At the top of a hlgh mountaln, the welght of an object Is:

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

Which statement correctly describe the nucleus of the atom

Elza [17]

Protons,neutrons,electrons

5 0

2 years ago

Electromagnetic radiation with a wavelength of 575 nm appears as yellow light to the human eye. The energy of one photon of this

Alecsey [184]

Answer:

9.12 * 10^20 photons

Explanation:

Given that;

E=n⋅h⋅ν

Where;

E= energy of the electromagnetic radiation

n = number of photons

h = Plank's constant

ν = frequency of electromagnetic radiation

Hence;

n = E/hν

n = 3.46 × 10 -19/6.6 * 10^-34 * 575 * 10^-9

n = 3.46 × 10 -19/3795 * 10^-43

n= 9.12 * 10^20 photons

7 0

2 years ago

2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g)a. Determine the volume (mL) of 15.0 M sulfuric acid needed to react with 45.0 g of

Ira Lisetskai [31]

Answer:

a. 167 mL.

b. 39.3 %.

Explanation:

Hello!

In this case, for the undergoing chemical reaction, since 45.0 g of aluminum react, based on the 2:3 mole ratio with sulfuric acid, we can compute the required moles as shown below:

$n_{H_2SO_4}=45.0gAl*\frac{1molAl}{27.0gAl} *\frac{3molH_2SO_4}{2molAl} =2.50molH_2SO_4$

Next, since the molarity of a solution is computed based on the moles and volume (M=n/V), we can compute the required volume of sulfuric acid as shown below:

$V=\frac{n}{M}=\frac{2.50mol}{15.0mol/L}=0.167L$

That in mL is 167 mL.

Moreover, for the percent yield, we compute the grams of aluminum sulfate that are produced based on the required 2.50 moles of sulfuric acid:

$m_{Al_2(SO_4)_3}=2.50molH_2SO_4*\frac{1molAl_2(SO_4)_3}{3molH_2SO_4}*\frac{342.15gAl_2(SO_4)_3}{1molAl_2(SO_4)_3} \\\\m_{Al_2(SO_4)_3}=285.13gAl_2(SO_4)_3$

Therefore the percent yield is:

$Y=\frac{112g}{285.13g}*100\%\\\\Y=39.3\%$

Best regards!

6 0

2 years ago

How does entropy change in gas reactions?

andre [41]

Answer:

where k is a proportionality constant equal to the ideal gas constant (R) divided by ... Isolated system - System in which neither heat nor work can be transferred between it and its surroundings. ... Gases particles are in a state of constant random motion. ... Does the entropy increase or decrease for the following reactions?

Explanation:

7 0

3 years ago

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