Find the oxidation number of each sulfur in the molecule H2S4, which has a linear arrangement of its atoms.

Answers

Answer 1
Answer: In the molecule H2S4, each sulfur atom has an oxidation number of -2.

Related Questions

What is scientific method?
When fossil fuels are burned, _____ is released into the atmosphere.A. oxygen B. nitrogen C. ammonia D. carbon dioxide
Which organic compound is most soluble in water? A) ethyne B) benzene C) butane D) ethanol
SELECT ALL THAT APPLYat STP conditions, 0.25 moles of CO2(g), H2(g), NH3(g)1. Will contain the same number of molecules 2. Will contain the same number of atoms3. Will occupy the same volume4. Will have the same g.f.w
Which compound is an unsaturated hydrocarbon

Which factors are needed for organisms to live earth

Answers

Answer: sunlight, water, air, habitat, and food.

Explanation: we are all living organisms and we all have our five basic necessities for survival; sunlight, water, air, habitat, and food.

What is a chemically active atom?

Answers

Is a stability relative to the elements surrounding it on the periodic table

What do hydrogen ions and hydroxide ions come together to form?a. a base
b. water
c.a tritant
d. an acid

Answers

They come together to form water. I hope this helps. Let me know if anything is unclear. That is the bases of a neutralization reaction.

Carbon-11 is used in medical imaging. The half-life of this radioisotope is 20.4 min. What percentage of a sample remains after 60.0 min?34.013.05.2871.22.94

Answers

Answer:

Percentage of a sample remains after 60.0 min is 13.03%.

Explanation:

  • It is known that the decay of isotopes of C-11 obeys first order kinetics.
  • Also, it is clear that in first order decay the half-life time is independent of the initial concentration.
  • Half-life time (t1/2) in first order reaction = 0.693/k, where k is the rate constant.

k = 0.693/(t1/2) = 0.693/(20.4 min) = 0.03397 min⁻¹.

  • The integrated law for first order reaction is:

kt = ln[A₀]/[A],

where, k is the rate constant (k = 0.03397 min⁻¹).

t is the time of the reaction (t = 60.0 min).

[A₀] is the initial concentration of C-11 ([A₀] = 100.0 %).

[A] is the remaining concentration of C-11 ([A] = ???%).

∵ kt = ln[A₀]/[A]

∴ (0.03397 min⁻¹)(60.0 min) = ln(100%)/[A]

∴ 2.038 = ln(100%)/[A]

  • Taking e for both sides:

∴ 7.677 = (100%)/[A]

∴ [A] = (100%)/(7.677) = 13.03%.

So, percentage of a sample remains after 60.0 min is 13.03%.

Final answer:

The half-life of Carbon-11 is 20.4 minutes, meaning the quantity reduces by half every 20.4 minutes. So, after approximately 60.0 minutes (roughly three half-lives), approximately 12.5% of the original sample would remain.

Explanation:

Carbon-11 is a radioisotope that's widely used in medical imaging due to its radioactive properties. It has a half-life of 20.4 minutes, which means that the quantity of Carbon-11 reduces to half in every 20.4 minutes.

If we denote one half-life as t, after three half-lives (60.0 min or 3t), the ratio of the remaining sample of Carbon-11 would be 1/2 * 1/2 * 1/2 = 1/8. In other words, approximately 12.5% of the original sample would be remaining. This percentage changes a bit due to the fact that 20.4 min is slightly less than 22.2 min (which would be the exact 3 half lives).

Learn more about Carbon-11 Half-Life here:

brainly.com/question/29112174

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10. How many g of Cu(OH)2 can be made from 9.1 x 1025 atoms of O?

Answers

Molar mass Cu(OH)₂ = 97.561 g/mol

97.561 g Cu(OH)₂ --------------- 6.02x10²³ atoms
  ? g Cu(OH)₂ -------------------- 9.1x10²⁵ atoms

mass = 9.1x10²⁵ * 97.561 / 6.02x10²³

mass = 8.87x10²⁷ / 6.02x10²³

mass = 14734.2 g

hope this helps!

When the equation H2S + HNO3 → S + NO + H2O is balanced, what is the coefficient for H2O?

Answers

It isn't balanced. You have 3 *H and 3 * O, so something in the first formula must be changed. Have been searching for 1 hour and can't find the answer.

the answer is not 3 but i am guessing 6 please let me know