Lewisite (2-chloroethenyldichloroarsine) was once manufactured as a chemical weapon, acting as a lung irritant and a blistering agent. During World War II, British biochemists developed an antidote which came to be known as British anti-Lewisite (BAL) (2,3-disulfanylpropan-1-ol). Today, BAL is used medically to treat toxic metal poisoning. Complete the reaction between Lewisite and BAL by giving the structure of the organic product and indicating the coefficient for the number of moles of HCl produced in the reaction.

Answers

Answer 1
Answer:

Answer:

2 HCl

Explanation:

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(NH4)2S(aq)+SrCl2(aq)→Express your answer as a chemical equation. Enter NOREACTION if no reaction occurs. Identify all of the phases in your answer.

Answers

The chemical equation will be;  

(NH4)2S(aq)+SrCl2(aq)→ 2 NH4Cl(aq) + SrSO4(s)

Further Explanation  

Chemical equation  

  • A chemical equation is an equation showing chemical symbols of reactants and those of products. They represent a chemical reactions between reactants to form products.
  • For example; (NH4)2S(aq)+SrCl2(aq)→ 2 NH4Cl(aq) + SrSO4(s), where (NH4)2S and SrCl2 are reactants while NH4Cl and SrSO4 are products.

Types of chemical reactions  

Precipitation reaction

  • Precipitation reactions are reactions which involves the formation of a precipitate as one of the products. A precipitate is a compound that is insoluble in water.
  • An example of a precipitation reaction is; (NH4)2S(aq)+SrCl2(aq)→ 2 NH4Cl(aq) + SrSO4(s), where the compound SrSO4 is the precipitate.  

Displacement reaction

  • Displacement reactions are reactions where ions replace other ions in their compounds.
  • For example; ; (NH4)2S(aq)+SrCl2(aq)→ 2 NH4Cl(aq) + SrSO4(s) is an example of a double displacement reaction where NH4+ takes the place of Sr ions in SrCl2 and Sr2+ takes the place of NH4+ in (NH4)2SO4.

Decomposition reaction  

  • Decomposition reactions are reactions which involves break down of a compound to its constituent’s elements or other compounds by use of a catalyst or heat.
  • For example; Decomposition of lead (II) nitrate using heat to get lead (ii) oxide, oxygen and nitrogen (IV) oxide.

Neutralization reaction  

  • Neutralization reactions are reactions that involve reacting a base or an alkali and an acid to form a salt and water as the only product.

Redox reactions

  • Redox reactions are reactions that involve both reduction and oxidation. Some species in reactions undergo reduction while others undergo oxidation.  

Keywords: Chemical reactions, precipitation reactions, chemical equations

Learn more about:

Level: High school  

Subject: Chemistry  

Topic: Chemical reactions  

Sub-topic: Precipitation reactions  

Final answer:

No reaction is expected when (NH4)2S(aq) and SrCl2(aq) are mixed, as solubility rules suggest no insoluble salts will form, leading to NOREACTION.

Explanation:

When (NH4)2S(aq) and SrCl2(aq) are mixed together, we expect a reaction where the cations (NH4+ and Sr2+) and anions (S2- and Cl-) exchange partners if any of them can form an insoluble salt. Looking at solubility rules, we know that most sulfides are insoluble except those of alkali metals and ammonium, and most chlorides are soluble except for Ag+, Pb2+, and Hg22+. Given that neither NH4+ nor Sr2+ forms an insoluble chloride and SrS is not listed as an insoluble sulfide, we can predict that no visible reaction will occur when these solutions are mixed. Therefore, the chemical equation to represent this mixture is NOREACTION.

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How do you find relative atomic mass

Answers

Final answer:

To find the relative atomic mass of an element, you need to consider the masses of its isotopes and their relative abundance.


Explanation:

To find the relative atomic mass of an element, you need to consider the masses of its isotopes and their relative abundance. The relative atomic mass is the weighted average of the masses of all the isotopes of the element. The formula for calculating relative atomic mass is:

Relative Atomic Mass = (Mass of Isotope1 * Abundance of Isotope1) + (Mass of Isotope2 * Abundance of Isotope2) + ...

For example, let's calculate the relative atomic mass of carbon, which has two isotopes: carbon-12 and carbon-13. The mass of carbon-12 is 12 amu and its abundance is about 98.9%. The mass of carbon-13 is 13.003 amu and its abundance is about 1.1%. We can use the formula:

(12 amu * 0.989) + (13.003 amu * 0.011) = 12.011 amu


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Under which of the following conditions of temperature and pressure will H2 gas be expected to behave most like an ideal gas? (1 Point) 50 K and 0.10 atm 50 K and 5.0 atm 500 K and 0.10 atm 500 K and 50 atm

Answers

Answer: 500K and 0.10atm

Explanation:

An important concept to remember is that gases behave most ideally under low pressure and high temperature. 500 K is a high temperature and 0.10 atm is a low pressure, which makes that the best answer.

Final answer:

The conditions under which H2 gas would behave most like an ideal gas are at a high temperature of 500 K and a low pressure of 0.10 atm. These are optimally suitable for a gas to behave ideally as per the ideal gas law.

Explanation:

Under the conditions of both temperature and pressure given in the question, H2 gas would behave most like an ideal gas at 500 K and 0.10 atm. The ideal gas law, which describes the relationship between the pressure, volume, and temperature of a gas, suggests that a gas behaves most ideally at low pressure and high temperature.

This is because at low pressures, the volume of individual gas molecules relative to the total volume of gas becomes negligible, and intermolecular forces become weak. Meanwhile, at high temperatures, the kinetic energy of the gas molecules becomes high enough to overpower any intermolecular forces of attraction. Hence, the gas behaves more ideally.

A good way to visualize this is to think of an ideal gas as perfectly 'free' – the particles move in straight lines until they hit the edge of their container, and they do not attract or repel each other. The closer we get to this scenario, the more 'ideal' the gas would behave.

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In living organisms, C-14 atoms disintegrate at a rate of 15.3 atoms per minute per gram of carbon. A charcoal sample from an archaeological site has a C-14 disintegration rate of 9.16 atoms per minute per gram of carbon. Estimate the age of this sample in years. The half-life of C-14 is 5730 years. (enter only the number of years in standard notation, not the unit years)

Answers

Rate of disintegration is defined as the time required by a sample or substance at which half of the radioactive substance disintegrates. It depends on the nature of disintegration and amount of substance.

The age of the sample is approximately 4241.17 years.

Given that:

C-14 atoms disintegration rates = 15.3 atom/ min-g

Rate of disintegration of the sample = 9.16 atom/ min-g

The digit proportion of carbon-14 is  = (9.16)/(15.3) = 0.5987

Now, also the half-life of carbon-14 is 5730 years.

Such that:

(1)/(2)^n = \text A

(1)/(2) = 0.5987

Taking log:

n log 2 = -log 0.5987

Thus, n = (0.227)/(0.3010)

n = 0.740

The age of the sample can be given by:

Age = n x half-life

Age = 0.740 x 5730

Age = 4241.17 years.

Therefore, the age of the substance is 4241.17 years.

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Answer:

The answer is "4,241 .17 years"

Explanation:

The disintegration rate, which shows in C-14 atoms =  15.3 (atoms)/(min-g)

Rate of sample disintegration =9.16 \frac{atoms} {min-gram}

The digit proportion of C-14 can be determined that is included in the sample = \frac {9.16}{15.3} \n\n = 0.5987

5730 years from half-life.  

The number with half-lives (n) which are repelled must be determined:

((1)/(2))^n= A\n\nA=  fraction of C-14, which is remaining \n\n((1)/(2))^n= 0.5987 \n\n n \log 2 = - \log 0.5987\n\n

\therefore \n\n \Rightarrow n= (0.227)/(0.3010) \n\n  = 0.740\n

So, the age of the sample is given by = n *\ half-life

                                                 = 0.740 * 5730 \ years \n\n=4241.17 \ years\n\n

Be sure to answer all parts. Consider both 5-methyl-1,3-cyclopentadiene (A) and 7-methyl-1,3,5-cycloheptatriene (B). Which labeled H atom is most acidic? Hb is most acidic because its conjugate base is aromatic. Hc is most acidic because its conjugate base is antiaromatic. Ha is most acidic because its conjugate base is antiaromatic. Hd is most acidic because its conjugate base is aromatic. Which labeled H atom is least acidic? Ha is least acidic because its conjugate base is aromatic. Hb is least acidic because its conjugate base is antiaromatic. Hd is least acidic because its conjugate base is aromatic. Hc is least acidic because its conjugate base is antiaromatic.

Answers

Due to the conjugate base of the hydrogen atom is aromatic, Hb is regarded as the most acidic. Because the conjugate base of the hydrogen atom Hc is anti-aromatic, it is the least acidic.

The correct options are:

(A) - (a)

(B) - (d)

What are the most and the least acidic hydrogen atom?

The hydrogen connected at the heptatriene's tertiary position (at the 7-methyl) would be particularly acidic, as its removal would leave a positive charge that could be transported around the ring via resonance.

The hydrogen connected to the pentadiene (5-methyl) at the tertiary position would not be acidic, as removing it would result in an anti-aromatic structure.

Thus, the least acidic H atom is Hc and the most acidic H atom is Hb.

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I don’t have a picture but I can describe it to you.

The hydrogen that is attached at the tertiary position on the heptatriene (at the 7-methyl) would be very acidic, as removal would leave a positive charge that could be moved throughout the ring through resonance. This would mean that the three double bonds would be participating in resonance, and the deprotonated structure would be aromatic, thus making this favorable.

The hydrogen that is attached at the tertiary position on the pentadiene (5-methyl) would NOT be acidic, as removal would cause an antiaromatic structure.

Any other hydrogens would NOT be acidic. Those vinylic to their respective double bonds would seriously destabilize the double bond if removed, and hydrogens attached to the methyl group jutting off the ring have no incentive to leave the carbon.

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1‑propanol ( nn ‑propanol) and 2‑propanol (isopropanol) form ideal solutions in all proportions. Calculate the partial pressure and the mole fraction ( yy ) of the vapor phase of each component in equilibrium with each of the given solutions at 25 °C. P∘prop=20.9 TorrPprop°=20.9 Torr and P∘iso=45.2 TorrPiso°=45.2 Torr at 25 °C. A solution with a mole fraction of xprop=0.243xprop=0.243 .

Answers

Answer:

Piso = 32.17 Torr

Pprop = 5.079 Torr

yprop = 0.1364

yiso = 0.8636

Explanation:

From the question; we can opine that :

  • The mole fraction of isopropanol in a mixture of isopropanol and propanol will be 1.
  • The partial pressure of isopropanol will be its mole fraction multiplied by vapor pressure of isopropanol
  • The partial pressure of propanol will be its mole fraction multiplied by vapor pressure of propanol
  • In the vapor, the mole fraction of propanol will be its partial pressure divided by the sum of the two partial pressures      

NOW;

When xprop = 0.243; xisopropanol will be 1- 0.243 = 0.757

P°iso = 45.2 Torr at 25 °C so

Piso will be 45.2 × 0.757 = 32.17 Torr

Pprop will be 20.9 × 0.243 = 5.079 Torr

yprop = 5.079/(5.079 +32.17) = 0.1364

yiso = 1-0.1364 = 0.8636