The swallowtail caterpillar is an example of a ?

Which of the following coefficients is needed to balance the combustion reaction of methane? a.3 b.2 c.4 d.equation is already b

Alla [95]

I might go with d it seems right to me hope this helps.

3 0

3 years ago

Nitrosyl bromide decomposes according to the chemical equation below. 2NOBr(g) 2NO(g) + Br2(g) 1.00 atm of NOBr is sealed in a f

Andrew [12]

Given :

2NOBr(g) - -> 2NO(g) + Br2(g)

Initial pressure of NOBr , 1 atm .

At equilibrium, the partial pressure of NOBr is 0.82 atm.

To Find :

The equilibrium constant for the reaction .

Solution :

2NOBr(g) - -> 2NO(g) + Br2(g)

t=0 s 1 atm 0 0

$t=t_{eqb}$ 1( 1-2x) 2x x

So ,

$1-2x=0.82\\\\x=0.09$

At equilibrium :

$K_{eq}=\dfrac{[NO]^2[br_2]}{[NOBr]^2}\\\\K_{eq}=\dfrac{0.18^2\times 0.9}{0.82^2}\\\\K_{eq}=0.043\ atm$

Hence , this is the required solution .

3 0

3 years ago

When light is shown on a mixture of chlorine and chloromethane, carbon tetrachloride is one of the components of the final react

velikii [3]

A free-radical substitution reaction is likely to be responsible for the observations. The reaction mechanism of a reaction like this can be grouped into three phases:

Initiation; the "light" on the mixture deliver sufficient amount of energy such that the halogen molecules undergo homologous fission. It typically takes ultraviolet radiation to initiate fissions of the bonds.
Propagation; free radicals react with molecules to produce new free radicals and molecules.
Termination; two free radicals combine and form covalent bonds to produce stable molecules. Note that it is possible for two carbon-containing free-radicals to combine, leading to the production of trace amounts of long carbon chains in the product.

Initiation

$\text{Cl}-\text{Cl} \stackrel{\text{UV}}{\to} \text{Cl}\bullet + \bullet\text{Cl}$

where the big black dot indicates unpaired electrons attached to the atom.

Propagation

$\text{CH}_3\text{Cl}+ \text{Cl}\bullet \to \bullet\text{CH}_2\text{Cl} + \text{HCl}$

$\bullet\text{CH}_2\text{Cl} + \text{Cl}_2 \to \text{CH}_2\text{Cl}_2 + \text{Cl}\bullet$

$\text{CH}_2\text{Cl}_2 + \text{Cl}\bullet \to \bullet\text{CHCl}_2 + \text{HCl}$

$\bullet\text{CHCl}_2+ \text{Cl}_2 \to \text{CHCl}_3 + \bullet \text{Cl}$

$\text{CHCl}_3 + \text{Cl}\bullet \to \bullet\text{CCl}_3 + \text{HCl}$

$\bullet\text{CCl}_3 + \text{Cl}_2 \to \text{CCl}_4 + \text{Cl}\bullet$

Termination

$\text{Cl}\bullet + \bullet\text{Cl} \to \text{Cl}-\text{Cl}$

8 0

3 years ago

Part complete When a 235 92U nucleus is bombarded by neutrons (10n) it undergoes a fission reaction, resulting in the formation

nikdorinn [45]

Answer: The isotopic symbol of barium is $_{56}^{138}\textrm{Ba}$ and that of strontium is $_{38}^{89}\textrm{Sr}$

Explanation:

Nuclear fission reactions are defined as the reactions in which a heavier nuclei breaks down in two or more smaller nuclei.

In a nuclear reaction, the total mass and total atomic number remains the same.

For the given fission reaction:

$^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^A_Z\textrm{Ba}+^{94}_{36}\textrm{Kr}+3^1_0\textrm{n}$

To calculate A:

Total mass on reactant side = total mass on product side

235 + 1 = A + 94 + 3

A = 139

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

92 + 0 = Z + 36 + 0

Z = 56

The isotopic symbol of barium is $_{56}^{139}\textrm{Ba}$

For the given fission reaction:

$^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^A_Z\textrm{Sr}+^{143}_{54}\textrm{Xe}+3^1_0\textrm{n}$

To calculate A:

Total mass on reactant side = total mass on product side

235 + 1 = A + 143 + 3

A = 90

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

92 + 0 = Z + 54 + 0

Z = 38

The isotopic symbol of strontium is $_{38}^{89}\textrm{Sr}$

Hence, the isotopic symbol of barium is $_{56}^{138}\textrm{Ba}$ and that of strontium is $_{38}^{89}\textrm{Sr}$

7 0

3 years ago

When a radioactive isotope releases an alpha particle, the atomic number of the atom is

Ipatiy [6.2K]

The radioactive isotope's atomic number decreases by two, and the alpha particle has an atomic number of 2. It wasnt clear which one you were asking for, so there is both.

5 0

3 years ago