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77julia77 [94]
2 years ago
10

What is the sum of the coefficients when the equation is balanced with the smallest whole numbers? __BaCl2 + __Fe2(SO4)3 → __FeC

l3 + __BaSO4 *?
Chemistry
1 answer:
aliina [53]2 years ago
8 0

Explanation:

Hphphphphphphphohoohohhpph

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What state of matter is the body mostly made up of
Alinara [238K]
Liquid makes up most of the body
5 0
3 years ago
Which one of the following would you expect to dissolve in water? limonene sucrose oil atmospheric oxygen none of the above
sdas [7]

-----------sucrose----------------

3 0
3 years ago
What is the theoretical yield of aluminum oxide if 3.00 mol of aluminum metal is exposed to 2.55 mol of oxygen?
lisabon 2012 [21]

Theoretical yield of Al₂O₃: 1.50 mol.

<h3>Explanation</h3>

2 \; \text{Al} + \dfrac{3}{2} \; \text{O}_2 \to {\bf 1} \; \text{Al}_2\text{O}_3;

4 \; \text{Al} + 3 \; \text{O}_2 \to 2 \; \text{Al}_2\text{O}_3 \; \textit{Balanced}.

How many moles of aluminum oxide formula units will be produced <em>if</em> aluminum is the limiting reactant?

Aluminum reacts to aluminum oxide at a two-to-one ratio.

3.00 \times \dfrac{1}{2} = 1.50 \; \text{mol}.

As a result, 3.00 moles of aluminum will give rise to 1.50 moles of aluminum oxide.

How many moles of aluminum oxide formula units will be produced <em>if</em> oxygen is the limiting reactant?

Oxygen reacts to produce aluminum oxide at a three-to-two ratio.

2.55 \times \dfrac{2}{3} = 1.70 \; \text{mol}

As a result, 2.55 moles of oxygen will give rise to 1.70 moles of aluminum oxide.

How many moles of aluminum oxide formula units will be produced?

Aluminum is the limiting reactant. Only 1.50 moles of aluminum oxide formula units will be produced. 1.70 moles isn't feasible since aluminum would run out by the time 1.50 moles was produced.

4 0
3 years ago
Mercury’s atomic emission spectrum is shown below. Estimate the wavelength of the orange line. What is its frequency? What is th
lions [1.4K]

The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

<em>"Your question is not complete, it seems to be missing the diagram of the emission spectrum"</em>

the diagram of the emission spectrum has been added.

<em>From the given</em><em> chart;</em>

The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m

The frequency of this emission is calculated as follows;

c = fλ

where;

  • <em>c is the speed of light = 3 x 10⁸ m/s</em>
  • <em>f is the frequency of the wave</em>
  • <em>λ is the wavelength</em>

f = \frac{c}{\lambda } \\\\f = \frac{3\times 10^8}{610 \times 10^{-9}} \\\\f = 4.92 \times 10^{14} \ Hz

The energy of the emitted photon corresponding to the orange line is calculated as follows;

E = hf

where;

  • <em>h is Planck's constant = 6.626 x 10⁻³⁴ Js</em>

<em />

E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)

E = 3.26 x 10⁻¹⁹ J.

Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

Learn more here:brainly.com/question/15962928

6 0
2 years ago
Why do electrons enter the 4s orbital before entering the 3D orbital
garri49 [273]

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. (:

8 0
3 years ago
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