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3 years ago

What would be the charge on an ion whose neutral atom has an electron configuration of 2.8.7?

Chemistry

1 answer:

kompoz [17]3 years ago

5 0

Answer:

–1

Explanation:

The electronic configuration of the neutral atom is given below:

2 8 7

From the electronic configuration of the neutral atom, we can see that the atom has 7 electrons in it's outermost shell. This implies that the element needs 1 more electron to complete it's octet configuration.

Therefore the charge on the ion of the element will be –1 indicating that the atom has received 1 electron to complete it's octet configuration.

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Black_prince [1.1K]

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8 0

3 years ago

Read 2 more answers

Given the density of iron (Fe) is 7.87 g/cm3, determine the mass of iron (in grams) in a rectangle block with the dimensions of

guapka [62]

Answer:

$m=14,105.71 g Fe$

Explanation:

Hello,

In this case, the first step is to compute the volume of the block considering the length, height and width:

$V=L \times W \times H =12.5 in\times 3.50 in \times 2.50 in =109.375 in^3$

Then, we compute the volume in cubic centimetres:

$V=109.375in^3\times \frac{16.3871 cm^3}{1in^3} =1792.34cm^3$

Finally, as the density is given by:

$\rho =\frac{m}{V}$

We solve for the mass:

$m=\rho \times V= 7.87\frac{g}{cm^3} \times 1792.34 cm^3\\\\m=14,105.71 g Fe$

Best regards.

8 0

3 years ago

t-Butyl alcohol (TBA) is an important octane enhancer that is used to replace lead additives in gasoline [Ind. Eng. Chem. Res.,

Andrew [12]

Complete Question:

t-Butyl alcohol (TBA) is an important octane enhancer that is used to replace lead additives in gasoline [Ind. Eng. Chem. Res., 27, 2224 (1988)]. TBA was produced by the liquid-phase hydration (W) of isobutene (I) over an Amberlyst-15 catalyst. The system is normally a multiphase mixture of hydrocarbon, water, and solid catalysts. However, the use of cosolvents or excess TBA can achieve reasonable miscibility. The reaction mechanism is believed to be

I + S ⇄ I*S

W + S ⇄ W*S

W*S + I*S ⇄ TBA * S * S

TBA * S ⇄ TBA + S

Derive a rate law assuming:

(a) The surface reaction is rate limiting

(b) The adsorption of isobutene is limiting

(c) The reaction follows Eley-rideal Kinetics

I*S+W ⇄ TBA * S

and surface reaction is limiting

(d) Isobutene (I) and water (W) are adsorbed on different sites

I + S₁ ⇄ I*S₁

W + S₂ ⇄ W*S₂

TBA is not on the surface, and the surface reaction is rate-limiting

$[Ans: r'_{TBA}=-r'_1=\frac{k[C_1C_w-C_{TBA/K_C}]}{(1+K_WC_W)(1+K_1C_1)} ]$

(e) What generalizations can you make by comparing rate laws derived from part (a) through (d)?

Answer and explanation:

The mechanism for the production of t-butyl alcohol is as follows:

the reaction and rate law for the adsorption of isobutene over the amberlyst-15 is as follows:

I + S ⇄ I * S $-r_{ADI} = k_I(C_1C_v-\frac{C_{I.S}}{K_I} )$

where $C_V$ is the concentration of vacant site

$K_I$ is the equilibrium constant of the adsorption

$k_I$ is the rate constant for forward

$C_I,C_{I.S}$ are concentration of isobutene and site filled with isobutene

The reaction and rate law for the adsorption of water (W) over the amberlyst-15 catalyst catalyst is as follows

W + S ⇄ W.S $-r_{ADW} = k_W(C_WC_V-\frac{C_{W.S}}{K_W} )$

The reaction and rate law for the surface reaction on the catalyst is as follows

W.S + I.S ⇄ TBA . S + Sn $-r_s = k_s(C_{W.S}C_{I.S}-\frac{C_{TBA.S}C_V}{K_s} )$

The reaction and rate law for the desorption of TBA from catalyst is as follows

TBA . S ⇄ TBA + S $-r_{D TBA} = k_{DTBA}(C_{TBA.S}-\frac{C_{TBA}C_V}{K_{DTBA}} )$

the attached image below gives the remaining steps

5 0

3 years ago

Calcule la normalidad de una solución

tensa zangetsu [6.8K]

Answer:

si por favor

Explanation:

5 0

4 years ago

Density is defined as mass per unit volume, or D = m/v. Solve the density

fgiga [73]

Answer:

v = m/D

Explanation:

Density is defined as the mass per unit volume, or D = m/v. Solve for Density equation for the variable v:

With this we are making v Which represents volume the subject so;

$d = \frac{m}{v}$

With this you multiply both sides by ;

$v$

$v(d) = ( \frac{m}{v})v$

And then the v will multiply the d making it

“vd” and the ‘v’ will cancel the v making it;

$vd = m$

So then we divide both sides by d to make v the subject.

$\frac{vd}{d} = \frac{m}{d}$

So the d cancels the d leaving;

$v = \frac{m}{d}$

5 0

3 years ago