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Charra [1.4K]
2 years ago
8

The temperature of the system is highest when it is a

Chemistry
1 answer:
xz_007 [3.2K]2 years ago
8 0

Answer:

gas

Explanation:

its either gas or plasma but gas should be it bc liquid and solid are cold

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RINLAOTITE<br> Un scramble
AlladinOne [14]

Answer:

Literation

Explanation:

4 0
3 years ago
What do you think happens when you add an emulsifier to oil and vinegar?
julsineya [31]
They separate because of different densities.
7 0
2 years ago
Read 2 more answers
Pls help me with this
Zina [86]

Answer:

[I_2]=[Br]=0.31M

Explanation:

Hello there!

In this case, according to the given information, it is possible for us to set up the following chemical equation at equilibrium:

I_2+Br_2\rightleftharpoons 2IBr

Now, we can set up the equilibrium expression in terms of x (reaction extent) whereas the initial concentration of both iodine and bromine is 0.5mol/0.250L=2.0M:

K=\frac{[IBr]^2}{[I_2][Br_2]} \\\\1.2x10^2=\frac{(2x)^2}{(2.0-x)^2}

Thus, we solve for x as show below:

\sqrt{1.2x10^2} =\sqrt{\frac{(2x)^2}{(2.0-x)^2}} \\\\10.95=\frac{2x}{2.0-x}\\\\21.91-10.95x=2x\\\\21.91=12.95x\\\\x=\frac{21.91}{12.95} \\\\x=1.69M

Therefore, the concentrations of both bromine and iodine are:

[I_2]=[Br]=2.0M-1.69M=0.31M

Regards!

8 0
3 years ago
How does silicon dioxide become a covalent bond. (explain)
tatuchka [14]
Silicon dioxide is SiO2.  Silicon has 4 valence electrons, while each oxygen has 6 valence electrons.  This can be shown as 
     **             *                   *
**  O *       *  Si   *        *   O  **
      *               *                 **
At points where there is one valence electron, represented by a lone *, the electrons will be 'shared' between the atoms.  This will make silicon dioxide appear as
   **           *--------------- *
**  O *--*  Si   * ------ *   O  **
      *------- *                 **
, as the lines with no arrows indicate that each electron moves between the atoms, and does not stay with one forever.
4 0
3 years ago
131i has a half-life of 8.04 days. assuming you start with a 1.53 mg sample of 131i, how many mg will remain after 13.0 days ___
inn [45]
For this problem we can use half-life formula and radioactive decay formula.

Half-life formula,
t1/2 = ln 2 / λ

where, t1/2 is half-life and λ is radioactive decay constant.
t1/2 = 8.04 days

Hence,         
8.04 days    = ln 2 / λ                         
λ   = ln 2 / 8.04 days

Radioactive decay law,
Nt = No e∧(-λt)

where, Nt is amount of compound at t time, No is amount of compound at  t = 0 time, t is time taken to decay and λ is radioactive decay constant.

Nt = ?
No = 1.53 mg
λ   = ln 2 / 8.04 days = 0.693 / 8.04 days
t    = 13.0 days 

By substituting,
Nt = 1.53 mg e∧((-0.693/8.04 days) x 13.0 days))
Nt = 0.4989 mg = 0.0.499 mg

Hence, mass of remaining sample after 13.0 days = 0.499 mg

The answer is "e"

8 0
2 years ago
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