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3 years ago

Compare the values of the spin quantum number for two electrons in the same orbital

1 answer:

3 0

If there are 2 electrons in the same orbital, the spin numbers would be different for both of these 2 electrons. One would have an up spin and the other a down spin.

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At constant volume, the heat of combustion of a particular compound, compound A, is − 3039.0 kJ / mol. When 1.697 g of compound

melisa1 [442]

Answer:

13.85 kJ/°C

-14.89 kJ/g

Explanation:

At constant volume, the heat of combustion of a particular compound, compound A, is − 3039.0 kJ/mol. When 1.697 g of compound A (molar mass = 101.67 g/mol) is burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 3.661 °C. What is the heat capacity (calorimeter constant) of the calorimeter?

The heat of combustion of A is − 3039.0 kJ/mol and its molar mass is 101.67 g/mol. The heat released by the combustion of 1.697g of A is:

$1.697g.\frac{1mol}{101.67g} .\frac{(-3039.0kJ)}{mol} =-50.72kJ$

According to the law of conservation of energy, the sum of the heat released by the combustion and the heat absorbed by the bomb calorimeter is zero.

Qcomb + Qcal = 0

Qcal = -Qcomb = -(-50.72 kJ) = 50.72 kJ

The heat capacity (C) of the calorimeter can be calculated using the following expression.

Qcal = C . ΔT

where,

ΔT is the change in the temperature

Qcal = C . ΔT

50.72 kJ = C . 3.661 °C

C = 13.85 kJ/°C

Suppose a 3.767 g sample of a second compound, compound B, is combusted in the same calorimeter, and the temperature rises from 23.23°C to 27.28 ∘ C. What is the heat of combustion per gram of compound B?

Qcomb = -Qcal = -C . ΔT = - (13.85 kJ/°C) . (27.28°C - 23.23°C) = -56.09 kJ

The heat of combustion per gram of B is:

$\frac{-56.09 kJ}{3.767g} =-14.89 kJ/g$

4 0

3 years ago

Zinc reacts with hydrochloric acid according to the reaction equation Zn ( s ) + 2 HCl ( aq ) ⟶ ZnCl 2 ( aq ) + H 2 ( g ) How ma

Katen [24]

Answer: 12.0 milliliters of 6.50 M HCl ( aq ) are required to react with 2.55 g Zn.

Explanation:

moles = $\frac{\text {given mass}}{\text {Molar mass}}$

moles of zinc = $\frac{2.55g}{65.38g/mol}=0.0390moles$

The balanced chemical equation is :

$Zn(s)+2HCl(aq)\rightarrow ZnCl_2(aq)+H_2(g)$

According to stoichiometry:

1 mole of zinc reacts with = 2 moles of HCl

Thus 0.0390 moles of zinc reacts with = $\frac{2}{1}\times 0.0390=0.0780$ moles of HCl

To calculate the volume for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution in ml}}$ .....(1)

Molarity of $HCl$ solution = 6.50 M

Volume of solution = ?

Putting values in equation 1, we get:

$6.50M=\frac{0.0780\times 1000}{\text{Volume of solution in ml}}$

${\text{Volume of solution in ml}}=12.0ml$

Thus 12.0 ml of 6.50 M HCl ( aq ) are required to react with 2.55 g Zn

3 0

3 years ago

Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom. What type of electr

Alecsey [184]

Answer:

The wavelength of the emitted photon will be approximately 655 nm, which corresponds to the visible spectrum.

Explanation:

In order to answer this question, we need to recall Bohr's formula for the energy of each of the orbitals in the hydrogen atom:

$E_{n} = -\frac{m_{e}e^{4}}{2(4\pi\epsilon_{0})^2\hbar^{2}}\frac{1}{n^2} = E_{1}\frac{1}{n^{2}}$ , where:

[tex]m_{e}[tex] = electron mass

e = electron charge

[tex]\epsilon_{0}[tex] = vacuum permittivity

[tex]\hbar[tex] = Planck's constant over 2pi

n = quantum number

[tex]E_{1}[tex] = hydrogen's ground state = -13.6 eV

Therefore, the energy of the emitted photon is given by the difference of the energy in the 3d orbital minus the energy in the 2nd orbital:

[tex]E_{3} - E_{2} = -13.6 eV(\frac{1}{3^{2}} - \frac{1}{2^{2}})=1.89 eV[tex]

Now, knowing the energy of the photon, we can calculate its wavelength using the equation:

[tex]E = \frac{hc}{\lambda}[tex], where:

E = Photon's energy

h = Planck's constant

c = speed of light in vacuum

[tex]\lambda[tex] = wavelength

Solving for [tex]\lambda[tex] and substituting the required values:

[tex]\lambda = \frac{hc}{E} = \frac{1.239 eV\mu m}{1.89 eV}=0.655\mu m = 655 nm[tex], which correspond to the visible spectrum (The visible spectrum includes wavelengths between 400 nm and 750 nm).

5 0

3 years ago

Gustave-Gaspard Coriolanus described this type of energy

Paha777 [63]

I believe it’s Chemical energy but please correct me if i’m wrong

8 0

3 years ago

How many liters is in a gallon

solong [7]

1 Gallon = 3.7854118 Liters

5 0

3 years ago

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