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Alex_Xolod [135]
3 years ago
8

In which orbital would the valence electrons for carbon be placed

Chemistry
1 answer:
Paha777 [63]3 years ago
4 0
Carbon have 4 valence electron


The electron configuration of C is : 1s2 2s2 2ps

the outer most s and p orbital contains 4 electrons and these are the valence electron of C atom


Key: to determine the valence electron of an element, you have to refer to the group it is located in, in the periodic table. (C is in group 4 (or 14) so VE = 4)

hope its clear enough 
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A tank in the shape of a rectangular prism measures 10 dm by 4 dm by 6 dm. the tank is completely filled with 8908.8 l of liquid
nevsk [136]
1) The question contains an unknown unit


The number 8908.8 has to be in units of mass: for example, kg or grams.


Here you indicated L.


I am going to work assuming that L is a mass unit. So you can see the way to solve the problem, but you have to verifiy the real unit of the statement and substitute with it.



With that in mind you can find the density of the liquid from:


density  = mass / volume


2) Calculate the volume.


The volume of the liquid is the volume of the vessel, because it is filled.


The volume of the vessel is calculated from the formula of volume for a rectantular prism.


Volume of a rectangular prism = area of the base * height = side * side * height


=> Volume = 10 dm * 4 dm * 6 dm = 240 dm^3 = 240 liter


3) Calculate the density:


density = mass /volume = 8,908.8 L / 240 liter   =  37.12 L / dm^3



Answer: 37.12 L / dm^3


 
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3 years ago
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valentina_108 [34]
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7 0
3 years ago
Calculate moles of oxygen atoms in 0.68mol of KMnO4
Dominik [7]
Alright, so that means we have 0.68 mol of the compound


For each 1 mol of the compound, we have 4*1 oxygens (because there are four oxygens in the formula)
Therefore for each 0.68 mol of the compound, we have 4*0.68 moles of oxygen!
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Unit: Chemical Quantities
Vaselesa [24]

Answer:

(See explanation for further details)

Explanation:

1) The quantity of moles of sulfur is:

n = \frac{1.20\times 10^{24}\,atoms}{6.022\times 10^{23}\,\frac{atoms}{mol} }

n = 1.993\,moles

2) The number of atoms of helium is:

x = (1.5\,moles)\cdot \left(6.022\times 10^{23}\,\frac{atoms}{mole} \right)

x = 9.033\times 10^{23}\,atoms

3) The quantity of moles of carbon monoxide is:

n = \frac{4.15\times 10^{23}\,molecules}{6.022\times 10^{23}\,\frac{molecules}{mol} }

n = 0.689\,moles

4) The number of molecules of sulfur dioxide is:

x = (2.25\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)

x = 1.355\times 10^{24}\,molecules

5) The quantity of moles of sodium chloride is:

n = \frac{2.4\times 10^{23}\,molecules}{6.022\times 10^{23}\,\frac{molecules}{mol} }

n = 0.399\,moles

6) The number of formula units of magnesium iodide is:

x = (1.8\,moles)\cdot \left(6.022\times 10^{23}\,\frac{f.u.}{mole} \right)

x = 1.084\times 10^{24}\,f.u.

7) The quantity of moles of potassium permanganate is:

n = \frac{3.67\times 10^{23}\,f.u.}{6.022\times 10^{23}\,\frac{f.u.}{mol} }

n = 1.214\,moles

8) The number of molecules of carbon tetrachloride is:

x = (0.25\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)

x = 1.506\times 10^{23}\,molecules

9) The quantity of moles of aluminium is:

n = \frac{3.67\times 10^{23}\,atoms}{6.022\times 10^{23}\,\frac{atoms}{mol} }

n = 0.609\,moles

10) The number of molecules of oxygen difluoride is:

x = (3.52\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)

x = 2.120\times 10^{24}\,molecules

3 0
4 years ago
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