For the balanced equation shown below how many grams of O2 reacted if 2420 of p4O10 are produced

12. Draw an energy curve for a chemical reaction that begins at 30 kj and has an activation energy of

Rufina [12.5K]

Answer:

<em>Since the activation energy is $50kJ$ , we need to supply it $50kJ$ from $30kJ$ to make it $80kJ$ .</em>

Diagram attached in the file .

8 0

3 years ago

100 points! Report on an element

Luden [163]

Hydrogen H weight: 81

Non-metal

Hydrogen is the simplest element; an atom consists of only one proton and one electron. It is also the most plentiful element in the universe. Despite its simplicity and abundance, hydrogen doesn't occur naturally as a gas on the Earth--it is always combined with other elements.

period 1 group 1

Hydrogen is easily the most abundant element in the universe. It is found in the sun and most of the stars, and the planet Jupiter is composed mostly of hydrogen. On Earth, hydrogen is found in the greatest quantities as water.

6 0

2 years ago

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Express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms

nirvana33 [79]

Answer:

$PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]$

B = b -a/RT

C = b^2

a = 1.263 atm*L^2/mol^2

b = 0.03464 L/mol

Explanation:

In the given question, we need to express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. Therefore:

Using the van deer Waals equation of state:

$P = \frac{RT}{V_{m}-b } - \frac{a}{V_{m} ^{2} }$

With further simplification, we have:

$P = RT[\frac{1}{V_{m}-b } - \frac{a}{RTV_{m} ^{2} }]$

Then, we have:

$P = \frac{RT}{V_{m} } [\frac{1}{1-\frac{b}{V_{m} } } - \frac{a}{RTV_{m} }]$

Therefore,

$PV_{m} = RT[(1-\frac{b}{V_{m} }) ^{-1} - \frac{a}{RTV_{m} }]$

Using the expansion:

$(1-x)^{-1} = 1 + x + x^{2} + ....$

Therefore,

$PV_{m} = RT[1+\frac{b}{V_{m} }+\frac{b^{2} }{V_{m} ^{2} } + ... -\frac{a}{RTV_{m} }]$

Thus:

$PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]$ equation (1)

Using the virial equation of state:

$P = RT[\frac{1}{V_{m} }+ \frac{B}{V_{m} ^{2}}+\frac{C}{V_{m} ^{3} }+ ...]$

Thus:

$PV_{m} = RT[1+ \frac{B}{V_{m} }+ \frac{C}{V_{m} ^{2} } + ...]$ equation (2)

Comparing equations (1) and (2), we have:

B = b -a/RT

C = b^2

Using the measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coefficients at 273 K.

$b = \sqrt{C} = \sqrt{1200} = 34.64[tex]cm^{3}/mol$ [/tex] = 0.03464 L/mol

a = (b-B)*RT = (34.64+21.7)*(1L/1000cm^3)*(0.0821)*(273) = 1.263 atm*L^2/mol^2

3 0

2 years ago

What causes the difference in bond angles in carbon dioxide and water?

egoroff_w [7]

Answer is: C) the fact that the number of lone pairs of electrons on the central atom is greater in the case of water.

Carbon(IV) oxide is nonpolar because CO₂ is linear molecule and the oxygen atoms are symmetrical (bond angles 180°).

Water is polar because of the bent shape of the molecule.

Oxygen atom in water molecule has sp3 hybridization. The bond angle between the two hydrogen atoms is approximately 104.45°.

Oxygen atom has atomic number 8, it means it has eight protons and eight electrons, so atom has neutral charge. Oxygen is a nonmetal.

Electron configuration of oxygen atom: ₈O 1s² 2s² 2p⁴.

Oxygen atom has six valence electrons , two lone pairs and two electrons that form two sigma bonds with hydrogen atoms.

Carbon is a chemical element with symbol C and atomic number 6, which means it has 6 protons and six electrons. Four valence electrons are in 2s and 2p orbitals.

Electron configuration of carbon atom: ₆C 1s² 2s² 2p².

In carbon dioxide, carban has sp hybridization with no lone pairs.

5 0

2 years ago

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4 and 5 pls and thank uuu

PtichkaEL [24]

Answer:

4. -ol

5. cyclic ketone

Explanation:

biology stuff

sorry it's hard to explain

5 0

1 year ago