I believe the correct answer from the choices listed above is option A. Fan blades would be an analogy for electron cloud model. Austrian physicist Erwin Schrödinger (1887-1961) developed an “Electron Cloud Model<span>” in 1926. It consisted of a dense nucleus surrounded by a cloud of electrons. Hope this helps.</span>

4 0

3 years ago

Lewis structure for 2O2

Mrrafil [7]

the big number describes the number ratio in a chemical equation

so for example,

2H2 + O2 --> 2H2O means

2 moles of hydrogen reacts with one mole of oxygen to form 2 moles of water

and as you know, the small (subscript) number determines the number of atoms of that element in one molecule of a compound

so I believe that drawing a normal lewis structure ( O=O ) should be correct

6 0

3 years ago

Read 2 more answers

47.0ml of a HBr solution were titrated with 37.5ml of a 0.215M LiOH solution to reach the equivalence point. what is the molarit

faltersainse [42]

Hello!

The molarity of the HBr solution is 0,172 M.

Why?

The neutralization reaction between LiOH and HBr is the following:

HBr(aq) + LiOH(aq) → LiBr(aq) + H₂O(l)

To solve this exercise, we are going to apply the common titration equation:

$M1*V1=M2*V2$

$M1=\frac{M2*V2}{V1}= \frac{0,215 M * 37,5 mL}{47 mL}=0,172 M$

Have a nice day!

4 0

4 years ago

The formation of SO3 from SO2 and O2 is an intermediate step in the manufacture of sulfuric acid, and it is also responsible for

jeka57 [31]

Answer:

118.22 atm

Explanation:

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

KP = 0.13 = $\frac{p(SO_{3})^{2}}{p(SO_{2})^{2}p(O_{2})}$

Where p(SO₃) is the partial pressure of SO₃, p(SO₂) is the partial pressure of SO₂ and p(O₂) is the partial pressure of O₂.

With 2.00 mol SO₂ and 2.00 mol O₂ if there was a 100% yield of SO₃, then 2 moles of SO₃ would be produced and 1.00 mol of O₂ would remain.
With a 71.0% yield, there are only 2*0.71 = 1.42 mol SO₃, the moles of SO₂ that didn't react would be 2 - 1.42 = 0.58; and the moles of O₂ that didn't react would be 2 - 1.42/2 = 1.29.

The total number of moles is 1.42 + 0.58 + 1.29 = 3.29. With that value we can calculate the molar fraction (X) of each component:

XSO₂ = 0.58/3.29 = 0.176

XO₂ = 1.29/3.29 = 0.392

XSO₃ = 1.42/3.29 = 0.432

The partial pressure of each gas is equal to the total pressure (PT) multiplied by the molar fraction of each component.

p(SO₂) = 0.176 * PT

p(O₂) = 0.392 * PT

p(SO₃) = 0.432 * PT

Rewriting KP and solving for PT:

$\frac{p(SO_{3})^{2}}{p(SO_{2})^{2}p(O_{2})}=0.13\\\frac{(0.432*P_{T})^{2}}{(0.176*P_{T})^{2}(0.392*P_{T})} =0.13\\\frac{0.1866*P_{T}^{2}}{0.0121*P_{T}^{3}} =0.13\\\frac{15.369}{P_{T}}=0.13\\P_{T}=118.22 atm$

5 0

3 years ago