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2 years ago

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Elements whose names end with âium are usually metals; sodium is one example. Identify a nonmetal whose name also ends with âium

.

1 answer:

Makovka662 [10]2 years ago

8 0

Answer: Helium (He) and Selenium (Se) both are non metals ends with ium sound

Explanation:

He is a inert gas and Se belongs to oxygen family (VI A group)

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Sulfur is composed of three isotopes: 32S, 33S, and 34S. The atomic masses of these isotopes are given below. 32S: 31.97207 amu

elena-14-01-66 [18.8K]

Answer:

Abundance of 32S is 94.41%

Explanation:

The average atomic mass is defined as the sum of the atomic masses of each isotope times its abundance:

Average atomic mass = ∑ Atomic mass istope*Abundance

For the sulfur:

32.07amu = 31.97207X + 32.97146Y + 33.96786*0.0422 <em>(1)</em>

<em>Where X is abundance of 32S and Y abundance of 33S</em>

Also we can write:

1 = X + Y + 0.0422 <em>(2)</em>

0.9578 - X = Y

Because the sum of the abundances = 1

Replacing (2) in (1):

32.07amu = 31.97207X + 32.97146(0.9578 - X) + 33.96786*0.0422

32.07 = 31.97207X + 31.58006 - 32.97146X + 1.43344

-0.9435 = -0.99939X

0.9441 =X

In percentage, abundance of 32S is 94.41%

3 0

1 year ago

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

2 years ago

As photosynthesis occurs in chloroplasts, O2 is produced from _____ via a series of reactions associated with _____. View Availa

Monica [59]

Answer:

H₂O, Photosystem II

Explanation:

Photosynthesis is the process that enables autotrophs such as green plants and algae to generate food using water, carbon dioxide and energy from the sun.
It occurs in two phases, that is, the light-dependent phase and the light-independent phase.
During photosystem II energy from the sun is used to break-down water molecules to yield oxygen and hydrogen ions. Oxygen is released away to the atmosphere while hydrogen ions are used in the next phase to generate ATP molecules.

6 0

2 years ago

How many moles of al2o3 can be produced from the reaction of 10.0 g of al and 19.0 of o2?

Advocard [28]

Answer:

0.185moles of Al₂O₃

Explanation:

Mass of Al = 10g

Mass of O₂ = 19g

Equation of the reaction: 4Al + 3O₂ → 2Al₂O₃

This is the balanced reaction equation.

Solution

From the given parameters, the reactant that would determine the extent of the reaction is Aluminium. It is called the limiting reagent. Oxygen is in excess and it is in an unlimited supply.

Working from the known mass to the unknown, we simply solve for the number of moles of Al using the mass given.

Then from the equation, we can relate the number of moles of Al to that of Al₂O₃ produced:

Number of moles of Al = $\frac{mass}{molar mass}$

= $\frac{10}{27}$

= 0.37mol

From the equation:

4 moles of Al produced 2 moles of Al₂O₃

0.37 mole will yield: $\frac{2 x 0.37}{4}$ = 0.185moles of Al₂O₃

8 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