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oee [108]
3 years ago
6

1

Chemistry
1 answer:
WINSTONCH [101]3 years ago
5 0

Answer:

The new substance will need more energy to form its chemical bonds than the old substance will release. ... More energy will be released from the old substance than the new substance will need to form its chemical bonds.

Explanation:

This is the answer I got. Hope it's really helpful

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Al comenzar la reacción: N2(g) + 2O2(g) ------> 2NO2(g) existe 1 mol de N2 y 2 moles de O2 y al
miss Akunina [59]

Answer:

65

Explanation:

estion

Al comenzar la reacción: N2(g) + 2O2(g) ------> 2NO2(g) existe 1 mol de N2 y 2 moles de O2 y al

finalizarla está presente una mezcla formada por 2,2 moles en total, ¿cuál es el rendimiento para la

reacción?

7 0
3 years ago
What type of reaction is this.... CH2(g)+02(g)
oee [108]

Combustion

Explanation:

This reaction is called a combustion reaction. It is very peculiar to reactions involving hydrocarbons.

In such reactants, oxygen gas is used as a reactant. It combines with fuel usually the hydrocarbon to produce heat and light energy.

For the combustion of most hydrocarbons, water and carbon dioxides are usually the product.

        2CH₂   +   3O₂  →   2CO₂   +  2 H₂O

learn more:

Combustion brainly.com/question/11089002

#learnwithBrainly

6 0
3 years ago
What two planets feud the most ?
Irina18 [472]
  <span>Fred and Ethel Mertz

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8 0
3 years ago
Read 2 more answers
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
Solid cesium iodide has the same kind of crystal structure as CsCl which is pictured below: If the edge length of the unit cell
Assoli18 [71]

The question is incomplete, here is the complete question:

Solid cesium iodide has the same kind of crystal structure as CsCl which is pictured below: If the edge length of the unit cell is 456.2 pm, what is the density of CsI in g/cm^3

The image is attached below.

<u>Answer:</u> The density of CsI is 9.09g/cm^3

<u>Explanation:</u>

To calculate the density of metal, we use the equation:

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density

Z = number of atom in unit cell = 2  (BCC)

M = atomic mass of CsI = 259.8 g/mol

N_{A} = Avogadro's number = 6.022\times 10^{23}

a = edge length of unit cell = 456.2pm=456.2\times 10^{-10}cm    (Conversion factor:  1cm=10^{10}pm  )

Putting values in above equation, we get:

\rho=\frac{1\times 259.8}{6.022\times 10^{23}\times (456.2\times 10^{-10})^3}\\\\\rho=9.09g/cm^3

Hence, the density of CsI is 9.09g/cm^3

3 0
3 years ago
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