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Katarina [22]
3 years ago
10

An atom of an element has 5 electrons in L-shell.

Chemistry
1 answer:
pishuonlain [190]3 years ago
6 0
I dont know your question but that is true
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Why did bohrs observations of gaseous elements not reveal a complete spectrum when the light was shone through a prism? explain
Vika [28.1K]
The Bohr model proposed that electrons could just have characterized vitality levels thus when rotting back to a lower vitality level discharge a specific measure of vitality. Since the measure of vitality could be changed over to a specific recurrence then particular emanation lines were found in the electromagnetic range. Alternate speculations couldn't clarify the discharge lines.
4 0
3 years ago
A 0.150-kg sample of a metal alloy is heated at 540 Celsius an then plunged into a 0.400-kg of water at 10.0 Celsius, which is c
Zarrin [17]

Answer:

C_{alloy}=0.497\frac{J}{g\°C}

Explanation:

Hello there!

In this case, according to this calorimetry problem on equilibrium temperature, it is possible for us to infer that the heat released by the metal allow is absorbed by the water for us to write:

Q_{allow}=-(Q_{water}+Q_{Al})

Thus, by writing the aforementioned in terms of mass, specific heat and temperature, we have:

m_{alloy}C_{alloy}(T_{eq}-T_{alloy})=-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})

Then, we solve for specific heat of the metallic alloy to obtain:

C_{alloy}=\frac{-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})}{m_{alloy}(T_{eq}-T_{alloy})}

Thereby, we plug in the given data to obtain:

C_{alloy}=\frac{-(400g*4.184\frac{J}{g\°C} (30.5\°C-10.0\°C)+200g*0.900\frac{J}{g\°C}(30.5\°C-10.0\°C)}{150g(30.5\°C-540\°C)} \\\\C_{alloy}=0.497\frac{J}{g\°C}

Regards!

3 0
3 years ago
Write the cell notation for an electrochemical cell consisting of an anode where Mn (s) is oxidized to Mn2 (aq) and a cathode wh
morpeh [17]

Answer:

Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)

Explanation:

In writing the cell notation for an electrochemical cell, the anode is written on the left hand side while the cathode is written on the right hand side. The two half cells are separated by two thick lines which represents the salt bridge.

For the cell discussed in the question; the Mn(s)/Mn^2+(aq) is the anode while the Co^2+(aq)/Co(s) half cell is the cathode.

Hence I can write; Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)

8 0
3 years ago
What does it means when your boyfriend doesnt talk to you ​
d1i1m1o1n [39]
Either he’s busy, or he is just trying to ignore you.

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7 0
4 years ago
How many molecules of NH3 are produced from 4.72x10 negative 4 power g of H2?
Delvig [45]
<span>9.40x10^19 molecules.
   The balanced equation for ammonia is:
 N2 + 3H2 ==> 2NH3
   So for every 3 moles of hydrogen gas, 2 moles of ammonia is produced. So let's calculate the molar mass of hydrogen and ammonia, starting with the respective atomic weights:
 Atomic weight nitrogen = 14.0067
  Atomic weight hydrogen = 1.00794

   Molar mass H2 = 2 * 1.00794 = 2.01588 g/mol
 Molar mass NH3 = 14.0067 + 3 * 1.00794 = 17.03052 g/mol

   Moles H2 = 4.72 x 10^-4 g / 2.01588 g/mol = 2.34140921086573x10^-4 mol
   Moles NH3 = 2.34140921086573x10^-4 mol * (2/3) = 1.56094x10^-4 mol

Now to convert from moles to molecules, just multiply by Avogadro's number: 1.56094x10^-4 * 6.0221409x10^23 = 9.400197448261x10^19
   Rounding to 3 significant figures gives 9.40x10^19 molecules.</span>
7 0
3 years ago
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