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4 years ago

What is the correct electron configuration for oxygen?

Chemistry

1 answer:

s344n2d4d5 [400]4 years ago

3 0

It would be: 1s2, 2s2, 2p4

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HELP PLSS!

Inessa05 [86]

. The empirical formula is the simplest whole number ratio.

The correct answer is 1. HCI

7 0

3 years ago

Difference betwifference between a mixture or compound

Hatshy [7]

The difference between a mixture and a compound is that a mixture can be easily separated like a salad where you can pick things out and a compounds you are usually not able to undo

7 0

3 years ago

PLEASEEE HELPPP!!!!!!

chubhunter [2.5K]

Answer:

Explanation:

Cs has a higher rate of reaction because it's easier to remove an electron from it, thereby leading to faster reactivity

5 0

3 years ago

Read 2 more answers

A galvanic (voltaic) cell consists of an electrode composed of nickel in a 1.0 M nickel(II) ion solution and another electrode c

Andre45 [30]

Answer: The standard potential of the cell is 0.77 V

Explanation:

We know that:

$E^o_{Ni^{2+}/Ni}=-0.25V\\E^o_{Cu^{+}/Cu}=0.52V$

The substance having highest positive $E^o$ reduction potential will always get reduced and will undergo reduction reaction.

The half reaction follows:

Oxidation half reaction: $Ni(s)\rightarrow Ni^{2+}(aq)+2e^-$

Reduction half reaction: $Cu^{+}(aq)+e^-\rightarrow Cu(s)$ ( × 2)

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

Putting values in above equation follows:

$E^o_{cell}=0.52-(-0.25)=0.77V$

Hence, the standard potential of the cell is 0.77 V

5 0

3 years ago

Wastewater discharged into a stream by a sugar refinery contains 3.40 g of sucrose (C12H22O11) per liter. A government-industry

Ipatiy [6.2K]

Answer: The pressure that must be applied to the apparatus is 0.239 atm

Explanation:

To calculate the osmotic pressure, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

or,

$\pi=i\times \frac{m_{solute}}{M_{solute}\times V_{solution}\text{ (in L)}}}\times RT$

where,

$\pi$ = osmotic pressure of the solution

i = Van't hoff factor = 1 (for non-electrolytes)

$m_{solute}$ = mass of sucrose = 3.40 g

$M_{solute}$ = molar mass of sucrose = 342.3 g/mol

$V_{solution}$ = Volume of solution = 1 L

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

T = temperature of the solution = $20^oC=[20+273]K=293K$

Putting values in above equation, we get:

$\pi =1\times \frac{3.40g}{342.3g/mol\times 1}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 293K\\\\\pi =0.239atm$

Hence, the pressure that must be applied to the apparatus is 0.239 atm

3 0

4 years ago