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PSYCHO15rus [73]
3 years ago
15

A 17g sample of H2O2 was decomposed to yield 1g of H2 and 16g of O2 An unknown sample containing only H and O was decomposed to

yeild 2g of H2 and 16 g of O2. The formula for the unknown sample is:
Chemistry
1 answer:
ludmilkaskok [199]3 years ago
5 0

Answer:

Explanation:

Our H_2O_2 sample yielded 1g of H_2 and 16g of O_2, but our unknown sample yielded 2 times as much H_2  for the same amount of O_2.

What does this mean? that the H:O proportion for the unknown sample is twice the H:O proportion for the H_2O_2 sample.

What is the  H:O proportion for the H_2O_2 sample? As we can see from its formula, it's 1:1, therefore the proportion for the unknown formula must be 2:1.

That means, two H atoms for every O atom. We could write that as: H_2O and you should recognize that formula, for it is one of the most common compounds on earth, Water.

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Ethanol (c2h5oh) melts at -114°c. the enthalpy of fusion is 5.02 kj/mol. the specific heats of solid and liquid ethanol are 0.9
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You'll want to add three amounts of heat. 

(1) Specific heat of lowering the temperature from -135°C to the melting point -114°C
(2) Latent heat of fusion/melting
(3) Specific heat of elevating the temperature from -114°C to -50°C

(1) E = mCΔT = (25 g)(0.97 J/g·°C)(1 kJ/1000 J)(-114 - -135) = 0.509 kJ
(2) E = mΔH = (25 g)(5.02 kJ/mol)(1 mol/46.07 g ethanol) = 2.724 kJ
(3) E = mCΔT = (25 g)(2.3 J/g·°C)(1 kJ/1000 J)(-50 - -114) = 3.68 kJ

<em>Summing up all energies, the answer is 6.913 kJ.</em>
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This type of chemist understands the structure of living systems and, in turn, their
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Scientists that study biochemistry are called Biochemists.

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3 0
2 years ago
51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b
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Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

a = \sqrt{8} r

a = \sqrt{8} \times 144 pm

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

density = \dfrac{mass}{volume}

density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}

density d = 19.41 g/cm³

Similarly; For a  body-centered cubic structure

r = \dfrac{\sqrt{3}}{4}a

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r = 144

144 = \dfrac{\sqrt{3}}{4}a

a = \dfrac{144 \times 4}{\sqrt{3}}

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V  3.68 × 10⁻²³   cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

density = \dfrac{mass}{volume}

density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

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