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weeeeeb [17]
3 years ago
14

A student wants to reclaim the iron from an 18.0 gram sample of iron (III) oxide, which occurs according to the reaction below:

how many grams of iron can be reclaimed.
Chemistry
1 answer:
geniusboy [140]3 years ago
4 0

Answer:

12.6 g

Explanation:

Step 1: Write the balanced decomposition reaction

2 Fe₂O₃ ⇒ 4 Fe + 3 O₂

Step 2: Calculate the moles corresponding to 18.0 g of Fe₂O₃

The molar mass of Fe₂O₃ is 159.69 g/mol.

18.0 g × 1 mol/159.69 g = 0.113 mol

Step 3: Calculate the moles of Fe formed from 0.113 moles of Fe₂O₃

The molar ratio of Fe₂O₃ to Fe is 2:4. The moles of Fe formed are 4/2 × 0.113 mol = 0.226 mol

Step 4: Calculate the mass corresponding to 0.226 moles of Fe

The molar mass of Fe is 55.85 g/mol.

0.226 mol × 55.85 g/mol = 12.6 g

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tatyana61 [14]
I think it’s 4 ........
4 0
3 years ago
Fill in the blanks.
ANTONII [103]

Answer:GASEOUS, LOWERING ,EMITTED AS HEAT, DOWN,DO NOT RELEASE,DECREASES,NEGATIVE.

Explanation: Filling the blanks gives

Lattice energy is associated with forming a crystalline lattice of alternating cations and anions from the _GASEOUS___________ ions.

Because the cations are positively charged and the anions are negatively charged, there is a LOWERING of potential- as described by Coulomb's law-when the ions come together to form a lattice energy.That energy is EMITTED AS HEAT when the lattice forms.

As the ionic radii increases as you move DOWN a group, ions cannot get as close to each other and therefore DO NOT RELEASE as much energy when the lattice forms. Thus the lattice energy DECREASES (becomes less) NEGATIVE  as the radius increases.

The above gives the  definition of Lattice  Energy and how it relates to atomic and ion charge.

6 0
3 years ago
Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.150 M acetic acid (Ka = 1.75x10-5) with 0.150
mylen [45]

Answer:

a) pH = 2.793

b) pH = 4.280

c) pH = 4.933

d) pH = 8.816

e) pH = 8.861

f) pH = 8.891

Explanation:

a) VNaOH = 0 mL

∴ CH3COOH ↔ CHECOO- + H3O+

⇒ Ka = 1.75 E-5 = [ H3O+ ] * [ CH3COO-] / [ CH3COOH ]

mass balance:

⇒ <em>C</em> CH3COOH = [ CH3COO- ] + [ CH3COOH ] = 0.150 M

charge balance:

⇒ [ H3O+ ] = [ CH3COO- ]

⇒ Ka = 1.75 E-5 = [ H3O+ ]² / ( 0.150 M - [ H3O+ ] )

⇒ [ H3O+ ]² + 1.75 E-5 [ H3O+ ] - 2.625 E-6 = 0

⇒ [ H3O+ ] = 1.61146 E-3 M

⇒ pH = - Log [ H3O+ ] = 2.793

b) after  5.0 mL NaOH:

∴ CH3COOH + NaOH ↔ CH3COONa + H2O

⇒ <em>C</em> NaOH = (5 E-3 L * 0.150 mol/L) / (0.025+0.01 ) = 0.02143 M

⇒ <em>C</em> CH3COOH = ((0.025*0.150) - (0.01*0.150)) / (0.025 + 0.01) = 0.0643 M

mass balance:

⇒ 0.02143 + 0.0643 = [ CH3COOH ] + [ CH3COO- ] = 0.086 M

charge balance:

⇒ [ H3O+ ] + [Na ] = [ CH3COO- ]

⇒ [ H3O+ ] + 0.02143 = [ CH3COO- ]

⇒ Ka = [ H3O+ ] * ( [ H3O+ ] + 0.150 ) / (0.086 - 0.02143 - [ H3O+ ]) = 1.75 E-5

⇒ [ H3O+ ]² + 0.02143 [ H3O+ ] = 1.13 E-6 - 1.75 E-5 [ H3O+ ]

⇒ [ H3O+ ]² + 0.02144 [ H3O+ ] - 1.13 E-6 = 0

⇒ [ H3O+ ] = 5.26 E-5 M

⇒ pH = 4.28

c) after 15 mL NaOH:

⇒ <em>C</em> CH3COOH = 0.0375 M

⇒ <em>C</em> NaOH = 0.05625 M

mass balance:

⇒ 0.09375 M = [ CH3COO- ] +[ CH3COOH ]

charge balance:

⇒ [ H3O+ ] + 0.05625 = [ CH3COO- ]

⇒ Ka = 1.75 E-5 = [ H3O+ ] * ([ H3O+ ] + 0.05625) / (0.09375 - 0.05625 - [H3O+])

⇒ [H3O+]² + 0.05625[H3O+] = 6.5625 E-7 - 1.75 E-5 [H3O+]

⇒ [ H3O+]² + 0.05626[H3O+] - 6.5625 E-7 = 0

⇒ [ H3O+ ] = 1.1662 E-5 M

⇒ pH = 4.933

d) after 25 mL NaOH:

⇒ <em>C </em>NaOH = 0.075 M

⇒ <em>C</em> CH3COOH = 0 M....equiv. point

⇒Kh = Kw/Ka = 1 E-14 / 1.75 E-8 = 5.7143 E-10 = [ OH-]² / ( 0.075 - [OH-])

⇒ [OH-]² + 5.7143 E-10[OH-] - 4.286 E-11 = 0

⇒ [ OH- ] = 6.5463 E-6 M

⇒ pOH = 5.184

⇒ pH = 8.816

e) after 40 mL NaOH:

⇒ <em>C </em>NaOH = 0.0923 M

⇒ [OH-]² + 5.7143 E-10 [OH-] - 5.275 E-11 = 0

⇒ [OH-] = 7.2624 E-6 M

⇒ pOH = 5.139

⇒ pH = 8.861

f) after 60 mL NaOH:

⇒ <em>C </em>NaOH = 0.106 M

⇒ [OH-]² + 5.7143 E-10 [OH-] - 6.05 E-11 = 0

⇒ [OH-] = 7.7782 E-6 M

⇒ pOH = 5.11

⇒ pH = 8.891

5 0
4 years ago
4 Recall What holds ionic compounds<br> together?
Kazeer [188]

Answer:

The opposite charges of the atoms hold the atoms together. One atom can have a positive charge while the other can have a negative charge and this causes an electrostatic attraction causing the atoms to stick to each other like glue. For example, two magnets, can only be attracted to each other when one side of the magnet has a "-" while the other magnet on one side has a "+". They are attracted to each other by these varying charges. The same applies to ionic compounds which are held together by the varying charges of the ions.

7 0
3 years ago
consider one glucose unit in glycogen. what is the overall or net reaction for the conversion of this unit into 2 pyruvate, star
liubo4ka [24]

This cycle is known as Glycolysis

Glycolysis is the 10 step process, which occurs in cytoplasm of cell and  is conversion of glucose to pyruvate.

There are several steps and enzymes that is required in glycolysis pathway.

STEP 1: PHOSPHORYLATION

This is irreversible reaction.

Here glucose is phosphorylated to glucose 6 phosphate with the help of enzyme hexokinase and 1 ATP is utilized.

STEP 2 : ISOMERISATION

The isomerization of Glucose-6-phosphate to Fructose-6-phosphate, done with the help of enzyme phosphoglucoisomerase.

STEP 3 : SECOND PHOSPHORYLATION

Fructose-1,6-bisphosphate is phosphorylated to Fructose 1,6 bisphosphate  which is catalyzed by  phosphofructokinase and cost another ATP.

STEP 4: BREAKDOWN

The fructose-1,6 bisphosphate  is breaken down too produce two 3carbon molecules -  Glyceraldehyde-3-phosphate, or GADP, and a molecule of Dihydroxyacetone phosphate or DHAP.

The reaction is catalyzed by aldolase.

STEP 5 : CONVERSION OF DHAP INTO GADP

DHAP is oxidized to form GADP.

The reaction is catalyzed by triose phosphate isomerase enzyme.

STEP 6: OXIDATION

Here 2 mol. of GADP are oxidized.The GDAP is converted to 1,3 bisphosphoglycerate with the help of  glyceraldehyde phosphate dehydrogenase. This requires NAD+ and a free phosphate.

STEP 7: DEPHOSPHORYLATION

First substrate level phosphorylation ( addition of phosphate to ADP to give ATP )

1,3 bisphosphoglycerate with the help of Phosphoglycerate kinase become 3-phosphoglycerate and will produce 1 ATP.

STEP 8: PHOSPHATE TRANSFER

The phosphate ester linkage in 3 phosphoglycerate is moved from 3 C to 2 , because of low free energy to form 2 phosphoglycerate  with the help of phosphoglycerate mutase .

STEP 9: DEHYDRATION

2 phopshoglycerate is dehydrated by enolase to form Phosphoenolpyruvate ( PEP)

This is reversible reaction.

STEP 10: SECOND DEPHOSPHORYLATION

2 substrate level phosphorylation which gives out ATP.

Non - oxidative phosphorylation.

Here Phosphoenolpyruvate ( PEP) is converted to last product of glycolysis pyruvate releasing ATP by pyruvate kinase.

The first five step is production of GADP, And usage of ATP and the next five steps are the formation of ATP and pyruvate. The net formation of ATP is 2 mol. of ATP and 2 mol. of NADH.  This pyruvate then move to the TCA cycle.

To know more about glycolysis -

brainly.com/question/26990754

#SPJ4

5 0
1 year ago
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