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3 years ago

Which is a property of both alkenes and alkynes?

Chemistry

2 answers:

musickatia [10]3 years ago

6 0

Their boiling points tend to increase with chain length.<span>
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kap26 [50]3 years ago

3 0

Answer: Option (c) is the correct answer.

Explanation:

An alkene molecule has a double bond whereas an alkyne molecule has a triple bond.

Properties of alkenes and alkynes are as follows.

Both are insoluble in water.
Their pH is less than 7.
Boiling point increases with increase in carbon atoms or chain length.
Their densities are less than the density of water.

Thus, we can conclude that out of the given options, their boiling points tend to increase with chain length is a property of both alkenes and alkynes.

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Calculate the formula for the following hydrate composed of 76.9% CaSO3 and 23.1%H2O

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At standard temperature and pressure (0 ∘C and 1.00 atm ), 1.00 mol of an ideal gas occupies a volume of 22.4 L . What volume wo

Nostrana [21]

Taking into account the Charles's law, the same amount of gas at the same pressure and 65 ∘C would occupy a volume of 27.73 L.

<h3>Charles's Law</h3>

Charles's Law consists of the relationship that exists between the volume and the temperature of a certain quantity of ideal gas, at a constant pressure.

Volume is directly proportional to the temperature of the gas: if the temperature increases, the volume of the gas increases, while if the temperature of the gas decreases, the volume decreases.

Mathematically, Charles's law is a law that says that the quotient that exists between the volume and the temperature will always have the same value:

V÷ T= k

Considering an initial state 1 and a final state 2, it is satisfied:

V1÷ T1= V2÷ T2

<h3>Volume at 65°C</h3>

In this case, you know:

V1= 22.4 L
T1= 0 C= 273 K
V2= ?
T2= 65 C= 338 K

Replacing in Charles's law:

22.4 L÷ 273 K= V2÷ 338 K

Solving:

(22.4 L÷ 273 K) ×338 K= V2

Finally, the same amount of gas at the same pressure and 65 ∘C would occupy a volume of 27.73 L.

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1 year ago

Which organ serves as the primary site of amino acid metabolism (i.e. oxidation, transamination, deamination, gluconeogenesis, e

GarryVolchara [31]

Answer:

The liver

Explanation:

The liver -

In human beings the location of liver is just in the upper right portion of the abdomen , exactly below the diaphragm .

Liver is present in specifically vertebrates , and is responsible for various metabolism , like producing hormones , decomposition of the red blood cells , regulating the storage of glycogen and the synthesis of proteins .

Liver is the main site for the process of gluconeogenesis , amino acid catabolism .

Liver is the very organ , that has full urea cycle .

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The correct term is the liver .

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3 years ago

The following mechanism has been suggested for the reaction between nitrogen monoxide and oxygen: NO(g) + NO(g) → N2O2(g) (fast)

Karo-lina-s [1.5K]

Answer:

b. Second order in NO and first order in O₂.

Explanation:

A. The mechanism

$\rm 2NO\xrightarrow[k_{-1}]{k_{1}}N_{2}O_{2} \, (fast)\\\rm N_{2}O_{2} + O_{2}\xrightarrow{k_{2}} 2NO_{2} \, (slow)$

B. The rate expressions

$-\dfrac{\text{d[NO]} }{\text{d}t} = k_{1}[\text{NO]}^{2} - k_{-1} [\text{N}_{2}\text{O}_{2}]^{2}\\\\\rm -\dfrac{\text{d[N$_{2}$O$_{2}$]}}{\text{d}t} = -\dfrac{\text{d[O$_{2}$]}}{\text{d}t} = k_{2}[ N_{2}O_{2}][O_{2}] - k_{1} [NO]^{2}\\\\\dfrac{\text{d[NO$_{2}$]}}{\text{d}t}= k_{2}[ N_{2}O_{2}][O_{2}]$

The last expression is the rate law for the slow step. However, it contains the intermediate N₂O₂, so it can't be the final answer.

C. Assume the first step is an equilibrium

If the first step is an equilibrium, the rates of the forward and reverse reactions are equal. The equilibrium is only slightly perturbed by the slow leaking away of N₂O₂ to form product.

$\rm k_{1}[NO]^{2} = k_{-1} [N_{2}O_{2}]\\\\\rm [N_{2}O_{2}] = \dfrac{k_{1}}{k_{-1}}[NO]^{2}$

D. Substitute this concentration into the rate law

$\rm \dfrac{\text{d[NO$_{2}$]}}{\text{d}t}= \dfrac{k_{2}k_{1}}{k_{-1}}[NO]^{2} [O_{2}] = k[NO]^{2} [O_{2}]$

The reaction is second order in NO and first order in O₂.

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