What is the density of sulfur dioxide (so2) at 1.2 atmospheres and 271 kelvin? show all of the work used to solve this problem?

Picric acid has been used in the leather industry and in etching copper. However, its laboratory use has been restricted because

olchik [2.2K]

Answer:

0.3023 M

Explanation:

Let Picric acid = $H_{picric}$

So, $H_{picric}$ + $H_2}O$ ⇄ $H_3}O^+$ + $Picric^-$

The ICE table can be given as:

$H_{picric}$ + $H_2}O$ ⇄ $H_3}O^+$ + $Picric^-$

Initial: 0.52 0 0

Change: - x + x + x

Equilibrium: 0.52 - x + x + x

Given that;

acid dissociation constant ( $K_a$ ) = 0.42

$K_a = \frac{[H_3O^+][Picric^-]}{H_{picric}}$

$0.42 = \frac{[x][x]}{0.52-x}}$

$0.42 = \frac{[x]^2}{0.52-x}}$

0.42(0.52-x) = x²

0.2184 - 0.42x = x²

x² + 0.42x - 0.2184 = 0 -------------------- (quadratic equation)

Using the quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$ ; ( where +/- represent ± )

= $\frac{-0.42+/-\sqrt{(0.42)^2-4(1)(-0.2184)} }{2*1}$

= $\frac{-0.42+/-\sqrt {0.1764+0.8736} }{2}$

= $\frac{-0.42+\sqrt {1.0496} }{2}$ OR $\frac{-0.42-\sqrt {1.0496} }{2}$

= $\frac{-0.42+1.0245}{2}$ OR $\frac{-0.42-1.0245}{2}$

= $\frac{0.6045}{2}$ OR $-\frac{1.4445}{2}$

= 0.30225 OR - 0.72225

So, we go by the +ve integer that says:

x = 0.30225

x = [ $H_3}O^+$ ] = [ $Picric^-$ ] = 0.3023 M

∴ the value of [H3O+] for an 0.52 M solution of picric acid = 0.3023 M (to 4 decimal places).

6 0

3 years ago

i have a balloon that can hold 100 liters of air. if i blow up this balloon with 3 moles of oxygen gas at a pressure of 1.0 atmo

Tcecarenko [31]

Baloon with 3 moles og oxygen at 1 atm.The temperature of the balloon is 4 Kelvin.

An ideal gas is a theoretical gas composed of many randomly transferring factor particles that aren't difficult to interparticle interactions. the best gasoline idea is beneficial because it obeys the precise gas law, a simplified equation of country, and is amenable to evaluation under statistical mechanics.

An ideal gas is described as one for which both the extent of molecules and forces between the molecules are so small that they have got no effect at the behavior of the gas. The real gas that acts almost like a really perfect gasoline is helium. that is due to the fact helium, in contrast to maximum gases, exists as an unmarried atom, which makes the van der Waals dispersion forces as low as viable

Using the ideal gas equation:-

Given;

P₁ = 1 atm

V₁ = 100 L

n = 3

r = 8.314

T = PV/nR

= 1 × 100 / 3 × 8.314

= 4 K

Learn more about ideal gas here:-brainly.com/question/20348074

#SPJ4

4 0

1 year ago

Colorful fireworks often involve the decomposition of barium nitrate and potassium chlorate and the reaction of the metals magne

Fynjy0 [20]

Fireworks owe their colors to reactions of combustion of the metals present. When Mg and Al burn, they emit a white bright light, whereas iron emits a gold light. Besides metals, oxygen is necesary for the combustion. The decomposition reactions of barium nitrate and potassium chlorate provide this element. At the same time, barium can burn emitting a green light.

(a) Barium nitrate is a salt formed by the cation barium Ba²⁺ and the anion nitrate NO₃⁻. Its formula is Ba(NO₃)₂. Potassium chlorate is a salt formed by the cation potassium K⁺ and the anion chlorate ClO₃⁻. Its formula is KClO₃.

(b) The balanced equation for the decomposition of potassium chloride is:

2KClO₃(s) ⇄ 2KCl(s) + 3O₂(g)

(c) The balanced equation for the decomposition of barium nitrate is:

Ba(NO₃)₂(s) ⇄ BaO(s) + N₂(g) + 3O₂(g)

(d) The balanced equations of metals with oxygen to form metal oxides are:

2 Mg(s) + O₂(g) ⇄ 2 MgO(s)

4 Al(s) + 3 O₂(g) ⇄ 2 Al₂O₃(s)

4 Fe(s) + 3 O₂(g) ⇄ 2 Fe₂O₃(s)

5 0

3 years ago

Given that a 12.00 g milk chocolate bar contains 8.000 g of sugar, calculate the percentage of sugar present in 12.00 g of milk

neonofarm [45]

So in order for us to know the percentage of sugar present in a 12.00 g of milk chocolate, what we are going to do is that, we just have to divide 8 by 12 and multiply in by 100 and we get 66.67. Therefore, the percentage of sugar present in 12.00 g of milk chocolate bar is 66.67%. Hope this answers your question. Have a great day!

5 0

3 years ago

Read 2 more answers

What mass of barium sulfate (233 g/mol) is produced when 125 mL of a 0.150 M solution of barium chloride is mixed with 125 mL of

Rzqust [24]

Answer:

4.37 g of barium sulphate

Explanation:

The reaction equation is;

3BaCl2(aq) + Fe2(SO4)3(aq) ---->3 BaSO4(s) + 2FeCl3(aq)

From the question, the number of moles of both barium chloride and FeSO4 = 125/1000 L × 0.150 M = 0.01875 moles

To find the limiting reactant;

3 moles of barium chloride yields 3 moles of barium sulphate

0.01875 moles of barium chloride yields 3 × 0.01875 moles/3 = 0.01875 moles of barium sulphate

1 mole of iron III sulphate yields 3 moles of barium sulphate

0.01875 molesof iron III sulphate yields 0.01875 moles ×3/1 = 0.05625 moles of barium sulphate

Hence,barium chloride is the limiting reactant

Amount of barium sulphate produced = 0.01875 moles × 233 g/mol = 4.37 g of barium sulphate

4 0

3 years ago