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Sunny_sXe [5.5K]
3 years ago
14

Why do electrons enter the 4s orbital before entering the 3D orbital

Chemistry
1 answer:
garri49 [273]3 years ago
8 0

The 3d sublevel is not filled until after the 4s sublevel, because the 3d sublevel has more energy than the 4s sublevel, and less energy than the 4p sublevel. (:

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If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi
zlopas [31]

Answer:

T_{eq}=28.9\°C

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

Q_{Cd}+Q_{W}=0

Thus, we insert mass, specific heat and temperatures to obtain:

m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}

Now, we plug in to obtain:

T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C

NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.

Best regards!

6 0
2 years ago
Im about to take first year chemistry for engineering. Should i review or will this be done in class? If yes what are some conce
sashaice [31]
You should always study. you can be above your class. thats what i do.. search google for ur course. they can tell u some key points for it. u should even study what you know and want to know as well.
6 0
3 years ago
How much faster does helium escape through a porous container than ozone?
otez555 [7]
According to Graham's Law ," the rates of effusion or diffusion of two gases are inversely proportional to the square root of their molecular masses at given pressure and temperature".

                                r₁ / r₂  =  \sqrt{M2 / M1}   ---- (1)

r₁    =  Rate of effusion of He

r₂    =  Rate of Effusion of O₃

M₁  =  Molecular Mass of He  =  4 g/mol

M₂  =  Molecular Mass of O₃  =  48 g/mol

Putting values in eq. 1,

                                r₁ / r₂  =  \sqrt{48 / 4}

                                r₁ / r₂  =  \sqrt{12}

                                r₁ / r₂  =  3.46

Result:
          Therefore, Helium will effuse 3.46 times more faster than Ozone.
4 0
3 years ago
carbon dioxide at 25 degrees celsius and 101.3 kpa has a density of 1.799 kg/m^3 determine gas constant
dusya [7]

Answer:

Explanation:

Temperature of gas in absolute scale T = 25 + 273 = 298 .

pressure of gas P = 1.013 x 10⁵ N / m²

density D = 1.799 kg / m³

= 1799 g / m³

From gas formula

PV / T = n R Where P is pressure , V is volume and T is absolute temperature , n is no of moles

P / T = n R / V

P / T = m R /M V where m is mass of gas and M is molecular weight .

m / V = D ( density )

P / T = DR/ M  

PM / DT = R

Putting the values

1.013 x 10⁵ x 44  /  (1799 x 298)

R = 8.314.09 J / K mole

5 0
3 years ago
Why did mendeleyev leave spaces in the periodic table
aalyn [17]

Answer:

Mendeleev left spaces in his periodic table because he predicted there were elements that weren't discovered yet that would fit in these spaces.

Explanation:

8 0
3 years ago
Read 2 more answers
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