Me i need to turn this in really !

1. a chemical equation is shown below. hgs + o2 → hgo + so2 what are the coefficients that should be added to balance this equation? use complete sentences to explain your answer. explain how this chemical reaction demonstrates the conservation of mass.

2. part 1: name the type of chemical reaction that occurs when aluminum (al) reacts with copper nitrate (cu(no3)2).

part 2: explain why aluminum does not react with potassium nitrate (kno3) although it reacts with copper nitrate.

3.
the table shows the amount of radioactive element remaining in a sample over a period of time.

radioactive decay rate

amount of radioactive sample
(grams) time
(hours)
186.0 0
147.0 7
117.0 14
93.0 21
23.3 63

part 1: what is the half-life of the element? explain how you determined this.
part 2: how long would it take 320 g of the sample to decay to 2.5 grams? show your work or explain your answer.

Question#1:

The given chemical equation is the following:

On the left side of the chemical equation, there are 1 Mercury atom, 1 Sulfur (or Sulphur) atom, and two oxygen atoms, whereas on the right side of the chemical equation, there are 1 Mercury atom, 1 Sulfur (or Sulphur) atom, but 3 oxygen atoms. Now let's balance the given equation!

The balanced equation will be:

The left side of the above chemical equation has the following configuration:

Number of Mercury atoms = 2*1 = 2.

Number of Sulfur (or Sulphur) atoms = 2*1 = 2.

Number of oxygen atoms = 3*2 = 6.

The right side of the above chemical equation has the following configuration:

Number of Mercury atoms = 2*1 = 2.

Number of Sulfur (or Sulphur) atoms = 2*1 = 2.

Number of oxygen atoms = 2 + (2*2) = 2+4 = 6.

Hence, on left hand side of the chemical equation, the coefficient 2 has been added to the Mercury sulphide (HgS) and the coefficient 3 has been added to oxygen molecule (O₂), whereas on the right hand side of the chemical equation, the coefficient 2 has been introduced with the Mercury oxide (aka Mercuric oxide)—HgO—and the coefficient 2 has been added with the Sulphur dioxide (SO₂).

In this context, the conservation of mass, in simple—non-bookish—words, means that there are equal number atoms on both sides of the equation. No matter is either created during or after the reaction not any matter was destroyed. As in the case of aforementioned balanced equation, there are 2 mercury, 2 suphur and 6 oxygen atoms on both sides of the equation, demostrating the conservation of mass after the reaction.

Question#2:

PART-1:

To tell you the type of the chemical reaction occuring between aluminum and copper nitrate, let me explain you the reaction first, which is, trust me, the saddest reaction you'd ever see. The (balanced) reaction is the following:

Why is this the saddest reaction? Well, according to the activity series (aka reactivity series), Aluminum (Al) is more reactive than Copper (Cu). Since Al is more reactive than Cu, it (Al) kicks out Cu from Cu(NO₃)₂ compound and takes its place, leaving Cu alone after the reaction. :(

Such kind of reaction, where one element takes the place of another, leaving the other alone is called single replacement reaction.

Hence, the correct answer is: Single replacement reaction.

PART-2: The reason why aluminum doesn't react with the potassium nitrate is that potassium is more reactive than aluminum. It means that it is impossible for the aluminum to break the bond between potassium and nitrate and takes the place of potassium. In this case, in informal way, you can say that potassium nitrate is a happily-together compound, and their bond cannot be broken by alumium, since aluminum is less reactive than potassium.

Question#3:

PART-1: In simple words, the half-life is the time the radioactive sample takes for half of its atoms to decay. In the table, as you can see, the sample at time t = 0 hours is 186 grams. Now find the time this radioactive sample will take for half of its grams to decay, which in this case is 186/2 = 93 grams...

Well, in the table, after time t = 21 hours, 93 grams of radioactive sample remains (which is half of its original/initial amount). Hence, the half life of this radioactive sample is:   hours.

PART-2: In the previous part, we have found the half life of the radioactive sample, which we shall use at the end of this problem. First let us find out number of half-lives that passes when 320 g of the sample will decay to 2.5 g.

Use the following formula:

--- (A)

Where,

n = number of half lives that passes = ?

= initial mass of the sample = 320 g

= Mass left after decaying (or after t amount of time) = 2.5 g

Plug the values in equation (A):

In Part-1, we have found the half life of this radioactive substance is 21. Now to get the time to decay t, all you need to do is to multiply n with the half life (21 hours).

Hence, it will take 147 hours to decay to 2.5 grams from 320 gram.

-i

o conteúdo de calor de um sistema é igual à entalpia apenas para um sistema que é constante