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Andrej [43]
4 years ago
5

What substance is reduced in the reaction 2Fe 2+ + CI2 = 2Fe 3+ + 2CI- ?

Chemistry
1 answer:
atroni [7]4 years ago
5 0

Answer:

$Cl_{2}$ is reduced in the reaction

Explanation:

The given reaction is

$2Fe^{2+}+Cl_{2} \to 2Fe^{3+}+2Cl^{-}$

The oxidation number of $Fe$ is changed from $+2\, \to \, +3$

$Fe^{2+} \to Fe^{3+}+e^{-}$

And The oxidation number of $Cl$ is changed from $0\, \to \, -1$

$Cl_{2}^{0} + 2e^{-} \to 2Cl^{-}$

Hence, $Fe^{2+}$ is oxidized and $Cl_{2}$ is reduced

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Answer:

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Explanation:

im also 15 , chill, takes jokes rlly well(even if it is sarcastic),bit shy sometimes, and I’ll support you all the way

6 0
2 years ago
Read 2 more answers
What are the two types of numbers in experimental calculations?
FinnZ [79.3K]
The logistics of a proposed larger study

Gain familiarity with the experimental material,

Ensure that treatments are not obviously excessively mild or severe

Check that staff are sufficiently well trained in the necessary procedures

Ensure that all steps in a proposed future experiment are feasible.

Gain some information on variability, although this will not usually be sufficiently reliable to form the basis of power analysis calculations of sample size.

Exploratory experiments can be used to generate data with which to develop hypotheses for future testing. They may “work” or “not work”. They may have no clearly stated hypothesis (“let’s see what happens if..” is not a valid hypothesis on which to base an experiment).

Often they will measure many outcomes (characters). Picking out “interesting looking differences” (known as data snooping) and then doing a hypothesis test to see if the differences are statistically significant will lead to serious overestimation of the magnitude of a response and excessive numbers of false positive results. Such differences should always be tested in a controlled experiment where the hypothesis is stated a priori before the results are published.

Depending on the nature of the data, statistical analysis will often be done using an analysis of variance (ANOVA)

Confirmatory experiments are used to test some relatively simple hypothesis stated a priori. This is the type of experiment mainly considered in this web site.

The basic principles are:

Experiments involve comparisons between two or more groups

Their aim is to test a “null hypothesis” that there is no difference among the groups for the specified outcome.

If the null hypothesis is rejected at a certain level of probability (often 5%) this means that the probability of getting a result as extreme as this or more extreme in the absence of a true effect is 5% (assuming also that the experiment has been properly conducted). So it is assumed that such a difference is likely to be the result of the treatment. But, it could be a false positive resulting from sampling variation.

Failure to reject the null hypothesis does not mean that the treatment has no effect, only that if there is a real effect this experiment failed to detect it. “Absence of evidence is not evidence of absence”.

Experimental subjects need to be independently replicated because individuals (of whatever type) vary. Two subjects can normally be regarded as being independent if they can theoretically receive different treatments.

Subjects need to be assigned to groups, held in the animal house and measured at random in order to minimise the chance of bias (a systematic difference between groups)

As far as possible the experimenter should be “blind” with respect to the treatment group in order to minimise bias.

The experiments need to be powerful, i.e. they should have a high probability of detecting an effect of clinical or scientific importance if it is present.

In many cases a formal experimental designsuch as a “completely randomised”, “randomised block”, “Latin square” etc. design will be used.

In most cases it is useful if the experiment has a wide range of applicability. In other words the results should hold true under a range of different conditions (different strains, both sexes, different diets, different environments etc.). At least some of these factors should be explored using factorial and randomised block designs.

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8 0
3 years ago
Why does it seem to us as if rocks remain the same and don't change
qaws [65]
Most rocks that we encounter in our normal everyday lives are sedimentary rocks.  Sedimentary rocks are rocks that have been worn down gradually over long periods of time.  Because it takes very long periods of time (couple decades) for these rocks to change, it often seems as if they don't change at all, when in reality the change is too small for us to realize it!
3 0
3 years ago
Consider a general reaction
choli [55]

Answer:

a) K = 5.3175

b) ΔG = 3.2694

Explanation:

a) ΔG° = - RT Ln K

∴ T = 25°C ≅ 298 K

∴ R = 8.314 E-3 KJ/K.mol

∴ ΔG° = - 4.140 KJ/mol

⇒ Ln K = - ( ΔG° ) / RT

⇒ Ln K = - ( -4.140 KJ/mol ) / (( 8.314 E-3 KJ/K.mol )( 298 K ))

⇒ Ln K = 1.671

⇒ K = 5.3175

b) A → B

∴ T = 37°C = 310 K

∴ [A] = 1.6 M

∴ [B] = 0.45 M

∴ K = [B] / [A]

⇒ K = (0.45 M)/(1.6 M)

⇒ K = 0.28125

⇒ Ln K = - 1.2685

∴ ΔG = - RT Ln K

⇒ ΔG = - ( 8.314 E-3 KJ/K.mol )( 310 K )( - 1.2685 )

⇒ ΔG = 3.2694

7 0
3 years ago
An oxide of aluminum contains 0.545 g of Al and 0.485 g of O. Find the empirical formula
uysha [10]

Answer:

The answer is: <u>Al2O3</u>

Explanation:

The data they give us is:

  • 0.545 gr Al
  • 0.485 gr O.

To find the empirical formula without knowing the grams of the compound, we find it per mole:

  • 0.545 g Al * 1 mol Al / 27 g Al = 0.02 mol Al
  • 0.485 g O * 1 mol O / 16 g O = 0.03 mol O

Then we must divide the results obtained by the lowest result, which in this case is 0.02:

  • 0.02 mol Al / 0.02 = 1  Al
  • 0.03 mol O / 0.02 = 1.5  O

Since both numbers have to give an integer, multiply by 2 until both remain integers:

  • 1Al * 2 = 2Al
  • 1.5O * 2 = 3O

Now the answer is given correctly:

  • Al2O3

8 0
3 years ago
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