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3 years ago

What is another way that diamonds are useful

1 answer:

7 0

Diamonds are common in the making of jewelry, but they are also used industrially because of their hardness. They are extremely useful in cutting, grinding, and drilling other materials.

You might be interested in

The two methods in the lab for measuring the volume of a geometric solid object are weighing it on the balance and measuring it

Yakvenalex [24]

Answer:

Measuring with a ruler and using final volume minus initial volume

Explanation:

You can measure the volume of a geometric object by measuring its sides with a ruler and calculating the volume according to the corresponding formula for each object. For example, for a rectangular prism it would be

$volume=length*width*height$

You can also measure the volume of an object by measuring how much water it displaces. To do this you have to fill a measuring cylinder with enough water for the object to be completely submerged and take note of the volume. Then, add the object and note again the volume of the water+object. The difference between both is the volume of the object.

$Volume of the object= volume of water and object - volume of water$

The advantage of the second method is that it can be used for objects with irregular shapes as long as they do not float.

8 0

3 years ago

Calculate the molarity of 0.289 moles of FeC13 dissolved in 120 ml of solution

yanalaym [24]

Answer:

2.41 M

Explanation:

The molarity is the moles of FeCl3 over the liters of solution. Since you're given mL you need to change it to L which is 0.12 L. 0.289 divided by 0.12 is your answer

5 0

3 years ago

Suppose you were told that the above reaction was a substitution reaction but you were not told the mechanism. Evaluate the foll

Zarrin [17]

Answer:

The alkyl halide is secondary

The nucleophile is a poor nucleophile

The solvent is a protic solvent

The product is racemic

Explanation:

The reaction is shown in the image attached.

Alkyl halides undergo nucleophilic substitution by two mechanisms; SN1 and SN2. The particular mechanism that applies depends on;

I) structure of the alkyl halide

ii) nature of the nucleophile

iii) nature of the solvent

Looking at the reaction under review, we can see from the structure that the alkyl halide is a secondary alkyl halide. A secondary alkyl halide may undergo substitution via SN1 or SN2 mechanism depending on the conditions of the reaction.

If the nucleophile is poor, and the solvent is protic, SN1 mechanism is favoured over SN2 mechanism. Since CH3CH2OH is a poor nucleophile and ethanol is a protic solvent, we expect the reaction to proceed via SN1 mechanism leading to the formation of a racemic product.

The organic product is also shown in the second image attached.

3 0

2 years ago

if 14.0 g of aluminium reacts with excess sulfuric acid to produce 75.26 g of aluminium sulfate, what is the percent yield?

vlada-n [284]

Taking into account definition of percent yield, the percent yield for the reaction is 84.88%.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 Al + 3 H₂SO₄ → Al₂(SO₄)₃ + 3 H₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Al: 2 moles
H₂SO₄: 3 moles
Al₂(SO₄)₃. 1 mole
H₂: 3 moles

The molar mass of the compounds is:

Al: 27 g/mole
H₂SO₄: 98 g/mole
Al₂(SO₄)₃: 342 g/mole
H₂: 2 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Al: 2 moles ×27 g/mole= 54 grams
H₂SO₄: 3 moles ×98 g/mole= 294 grams
Al₂(SO₄)₃: 1 mole ×342 g/mole= 342 grams
H₂: 3 moles ×2 g/mole= 6 grams

<h3>Mass of aluminium sulfate formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 54 grams of aluminium form 342 grams of aluminium sulfate, 14 grams of aluminium form how much mass of aluminium sulfate?

$mass of aluminium sulfate=\frac{14 grams of aluminium x342 grams of aluminium sulfate}{54 grams of aluminium}$

<u><em>mass of aluminium sulfate= 88.67 grams</em></u>

Then, 88.67 grams of aluminium sulfate can be produced if 14.0 g of aluminium reacts with excess sulfuric acid.

<h3>Percent yield</h3>

The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.

The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:

$percent yield=\frac{actual yield}{theorical yield} x100$

where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

<h3>Percent yield for the reaction in this case</h3>

In this case, you know: