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3 years ago

How to assign oxidation and reduction numbers

1 answer:

4 0

Answer:Rule 1: The oxidation number of an element in its free (uncombined) state is zero — for example, Al(s) or Zn(s). This is also true for elements found in nature as diatomic (two-atom) elements

Explanation:

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For an organism’s scientific names, the first part is the ________ and the second is the ________.

kari74 [83]

Answer:

Explanation:

For an organism scientific names the first part is the genus and the second is the species

3 0

2 years ago

Read 2 more answers

If the energy of photon emitted from the hydrogen atom is 4.09 x 10-19 J, what is

Aliun [14]

Answer:

486 nm

Explanation:

From the question given above, the following data were obtained:

Energy (E) = 4.09×10¯¹⁹ J

Wavelength (λ) =?

Next, we shall determine the frequency of the photon. This can be obtained as follow:

Energy (E) = 4.09×10¯¹⁹ J

Planck's constant (h) = 6.63×10¯³⁴ Js

Frequency (f) =?

E = hf

4.09×10¯¹⁹ = 6.63×10¯³⁴ × f

Divide both side by 6.63×10¯³⁴

f = 4.09×10¯¹⁹ / 6.63×10¯³⁴

f = 6.17×10¹⁴ Hz

Next, we shall determine the wavelength of the photon. This can be obtained as follow:

Frequency (f) = 6.17×10¹⁴ Hz

Velocity of photon (v) = 3×10⁸ m/s

Wavelength (λ) =?

v = λf

3×10⁸ = λ × 6.17×10¹⁴

Divide both side by 6.17×10¹⁴

λ = 3×10⁸ / 6.17×10¹⁴

λ = 4.86×10¯⁷ m

Finally, we shall convert 4.86×10¯⁷ m to nm. This can be obtained as follow:

1 m = 1×10⁹ nm

Therefore,

4.86×10¯⁷ m = 4.86×10¯⁷ m × 1×10⁹ nm / 1 m

4.86×10¯⁷ m = 486 nm

Therefore, the wavelength of the photon is 486 nm

7 0

3 years ago

How many grams of calcium phosphate (Ca3(PO4)2) re theoretically produced if we start with 3.40 moles of Ca(NO3)2 and 2.40moles

sattari [20]

1) Balance the chemical equation.

$3Ca(NO_3)_2+2Li_3PO_4\rightarrow6LiNO_3+Ca_3(PO_4)_2$

2) List the known and unknown quantities.

Reactant 1: Ca(NO3)2.

Amount of substance: 3.40 mol.

Reactant 2: Li3PO4.

Amount of substance: 2.40 mol.

Product: Ca3(PO4)2

Mass: unknown.

3) Which is the limiting reactant?

3.1-How many moles of Li3PO4 do we need to use all of the Ca(NO3)2?

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

$mol\text{ }Li_3PO_4=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{2\text{ }mol\text{ }Li_3PO_4}{3\text{ }mol\text{ }Ca(NO_3)_2}=2.2667\text{ }mol\text{ }Li_3PO_4$

We need 2.2667 mol Li3PO4 and we have 2.40 mol Li3PO4. We have enough Li3PO4. This is the excess reactant.

3.2-How many moles of Ca(NO3)2 do we need to use all of the Li3PO4?

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

$mol\text{ }Ca(NO_3)_2=2.40\text{ }mol\text{ }Li_3PO_4*\frac{3\text{ }mol\text{ }Ca(NO_3)_2}{2\text{ }mol\text{ }Li_3PO_4}=3.60\text{ }mol\text{ }Ca(NO_3)_2$

We need 3.60 mol Ca(NO3)2 and we have 3.40 mol Ca(NO3)2. We do not have enough Ca(NO3)2. This is the limiting reactant.

4) Moles of Ca3(PO4)2 produced from the limiting reactant.

We have 3.40 mol Ca(NO3)2 of the limiting reactant.

The molar ratio between Ca(NO3)2 and Ca3(PO4)2 is 3 mol Ca(NO3)2: 1 mol Ca3(PO4)2.

$mol\text{ }Ca_3(PO_4)_2=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{1\text{ }mol\text{ }Ca_3(PO_4)_2}{3\text{ }mol\text{ }Ca(NO_3)_2}=1.1313\text{ }mol\text{ }Ca_3(PO_4)_2$

5) Mass of Ca3(PO4)2 produced.

The molar mass of Ca3(PO4)2 is 310.1767 g/mol.

$g\text{ }Ca_3(PO_4)_2=1.1333\text{ }mol\text{ }Ca_3(PO_4)_2*\frac{310.1767\text{ }g\text{ }Ca_3(PO_4)_2}{1\text{ }mol\text{ }Ca_3(PO_4)_2}$ $g\text{ }Ca_3(PO_4)_2=351.526\text{ }g\text{ }Ca_3(PO_4)_2$

The mass of Ca3(PO4)2 produced is 351 g Ca3(PO4)2.

Option D.

8 0

1 year ago

The osmotic pressure exerted by a solution is equal to the molarity multiplied by the absolute temperature and the gas constant

enot [183]

Answer: $C=\frac{\pi}{R\times T}$