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Lubov Fominskaja [6]
3 years ago
12

Where should you place the title when constructing a graph?

Chemistry
2 answers:
Goshia [24]3 years ago
7 0
It should be on the top middle where you work and be on the side and bottom
Aliun [14]3 years ago
3 0
At the top since the numbers/data would be on the side and the bottom
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Carbon tetrachloride can be produced by the following reaction: Suppose 1.20 mol of and 3.60 mol of were placed in a 1.00-L flas
hjlf

The given question is incomplete. The complete question is :

Carbon tetrachloride can be produced by the following reaction:

CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)

Suppose 1.20 mol CS_2(g) of and 3.60 mol of Cl_2(g)  were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol  of CCl_4. Calculate equilibrium constant at the unknown temperature.

Answer: The equilibrium constant at unknown temperature is 0.36

Explanation:

Moles of  CS_2 = 1.20 mole

Moles of  Cl_2 = 3.60 mole

Volume of solution = 1.00  L

Initial concentration of CS_2 = \frac{moles}{volume}=\frac{1.20mol}{1L}=1.20M

Initial concentration of Cl_2 = \frac{moles}{volume}=\frac{3.60mol}{1L}=3.60M

The given balanced equilibrium reaction is,

                 CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)

Initial conc.         1.20 M        3.60 M                  0                  0

At eqm. conc.     (1.20-x) M   (3.60-3x) M   (x) M        (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[S_2Cl_2]\times [CCl_4]}{[Cl_2]^3[CS_2]}

Now put all the given values in this expression, we get :

K_c=\frac{(x)\times (x)}{(3.60-3x)^3\times (1.20-x)}

Given :Equilibrium concentration of CCl_4 , x = \frac{moles}{volume}=\frac{0.72mol}{1L}=0.72M

K_c=\frac{(0.72)\times (0.72)}{(3.60-3\times 0.72)^3\times (1.20-0.72)}

K_c=0.36

Thus equilibrium constant at unknown temperature is 0.36

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3 years ago
He orbital period of an object is 2 × 107 s and its total radius is 4 × 1010 m
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The answer is 12,560  

Explanation:

The orbital period is the time a given cosmic question takes to finish one circle around another protest and applies in space science as a rule to planets or space rocks circling the Sun, moons circling planets, exoplanets circling different stars, or double stars. Mercury, for instance, has an orbital time of 88 days while it takes Jupiter around 11.86 years. The time of the Earth's circle is generally thought to be 365 days as timetables appear.

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In a physical change of matter,​
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Answer: C. no new substances are formed<span>
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<span>In the physical change of matter, there is no new substance that is formed. It is only the appearance of the matter that is being changed and not its chemical composition. Cutting, tearing and grinding are only some of the examples that exhibit physical change. </span></span>

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Calculate the mass percent composition of carbon in C4F8
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My work to your question

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