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2 years ago

A compound's molecular formula must always be different than the compound's empirical formula. TRUE FALSE

Chemistry

1 answer:

Juli2301 [7.4K]2 years ago

5 0

Answer: False

Explanation:

Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.

Example: $CH_4$ has similar molecular formula and empirical formula as the elements are already present in simplest of the ratios.

$C_2H_2$ has molecular formula of $C_2H_2$ but $CH$ as the empirical formula.

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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

How many grams are in 2.3 moles of N?

jok3333 [9.3K]

The amount of grams that are in 2.3 moles of N = 32.223 or 32/100
Because there are 14.01 grams per mile of nitrogen atoms.
So…
14.01 x 2.3= 32.223
Hope this helps :)

6 0

2 years ago

Read 2 more answers

Having a child will change your financial obligations, and can lead to a career change.

MrRa [10]

Yes, that's a true statement?

3 0

3 years ago

Read 2 more answers

For question numbers 1 and 2, two statements are given - one labelled

Svetllana [295]

Answer:

(iv) (A) is false, but (R) is true.

Explanation:

It is not true that carbon has a strong tendency to either lose or gain electrons to attain noble gas configuration. Carbon is a member of group 14, it is the first member of the group and carbon is purely a non metal. Only metals metals can loose electrons to attain the noble gas configuration. Moreover, carbon does not participate in ionic bonding so it does not gain electrons to attain the noble gas configuration.

However, carbon participates in covalent bonding where it is covalently bonded to four other chemical species using its four outermost electrons. Carbon forms covalent bonds in which four electrons are shared with other chemical species.

5 0

3 years ago

How many moles of co2 must enter the calvin cycle for the synthesis of one mole of glucose?

11111nata11111 [884]

Answer:3 moles

Explanation:

For every three molecules of CO2 that enters the Calvin cycle, one molecule of the three carbon glyceraldehyde 3-phosphate (G3P) is produced. Two molecules of G-3-P are required to produce one molecule of glucose. Therefore, the Calvin cycle needs to make a total of 6 turns to produce two molecules of G-3-P.

7 0

3 years ago