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2 years ago

At a cost of $1600/oz, how much would you have to pay for a solid cubic foot of gold?

Chemistry

1 answer:

Alex777 [14]2 years ago

4 0

We'll begin by calculating the mass in ounce (oz) of a cube foot (ft³) of gold. This can be obtained as follow:

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Density of gold = 19298 oz/ft³

Volume of gold = 1 ft³

Density = mass /volume

19298 = mass / 1

Finally, we shall determine the cost of 19298 oz of gold. This can be obtained as follow:

1 oz = $ 1600

Therefore,

19298 oz = 19298 × 1600

19298 oz = $ 30876800

Therefore, a solid cube foot of gold (i.e 19298 oz) will cost $ 30876800

Learn more: brainly.com/question/15407624

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Propose a mechanism for the following reaction.

raketka [301]

First, recognize that this is an elimination reaction in which hydroxide must leave and a double bond must form in its place. It is likely an E2 reaction. Here is an efficient mechanism:
1) Pre-reaction: Protonate the -OH to make it a good leaving group, water. H2SO4 or any strong H+ donor works. The water is positively charged but still connected to the compound.
2) E2: Use a sterically hindered base, such as tert-butoxide (tButO-) to abstract the hydrogen from the secondary carbon. [You want a sterically hindered base because a strong, non-sterically hindered base could also abstract a hydrogen from one of the two methyl groups on the tertiary carbon, and that leads to unwanted products, which is not efficient]. As the proton of hydrogen is abstracted, water leaves at the same time, creating an intermediate tertiary carbocation, and the 2 electrons in the C-H bond immediately are used to make a double bond towards the partial positive charge.
In the products we see the major product and water, as expected. Even though you have an intermediate, remember that an E2 mechanism technically happens in one step after -OH protonation.

5 0

3 years ago

Compound X has a molar mass of 416.48 g mol

bazaltina [42]

Answer:

P₂Cl₁₀

Explanation:

From the question given above, the following data were obtained:

Molar mass of compound X = 416.48 g/mol

Percentage of phosphorus (P) = 14.87%

Percentage of Chlorine (Cl) = 85.13%

Molecular formula of X =?

Next, we shall determine the empirical formula of compound X. This can be obtained as follow:

P = 14.87%

Cl = 85.13%

Divide by their molar mass

P = 14.87 / 31 = 0.480

Cl = 85.13 / 35.5 = 2.398

Divide by the smallest

P = 0.480 / 0.480 = 1

Cl = 2.398 / 0.480 = 5

Empirical formula of compound X is PCl₅

Finally, we shall determine the molecular formula of compound X. This can be obtained as follow:

Molar mass of compound X = 416.48 g/mol

Empirical formula = PCl₅

Molecular formula =?

Molecular formula= [Empirical formula]ₙ

[PCl₅]ₙ = 416.48

[31 + (35.5 × 5)]ₙ = 416.48

[31 + 177.5]n = 416.48

208.5n = 416.48

Divide both side by 208.5

n = 416.48 / 208.5

n = 2

Molecular formula = [PCl₅]ₙ

Molecular formula = [PCl₅]₂

Molecular formula = P₂Cl₁₀

Therefore, the molecular formula of compound X is P₂Cl₁₀

4 0

3 years ago

Which group on the periodic table would have zero electronegativity because they have a full octet?

bija089 [108]

The group on the periodic table that would have 0 electronegativity due to the fact that their valence shell is full, i.e, have a full octet would be the inert or noble gases. They have a total of 8 electrons in their valence shell and are thus inert and cannot strongly attract electrons toward itself, from neighbouring atom electrons as it does not need to.

3 0

3 years ago

The terms Q and K refer to reaction components at non-equilibrium and equilibrium conditions, respectively. For a forward reacti

daser333 [38]

Answer:

The value of Q must be less than that of K.

Explanation:

The difference of K and Q can be understood with the help of an example as follows

A ⇄ B

In this reaction A is converted into B but after some A is converted , forward reaction stops At this point , let equilibrium concentration of B be [B] and let equilibrium concentration of A be [A]

In this case ratio of [B] and [A] that is

K = [B] / [A] which is called equilibrium constant.

But if we measure the concentration of A and B ,before equilibrium is reached , then the ratio of the concentration of A and B will be called Q. As reaction continues concentration of A increases and concentration of B decreases. Hence Q tends to be equal to K.

Q = [B] / [A] . It is clear that Q < K before equilibrium.

If Q < K , reaction will proceed towards equilibrium or forward reaction will

proceed .

8 0

3 years ago

Hi, can someone help me balance this chemistry equation:<br> H2SO4 + RbOH -> Rb2SO4 + H2O

zloy xaker [14]

H2SO4 + 2RbOH -> Rb2SO4 + 2H2O

If you want an explanation, keep reading.

In the first portion, there are two hydrogen ions and four sulfate ions.

The second portion has one rubidium ions and one hydroxide ion.

On the other side of the equation, in order to keep those two rubidiums balanced, you'll need to add a two at the beginning of the second portion, but in that process you are giving a second hydroxide value.

Back to the right side, there is there is water (H2O).

On the first portion, there were two hydrogen ions. The second portion also has two hydroxides because of the value change (adding the two to the front).

So on the fourth portion, you'd have to add another two so you could balance the four hydrogen ions (H2 and 2OH) and the two oxygen ions (2OH).

I hope this was easy to understand.

6 0

3 years ago