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2 years ago

B) Quelle est la masse de tétraoxyde de trifer (Fe3O4) produite si 3,60 moles de trioxyde de

Chemistry

1 answer:

hram777 [196]2 years ago

7 0

Answer:

626,4 g de Fe₃O₄

Explanation:

Nous commencerons par écrire l'équation équilibrée de la réaction entre le fer (Fe) et le trioxyde d'aluminium. Ceci est donné ci-dessous:

9Fe + 4Al₂O₃ -> 3Fe₃O₄ + 8Al

De l'équation équilibrée ci-dessus,

4 moles d'Al₂O₃ ont réagi pour produire 3 moles de Fe₃O₄.

Par conséquent, 3,6 moles d'Al₂O₃ réagiront pour produire = (3,6 × 3) / 4 = 2,7 moles de Fe₃O₄

Ainsi, 2,7 moles de Fe₃O₄ sont produites à partir de la réaction.

Enfin, nous déterminerons la masse massique de Fe₃O₄ produite par la réaction. Ceci peut être obtenu comme suit:

Mole de Fe₃O₄ = 2,7 moles

Masse molaire de Fe₃O₄ = (3 × 56) + (4 × 16)

= 168 + 64

= 232 g / mol

Masse de Fe₃O₄ =?

Mole = masse / masse molaire

2,7 = Masse de Fe₃O₄ / 232

Croiser multiplier

Masse de Fe₃O₄ = 2,7 × 232

Masse de Fe₃O₄ = 626,4 g

Par conséquent, 626,4 g de Fe₃O₄ sont produits à partir de la réaction

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A gas balloon has a volume of 80.0 mL at 300K, and a pressure of 50.0 kPa. If the pressure changes to 80 kPa and the temperature

fomenos

Answer: The new volume is 53.3 ml

Explanation:

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 50.0 kPa

$P_2$ = final pressure of gas = 80.0 kPa

$V_1$ = initial volume of gas = 80.0 ml

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $300K$

$T_2$ = final temperature of gas = $320K$

Now put all the given values in the above equation, we get:

$\frac{50.0\times 80.0}{300}=\frac{80.0\times V_2}{320}$

$V_2=53.3ml$

The new volume is 53.3 ml

4 0

2 years ago

Consider the reaction N2(g) + 2O2(g)2NO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

zysi [14]

<u>Answer:</u> The value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$N_2+2O_2\rightarrow 2NO_2$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(NO_2(g))})]-[(1\times \Delta S^o_{(N_2(g))})+(2\times \Delta S^o_{(O_2(g))})]$

We are given:

$\Delta S^o_{(NO_2(g))}=240.06J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(N_2)}=191.61J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (240.06))]-[(1\times (191.61))+(2\times (205.14))]\\\\\Delta S^o_{rxn}=-121.77J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-121.77) J/K = 121.77 J/K

We are given:

Moles of nitrogen gas reacted = 1.90 moles

By Stoichiometry of the reaction:

When 1 mole of nitrogen gas is reacted, the entropy change of the surrounding will be 121.77 J/K

So, when 1.90 moles of nitrogen gas is reacted, the entropy change of the surrounding will be = $\frac{121.77}{1}\times 1.90=231.36 J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

7 0

3 years ago

Determine the distance of closest approach (in fm) before the alpha particle reverses direction. Assume the lead nucleus remains

kiruha [24]

Answer:

Take E(alpha particle energy) = 5.5 MeV (5.5x106x1.6x10-19)

If the charge on the lead nucleus is +82e(atomic number of lead is 82) = +82x1.6x10-19 C and the charge on the alpha particle is +2e = 2x1.6x10-19 C

Using dc = (1/4πεo)qQ/Eα we have

dc = [9x10^9x(2x1.6x10-19x82x1.6x10-19)]/5.5x10-13 = 6.67x10^-13m. = 6.67 x 10^-13 x 10^15 = 6.67 x 10^2fm

Note: 1meter = 10^15fentometer

Explanation:

This is well inside the atom but some eight nuclear diameters from the centre of the lead nucleus.

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3 years ago

When the concentration of A in the reaction A ..... B was changed from 1.20 M to 0.60 M, the half-life increased from 2.0 min to

Reika [66]

Answer:

$0.4167\ \text{M}^{-1}\text{min}^{-1}$

Explanation:

Half-life

${t_{1/2}}A=2\ \text{min}$

${t_{1/2}}B=4\ \text{min}$

Concentration

${[A]_0}_A=1.2\ \text{M}$

${[A]_0}_B=0.6\ \text{M}$

We have the relation

$t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}$

$\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}$

Comparing the exponents we get

$1=n-1\\\Rightarrow n=2$

The order of the reaction is 2.

$t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}$

The rate constant is $0.4167\ \text{M}^{-1}\text{min}^{-1}$

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2 years ago

What does a horizontal row of elements in the periodic table represents

xxTIMURxx [149]

The periods or energy levels

8 0

3 years ago

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