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3 years ago

All members of the alkene series of hydrocarbons have the general formula:

Chemistry

1 answer:

LenaWriter [7]3 years ago

7 0

CnH2n since the equivalent unsaturation is equal with 1

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Identify the number and kinds of atoms present in a molecule of each compound butane

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The butane is a kind of alkane. And there is only carbon and hydrogen and single bonds. The formula of butane is C4H10. There are four carbon atoms and ten hydrogen atoms present in one butane molecule.

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4 years ago

Yusef adds all of the values in his data set and then divides by the number of values in the set. What is Yusef most likely find

Zanzabum

He is most likely finding the mean of the data.

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3 years ago

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A weather balloon is filled with helium that occupies a volume of 5.37 104 L at 0.995 atm and 32.0°C. After it is released, it r

svp [43]

Answer:

The new volume is 63583 L

Explanation:

Step 1: Data given

The initial volume of the balloon = 5.37 * 10^4 L

The initial pressure = 0.995 atm

The initial temperature = 32.0 °C = 305.15 K

The pressure decreased to 0.720 atm

The temperature decreased to -11.7 °C = 261.45 K

Step 2: Calculate the new volume

P1*V1 / T1 = P2*V2/T2

⇒with P1 = the initial pressure = 0.995 atm

⇒with V1 = the initial volume = 5.37 *10^4 L

⇒with T1 = the initial temperature = 305.15 K

⇒with P2 = the decreased pressure = 0.720 atm

⇒with V2 = the new volume = TO BE DETERMINED

⇒with T2 = the decreased temperature : 261.45 K

(0.995 * 5.37*10^4)/305.15 = (0.720 * V2) / 261.45

V2 = 63583 L

The new volume is 63583 L

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3 years ago

a chemist uses hot hydrogen gas to convert chromium (iii) oxide to pure chromium. how many grams of hydrogen are needed to produ

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Answer:

Explanation:

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3 years ago

He rate constant of a reaction is 4.55 × 10−5 l/mol·s at 195°c and 8.75 × 10−3 l/mol·s at 258°c. what is the activation energy o

Xelga [282]

Answer : The activation energy of the reaction is, $17.285\times 10^4kJ/mole$

Solution :

The relation between the rate constant the activation energy is,

$\log \frac{K_2}{K_1}=\frac{Ea}{2.303\times R}\times [\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = initial rate constant = $4.55\times 10^{-5}L/mole\text{ s}$

$K_2$ = final rate constant = $8.75\times 10^{-3}L/mole\text{ s}$

$T_1$ = initial temperature = $195^oC=273+195=468K$

$T_2$ = final temperature = $258^oC=273+258=531K$

R = gas constant = 8.314 kJ/moleK

Ea = activation energy

Now put all the given values in the above formula, we get the activation energy.

$\log \frac{8.75\times 10^{-3}L/mole\text{ s}}{4.55\times 10^{-5}L/mole\text{ s}}=\frac{Ea}{2.303\times (8.314kJ/moleK)}\times [\frac{1}{468K}-\frac{1}{531K}]$

$Ea=17.285\times 10^4kJ/mole$

Therefore, the activation energy of the reaction is, $17.285\times 10^4kJ/mole$

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3 years ago

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