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Luda [366]
3 years ago
5

For the following reaction, 2 SO2(g) + O:(g) 2 SO,(g) the equilibrium constant, K, is 1.32 at 627°C. What is the equilibrium con

stant for the reaction below? SO;(g)+ 1/2 O:(g) SO,(g)
Chemistry
1 answer:
cupoosta [38]3 years ago
8 0

Answer:

1.15  

Explanation:

2SO₂ + O₂ ⟶ 2SO₃; K =1.32

SO₂ + ½O₂ ⟶ SO₃;    K₁ = ?

When you divide an equation by 2, you take the square root of its equilibrium constant.

K₁ = √1.32 = 1.15

The equilibrium constant is 1.15.

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The mass spectrum of an unknown compound has a molecular ion peak with a relative intensity of 57.10% and an M+1 peak of 6.83%.
max2010maxim [7]

There are 11 Carbon atoms in the compound.

<u>Solution:</u>

Carbon atom count is the ratio of the M peak to the M+1 peak.

\text{ Number of Carbon atoms }=\frac{\text { Relative intensity of } M+1 \text { peak }}{0.011 \times \text { Relative tntensity of } M \text { peak }}

Here M peak is 57.10% and M+1 peak is 6.83%. On applying the values in the formula we get,

\frac{0.0683}{0.011\times0.571g} = 10.87\approx11

Therefore, the number of Carbon atoms in the compound are 11.

Refer the image attached below for a better understanding of M peak and M+1 peak.

The heaviest ion that has the greatest m/z value is said to be the molecular ion peak in mass spectrum.

6 0
3 years ago
Which two formulas represent compounds
SVEN [57.7K]

CO2 ; H20- They are the only ones that, on both sides, combined with another element and bonding of atoms

4 0
3 years ago
Choose the isotope that would be most stable.<br> carbon-14<br> barium-56<br> lead-82<br> radon-85
shutvik [7]
The most stable isotope would be lead-82.
4 0
3 years ago
What is the mass percentage for copper ?
RideAnS [48]

Answer:

Explanation:

To find the mass percent composition of an element, divide the mass contribution of the element by the total molecular mass. This number must then be multiplied by 100% to be expressed as a percent.

3 0
2 years ago
A solution may contain Ag+, Pb2+, and/or Hg22+. A white precipitate forms when 6 M HCl is added. The precipitate is partially so
omeli [17]

Answer:

All three are present

Explanation:

Addition of 6 M HCl would form precipitates of all the three cations, since the chlorides of these cations are insoluble: AgCl (s), PbCl_2 (s), Hg_2Cl_2 (s).

  • Firstly, the solid produced is partially soluble in hot water. Remember that out of all the three solids, lead(II) choride is the most soluble. It would easily completely dissolve in hot water. This is how we separate it from the remaining precipitate. Therefore, we know that we have lead(II) cations present, as the two remaining chlorides are insoluble even at high temperatures.
  • Secondly, addition of liquid ammonia would form a precipitate with silver: AgCl (s) + 2 NH_3 (aq) + H_2O (l)\rightarrow [Ag(NH_3)_2]OH (s) + HCl (aq); Silver hydroxide at higher temperatures decomposes into black silver oxide: 2 [Ag(NH_3)_2]OH (s)\rightarrow Ag_2O (s) + H_2O (l) + 4 NH_3 (g).
  • Thirdly, we also know we have Hg_2^{2+} in the mixture, since addition of potassium chromate produces a yellow precipitate: Hg_2^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow Hg_2CrO_4 (s). The latter precipitate is yellow.
3 0
3 years ago
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