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sergey [27]
3 years ago
10

PLZ HURRY RUNNING OUT OF TIME

Chemistry
2 answers:
horrorfan [7]3 years ago
4 0

Question:

What title belongs in the circle labeled X in the diagram below?

Answer:

C. Solar System

The solar system carries all the planet and moons and meteors and all that stuff that the solar system has.

Vilka [71]3 years ago
3 0

Solar system makes the most sense of all of these, as a Solar system has all of the things within it, and none of the others make sense

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Consider the reaction of peroxydisulfate ion (S2O2−8) with iodide ion (I−) in aqueous solution: S2O2−8(aq)+3I−(aq)→2SO2−4(aq)+I−
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Answer:

r = 3.61x10^{-6} M/s

Explanation:

The rate of disappearance (r) is given by the multiplication of the concentrations of the reagents, each one raised of the coefficient of the reaction.

r = k.[S2O2^{-8} ]^{x} x [I^{-} ]^{y}

K is the constant of the reaction, and doesn't depends on the concentrations. First, let's find the coefficients x and y. Let's use the first and the second experiments, and lets divide 1º by 2º :

\frac{r1}{r2} = \frac{0.018^{x} x0.036^{y} }{0.027^xx0.036^y}

\frac{2.6x10^{-6}}{3.9x10^{-6}} = (\frac{0.018}{0.027})^xx(\frac{0.036}{0.036})^y

0.67 = 0.67^x

x = 1

Now, to find the coefficient y let's do the same for the experiments 1 and 3:

\frac{r1}{r3} = \frac{0.018x0.036^y}{0.036x0.054^y}

\frac{2.6x10^{-6}}{7.8x10^{-6}} = (\frac{0.018}{0.036})x(\frac{0.036}{0.054})^y

0.33 = 0.5x 0.67^y

0.67 = 0.67^y

y = 1

Now, we need to calculate the constant k in whatever experiment. Using the first :

2.6x10^{-6} = kx0.018x0.036kx6.48x10^{-4} = 2.6x10^{-6}

k = 4.01x10^{-3} M^{-1}s^{-1}[/tex]

Using the data given,

r = 4.01x10^{-3}x1.8x10^{-2}x5.0x10^{-2}

r = 3.61x10^{-6} M/s

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If kerosene has a specific gravity of 0.820, what force will be exerted on the circular bottom of a cylindrical kerosene tank th
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Answer:

F = 774146.534\,N

Explanation:

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P_{bottom} = (0.820)\cdot (1000\,\frac{kg}{m^{3}})\cdot (9.807\,\frac{m}{s^{2}})\cdot (30\,ft)\cdot (\frac{0.305\,m}{1\,ft} )

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The force exerted on the circular bottom is:

F=(73581.921\,Pa)\cdot (\frac{\pi}{4} )\cdot [(12\,ft)\cdot (\frac{0.305\,m}{1\,ft} )]^{2}

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