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larisa [96]
3 years ago
6

Help!: Determine the mass of oxygen that can be removed from 25.0 grams of Fe2O3?

Chemistry
1 answer:
Lorico [155]3 years ago
3 0

7.5g

Explanation:

Given parameters:

Mass of Fe₂O₃ = 25g

Unknown:

Mass of oxygen that can be removed  = ?

Solution:

Find the ratio of the oxygen atoms to that of the compound using the molar mass. Then multiply with the given mass.

   Mass of oxygen = \frac{molar mass of oxgygen}{molar mass of [tex]Fe_{2}O_{3}}[/tex]  x given mass

  Mass of oxygen = \frac{3 x 16}{160} x 25 = 7.5g

Learn more:

Percentage composition brainly.com/question/8654167

#learnwithBrainly

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stellarik [79]

Answer:

when an electron move closer to the nucleus the magnitude of energy of the nucleus increases

8 0
2 years ago
In an experiment, 34.8243g of copper (II) nitrate hydrate, Cu(NO3)2•zH2O was heated to a constant mass of 27.0351g. Calculate th
Leto [7]

Answer:

1) The mass of water lost = 7.7892 grams

2) Z = 3: Cu(NO3)2*3H2O

Explanation:

<u>Step 1: </u>Data given

Mass of copper (II) nitrate hydrate, Cu(NO3)2•zH2O = 34.8243 grams

Mass of substance after heating = 27.0351 grams

Molar mass of Cu(NO3)2 = 187.56 g/mol

Molar mass of H2O = 18.02 g/mol

<u>Step 2:</u> Calculate mass of water

The mass of water is the mass lost after heating.

Mass water = 34.8243 - 27.0351 = 7.7892 grams of water

<u>Step 3:</u> Calculate moles of Cu(NO3)2

Moles Cu(NO3)2 = Mass Cu(NO3)2 / Molar mass Cu(NO3)2

Moles Cu(NO3)2 = 27.0351 grams / 187.56 g/mol

Moles Cu(NO3)2 = 0.144 moles

<u>Step 4:</u> Calculate moles of H2O

Moles H2O = 7.7892 grams / 18.02 g/mol

Moles H2O = 0.432 moles

<u>Step 5:</u> Calculate Z

z = moles H2O / moles Cu(NO3)2

Z = 0.432/0.144

Z = 3

This means we have 3 water molecules in the formula. This makes the formula ofthe hydrate: Cu(NO3)2*3H2O

5 0
3 years ago
Consider the resonance structures for the carbonate ion. carbonate is a carbon double bonded to an oxygen with two lone pairs an
Bumek [7]

Answer:

The correct answers are: <u>Each oxygen of carbonate ion has -2/3 or -0.67 charge.</u>

<u>Bond order of each carbon‑oxygen bond in the carbonate ion</u> = <u>1.33</u>

Explanation:

The carbonate ion (CO₃²⁻) is an organic compound, in which a carbon atom is covalently bonded to three oxygen atoms. The net formal charge on a carbonate ion is −2.

The carbonate ion is <u>resonance stabilized</u> and has three equivalent resonating structures, which exhibits that all the three carbon-oxygen bonds in a carbonate ion are equivalent.

In the resonance hybrid of carbonate ion,<u> the negative charge is equally delocalized on all the three oxygen atoms. </u>

<u>Thus, each bonded oxygen has -2/3 or -0.67 charge.</u>

<u />

In a carbonate ion there is one double bond oxygen (C=O) and two single bonded oxygen (C-O). Bond order of 1 C=O is 2 and bond order of C-O is 1.

∴ <u>Bond order</u> = sum of all bond orders ÷ number of bonding groups = (2+1+1) ÷ 3 = <u>1.33</u>

8 0
3 years ago
Predict the products of the double replacement reactions given. Check to see that the equations are
Angelina_Jolie [31]

Answer:

Option (2)

Explanation:

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We can see that this will result in a balanced equation, so the answer is Option (2).

4 0
2 years ago
Able 1 Cell Type Operating Cell Potential for Commercial Batteries, E (V) Lithium-iodine Zinc-mercury +2.80 +1.35 Table 2 Standa
IrinaK [193]

Answer:

During the initial cell operation, each reaction is thermodynamically favorable, but the larger operating potential of the lithium-iodine cell indicates that its cell reaction is more thermodynamically favorable.  ( B )

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Explanation:

<u>The major Differences between The Zinc mercury cell and Lithium-iodine cell are :</u>

During the initial cell operation, each reaction is thermodynamically favorable, but the larger operating potential of the lithium-iodine cell indicates that its cell reaction is more thermodynamically favorable.   and

During the initial cell operation, the oxidation of iodine is thermodynamically favorable but the oxidation of mercury is not.

Given the relationship below,

Δ G = -nFE

E = emf of cell ,  G = free energy.

This relationship shows that if E is positive the reaction will be thermodynamically favorable also if E is large it will increase the negativity of free energy  also From the question we can see that with the reduction of mercury the value of E is more positive and this shows that Mercury is thermodynamically unfavorable

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