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nydimaria [60]
3 years ago
8

Use the periodic table to classify each of the elements below.

Chemistry
2 answers:
Alika [10]3 years ago
5 0

Answer:

Explanation:

Cadmium:(Cd)

Cadmium is transition metal present in group twelve. It is soft metal and properties are similar to the other group members like zinc and mercury. Its atomic number is forty eight and have two valance electrons.

Electronic configuration:

Cd₄₈ = [Kr] 4d¹⁰ 5s²

Vanadium: (V)

It is present in group five. It is malleable and ductile transition metal. Its atomic number is twenty three. Vanadium have five valance electrons.

Electronic configuration:

V₂₃ =[Ar] 3d³ 4s²

Xenon: Xe

Xenon is present in group eighteen. It is noble gas. Its outer most valance shell is complete that's why it is inert. its atomic number is fifty four. Xenon have eight valance electrons.

Electronic configuration:

Xe₅₄ = [Kr] 4d¹⁰ 5s² 5p⁶

Iodine: (I)

Iodine is present in group seventeen. Its outer most valance shell have seven electrons. Iodine is the member of halogen family. It gain one electron to complete the octet. its atomic number is fifty three.

Electronic configuration:

I₅₃ = [Kr] 4d¹⁰ 5s² 5p⁵

Potassium: (K)

Potassium is present in group one. it is alkali metal. Its atomic number is nineteen. Its valance shell has one electron. Potassium loses its one valance electron and gets stable electronic configuration.

Electronic configuration:

K₁₉ = [Ar] 4s¹

Strontium: Sr

Strontium is present in group two. it is alkaline earth metal. its atomic number is thirty eight and have two valance electrons.

Electronic configuration:

Sr₃₈ = [Kr] 5s²

dalvyx [7]3 years ago
5 0

Answer:

cadmium (Cd): transition metal

vanadium (V): transition metal

xenon (Xe): noble gas

iodine (I):  halogen

potassium (K): alkali metal

strontium (Sr): alkaline earth metal

Explanation:

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1. (8pt) Using dimensional analysis convert 600.0 calories into kilojoules
Ivanshal [37]

Answer:

1. 2.510kJ  

2. Q = 1.5 kJ

Explanation:

Hello there!

In this case, according to the given information for this calorimetry problem, we can proceed as follows:

1. Here, we consider the following equivalence statement for converting from calories to joules and from joules to kilojoules:

1cal=4.184J\\\\1kJ=1000J

Then, we perform the conversion as follows:

600.0cal*\frac{4.184J}{1cal}*\frac{1kJ}{1000J}=2.510kJ

2. Here, we use the general heat equation:

Q=mC(T_2-T_1)

And we plug in the given mass, specific heat and initial and final temperature to obtain:

Q=236g*0.24\frac{J}{g\°C} (34.9\°C-8.5\°C)\\\\Q=1495.3J*\frac{1kJ}{1000J} \\\\Q=1.5kJ

Regards!

7 0
2 years ago
The term El Nino/Southern Oscillation refers to the occurrence of warmer than normal sea surface temperatures in the Eastern Pac
Amiraneli [1.4K]

Answer:

True

Explanation:

The  El Nino/Southern Oscillation occurs in between the three phases Neutral, La Nina or El Nino.

The humboldt current takes cold water to South America's west coast to the tropics from the  Southern Ocean. Because of the welling which takes place in Peru this process is enhanced. Due to the trade winds near the equator the cold water travels west where it heats up due to the sun. This makes the water of the western Pacific warmer.

6 0
3 years ago
A gas occupies 4.30 Liters at STP. What volume does it occupy at 2.56 atm and 302
laiz [17]

Answer:

V₂ = 1.86 L

Explanation:

Given data:

Initial volume = 4.30 L

Initial pressure = 1 atm

Initial temperature = 273.15 K

Final temperature = 302 K

Final volume =  ?

Final pressure = 2.56 atm

Solution:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁T₂ /T₁ P₂

V₂ = 1 atm ×4.30 L × 302 K / 273.15 K × 2.56 atm

V₂ = 1298.6 atm.L.K / 699.26 K.atm

V₂ = 1.86 L

4 0
3 years ago
When the paramagnetic [co(cn)6] 4– ion is oxidized to [co(cn)6] 3– , the ion becomes diamagnetic. however, when the paramagnetic
S_A_V [24]
Answer:  
Basically, paramagnetic and diamagnetic refer to the way a chemical species interacts with a magnetic field. More specifically, it refers to whether or not a chemical species has any unpaired electrons or not. 
 A diamagnetic species has no unpaired electrons, while a paramagnetic species has one or more unpaired electrons. 
 Now, I won't go into too much detail about crystal field theory in general, since I assume that you're familiar with it. 
 So, you're dealing with the hexafluorocobaltate(III) ion, [CoF6]3â’, and the hexacyanocobaltate(III) ion, [Co(CN)6]3â’. 
 You know that [CoF6]3â’ is paramagnetic and that [Co(CN)6]3â’ is diamagnetic, which means that you're going to have to determine why the former ion has unpaired electrons and the latter does not. 
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  Co3+:1s22s22p63s23p63d6 
 For an isolated cobalt(III) cation, all these five 3d-orbitals are degenerate. The thing to remember now is that the position of the ligand on the spectrochemical series will determine how these d-orbtals will split. 
 More specifically, you can say that 
  a strong field ligand will produce a more significant splitting energy, Δ  a weak field ligand will produce a less significant splitting energy, Δ 
 Now, the spectrochemical series looks like this 
 http://chemedu.pu.edu.tw/genchem/delement/9.htmhttp://chemedu.pu.edu.tw/genchem/delement/9.htm 
 Notice that the cyanide ion, CNâ’, is higher on the spectrochemical series than the fluoride ion, Fâ’. This means that the cyanide ion ligands will cause a more significant energy gap between the eg and t2g orbitals when compared with the fluoride ion ligands. 
 http://wps.prenhall.com/wps/media/objects/3313/3393071/blb2405.htmlhttp://wps.prenhall.com/wps/media... 
 In the case of the hexafluorocobaltate(III) ion, the splitting energy is smaller than the electron pairing energy, and so it is energetically favorable to promote two electrons from the t2g orbitals to the eg orbitals → a high spin complex will be formed. 
 This will ensure that the hexafluorocobaltate(III) ion will have unpaired electrons, and thus be paramagnetic. 
 On the other hand, in the case of the hexacyanocobaltate(III) ion, the splitting energy is higher than the electron pairing energy, and so it is energetically favorable to pair up those four electrons in the t2g orbitals → a low spin complex is formed. 
 Since it has no unpaired electrons, the hexacyanocobaltate(III) ion will be diamagnetic.
6 0
3 years ago
What types of volcanic landforms occur at divergent plate boundaries
DochEvi [55]
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