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weqwewe [10]
3 years ago
13

The number of moles of aluminum oxide. Urgent!!!! Please.

Chemistry
1 answer:
Nimfa-mama [501]3 years ago
5 0
<h2>Answer: 0.035 mol</h2>

<h3>Explanation:</h3>

Based on the balanced equation, for every mole of iron (II) oxide reacted, ¹/₃ the number of moles of aluminum oxide is produced. This is because the mole ratio of FeO : Al₂O₃ is 3 : 1.

∴ the moles of Al₂O₃ produced = 0.106 mol × ¹/₃

                                                    = 0.035 mol

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Identify the covalent compounds based on the names of the compounds. barium nitrate dinitrogen tetroxide boron trifluoride ammon
noname [10]

Covalent compounds are composed of atoms that are linked via covalent bonds i.e. bonds formed by mutual sharing of electrons. This is in complete contrast to ionic compounds which are held together by ionic bonds, i.e. bonds formed by complete transfer of electrons from one atom to the other.

In the given examples we have:

Barium nitrate: Ba(NO3)2 - Ionic

Dinitrogen tetroxide: N2O4- Covalent

Boron trifluoride: BF3-Covalent

Ammonium sulfate: (NH4)2SO4- Ionic

Carbon tetrachloride: CCl4- Covalent

Barium chloride: BaCl2 - Ionic



5 0
3 years ago
Read 2 more answers
In a room full of air, the air is mainly composed of Nitrogen and Oxygen molecules (both at room temperature). Find (to two sign
zloy xaker [14]

Explanation:

Expression for the v_{rms} speed is as follows.

            v_{rms} = \sqrt{\frac{3kT}{M}}

where,   v_{rms} = root mean square speed

                     k = Boltzmann constant

                    T = temperature

                    M = molecular mass

As the molecular weight of oxygen is 0.031 kg/mol and R = 8.314 J/mol K. Hence, we will calculate the value of v_{rms} as follows.

               v_{rms} = \sqrt{\frac{3kT}{M}}

                            = \sqrt{\frac{3 \times 8.314 J/mol K \times 309.02 K}{0.031 kg/mol}}

                            = 498.5 m/s

Hence, v_{rms} for oxygen atom is 498.5 m/s.

For nitrogen atom, the molecular weight is 0.028 kg/mol. Hence, we will calculate its v_{rms} speed as follows.

                v_{rms} = \sqrt{\frac{3kT}{M}}

                              = \sqrt{\frac{3 \times 8.314 J/mol K \times 309.92 K}{0.028 kg/mol}}

                              = 524.5 m/s

Therefore, v_{rms} speed for nitrogen is 524.5 m/s.

3 0
3 years ago
Olive oil has a density of 0.92 g/mL. How much would 1 liter of olive oil weigh in grams?
svp [43]
Dimension analysis is to be used to solve this problem. First convert 1L to milliliters. That is equivalent to 1000 ml. Then by dimension analysis, multiply the volume ( 1000ml) to the density of oil (0.92 g/ml) resulting to the answer: 920 grams. 
4 0
3 years ago
What kind of glacier has pieces that breaks off as icebergs
forsale [732]
There is no specific name for a glacier that break off as an iceberg. However, the part of the glacier in which this happens is called the "zone of wastage". Chunks break off in a process called "calving".
5 0
3 years ago
A site in Pennsylvania receives a total annual deposition of 2.688 g/mof sulfate from fertilizer and acid rain. The ratio by mas
sertanlavr [38]

According to the statement

2.12 x 10^4 lbs pounds of CaCO₃ are needed to neutralize this acid

<h3>What is neutralization?</h3>

A chemical reaction in which an acid and a base react quantitatively with each other is known as neutralization or neutralization. In a water reaction, neutralization ensures that there is no excess of hydrogen or hydroxide ions in the solution.

<h3>According to the given information:</h3>

The equation of the neutralization reaction between H2SO4 and CaCO3.

CaCO3 + H2SO4 → CaSO4 + H2CO3

H2CO3 dissociate to water and carbon dioxide.

        CaCO3 + H2SO4 → CaSO4  + H2O + CO2

Now solving for the mass of CaCO3 needed to neutralize the acid.

mass of CaCO3 = 9460 Kg H2SO4  × \frac{1000 \mathrm{~g}}{1 \mathrm{~kg}} \times \frac{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO} 4}{98.1 \mathrm{gH}_2 \mathrm{SO}_4} \times \frac{1 \mathrm{~mol} \mathrm{CaCO}\left(\mathrm{O}_3\right.}{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO}_4}\times \frac{100.1 \mathrm{~g} \mathrm{CaCO}_3}{1 \mathrm{~mol} \mathrm{CaCO}_3} \times \frac{2.205 \mathrm{lb}}{1000 \mathrm{~g}}

= 21284.56606

mass of CaCO3 =  2.12 x 10^4 lbs

2.12 x 10^4 lbs pounds of CaCO₃ are needed to neutralize this acid.

To know more about neutralization visit:

brainly.com/question/12498769

#SPJ4

4 0
2 years ago
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