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Elanso [62]
2 years ago
11

Addition reactions of alkenes are characterized by _________. A) formation of a bond B) addition of two groups across a double b

ond C) breaking of a bond A and B B and C
Chemistry
1 answer:
mr_godi [17]2 years ago
7 0

Answer:

B. ADDITION OF TWO GROUPS ACROSS A DOUBLE BOND

Explanation:

Addition reaction of alkenes involves the conversion of the double bond in alkenes Inyo single bonds by the addition of two groups of atoms or radicals.

During this addition reaction, two substances, an unsaturated compound(e.g. ethane) and an attacking reagent (hydrogen, halogens, hydrogen halides, chlorine and bromine water) combines to form a single new compound without forming any other products. So a saturated product or one in which is an increase in degree of saturation is formed.

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Si tuvieras que definir los atomos segun los postulados de Dalton, dirias que los atomos son particulas:
Arada [10]

Answer:

Según el científico inglés John Dalton, los átomos son esferas elásticas e indivisibles. Así, según él, el átomo es el bloque de construcción más pequeño de la materia. Es homogéneo e indivisible, y todos los átomos de un elemento químico dado son idénticos (es decir, tienen el mismo conjunto de propiedades).

Aunque se descubrió a finales del siglo XIX que los átomos están hechos de partículas aún más pequeñas y pueden sufrir transformaciones, y que los átomos de un elemento dado pueden diferir ligeramente entre sí (isótopos), la teoría de Dalton fue la base para el desarrollo de la tecnología química moderna.

3 0
2 years ago
Which is a difference between global and local winds?
valentina_108 [34]

Explanation:

local winds are considered breezes while global winds produce mostly storms

3 0
3 years ago
An atom has an average atomic mass of about 63.5 amu. What is the chemical symbol for the atom? Consult the periodic table.
Mama L [17]
The best and most correct answer among the choices provided by your question is the fourth choice.

Copper <span>has an average atomic mass of about 63.5 </span>amu<span>.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
8 0
3 years ago
Read 2 more answers
Hydrogen gas is filled in a 224 ml glass container at 0 degree celcius and 1 atmospheric pressure. What are the number of molecu
Step2247 [10]

Answer:

\boxed{\text{6.02 $\math{\times 10^{21}}$ molecules}}

Explanation:

A pressure of 1 atm and a temperature of 0 °C is the old definition of STP. Under these conditions, 1 mol of a gas occupies 22.4 L.

1. Calculate the moles of hydrogen.

n = \text{0.224 L} \times \dfrac{\text{1 mol}}{\text{22.4 L}} = \text{0.0100 mol}

2. Calculate the number of molecules

\text{No. of molecules} = \text{0.0100 mol} \times \dfrac{\text{6.022 $\times 10^{23}$ molecules}}{\text{1 mol}}\\\\= \textbf{6.02 $\mathbf{\times 10^{21}}$ molecules}\\\\\text{The sample contains }\boxed{\textbf{6.02 $\mathbf{\times 10^{21}}$ molecules}}

8 0
3 years ago
Read 2 more answers
Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =
Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Wherem  

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314 J / K mol  

Thus, given that, E_a = ?

k_2=4.0\times 10^3s^{-1}

k_1=1.5\times 10^3s^{-1}  

T_1=5\ ^0C  

T_2=25\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (5 + 273.15) K = 278.15 K  

T = (25 + 273.15) K = 298.15 K  

T_1=278.15\ K

T_2=298.15\ K

So,

\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)

E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}

E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}

E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}

E_a=33813.28\ J/mol

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

8 0
2 years ago
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