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3 years ago

Select all the correct answers.

Chemistry

2 answers:

Iteru [2.4K]3 years ago

8 0

Answer:

An experiment may have had errors that the scientists didn’t recognize.

The results of individual scientists may be influenced by bias.

Explanation: PLATO

Brut [27]3 years ago

5 0

Hi!

The reason for reviewing and replicating work of other scientists is influenced by two factors listed in the answer:

1. An experiment may have had errors that the scientist did not recognize. (unintentional)

2. The results of individual scientists may be influenced by bias. (intentional)

Personal motivations may often influence a scientist to manipulate the results to best suit his/her hypothesis, consequently leading to scientific fraud.

While other options listed are elements of the scientific method, they are not the reasons behind the review and replication of a scientist's work by others.

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An organic compound that contains only carbon and hydrogen and a triple bond (all the other bonds are single bonds) is classifie

Basile [38]

Answer:

c. alkyne.

Explanation:

Hello there!

In this case, according to the attached file, it turns out possible for us to say that alkanes have only single-bonded carbon atoms, alkenes have two double-bonded carbon atoms and alkynes have two triple-bonded carbon atoms.

In such a way, according to the aforementioned definition, we infer that that an organic compound that contains only carbon and hydrogen and a triple bond (all the other bonds are single bonds) is classified as c. alkyne.

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3 years ago

A 50.0 mL solution of 0.133 M KOH is titrated with 0.266 M HCl . Calculate the pH of the solution after the addition of each of

shtirl [24]

Answer:Calculate The PH Of The Solution After The Addition Of The Following Amounts Of HCl. PLEASE HELP! SHOW ALL STEPS!! This problem has been solved!

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3 years ago

Sarah added sugar to her iced tea but it is not dissolveing very quickly which can Sarah do to make the sugar dissolve faster ?

Aliun [14]

Stir it,
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7 0

3 years ago

At sea level, where the pressure was 104 kPa and temperature 21.1 ºC, a certain mass of air occupies 2.0 m3 . To what volume wil

Romashka [77]

Answer:

The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is $3.64 m^3$ .

Explanation:

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 104 kPa

$P_2$ = final pressure of gas = 52 kPa

$V_1$ = initial volume of gas = $2.0m^3$

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $21.1^oC=273+21.1=294.1K$

$T_2$ = final temperature of gas = $-5.0^oC=273+(-5.0)=268 K$

Now put all the given values in the above equation, we get:

$\frac{104 kPa\times 2.0m^3}{294.1 K}=\frac{52 kPa\times V_2}{268 K}$

$V_2=3.64 m^3$

The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is $3.64 m^3$ .

3 0

3 years ago

A student runs an experiment in the lab and then uses the data to prepare an Arrhenius plot of the natural log of the rate const

Thepotemich [5.8K]

Answer:

21.86582KJ

Explanation:

The graphical form of the Arrhenius equation is shown on the image attached. Remember that in the Arrhenius equation, we plot the rate constant against the inverse of temperature. The slope of this graph is the activation energy and its y intercept is the frequency factor.

Applying the equation if a straight line, y=mx +c, and comparing the given equation with the graphical form of the Arrhenius equation shown in the image attached, we obtain the activation energy of the reaction as shown.

5 0

3 years ago