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2 years ago

What determines the temperature of an object at the molecular level?

1 answer:

7 0

At the molecular level, temperature is related to the random motions of the particles (atoms and molecules) in matter. Because there are different types of motion, the particles' kinetic energy (energy of motion) can take different forms, and each form contributes to the total kinetic energy of the particles.

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Which is NOT a type of fracture?

Andrews [41]

B. Perfect ---------

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2 years ago

What heat is liberated in the formation of 10.0 grams of sulfur hexafluoride, SF6, from the elements sulfur and fluorine?

tensa zangetsu [6.8K]

Answer:

$Q=-76.7kJ$

Explanation:

Hello,

In this case, for such formation of sulfur hexafluoride, the standard enthalpy of formation is -1220.47 kJ/mol (data extracted from NIST database). Next, we compute the moles in 10.0 grams of sulfur hexafluoride as shown below:

$n_{SF_6}=10.0gSF_6*\frac{1molSF_6}{146.06 gSF_6}=0.0685mol$

Next, for the given energy, we compute the total heat that is liberated:

$Q=-1220.47\frac{kJ}{mol}*0.0685 mol\\\\Q=-76.7kJ$

Finally, we conclude such symbol has sense since negative heat is related with liberated heat.

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2 years ago

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Olenka [21]

Answer: What are you german

Explanation:

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2 years ago

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Write equation to represent the Hydrogenation reactions

garik1379 [7]

Explanation:

vegetable oils have long unsaturated carbon chains which can converted into vegetable ghee

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2 years ago

What is the composition, in atom percent, of an alloy that consists of a) 5.5 wt% Pb and b) 94.5 wt% of Sn? Assume that the atom

Anastaziya [24]

Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

Explanation :

First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

As, 207.2 g of lead contains $6.022\times 10^{23}$ atoms

So, 5.5 g of lead contains $\frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22}$ atoms

and,

As, 118.71 g of lead contains $6.022\times 10^{23}$ atoms

So, 94.5 g of lead contains $\frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23}$ atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

$\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%$

and,

$\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%$

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

6 0

2 years ago