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2 years ago

14

Draw the product(s) obtained when benzoquinone is treated with excess butadiene. using wedges and dashes, indicate the stereoche

mistry.

1 answer:

Alex17521 [72]2 years ago

6 0

Answer:

The structure can be found on the attached documents

Explanation:

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(a) The mass density of a gaseous compound was found to be 1.23 kg m^−3 at 330 K and 20 kPa. What is the molar mass of the compo

Luden [163]

Answer:

The molar mass of the compound is:- 168.82 g/mol

The molar mass of the gas is:- 16.38 g/mol

Explanation:

(a)

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Also,

Moles = mass (m) / Molar mass (M)

Density (d) = Mass (m) / Volume (V)

So, the ideal gas equation can be written as:

$PM=dRt$

Given that:-

Pressure = 20 kPa = 20000 Pa

The expression for the conversion of pressure in Pascal to pressure in atm is shown below:

P (Pa) = $\frac {1}{101325}$ P (atm)

20000 Pa = $\frac {20000}{101325}$ atm

Pressure = 0.1974 atm

Temperature = 330 K

d = 1.23 kg/m³ = 1.23 g/L

Molar mass = ?

Applying the equation as:

0.1974 atm × M = 1.23 g/L × 0.0821 L.atm/K.mol × 330 K

⇒M = 168.82 g/mol

<u>The molar mass of the compound is:- 168.82 g/mol</u>

(b)

Given that:

Pressure = 152 Torr

Temperature = 298 K

Volume = 250 cm³ = 0.25 L

Using ideal gas equation as:

$PV=nRT$

R = $62.3637\text{torr}mol^{-1}K^{-1}$

Applying the equation as:

152 Torr × 0.25 L = n × 62.3637 L.torr/K.mol × 298 K

⇒n = 0.002045 moles

Given that :

Mass of the gas = 33.5 mg = 0.0335 g

Molar mass = ?

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.002045\ moles
= \frac{0.0335\ g}{Molar\ mass}$

<u>The molar mass of the gas is:- 16.38 g/mol</u>

5 0

2 years ago

A solution is prepared by dissolving 91.7 g fructose in 545 g of water. Determine the mole fraction of fructose if the final vol

oee [108]

Answer:

Mole fraction = 0,0166

Explanation:

Mole fraction is defined as mole of a compound per total moles of the mixture. In the solution, the solute is fructose and the solvent is water. That means you need to find moles of fructose and moles of water.

The molecular mass of fructose is 180,16g/mol and mass of water is 18,02 g/mol. Using these values:

91,7g fructose × (1mol / 180,16g) = <em>0,509 moles of fructose</em>

545g water × (1mol / 18,02g) = <em>30,24 moles of water</em>

Thus, mole fraction of fructose is:

$\frac{0,509 moles}{0,509mol + 30,24mol} = 0,0166$

<em>Mole fraction = 0,0166</em>

I hope it helps!

5 0

2 years ago

What effect do tube dwelling worms have on mudflat ecosystems?

pentagon [3]

Earthworms influence (and benefit) the soil ecosystem in a number of ways: Recycling organic material: Earthworms, along with bacteria and fungi, decompose organic material. ... Improving soil structure: Earthworm burrows alter the physical structure of the soil. They open up small spaces, known as pores, within the soil.

8 0

2 years ago

Suppose you had a balloon containing 1 mole of helium at STP and a balloon containing 1 mole of oxygen at STP. Which statement(s

AleksAgata [21]

Answer:

The true statement is option A.

Explanation:

Using ideal gas equation:

PV = nRT

where,

P = Pressure of gas = 1 atm

V = Volume of gas = ?

n = number of moles of gas = 1 mol

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 273.15 K

$V=\frac{nRT}{P}=\frac{1 mol\times 0.0821 atm L/mol K\times 273.15 K}{1 atm}$

V = 22.42 L

This means that 1 mole of an ideal gas at STP occupies 22.42 liters of volume.

So, 1 mole of helium gas and 1 mole of oxygen gas will have same value of volume in their respective balloons at STP.

7 0

2 years ago

A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (

djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

$Kb = \frac{[ HSO3-][0H-]}{SO3-2}$

$Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7$

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation $Kb=\frac{[HSO3-][OH-]}{[SO3-2]}$

We noe have:

$1.485*10^-^7=\frac{x*x}{(0.056-x)}$

$x = [OH-] = 9.11*10^-^5$

$pOH = -log(OH) = -log(9.11*10^-^5)$

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0

2 years ago