Identify which location in the periodic table you would have the largest atomic radii.

The compound 3-methylpentane is an example of

Cerrena [4.2K]

Answer:

A) An Alkaline

Explanation:

An alkane that is pentane which is substituted by a methyl group at position 3. It is used as a solvent in organic synthesis, as a lubricant and as a raw material for producing carbon black.

3 0

2 years ago

The reaction 2NO(g)+O2(g)−→−2NO2(g) is second order in NO and first order in O2. When [NO]=0.040M, and [O2]=0.035M, the observed

Oksanka [162]

Answer:

(a) The rate of disappearance of $O_{2}$ is: $4.65*10^{-5}$ M/s

(b) The value of rate constant is: 0.83036 $M^{-2}s^{-1}$

(c) The units of rate constant is: $M^{-2}s^{-1}$

(d) The rate will increase by a factor of 3.24

Explanation:

The rate of a reaction can be expressed in terms of the concentrations of the reactants and products in accordance with the balanced equation.

For the given reaction:

$2NO(g)+O_{2}->2NO_{2}$

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = $-\frac{d}{dt}[O_{2}]$ = $\frac{1}{2}\frac{d}{dt}[NO_{2}]$ -----(1)

According to the question, the reaction is second order in NO and first order in $O_{2}$ .

Then we can say that, rate = k $[NO]^{2}[O_{2}]$ -----(2)

where k is the rate constant.

The rate of disappearance of NO is given:

$-\frac{d}{dt}[NO]$ = $9.3*10^{-5}$ M/s.

(a) From (1), we can get the rate of disappearance of $O_{2}$ .

Rate of disappearance of $O_{2}$ = $-\frac{d}{dt}[O_{2}]$ = (0.5)*( $9.3*10^{-5}$ ) M/s = $4.65*10^{-5}$ M/s.

(b) The rate of the reaction can be obtained from (1).

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = (0.5)*( $9.3*10^{-5}$ )

rate = $4.65*10^{-5}$ M/s

The value of rate constant can be obtained by using (2).

rate constant = k = $\frac{rate}{[NO]^{2}[O_{2}]}$

k = $\frac{4.65*10^{-5}}{(0.040)^{2}(0.035)}$ = 0.83036 $M^{-2}s^{-1}$

(c) The units of the rate constant can be obtained from (2).

k = $\frac{rate}{[NO]^{2}[O_{2}]}$

Substituting the units of rate as M/s and concentrations as M, we get:

$\frac{Ms^{-1} }{M^{3}}$ = $M^{-2}s^{-1}$

(d) The reaction is second order in NO. Rate is proportional to square of the concentration of NO.

$rate\alpha [NO]^{2}$

If the concentration of NO increases by a factor of 1.8, the rate will increase by a factor of $(1.8)^{2}$ = 3.24

5 0

2 years ago

Find the quantinum numbers n,m,l,s for the last of potassium layer pleasee help explain correctly all

Fantom [35]

Answer:

Quantum numbers of the outermost electron in potassium:

$n = 4$ .
$l = 1$ .
$m_l = 0$ .
Either $m_s = 1/2$ .

Explanation:

Refer to the electron configuration of a potassium atom. The outermost electron in a ground-state potassium atom is in the $4s$ orbital (fourth $s$ orbital.)

The quantum number $n$ (the principal quantum number) specifies the main energy shell of an electron. This electron is in the fourth main energy shell (as seen in the number four in the orbital.) Hence, $n = 4$ for this electron.

The quantum number $l$ (the angular momentum quantum number) specifies the shape ( $s$ , $p$ , $d$ , etc.) of an electron. $l = 1$ for $s\!$ orbitals (such as the one that contains this electron.

Quantum numbers $n$ and $l$ specify the shape of an orbital. On the other hand, the magnetic quantum number $m_l$ specifies the orientation of these orbitals in space.

However, $s$ orbitals are spherical. Regardless of the value of $n$ , the only possible $m_l$ value for electrons in $s\!$ orbitals is $m_l = 0$ .

The spin quantum number $m_s$ distinguishes between the two electrons in an orbital. The two possible values of $m_s \!$ are $(+1/2)$ and $(-1/2)$ . Typically, the first electron in an orbital is assigned an upward ( $\uparrow$ ) spin, which corresponds to $m_s = (+1/2)$ .

5 0

2 years ago

1. Describe the correlation between acid concentration and reaction rate and use ideas about particles to explain the correlatio

Nostrana [21]

Explanation:

an increase in concentration increases the rate of the reaction. This is because there are more reactant particles available which allows for more effective collisions between reactant particles in a given period of time. More effective collisions bring about a faster rate of reaction.

5 0

2 years ago

Can someone help me

noname [10]

From Earth's density we can estimate what elements must compose the Earth; an iron core just happens to estimate Earth's mass the best. Now from energy waves, geologists use seismometers to measure movements in Earth's interior (e.g. Earthquakes), These energy-waves form compressional and shear waves

4 0

2 years ago