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Oduvanchick [21]

4 years ago

11

What mass of iron(II) oxide must be used in the reaction given by the equation below to release 44.7 kJ? 6FeO(s) + O2(g) => 2

Fe3O4(s) ΔH° = -635 kJ Calculate your answer in g. Enter it with two decimal places and no units.

1 answer:

zavuch27 [327]4 years ago

7 0

<u>Answer:</u> The mass of iron (II) oxide that must be used in the reaction is 30.37

<u>Explanation:</u>

The given chemical reaction follows:

$6FeO(s)+O_2(g)\rightarrow 2Fe_3O_4(s);\Delta H^o=-635kJ$

By Stoichiometry of the reaction:

When 635 kJ of energy is released, 6 moles of iron (II) oxide is reacted.

So, when 44.7 kJ of energy is released, $\frac{6}{635}\times 44.7=0.423mol$ of iron (II) oxide is reacted.

Now, calculating the mass of iron (II) oxide by using the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of iron (II) oxide = 0.423 moles

Molar mass of iron (II) oxide = 71.8 g/mol

Putting values in above equation, we get:

$0.423mol=\frac{\text{Mass of FeO}}{71.8g/mol}\\\\\text{Mass of FeO}=(0.423mol\times 71.8g/mol)=30.37g$

Hence, the mass of iron (II) oxide that must be used in the reaction is 30.37

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Please someone help or tell me the answer for number 2 its a quiz grade

zubka84 [21]

Answer:

1 rearrange the equation

Explanation:

3c + 2h 2 = 2 c3h4

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So divide 40 by 1.5

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2 years ago

It takes to break an carbon-chlorine single bond. Calculate the maximum wavelength of light for which an carbon-chlorine single

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Answer:

It takes approximately $5.43\times 10^{-19}\; \rm J$ of energy to break one $\rm C-Cl$ single bond.

The maximum wavelength of a photon that can break one such bond is approximately $3.66\times 10^{-7}\; \rm m$ (in vacuum.) That's the same as $3.66 \times 10^{2}\; \rm nm$ (rounded to three significant figures.)

Explanation:

<h3>Energy per bond</h3>

The standard bond enthalpy of $\rm C-Cl$ single bonds is approximately $\rm 327\; \rm kJ \cdot mol^{-1}$ (note that the exact value can varies across sources.) In other words, it would take approximately $327\; \rm kJ$ of energy to break one mole of these bonds.

The Avogadro Constant $N_A \approx 6.023\times 10^{23}\; \rm mol^{-1}$ gives the number of $\rm C-Cl$ bonds in one mole of these bonds. Based on these information, calculate the energy of one such bond:

$\begin{aligned}& E(\text{one $\mathrm{C-Cl}$ bond}) \\ &= \frac{E(\text{one mole of $\mathrm{C-Cl}$ bonds})}{N_A} = \frac{327\; \rm kJ\cdot mol^{-1}}{6.023\times 10^{23}\; \rm mol^{-1}} \\ &\approx 5.429\times 10^{-22}\; \rm kJ = 5.429\times 10^{-19}\; \rm J \end{aligned}$ .

Therefore, it would take approximately $5.43\times 10^{-19}\; \rm J$ of energy to break one $\rm C-Cl$ single bond.

<h3>Minimum frequency and maximum wavelength </h3>

The Einstein-Planck Relation relates the frequency $f$ of a photon to its energy $E$ :

$E = h \cdot f$ .

The $h$ here represents the Planck Constant:

$h \approx 6.63 \times 10^{-34}\; \rm J \cdot s$ .

A photon that can break one $\rm C-Cl$ single bond should have more than $5.43\times 10^{-19}\; \rm J$ of energy. Apply the Einstein-Planck Relation to find the frequency of a photon with exactly that much energy:

$\begin{aligned}f &= \frac{E}{h}\\ &\approx \frac{5.43\times 10^{-19}\; \rm J}{6.63 \times 10^{-34}\; \rm J\cdot s} \\ &\approx 8.19 \times 10^{14}\; \rm s^{-1} = 8.19 \times 10^{14}\; \rm Hz\end{aligned}$ .

What would be the wavelength $\lambda$ of a photon with a frequency of approximately $8.19 \times 10^{14}\; \rm Hz$ ? The exact answer to that depends on the medium that this photon is travelling through. To be precise, the exact answer depends on the speed of light in that medium:

$\displaystyle \lambda = \frac{(\text{speed of light})}{f}$ .

In vacuum, the speed of light is $c \approx 2.998\times 10^{8}\; \rm m \cdot s^{-1}$ . Therefore, the wavelength of that $8.19 \times 10^{14}\; \rm Hz$ photon in vacuum would be:

$\begin{aligned} \lambda &= \frac{c}{f} \\ & \approx \frac{2.998\times 10^{8}\; \rm m \cdot s^{-1}}{8.19\times 10^{14}\; \rm s^{-1}} \\ &\approx 3.66 \times 10^{-7}\; \rm m = 3.66 \times 10^{2}\; \rm nm\end{aligned}$ .

(Side note: that wavelength corresponds to a photon in the ultraviolet region of the electromagnetic spectrum.)

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