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jonny [76]
2 years ago
15

Write L values for the following energy levels. What orbital shapes exist in each of

Chemistry
1 answer:
den301095 [7]2 years ago
4 0

Answer:

n   l    m  

����������������������������������

1   0    0 1s 1 2 2

����������������������������������

2   0    0 2s 1 2  

2   1    1,0,-1 2p 3 6 8

����������������������������������

3   0    0 3s 1 2  

3   1    1,0,-1 3p 3 6  

3   2    2,1,0,-1,-2 3d 5 10 18

����������������������������������

4   0    0 4s 1 2  

4   1    1,0,-1 4p 3 6  

4   2    2,1,0,-1,-2 4d 5 10  

4   3    3,2,1,0,-1,-2,-3 4f 7 14 32

Explanation:

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II) Objects made of silver become tarnished.

Explanation:

Physical properties are the properties which can be observed without changing identity of substance.

Chemical properties are the properties which describe how the substance changes into the different substance completely.

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Read 2 more answers
How many moles, kmols in: 100 g of CO2, 1 litre of ethyl alcohol of density 0.789 g/cm3 and a) 1.5m3 of O2 at 25°C and 1 atm. b)
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Explanation:

1) Mass of carbon dioxide = 100 g

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide =\frac{100 g}{44 g/mol}=2.273 moles

1 mol = 0.001 kmol

2.273 moles= 2.273 × 0.001 kmol = 2.273\times 10^{-3} kmol

2) 1 liter of ethyl alcohol of density 0.789 g/cm^3

Volume of ethyl alcohol ,V= 1 L = 1000 mL

Density of ethyl alcohol =d = 0.789 g/cm^3

1 cm^3=1 mL

Mass of ethyl alcohol = m

m=d\times V=0.789 g/cm^3\times 1000 mL=789 g

Molar mass of  ethyl alcohol = 46 g/mol

Moles of ethyl alcohol = \frac{789 g}{46 g/mol}=17.152 mol

17.152 mol=17.152\times 0.001 kmol=1.7152\times 10^{-4} kmol

3) Volume of oxygen gas,V =1.5 m^3=1500 L

1 m^3= 1000 L

Temperature of the gas = T= 25°C = 298.15 K

Pressure of the gas ,P= 1 atm

Moles of oxygen gas = n

PV=nRT

n=\frac{RT}{PV}=\frac{0.0821 atm L/mol K\times 298.15 K}{1 atm\times 1500 L}=0.01632 mol

0.01632 mol = 0.01632 × 0.001 kmol=1.632\times 10^{-5} kmol

4) Volume water in mixture = 1 L

Density of water =  1000 kg/m^3=\frac{1,000,000 g}{1000 L}=1000 g/L

Mass of water = 1000 g/L\times 1 L = 1000 g

Volume of alcohol = 2.5 L

Density of alcohol =  789 kg/m^3=\frac{789000 g}{1000 L}=789 g/L

Mass of alcohol = 789 g/L\times 2.5 L = 1972.5 g

Mass of mixture = 1000 g + 1972.5 g = 2972.5 g

Mass percentage of water :

\frac{1000 g}{2972.5 g}\times 100=33.64\%

Mass percentage of alcohol :

\frac{1972.5 g}{2972.5 g}\times 100=66.36\%

Moles of water :

n_1=\frac{1000 g}{18 g/mol}=55.55 mol

Moles of alcohol =

n_2=\frac{1972.5 g}{46 g/mol}=42.88 mol

Mole fraction of water :

\chi_1=\frac{n_1}{n_1+n_2}=\frac{55.55 mol}{55.55 mol+42.88 mol}=0.5644

Mole fraction of alcohol :

\chi_2=\frac{n_2}{n_1+n_2}=\frac{42.88 mol}{55.55 mol+42.88 mol}=0.4356

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