Consider the equation: 4AI+30² → 2AI²O³ Is this equation balanced? Why or why not?

As you may know, ethyl alcohol, C2H5OH, can be produced by the fermentation of grains, which contain glucose, C6H12O6 → +2C2H5OH

fredd [130]

Answer:

a. 510.6 g of C₂H₅OH are produced from 1kg of glucose

b. 171.1 g of glucose are required

Explanation:

Chemist reaction is this:

C₆H₁₂O₆ → 2C₂H₅OH(l) + 2CO₂(g)

So 1 mol of glucose can produce 1 mol of ethyl alcohol.

First of all, we should convert the mass to g, afterwards to moles

1 kg . 1000 g/ 1kg = 1000 g . 1 mol/180 g = 5.55 moles

Then we can think, this rule of three

1 mol of glucose can produce 2 moles of ethyl alcohol

Then 5.55 moles of glucose may produce the double of moles of C₂H₅OH

(5.55 .2)/1 = 11.1 moles.

Let's convert the moles to mass → 11.1 mol . 46g /1mol = 510.6 g

b. Let's determine the liters of ethyl alcohol we need.

1 gasohol is 10 mL C₂H₅OH / 90 ml of gasoline. We should make a rule of three.

In 90 mL of gasoline we have 10 mL of C₂H₅OH

In 1000 mL (1L) we would have (1000 . 10)/ 90 = 111.1 mL

Now we have to determine the mass of C₂H₅OH that is contained in the volume we have calculated. We must use the density.

Density = Mass /Volume

0.79 g/mL = Mass / 111.1 mL

0.79 g/mL . 111.1 mL = 87.7 g

Now, we convert the mass to moles → 87.7 g . 1mol/ 46g = 1.91 mol

Ratio is 2:1 so 2 moles of C₂H₅OH are produced by 1 mol of glucose

Therefore 1.91 mol would be produced by (1.91 .1)/2 = 0.954 moles

Finally we convert the moles of glucose to mass:

0.954 mol . 180 g/ 1mol = 171.7 grams.

5 0

1 year ago

A compound of barium and oxygen was dissolved in hydrochloric acid to give a solution of barium ion, which was then precipitated

n200080 [17]

Answer:

The empiricial formula of the compound is BaO2

Explanation:

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

Barium and oxygen dissolved in hydrochloric acid gives a solution of barium ion. This was precipitated with an excess of potassium chromate and gives barium chromate.

The original compound weighs 1.345g and gives 2.012g of BaCrO4

<u>Step 2: </u>Calculate moles of BaCrO4

moles of BaCrO4 = mass of BaCrO4 / molar mass of BaCrO4

moles of BaCrO4 = 2.012g / 253.37 g/mol = 0.0079 moles

<u>Step 3</u>: Calculate moles of Ba

Mole ratio for Ba and BaCrO4 is 1:1 so this means for 0.0079 moles of BaCrO4, there are 0.0079 moles of Ba-ion

<u>Step 4:</u> Calculate mass of Ba-ion

Mass of Ba = Moles of Ba / Molar mass of Ba

Mass of Ba = 0.0079 moles * 137.327g/mole = 1.085 g Ba

<u>Step 5:</u> Mass of oxygen

Since the original compound has barium and oxygen, the mass of oxygen is the difference between the original mass and the mass of the Ba-ion

1.345 g - (1.085 g Ba) = 0.260 g O

<u>Step 6:</u> Calculate moles of Oxygen

moles oxygen = mass of oxygen / Molar mass of oxygen

moles oxygen = 0.260g / 16g/mole = 0.01625 moles O

We divide the number of moles by the smallest number of moles which is 0.0079

Ba → 0.0079/0.0079 = 1

O → 0.01625 / 0.0079 ≈ 2

(1.091 g Ba) / (137.3277 g Ba/mol) = 0.00794450 mol Ba

(0.254 g O) / (15.99943 g O/mol) = 0.0158756 mol O

This gives us the empirical formula of BaO2 for this compound

7 0

2 years ago

A student has two compounds in two separate bottles but with no labels on either one. One is an unbranched alkane, octane, C8H18

Aleks [24]

Answer:

a) both substances are insoluble in water

b) both substances are soluble in ligroin

c) both substances suffer combustion, octane produces more CO₂ than hexene.

d) both substances are less dense than waterl, with hexene having the lowest density.

e) only hexene would react with bromine

f) only hexene would react with permanganate

Explanation:

a) both substances are non-polar and water is polar

b) both substances are non-polar and lingroin is non-polar

c) C₈H₁₈ + 17.5O₂ → 8CO₂ + 9H₂O

C₆H₁₂ + 9O₂ → 6CO₂ + 6H₂O

d) water = 997 kg/m³

ocatne = 703 kg/m³

hexene = 673 kg/m³

e) bromine test is used to detect unsaturations

f) permanganate test is used to detect unsaturations

7 0

2 years ago

If Mary drove 525miles in 7hours at an average speed of 75 miles per hour how much faster would her average speed need to be to

Makovka662 [10]

<h3>Answer:</h3>

30 miles per hour faster

<h3>Explanation:</h3>

We are given;

Distance that Mary drove as 525 miles
Time she took as 7 hours
Average speed as 75 miles per hour

We are supposed to determine how fast the average speed would be if she made the trip in 5 hours.

We know that speed is given by dividing distance by time taken.
In this case distance remained constant

Therefore, we can determine the speed when she took 5 hours

Speed = Distance ÷ time

= 525 miles ÷ 5 hours

= 105 hours

Thus, the new speed would be 105 miles per hours

But, initial speed was 75 miles per hour

Therefore, she would travel 30 miles per hour faster to cover the journey in 5 hours.

6 0

2 years ago

Citric acid is a naturally occurring compound. what orbitals are used to form each indicated bond? be sure to answer all parts.2

Rina8888 [55]

Answer : The orbitals that are used to form each indicated bond in citric acid is given below as per the attachment.

Answer 1) σ Bond a : C has $SP^{2}$ O has $SP^{2}$ .

Explanation : The orbitals of oxygen and carbon which are involved in $SP^{2}$ hybridization to form sigma bonds. This is observed at 'a' position in the citric acid molecule.

Answer 2) π Bond a: C has π orbitals and O also has π orbitals.

Explanation : The pi-bond at the 'a'position has carbon and oxygen atoms which undergoes pi-bond formation. And has pi orbitals of oxygen and carbon involved in the bonding process.

Answer 3) Bond b: O $SP^{3}$ H has only S orbital involved in bonding.

Explanation : The bonding at 'b' position involves oxygen $SP^{3}$ hydrogen atoms in it. It has $SP^{3}$ hybridized orbitals and S orbital of hydrogen involved in the bonding.

Answer 4) Bond c: C is $SP^{3}$ O is also $SP^{3}$

Explanation : The bonding involved at 'c' position has carbon and oxygen atoms involved in it. Both the atoms involves the orbitals of $SP^{3}$ hybridized bonds.

Answer 5) Bond d: C atom has $SP^{3}$ C atom has $SP^{3}$

Explanation : At the position of 'd' the bonding between two carbon atoms is found to be $SP^{3}$ . Therefore, the orbitals that undergo $SP^{3}$ hybridization are $SP^{3}$ .

Answer 6) Bond e : C1 containing O $SP^{2}$

C2 is $SP^{3}$

Explanation : The carbon atom which contains oxygen along with a double bond has $SP^{2}$ hybridized orbitals involved in the bonding process; whereas the carbon at C2 has $SP^{3}$ hybridized orbitals involved during the bonding. This is for the 'e' position.

7 0

2 years ago