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3 years ago

Define "Dispersed System", "Dispersion Medium", and "Dispersed Particles

Chemistry

1 answer:

ollegr [7]3 years ago

4 0

Answer:

Dispersion system is a system in which certain particles are scattered in a continuous liquid or solid medium. The two phases present in this system are the dispersed particles and dispersion medium. These phases may or may not be present in the same state.

In a dispersion system, the particles that are dispersed are known as the dispersed particles and the medium in which the particles are dispersed is known as the dispersion medium.

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What is the pH of a 0.1 M phosphate buffer (pKa = 6.86) that contains equal amounts of acid and conjugate base?

Anvisha [2.4K]

Answer : The pH of a 0.1 M phosphate buffer is, 6.86

Explanation : Given,

$pK_a=6.86$

Concentration of acid = 0.1 M

Concentration of conjugate base (salt) = 0.1 M

Now we have to calculate the pH of buffer.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

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

$pH=6.86+\log (\frac{0.1}{0.1})$

$pH=6.86$

Therefore, the pH of a 0.1 M phosphate buffer is, 6.86

4 0

3 years ago

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.

Experiments to explore relationships between variables. A typical example would be a growth curve or a dose-response relationship. In these experiments the aim is often to test whether the two variables are associated, and if so, what is the nature of that relationship. The typical statistical analysis involves correlation and/or regression.

8 0

3 years ago

Carbon-14 (14C) dating assumes that the carbon dioxide on the Earth today has the same radioactive content as it did centuries a

Nataly [62]

Answer: The tree was burned 16846.4 years ago to make the ancient charcoal

Explanation:

The equation used to calculate rate constant from given half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

where,

$t_{1/2}$ = half life of the reaction = 5715 years

Putting values in above equation, we get:

$k=\frac{0.693}{5715yrs}=1.21\times 10^{-4}yrs^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $1.21\times 10^{-4}yr^{-1}$

t = time taken for decay process = ? yr

$[A_o]$ = initial amount of the sample = 100 grams

[A] = amount left after decay process = 13 grams

Putting values in above equation, we get:

$1.21\times 10^{-4}=\frac{2.303}{t}\log\frac{100}{13}\\\\t=16864.4yrs$

Hence, the tree was burned 16846.4 years ago to make the ancient charcoal

8 0

3 years ago

A gas occupies 18.5L at stp. what volume will it occupy at 735 torr and 57°c?

olga nikolaevna [1]

The volume that will be occupied at 735 torr and 57 c is 23.12 L

calculation

At STP temperature=273 k and pressure=760 torr

by use of combined gas formula

that is P1V1/T1= P2V2/T2

where; P1 =760 torr

T1= 273 K

V1= 18.5 L

P2= 735 torr

T2= 57+273= 330 K

V2=?

by making V2 the formula of subject

V2= T2P1V1/P2T1

V2= [(18.5L x 330 k x 760 torr)/(735 torr x 273 k)]= 23.12 L

5 0

3 years ago

The number of glyceraldehyde-3-phosphate molecules that would be produced from 24 turns of the calvin cycle would be

SVEN [57.7K]

The Calvin cycle, also called the light-independent or carbon fixation reactions, is the second stage of photosynthesis where water, and carbon dioxide (CO2) from air, are converted into organic compounds (i.e. sugars) using the energy from short-lived electronically excited carriers (ATP and NADPH) for the reactions. These organic compounds can then be used by the producing organism (i.e. plants) and the animals that feed on it.

One product of the Calvin cycle is the glyceraldehyde-3-phosphate (G3P), which is later on used in the production of glucose and in the regeneration of Ribulose 1,5-bisphosphate (RuBP), which is an organic compound essential to the reactions in the cycle.

One turn of Calvin Cycle produces 2 G3P molecules, each comprising of 3 carbons. This gives a total of 6 carbons. Five (5) of these carbons will be used to regenerate RuBP and only 1 will be available to form a surplus G3P later on. This surplus G3P will be used for the production of glucose (a 6-carbon sugar).

Thus, 3 turns of the carbon cycle will produce 1 surplus G3P. There are 8 sets of 3-turns in 24 cycles, therefore,

1 net G3P molecule * 8 sets of 3-turns = 8 G3P molecules

Therefore, there are 8 net or surplus G3P molecules produced for 24 cycles of the Calvin Cycle. The total G3P molecules produced, including the ones that participated in the regeneration of RuBP would be 48 G3Ps.

For every 3 turns, 6 G3P molecules are produced, 5 of which will be used in the regeneration of RuBP and 1 will be the net or surplus, to be used for the production of glucose. The 48 G3Ps then come from the calculation,

6 total G3P molecules * 8 sets of 3-turns = 48 G3P molecules

The figure below shows the products of the cycle after 3 turns (Source: https://ka-perseus-images.s3.amazonaws.com/2f4bdc8f8275834d3f5ef434d93bf16b991b2357.png).

7 0

3 years ago