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3 years ago

An unsaturated hydrocarbon us a hydrogen-carbon compound with ___

2 answers:

8 0

An unsaturated hydrocarbon is a hydrogen-carbon compound with one or more double or triple bonds of carbon-carbon atoms.

Explanation: 

Generally hydrocarbons are compounds formed entirely by combination of carbon and hydrogen molecules. Hydrocarbons are classified as saturated and unsaturated hydrocarbons. A saturated hydrocarbons are those where the hydrogen-carbon molecules are formed with carbon-hydrogen single bond.

They are the simplest of hydrocarbons and the hydrogen atoms are dominant in them. While in unsaturated hydrocarbons, the number count of hydrogen atoms will be less to form bonds with carbon. So in unsaturated hydrocarbons, more number of carbon-carbon double or triple bonds are observed and less number of single bonds of C-H will be present.

irina [24]3 years ago

3 0

Answer:with double or triple carbon carbon bond between it's corresponding hydrocarbon or organic compound thus gives rise to alkene and alkyne

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See if you can think of a way to prove that it is the oxygen in air, not nitrogen, that causes rusting.

allsm [11]

Answer:

rusting occur in oxygen and water coz when you put a nail into the water and you leave it three days you will the nail become brownish and that is due to oxygen and water

8 0

4 years ago

The successive ionization energies for a given element are listed. How many valence electrons does it have?

damaskus [11]

Answer:

3 valence electrons

Explanation:

Let us first find the amount increased for everything.

From 1st to 2nd, 1048 - 620 = 428

428 increased from 1st to 2nd.

From 2nd to 3d, 2144 - 1048 = 1096

1096 increased from 2nd to 3rd.

From 3rd to 4th, 8724 - 2144 = 6580

6580 increased from 3rd to 4th.

From 4th to 5th, 10141 - 8724 = 1417

From 3rd to 4th, it increased the most. Therefore, we assume that those are the beginning core electrons and not valence electrons. Also, starting from 3rd, they are all core electrons.

For the 1st and 2nd, we can assume they are valence electrons because the kJ/mol didn't increase that much.

Since the core electrons started from the 3rd, we most likely have 3 valence electrons.

So the final answer is 3.

Hope it helped!

5 0

3 years ago

Is the following equation balanced? SO3 + 2H2O H2SO4 no yes

postnew [5]

Answer: No

Explanation: It's not balanced because four oxygen atoms in H2SO4, whereas there are 5 oxygen atoms in the reactants side. Also, there's more hydrogen atoms on the reactants side.

I hope this helps!

7 0

3 years ago

Order the follow processes from (1) the least work done by the system to (5) the most work done by one mole of an ideal gas at 2

quester [9]

Answer : The order of process from (1) the least work done by the system to (5) the most work done by the system will be:

(1) < (5) < (3) < (4) < (2)

Explanation :

The formula used for isothermally irreversible expansion is :

$w=-p_{ext}dV\\\\w=-p_{ext}(V_2-V_1)$

where,

w = work done

$p_{ext}$ = external pressure

$V_1$ = initial volume of gas

$V_2$ = final volume of gas

The expression used for work done in reversible isothermal expansion will be,

$w=-nRT\ln (\frac{V_2}{V_1})$

where,

w = work done = ?

n = number of moles of gas = 1 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas = $25^oC=273+25=298K$

$V_1$ = initial volume of gas

$V_2$ = final volume of gas

First we have to determine the work done for the following process.

(1) An isothermal expansion from 1 L to 10 L at an external pressure of 2.5 atm.

$w=-p_{ext}(V_2-V_1)$

$w=-(2.5atm)\times (10-1)L$

$w=-22.5L.atm=-22.5\times 101.3J=-2279.25J$

(2) A free isothermal expansion from 1 L to 100 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{100L}{1L})$

$w=-11409.6J$

(3) A reversible isothermal expansion from 0.5 L to 4 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{4L}{0.5L})$

$w=-5151.97J$

(4) A reversible isothermal expansion from 0.5 L to 40 L.

$w=-nRT\ln (\frac{V_2}{V_1})$

$w=-1mole\times 8.314J/moleK\times 298K\times \ln (\frac{40L}{0.5L})$

$w=-10856.8J$

(5) An isothermal expansion from 1 L to 100 L at an external pressure of 0.5 atm.

$w=-p_{ext}(V_2-V_1)$

$w=-(0.5atm)\times (100-1)L$

$w=-49.5L.atm=-49.5\times 101.3J=-5014.35J$

Thus, the order of process from (1) the least work done by the system to (5) the most work done by the system will be:

(1) < (5) < (3) < (4) < (2)

8 0

3 years ago

Butane (C4H10) has a heat of vaporization of 22.44 kJ/mol and a normal boiling point of -0.4 ∘C. A 250 mL sealed flask contains

erastova [34]

Given that:

The heat of vaporization = 22.44 kJ/mol = 22440 J/mol
normal boiling point which is the initial temperature = 0.4° C = (273 + (-0.4))K = 272.6 K
volume = 250 mL = 0.250 L
Mass of butane = 0.8 g
the final temperature = -22° C = (273 + (-22)) K = 251 K

The first step is to determine the vapor pressure at the final temperature of 251K by using the Clausius-Clapeyron equation. This is following by using the ideal gas equation to determine the numbers of moles of butane gas. After that, the mass of butane present in the liquid is determined by using the relation for the number of moles.

Using Clausius-Clapeyron Equation:

$\mathbf{In (\dfrac{P_2}{P_1} )= -\dfrac{\Delta H_{vap}}{R}(\dfrac{1}{T_2} - \dfrac{1}{T_1})}$