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NARA [144]
2 years ago
5

A 283 g silver figure of a polar bear is dropped into the 217 g aluminum cup of a well insulated calorimeter Containing 263 g of

liquid water at 23.7°C period the Bears initial temperature is 96.3°C period what is the final temperature of the water, cup, and beer when they reach thermal equilibrium Librium? The specific heat of silver, aluminum, and liquid water are, respect of Lee, 234J/(KJ times K), 910 J/(KJ times K), and 4190J/(kg*K)
Chemistry
1 answer:
notka56 [123]2 years ago
3 0

Answer:

I'm Not Sure Sorry

Explanation:

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During photosynthesis,blank energy is transformed into blank energy
11Alexandr11 [23.1K]

Answer:

I am also beginners I don't know I am only completing this answer and sorry!!!!!!

8 0
2 years ago
Read 2 more answers
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

5 0
3 years ago
Which molecule has a dipole moment of 0 d?
fiasKO [112]
The answer is 

a symmetrical molecule.


5 0
3 years ago
Part 1: Name the type of chemical reaction that occurs when calcium hydroxide (Ca(OH)2) reacts with nitric acid (HNO3).
liberstina [14]

<u>Answer:</u>

<u>For 1:</u> Neutralization reaction

<u>For 2: </u>Zinc is more reactive than lead and less reactive than calcium.

<u>Explanation:</u>

  • <u>For (1):</u>

When a base reacts with an acid to form a salt and water molecule, it is known as a neutralization reaction. The general equation follows:

HX+BOH\rightarrow BX+H_2O

The chemical equation for the reaction of calcium hydroxide and nitric acid follows:

Ca(OH)_2(aq)+2HNO_3(aq)\rightarrow Ca(NO_3)_2(aq)+2H_2O(l)

  • <u>For (2):</u>

A single displacement reaction is defined as the reaction in which a more reactive metal displaces a less reactive metal from its salt solution. The general chemical equation follows:

A+BX\rightarrow AX+B

where,

Metal A is more reactive than metal B

The reactivity of metals is judged by the reactivity series where a metal lying above in the series is more reactive than the metal lying below it.

From the reactivity series below,

Zinc lies above in the series than lead thus is more reactive and will easily replace lead from its aqueous solution.

While zinc lies below in the series than calcium thus is less reactive and will not easily replace calcium from its aqueous solution.

Zn(s)+Pb(NO_3)_2(aq)\rightarrow Zn(NO_3)_2(aq)+Pb(s)

Zn(s)+CaCl_2(aq)\rightarrow \text{No reaction}

5 0
3 years ago
To what characteristics of the group 1 metallic elements can the ease with which their sons form +1 monatomic ions be attributed
ArbitrLikvidat [17]
Group 1 elements (usually called alkali metals) are not very electronegative and have small ionization energies due to that.  The reason why they are not very electronegative is that they really want to loose their one valence electron so that they can have a noble gas electron configuration (completed octet).

I hope this helps.
3 0
3 years ago
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