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3 years ago

2. what is nappening in this diagram:

Chemistry

1 answer:

natka813 [3]3 years ago

7 0

Answer:

I'm sure but send thru this picture for the question so I can help.

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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

gThe mole fraction of potassium nitrate in an aqueous solution is 0.0194. The solution's density is 1.0627 g/mL. What is the mol

xxMikexx [17]

Answer:

0.595 M

Explanation:

The number of moles of water in 1L = 1000g/18g/mol = 55.6 moles of water.

Mole fraction = number of moles of KNO3/number of moles of KNO3 + number of moles of water

0.0194 = x/x + 55.6

0.0194(x + 55.6) = x

0.0194x + 1.08 = x

x - 0.0194x = 1.08

0.9806x= 1.08

x= 1.08/0.9806

x= 1.1 moles of KNO3

Mole fraction of water= 55.6/1.1 + 55.6 = 0.981

xA= mole fraction of solvent

xB= mole fraction of solute

nA= number of moles of solvent

nB = number of moles of solute

MA= molar mass of solvent

MB = molar mass of solute

d= density of solution

Molarity = xBd × 1000/xAMA ×xBMB

Molarity= 0.0194 × 1.0627 × 1000/0.981 × 18 × 0.0194×101

Molarity= 20.6/34.6

Molarity of KNO3= 0.595 M

8 0

3 years ago

Which of the following is not an example of a molecule?

9966 [12]

D. F

Molecules are a group of bonded atoms but Fluorine stands on its own

4 0

4 years ago

Read 2 more answers

Please help with these two (last question is nwse)

Vlad1618 [11]

Answer:

your answer gonna be 20 miles

5 0

3 years ago

Think about water and how it rolls up on the beach. Think of all its qualities. Is it alive according to the characteristics of

gulaghasi [49]

Water is not alive because it has no living organisms.

6 0

3 years ago