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3 years ago

In which direction does the electric field point a position directly south of a positive charge?

Chemistry

1 answer:

Stella [2.4K]3 years ago

7 0

Answer: Outward

Explanation: :The direction of electric field is radially outward for a positive charge and radially inward for a negative charge. Thus, for the electric field points toward south at a position directly south of a positive charge.

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Pulverized coal pellets, which may be approximated as carbon spheres of radius ro= 1 mm, are burned in a pure oxygen atmosphere

Ganezh [65]

Answer:

Explanation:

SO; If we assume that:

P should be the diffusion of oxygen towards the surface ; &

Q should be the diffusion of carbondioxide away from the surface.

Then the total molar flux of oxygen is illustrated by :

$Na,x = - cD_{PQ}\frac{dy_P}{dr} +y_P(NP,x + N_Q,x)$

where;

r is the radial distance from the center of the carbon particle.

Since ;

$N_P,x = - N_Q, x$ ; we have:

$Na,x = - cD_{PQ}\frac{dy_P}{dr}$

The system is not steady state and the molar flux is not independent of r because the area of mass transfer $4\pi r^{2}$ is not a constant term.

Therefore, using quasi steady state assumption, the mass transfer rate $4\pi r^{2}N_{P,x}$ is assumed to be independent of r at any instant of time.

$W_{P}=4\pi r^{2}N_{P,x}$

$W_{P}=-4\pi r^{2}cD_{PQ}\frac{dy_{P}}{dr}$

= constant

The oxygen concentration at the surface of the coal particle $yP,R$ will be calculated from the reaction at the surface.

The mole fraction of oxygen at a location far from pellet is 1.

Thus, separating the variables and integrating result into the following:

$W_{P}\int_{R}^{\infty} \frac{dr}{r^{2}}=-4\pi$

$r^{2}cD_{PQ}\int_{y_{P,R}}^{y_{P,\infty }}dy_{P}$

$-W_{P}\frac{1}{r}\mid ^{\infty }_{R}= -4\pi cD_{PQ}(y_{P,\infty }-y_{P,R})$

$=> W_{P}= - 4\pi cD_{PQ}(1-y_{P,R})R$

The mole of oxygen arrived at the carbon surface is equal to the mole of oxygen consumed by the chemical reaction.

$W_{P} = 4 \pi R^2R"$

$W_{P}= 4\pi R^{2}k_{1}"C_{O_{2}}\mid _{R}$

$W_{P}= 4\pi R^{2}k_{1}"c y _{P,R}$

$-4\pi cD_{PQ}(1-y_{P,R})R= - 4\pi R^{2}k_{1}"c y _{P,R}$

$y_{P,R}=\frac{D_{PQ}}{D_{PQ}+Rk_{1}}$

$y_{P,R}=\frac{1.7 \times 10^{-4}}{1.7\times 10^{-4}+10^{-3}\times 0.1}$

$\mathbf{= 0.631}$

Obtaining the total gas concentration from the ideal gas law; we have the following:

where;

R= $0.082m^3atm/kmolK$

$c=\frac{P}{RT} \\ \\ c=\frac{1}{0.082\times 1450} \\ \\ = 0.008405kmol/m^3$

The steady state $O_2$ molar consumption rate is:

$W_{P}= -4\pi cD_{PQ}(1-y_{P,R})R$

$W_{P}= -4\pi (0.008405)(1.7\times 10^{-4})(1-0.631)(10^{-3})$

$W_{P}= - 6.66\times 10^{-9}kmol/s$

5 0

3 years ago

In an eluent solution of 70% hexanes and 30% ethyl acetate, compound A moves farther on the TLC plate than compound B. Compound

8090 [49]

Answer:

it means compound A is <u>less</u> polar than compound B

Explanation:

It is sure that the more polar the solvent, the faster compounds elute, regardless of the compounds polarity. This means changing the solvent polarity cannot change the order compounds elute from a TLC or column. This may seem non intuitive, as it would seem that a polar solvent would move a polar compound farther than a nonpolar compound.

6 0

3 years ago

Is sponge cake a heterogeneous mixture

maxonik [38]

Answer:

though many may argue that cake batter is a homogeneous mixture because of its observable uniformity, we know it is heterogeneous. ... In other words, the mixture of ingredients from cake batter results in some chemical reactions here and there, but we are still left with a mixture of things that can be separated.

Explanation:

5 0

3 years ago

A 5.20 mol sample of solid A was placed in a sealed 1.00 L container and allowed to decompose into gaseous B and C. The concentr

ELEN [110]

Answer:

2.60 moles of A remaining.

Explanation:

According to Le Chatelier's principle, the equilibrium would shift if the volume, concentration, pressure, or temperature changes.

In this question, we were told that the volume doubles, that implies that we would have to double the molarity of B/ C (since B=C.)

However, it is obvious and clear from the given equation of the reaction that A is solid in it's activity = 1. Hence, it is then ignored.

So doubling B would be 1.30 M × 2 = 2.60 M

i.e 2.60 M moles of A was consumed.

Now; the number of moles of A remaining is 5.20 - 2.60 = 2.60 moles of A remaining.

5 0

3 years ago

1.25mole Nocl was placed in a2.50l reaction chamber at 427c. After equilibrium was reached .1.10moles of Nocl remained.Calculate

Dmitry [639]

Answer:

Kc → 5.58×10⁻⁴

Explanation:

Equilibrium reaction is:

2NOCl (g) ⇄ 2NO (g) + Cl₂(g)

Initially we have 1.25 moles of NOCl

After the equilibrium, we have 1.10 moles. So, during the process:

(1.25 mol - 1.1 mol) = 0.15 moles have reacted.

As ratio are 2:2, and 2:1, 0.15 moles of NO and (0.15 /2) = 0.075 moles of chlorine, were produced in the equilibrium.

Finally in equilibrium we have: 1.10 moles of NOCl, 0.15 moles of NO and 0.075 moles of Cl₂. But these amount are not molar, so we need molar concentration in order to determine Kc:

1.10 mol /2.50L = 0.44 M

0.15 mol / /2.50L = 0.06 M

0.075 mol /2.50L = 0.03 M

Let's make expression for Kc → [Cl₂] . [NO]² / [NOCl]²

Kc = (0.03 . 0.06²) / 0.44² → 5.58×10⁻⁴

4 0

3 years ago