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Zielflug [23.3K]
3 years ago
12

Which of the following statements is a testable hypothesis for an experiment

Chemistry
2 answers:
Arlecino [84]3 years ago
6 0

Answer:

Explanation:

There are many different testable hypothesis and experiments that can be made in order to prove this. Since there are no available statements let me provide one.

Testing and Comparing the excrement of wild and tamed horses.

This is because the excrement of tame horses will show higher levels of man-made chemicals and additives due to the different man-made horse food that the owners are giving them. Wild horses on the other hand will not since they nourish themselves on naturally grown grass.

valentinak56 [21]3 years ago
3 0

The teeth of wild horses wear down faster than the teeth of a tame horse.

5 stars?

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Use one of your experimentally determined values of k, the activation energy you determined, and the Arrhenius equation to calcu
stealth61 [152]

Answer:

Explanation:

Use one of your experimentally determined values of k, the activation energy you determined, and the Arrhenius equation to calculate the value of the rate constant at 25 °C. Alternatively, you can simply extrapolate the straight line plot of ln(k) vs. 1/T in your notebook to 1/298 , read off the value of ln(k), and determine the value of k. Please put your answer in scientific notation. slope=-12070, Ea=100kJ/mol, k= 0.000717(45C), 0.00284(55C), 0.00492(65C), 0.0165(75C), 0.0396(85C)

Explanation;

According to Arrhenius equation:

i.e. ln(k2/k1) = -Ea/R (1/T2 - 1/T1)

Where, k1 = 0.000717, T1 = 45 oC = (45+273) K = 318 K

T2 = 25 oC = (25 + 273) K = 298 K

i.e. ln(k2/0.000717) = -12070 (1/298 - 1/318)

i.e. ln(k2/0.000717) = -2.54738

i.e. k2/0.000717 = e^{-2.54738}

= 0.078286

Therefore, the required constant (k2) = 0.078286 * 0.000717 = 5.61*10^-^5

6 0
3 years ago
What kind of non-flowering plant produces seeds instead of spores?
Pie

Answer:

The answer to your question is Ferns

Explanation:

Ferns are called nonflowering plants and produce spores instead of seeds.

6 0
3 years ago
(( PLEASE HELP QUICK ))Which expression can be used to calculate the ratio of the rate of effusion of gas A to the rate of effus
borishaifa [10]

Answer:

The square root of the molar mass of B ÷ the square root of the molar mass of A  

Explanation:

Graham’s Law applies to the effusion of gases:

The rate of effusion (r) of a gas is inversely proportional to the square root of its molar mass (M).

r \propto \dfrac{1}{\sqrt{M}}

If you have two gases A and B, the ratio of their rates of effusion is

\dfrac{r_{\text{A}}}{r_{\text{B}}} = \sqrt{\dfrac{M_{\text{B}}}{M_{\text{A}}}}

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Bronze alloy and porcelain dentures
5 0
3 years ago
What’s the correct answer
butalik [34]
He ^ 2+

Helium has two electrons the 2+ means that it has lost its two electrons leaving it with none.
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3 years ago
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