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3 years ago

8.) If 396 g of Carbon Dioxide (CO2) are produced, what mass of Oxygen

Chemistry

1 answer:

KengaRu [80]3 years ago

3 0

Answer:

480 g of oxygen.

Explanation:

C3H8 + 5O2 ---> 3CO2 + 4H2O

Using the molar masses:

3*12 + 6*16 g of CO2 were formed from 10*16 g O2

132g g CO2 from 160 g O2

1g CO2 from (160/132) g O2

396 g from (160/132) * 396

= 480 g of oxygen.

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Add oxidation numbers to the following reaction: 2 H3PO4 (aq) + 2 Cr(s) → 2 CrPO4 (aq) + 3 H2(g). Identify the atom that is oxid

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The atom that is oxidized : Cr

The oxidizing agent : H₃PO₄

<h3>Further explanation</h3>

Reaction

2 H₃PO₄ (aq) + 2Cr(s) → 2 CrPO₄ (aq) + 3H₂(g)

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The atomic mass is greater than the atomic number.

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4 years ago

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When metallic sodium is dissolved in liquid sodium chloride, electrons are released into the liquid. These dissolved electrons a

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Answer:

The edge of the length is $\mathbf{L = 8.54 \times 10^{-10} \ m}$

Explanation:

From the given information:

The associated energy for a particle in three - dimensional box can be expressed as:

$E_n = \dfrac{h^2}{8mL^2}(n_x^2+n_y^2+n_z^2)$

here;

h = planck's constant = $6.626 \times 10^{-34} \ Js$

$n_i$ = the quantum no in a specified direction

m = mass (of particle)

L = length of the box

At the ground state $n_x = n_y = n_z=1$

The energy at the ground state can be calculated by using the formula:

$E_1 =\dfrac{3h^2}{8mL^2}$

At first excited energy level, one of the quantum values will be 2 and the others will be 1.

Thus, the first excited energy will be: 2,1,1

∴

$E_2 =\dfrac{(2^2+1^2+1^2)h^2}{8mL^2}$

$E_2 =\dfrac{(4+1+1)h^2}{8mL^2}$

$E_2 =\dfrac{(6)h^2}{8mL^2}$

The transition energy needed to move from the ground to the excited state is:

$\Delta E= E_2 - E_1$

$\Delta E= \dfrac{6h^2}{8mL^2}- \dfrac{3h^2}{8mL^2}$

$\Delta E= \dfrac{3h^2}{8mL^2}}$ ----- (1)

Recall that:

the wavelength identified with the electronic transition is: 800 nm

800 nm = 8.0 × 10⁻⁷ m

However, the energy-related with the electronic transition is:

$\Delta E =\dfrac{hc}{\lambda}$

$\Delta E =\dfrac{6.626 \times 10^{-34} \times 2.99 \times 10^8}{8.0 \times 10^{-7} }$

$\Delta E =2.48 \times 10^{-19} \ J$

Replacing the value of $\Delta E$ in (1); then:

$2.48 \times 10^{-19}= \dfrac{3h^2}{8mL^2}}$

Making the edge length L the subject of the formula; we have:

$L = \sqrt{\dfrac{3h^2}{8m \times2.48 \times 10^{-19}} }$

$L = \sqrt{\dfrac{3\times (6.626 \times 10^{-34})^2}{8(9.1 \times 10^{-31} ) \times2.48 \times 10^{-19}} }$

$\mathbf{L = 8.54 \times 10^{-10} \ m}$

Thus, the edge of the length is $\mathbf{L = 8.54 \times 10^{-10} \ m}$

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