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Nina [5.8K]
3 years ago
9

Read the following chemical equations. Reaction 1: H2S + Cl2 → 2HCl + S Reaction 2: Fe + S → FeS Which of the following statemen

ts is true for both the chemical equations? A. Chlorine is reduced in reaction 1 and iron is reduced in reaction 2. B. Chlorine is oxidized in reaction 1 and iron is oxidized in reaction 2. C. Chlorine is oxidized in reaction 1 and iron is reduced in reaction 2. D. Chlorine is reduced in reaction 1 and iron is oxidized in reaction 2.
Chemistry
1 answer:
DIA [1.3K]3 years ago
8 0

Answer:

D. Chlorine is reduced in reaction 1 and iron is oxidized in reaction 2.

Explanation:

  • Reduction is the process where by the Oxidation number or state of an element or substance is reduce.
  • Oxidation is the increase in oxidation number or state of any element or substance
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The protein lysozyme unfolds at a transition temperature of 75.5°C, and the standard enthalpy of transition is 509 kJ mol-1. Cal
spin [16.1K]

Answer:

0.4774 KJ/K.mol

Explanation:

We are told that the transition at 25.0°C occurs in three steps. Steps i, ii and iii.

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii

Now,

C_p,m(unfolded protein) = C_p,m(folded protein) + 6.28 kJ/K.mol

Now, for the first process, ΔS_i is given as;

ΔS_i = C_p,m × In(T2/T1)

We are given;

T1 = 25°C = 25 + 273.15K = 298.15 K

T2 = 75.5°C = 75.5 + 273.15 K=348.65 K

Thus;

ΔS_i = C_p,m × In(348.65/298.15)

Now, for the third process, ΔS_iii is given as;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(T1/T2)

Thus;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)

Now, we don't know C_pm. So, we have to find a way to eliminate it. We will do it by rewriting In(298.15/348.65) in such a way that when ΔS_iii is added to ΔS_i, C_p,m will cancel out. Thus;

In(298.15/348.65) can also be written as;

In(348.65/298.15)^(-1) or

- In(348.65/298.15)

Thus;

ΔS_iii = - [(C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)]

Now, let's add ΔS_iii to ΔS_i to get;

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] + [(-C_p,m - 6.28 kJ/K.mol) × In(348.65/298.15)]

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] - [C_p,m × In(348.65/298.15)] - [6.28In(348.65/298.15)]

First 2 terms will cancel out to give;

ΔS_i + ΔS_iii = -6.28In(348.65/298.15)

ΔS_i + ΔS_iii = -0.9826 KJ/K.mol

Now,for process ii;

ΔS_ii = standard enthalpy of transition/Transition Temperature

Thus;

ΔS_ii = (509 KJ/K.mol)/348.65

ΔS_ii = 1.46 KJ/K.mol

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii = -0.9826 + 1.46 = 0.4774 KJ/K.mol

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3 years ago
Maple syrup, which comes from the sap of maple trees, is a mixture of water and natural sugars. It's a clear, brown liquid. It's
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C.) It's a Solution ..............
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Metals react with oxygen to give oxides with the general formula MxOy. What is a balanced chemical equation for the reaction of
katovenus [111]

Answer: 4Fe+3O_2\rightarrow 2Fe_2O_3

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.  

Here iron is having an oxidation state of +3 called as Fe^{3+} cation and oxide O^{2-} is an anion with oxidation state of -2. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral Fe_2O_3.

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

2Fe+3O_2\rightarrow 2Fe_2O_3

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Answer:

Yes, Pb3(PO4)2.

Explanation:

Hello there!

In this case, according to the given balanced chemical reaction, it is possible to use the attached solubility series, it is possible to see that NaNO3 is soluble for the Na^+ and NO3^- ions intercept but insoluble for the Pb^3+ and PO4^2- when intercepting these two. In such a way, we infer that such reaction forms a precipitate of Pb3(PO4)2, lead (II) phosphate.

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