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3 years ago

How many unpaired electrons are in sulfur

2 answers:

6 0

Distribute the 16 electrons into the boxes representing the orbitals, allowing a maximum of 2 electrons per orbital ad remembering Hund's rule. You can see from the diagram that sulfur has 2 unpaired electrons.

12345 [234]3 years ago

5 0

There are 2 unpaired electrons in sulphur orbital...

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If the vapor pressure of ethanol at 34.7degree C is 100

saw5 [17]

Answer:

we will use the Clausius-Clapeyron equation to estimate the vapour pressures of the boiling ethanol at sea level pressure of 760mmHg:

ln (P2/P1) = $\frac{ΔvapH}{R}([tex]\frac{1}{T1}$ - $\frac{1}{T2}$ )

where

P1 and P2 are the vapour pressures at temperatures T1 and T2

Δ vapH = the enthalpy of vaporization of the ETHANOL

R = the Universal Gas Constant

In this problem,

P 1 = 100 mmHg

; T 1 = 34.7 °C = 307.07 K

P 2 = 760mmHg

T 2 =T⁻²=?

Δ vap H = 38.6 kJ/mol

R = 0.008314 kJ⋅K -1 mol -1

ln ( 760/10)=(0.00325 - T⁻²) (38.6kJ⋅mol-1 /0.008314 )

0.0004368=(0.00325 - T⁻²)

T⁻²=0.002813

T² = 355.47K

6 0

3 years ago

You are given a cube of pure copper when you calculate the density using

lara [203]

Answer:

2.01% to the nearest hundredth.

Explanation:

Percent error =[ (8.96-8.78) / 8.96]* 100

= 0.020089 * 100

= 2.0089 %

5 0

3 years ago

Zinc metal and aqueous silver nitrate react to give Zn(NO3)2(aq) plus silver metal. When 5.00 g of Zn(s) and solution containing

Fittoniya [83]

Answer : The percent yield is, 83.51 %

Solution : Given,

Mass of Zn = 5.00 g

Mass of $AgNO_3$ = 25.00 g

Molar mass of Zn = 65.38 g/mole

Molar mass of $AgNO_3$ = 168.97 g/mole

Molar mass of Ag = 107.87 g/mole

First we have to calculate the moles of Zn and $AgNO_3$ .

$\text{ Moles of }Zn=\frac{\text{ Mass of }Zn}{\text{ Molar mass of }Zn}=\frac{5.00g}{65.38g/mole}=0.0765moles$

$\text{ Moles of }AgNO_3=\frac{\text{ Mass of }AgNO_3}{\text{ Molar mass of }AgNO_3}=\frac{25.00g}{168.97g/mole}=0.1479moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag$

From the balanced reaction we conclude that

As, 2 mole of $AgNO_3$ react with 1 mole of $Zn$

So, 0.1479 moles of $AgNO_3$ react with $\frac{0.1479}{2}=0.07395$ moles of $Zn$

From this we conclude that, $Zn$ is an excess reagent because the given moles are greater than the required moles and $AgNO_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Ag$

From the reaction, we conclude that

As, 2 mole of $AgNO_3$ react to give 2 mole of $Ag$

So, 0.1479 moles of $AgNO_3$ react to give 0.1479 moles of $Ag$

Now we have to calculate the mass of $Ag$

$\text{ Mass of }Ag=\text{ Moles of }Ag\times \text{ Molar mass of }Ag$

$\text{ Mass of }Ag=(0.1479moles)\times (107.87g/mole)=15.95g$

Theoretical yield of $Ag$ = 15.95 g

Experimental yield of $Ag$ = 13.32 g

Now we have to calculate the percent yield.

$\% \text{ yield}=\frac{\text{ Experimental yield of }Ag}{\text{ Theretical yield of }Ag}\times 100$

$\% \text{ yield}=\frac{13.32g}{15.95g}\times 100=83.51\%$

Therefore, the percent yield is, 83.51 %

7 0

3 years ago

What mass of precipitate forms when 185.5 ml of 0.533 m naoh is added to 627 ml of a solution that contains 15.8 g of aluminum s

Law Incorporation [45]

Answer:

2,57 g of precipitate.

Explanation:

For the reaction:

6 NaOH + Al₂(SO₄)₃ → 2 Al(OH)₃ + 3 Na₂SO₄

The precipitate is Al(OH)₃.

185,5mL of 0,533M NaOH are:

0,1855L × 0,533M = 0,0989 moles NaOH

Moles of Al₂(SO₄)₃ are:

15,8g × $\frac{1mol}{342,15g}$ = 0,0462 moles Al₂(SO₄)₃

For the total reaction of 0,0989 moles NaOH with Al₂(SO₄)₃ you need:

0,0989moles NaOH × $\frac{1molAl_{2}(SO_{4})_{3}}{6 moles NaOH}$ = 0,0165 moles Al₂(SO₄)₃

As you have 0,0462 moles Al₂(SO₄)₃ the limiting reactant is NaOH.

0,0989 moles of NaOH produce:

0,0989moles NaOH × $\frac{2molAl(OH)_{3}}{6 moles NaOH}$ = 0,0330 moles of Al(OH)₃

These moles are:

0,0330 moles of Al(OH)₃ × (78 g/mol) = 2,57 g of Al(OH)₃ ≡ mass of precipitate

I hope it helps!

3 0

3 years ago

What is the affect of viscosity on the shape of a volcano?

lions [1.4K]

Answer:

the lava viscosity defines the size and shape of a volcano. Even though lava is 100,000 times more viscous than water, it can still flow great distances. When lava has low viscosity, it can flow very easily over long distances. This creates the classic rivers of lava, with channels, puddles and fountains.

3 0

3 years ago