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3 years ago

1 pts

Chemistry

1 answer:

stealth61 [152]3 years ago

3 0

Its true . air particles move faster when heated and collide with each other

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How would you calculate the difference in enthalpy?

Bad White [126]

(Biology)
Equation: (delta)G = (delta)H - T (delta)S
(delta)G stand for change in free energy. When it is positive, it is endergonic. When it is negative, it is exergonic.
H stands for bond energy and is enthalpy
T is the absolute temperature expressed in the Kelvin scale
S is entropy or disorder in the system

7 0

3 years ago

Please help!!! Bob puts several pieces of dry ice inside an empty water bottle and covers the spout with a balloon. After 5 minu

saw5 [17]

The answer would be A.

7 0

3 years ago

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xxTIMURxx [149]

Answer:

Explanation:

5 0

3 years ago

Consider the reaction 2CO(g) + O2(g)2CO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

Marizza181 [45]

<u>Answer:</u> The value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$2CO(g)+O_2(g)\rightarrow 2CO_2(g)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(CO_2(g))})]-[(1\times \Delta S^o_{(O_2(g))})+(2\times \Delta S^o_{(CO(g))})]$

We are given:

$\Delta S^o_{(CO_2(g))}=213.74J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(CO)}=197.674J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (213.74))]-[(1\times (205.14))+(2\times (197.674))]\\\\\Delta S^o_{rxn}=-173.008J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-173.008) J/K = 173.008 J/K

We are given:

Moles of CO gas reacted = 2.25 moles

By Stoichiometry of the reaction:

When 2 moles of CO is reacted, the entropy change of the surrounding will be 173.008 J/K

So, when 2.25 moles of CO is reacted, the entropy change of the surrounding will be = $\frac{173.008}{1}\times 2.25=432.52J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 J/K

3 0

3 years ago

Another chemistry question i’m not good at this at all:( has me stressed

adoni [48]

Answer:

Average atomic mass = 15.86 amu.

Explanation:

Given data:

Number of atoms of Z-16.000 amu = 205

Number of atoms of Z-14.000 amu = 15

Average atomic mass = ?

Solution:

Total number of atoms = 205 + 15 = 220

Percentage of Z-16.000 = 205/220 ×100 = 93.18%

Percentage of Z-14.000 = 15/220 ×100 = 6.82 %

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100

Average atomic mass = (93.18×16.000)+(6.82×14.000) /100

Average atomic mass = 1490.88 + 95.48 / 100

Average atomic mass = 1586.36 / 100

Average atomic mass = 15.86 amu.

5 0

3 years ago