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AURORKA [14]
3 years ago
15

8. What behavior do all liquid or plastic solids near a heat source exhibit? L

Chemistry
1 answer:
LuckyWell [14K]3 years ago
7 0

Answer:

Explanation:

One of the properties of a liquid is that, <u>it's particles move freely (not tightly packed)</u> hence the reason for <u>it's free flowing (no definite shape)</u> when shaken in a container, unlike a solid whose <u>particles are tightly packed with restricted/no movement</u> and hence the reason for it's compactness and it's definite shape.

When a plastic solid (whose particle is tightly packed and have a restricted movement/no movement) is placed near a heat source, <u>it's particles gains energy in the process and starts to move (though slightly free) and become less tightly packed</u> hence the reason it is observed that plastic solids near a heat source melts.

From the above, it can be deduced that a liquid and a plastic solid near a heat source <u>have there particles move freely (and not tightly packed) hence making the two substances flow freely with no definite shape.</u>

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Notice that "SO4" appears in two different places in this chemical equation. SO42− is a polyatomic ion called "sulfate." What nu
vichka [17]

Answer:

3.

Explanation:

Hello,

In this case, it is convenient to write the chemical reaction as:

CaSO_4+AlCl_3\rightarrow CaCl_2+Al_2(SO_4)_3

Which balanced turns out:

3CaSO_4+2AlCl_3\rightarrow 3CaCl_2+Al_2(SO_4)_3

Thus the number that should be in front of the calcium sulfate is 3 in order to balance the reaction.

Best regards.

8 0
3 years ago
(I00p)A chemical equation example of a replacement of 1 ion for another in a compound! I need an example
snow_tiger [21]
A single replacement reaction could look like this:

2FeCl3 + 3Ba ➡️ 3BaCl2 + 2Fe

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8 0
3 years ago
Calculate the unit cell edge length for an 85 wt% fe-15 wt% v alloy. All of the vanadium is in solid solution, and, at room temp
Lady bird [3.3K]

Answer is 0.289nm.

Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.

wt % of Fe in Fe-V alloy = 85%

wt % of V in Fe-V alloy = 15%

We need to calculate edge length of the unit cell having bcc structure.

Using density formula,

\rho_{ave}=\frac{Z\times M_{ave}}{a^3\times N_A}

For calculating edge length,

a=(\frac{Z\times M_{ave}}{\rho_{ave}\times N_A})^{1/3}

For calculating M_{ave}, we use the formula

M_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{V}}{M_V}}

Similarly for calculating (\rho)_{ave}, we use the formula

\rho_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{\rho_{Fe}}+\frac{(wt\%)_{V}}{\rho_V}}

From the periodic table, masses of the two elements can be written

M_{Fe}= 55.85g/mol

M_{V}=50.941g/mol

Specific density of both the elements are

(\rho)_{Fe}=7.874g/cm^3\\(\rho)_{V}=6.10g/cm^3

Putting  M_{ave} and \rho_{ave} formula's in edge length formula, we get

a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}}  \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}}  \right )}  \right ]^{1/3}

a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}}  \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}}  \right )}  \right ]^{1/3}

By calculating, we get

a=2.89\times10^{-8}cm=0.289nm

7 0
3 years ago
Can anyone help me on this please?
Bad White [126]
This question is testing to see how well you understand the "half-life" of radioactive elements, and how well you can manipulate and dance around them.  This is not an easy question.

The idea is that the "half-life" is a certain amount of time.  It's the time it takes for 'half' of the atoms in any sample of that particular unstable element to 'decay' ... their nuclei die, fall apart, and turn into nuclei of other elements.

Look over the table.  There are 4,500 atoms of this radioactive substance when the time is 12,000 seconds, and there are 2,250 atoms of it left when the time is ' y ' seconds.  Gosh ... 2,250 is exactly half of 4,500 !  So the length of time from 12,000 seconds until ' y ' is the half life of this substance !  But how can we find the length of the half-life ? ? ?

Maybe we can figure it out from other information in the table !

Here's what I found:

Do you see the time when there were 3,600 atoms of it ? 
That's 20,000 seconds.

... After one half-life, there were 1,800 atoms left.
... After another half-life, there were 900 atoms left.
... After another half-life, there were 450 atoms left. 

==>  450 is in the table !  That's at 95,000 seconds.

So the length of time from 20,000 seconds until 95,000 seconds
is three half-lifes.

The length of time is (95,000 - 20,000) = 75,000 sec

                                     3 half lifes = 75,000 sec

Divide each side by 3 :   1 half life = 25,000 seconds

There it is !  THAT's the number we need.  We can answer the question now.

==> 2,250 atoms is half of 4,500 atoms.

==> ' y ' is one half-life later than 12,000 seconds

==> ' y ' = 12,000 + 25,000

         y   = 37,000 seconds  .

Check: 
Look how nicely 37,000sec fits in between 20,000 and 60,000 in the table.

As I said earlier, this is not the simplest half-life problem I've seen.
You really have to know what you're doing on this one.  You can't
bluff through it.


7 0
3 years ago
Compare and contrast an electric generator and a battery??
anygoal [31]
Both generators and batteries both convert a form of energy into electrical energy. In a battery, a chemical reaction takes place which converts chemical energy into electrical energy. In a generator however, many times mechanical energy is being converted into electrical energy. A process called electromagnetic induction can take place in some generator which is where an electromagnet is used to help conduct electricity. hope this helped!!! 
3 0
3 years ago
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