoms of the element when the valence electrons are emitted and that depended on the electrons configuration.

7 0

3 years ago