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3 years ago

What is the predicted outcome for very low mass stars (those that are smaller than about the mass of the sun)?

Chemistry

1 answer:

White raven [17]3 years ago

4 0

Answer:

In case of low-mass stars,the outer layers of the low mass stars are expelled as the core collapses such that the outer layers form a planetary nebula.

Explanation:

In case of low-mass stars,the outer layers of the low mass stars are expelled as the core collapses such that the outer layers form a planetary nebula. The core remains as a white dwarf and finally become a black dwarf as it cools down. A low mass star consumes its core hydrogen and turns it into helium over its lifetime.

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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

Answer plzzzz im begging

lisabon 2012 [21]

¿es demasiado tarde para ayudarte?

4 0

3 years ago

10. Hydrogen peroxide (H2O2: M = 34 g mol-1) is a powerful oxidising agent that is used in concentrated solution in rocket fuel

Lubov Fominskaja [6]

Answer:

The molality of solution=12.605 m

Explanation:

We are given that

Molar mass of Hydrogen peroxide, M=34 g/mol

Density of solution, $\rho=1.11gcm^{-3}$

30% Means mass of solute (Hydrogen peroxide)=30 g

Mass of solvent =100-30=70 g

Total mass of solution, m=100 g

Number of moles of solute= $\frac{given\;mass}{molar\;mass}$

Using the formula

Number of moles of hydrogen peroxide= $\frac{30}{34}$

Now, molality of solution

$m=\frac{number\;of\;moles\;of\;solute}{mass\;of\;solvent}\times 1000$

$m=\frac{30}{70\times 34}\times 1000$

$m=12.605 m$

Hence, the molality of solution=12.605 m

6 0

4 years ago

15.0 mL of 0.050 M Ba(NO3)2 M and 100.0 mL of 0.10 M KIO3 are added together in a 250 mL erlenmeyer flask. In this problem, igno

timama [110]

Answer:

Yes, precipitation of barium iodate will occur.

Explanation:

Molarity of barium nitrate solution = 0.050 M

Volume of barium nitrate solution =15.0 mL = 0.0150 L

1 mL = 0.001 L

Moles of barium nitrate = n

$n=0.050 M\times 0.0150L=0.00075 mol$

$Ba(NO_3)_2(aq)\rightarrow Ba^{2+}(aq)+2NO_3^{-}(aq)$

Moles of barium ions: $1\times 0.00075 mol=0.00075 mol$

Molarity of potassium iodate solution = 0.10 M

Volume of potassium iodate solution =100.0 mL = 0.1000 L

1 mL = 0.001 L

Moles of potassium iodate = n'

$n'=0.10 M\times 0.1000 L=0.01 mol$

$KIO_3(aq)\rightarrow K^{+}(aq)+IO_3^{-}(aq)$

Moles of iodate ions = $1\times 0.01 mol=0.01 mol$

After mixing of both solution in 250 mL in erlenmeyer flask

Volume of the final solution = 250 mL = 0.250 L

Concentration of barium ions in 250 mL solution :

$[Ba^{2+}]=\frac{0.00075 mol}{0.250 L}=0.003 M$

Concentration of iodate ions:

$[IO_3^{-}]=\frac{0.01 mol}{0.250 L}=0.04 M$

Solubility product of barium iodate, $K_{sp}=4.01\times 10^{-9}$

Ionic product of the barium iodate in solution : $K_i$

$Ba(IO_3)_2\rightleftahrpoons Ba^{2+}+2IO_3^{-}$

$K_i=[Ba^{2+}][IO_2^{-}]^2$

$K_i=0.003 M\times (0.04 M)^2=4.8\times 10^{-6}$

$K_{sp}$ ( precipitation)

As we can see, the ionic product of the barium iodate is greater than the solubility product of the barium iodate precipitation of barium iodiate will occur in 250 mL of final solution.

5 0

3 years ago

A chemist reacted 57.50 grams of sodium metal with an excess amount of chlorine gas. The chemical reaction that occurred is show

Readme [11.4K]

Answer:

actual yield=125.6 g Na + CI 2---->NaCI mass

Explanation:

6 0

2 years ago