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anastassius [24]
4 years ago
8

The formulas for ethane, ethene, and ethyne are C 2 H 6 C2H6 , C 2 H 4 C2H4 , and C 2 H 2 C2H2 , respectively. Rank these compou

nds by the length of the carbon–carbon bond.
Chemistry
1 answer:
Anuta_ua [19.1K]4 years ago
6 0

Answer:

Ethane (1.09 A) > Ethene(1.076 A) > Ethyne(1.06 A)

Explanation:

The types of bonds present are:

a) Ethane : single bond

b) Ethene : double bond (one sigma one pi)

c) Ethyne: triple bond (one sigma and two pi bonds)

More the number of bonds lesser the bond length.

Thus the order of bond length of carbon-carbon bond will be:

Ethane (1.09 A) > Ethene(1.076 A) > Ethyne(1.06 A)

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3. A PS6 plugged into a 120 V outlet uses 3 Amps. What is the Resistance rating of
blondinia [14]

Answer:

R = 40 ohms

Explanation:

Given that,

The voltage of outlet, V = 120 V

Current flowing through the device, I = 3 A

We need to find the Resistance rating of  the power cord. Let the resistance is R. We know that,

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R=\dfrac{V}{I}\\\\R=\dfrac{120}{3}\\\\R=40\ \Omega

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8 0
3 years ago
one of the isotopes of sulfur is represented as sulfur -35 or S-35. what does the number 35 signify?
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4 years ago
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Use the data provided to calculate benzaldehyde's heat of vaporization smartwork5
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This problem is incomplete. Luckily, I found a similar problem from another website shown in the attached picture. The data given can be made to use through the Clausius-Clapeyron equation:

ln(P₂/P₁) = (-ΔHvap/R)(1/T₂ - 1/T₁)

where
P₁ = 14 Torr * 101325 Pa/760 torr = 1866.51 Pa
T₁ = 345 K
P₂ = 567 Torr * 101325 Pa/760 torr = 75593.78 Pa
T₂ = 441 K

ln(75593.78 Pa/1866.51 Pa) = (-ΔHvap/8.314 J/mol·K)(1/441 K - 1/345 K)
Solving for ΔHvap,
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6 0
3 years ago
A balloon occupies 1.50 L with 0.205 mol of carbon dioxide. How many moles would be required to increase the size of the balloon
Gekata [30.6K]

Answer:

0.683 moles of the gas are required

Explanation:

Avogadro's law relates the moles of a gas with its volume. The volume of a gas is directely proportional to its moles when temperature and pressure of the gas remains constant. The law is:

V₁n₂ = V₂n₁

<em>Where V is volume and n are moles of 1, initial state and 2, final state of the gas.</em>

<em />

Computing the values of the problem:

1.50Ln₂ = 5L*0.205mol

n₂ = 0.683 moles of the gas are required

<em />

8 0
3 years ago
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