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3 years ago

WORTH 20 POINTS Confirmation bias is where a scientist may change the data to support a certain bias.

Chemistry

1 answer:

Orlov [11]3 years ago

8 0

Answer:

True

Explanation:

Confirmation Bias is the tendency to look for information that supports, rather than rejects, one’s preconceptions, typically by interpreting evidence to confirm existing beliefs while rejecting or ignoring any conflicting data

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The bond enthalpy of the oxygen-oxygen bond in O2 is 498 kJ/mol. Based on the enthalpy of the reaction represented above, what i

kirill115 [55]

The given question is incomplete. The complete question is as follows.

The enthalpy of formation of ozone is 142.7 kJ / mol. The bond energy of $O_{2}$ is 498 kJ / mol. What is the average O=O bond energy of the bent ozone molecule O=O=O?

Explanation:

The given reaction is as follows.

$3O_{2}(g) \rightarrows 2O_{3}$

The value of $\Delta H$ for 2 moles of ozone is $2 \times 142.7 kJ/mol$ = 285.4 kJ/mol.

So, in this reaction three O=O bonds are broken down and four O-O bonds of ozone are formed.

Expression for the enthalpy of reaction is as follows.

$\Delta H = B.E_{reactants} - B.E_{products}$

285.4 kJ/mol = $(3 \times 498) - (2 \times O_{3})$

285.4 kJ/mol = $1494 - 2 \times O_{3})$

-1208.6 kJ/mol = $-2O_{3}$

$O_{3}$ = 604.3 kJ/mol

Now, the average bond enthalpy of O-O bond in $O_{3}$ is as follows.

= $\frac{604.3}{3}$ kJ/mol

= 201.43 kJ/mol

Thus, we can conclude that for the given reaction average bond enthalpy of an oxygen-oxygen bond in $O_{3}$ is 201.43 kJ/mol.

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4 years ago

How is stoicheometry used to keep camels alive

Novay_Z [31]

Answer:

by heart beat camel lives

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3 years ago

Classify these substances. More than one answer may apply in each case.

enyata [817]

Answer: $H_2$ : Pure substance, Element

$O_2$ : Pure substance , Element

$H_2O$ : Pure substance, Compound

Explanation:

Element is a pure substance which is composed of atoms of similar elements.It can not be decomposed into simpler constituents using chemical reactions.Example: $H-2$ and $O_2$

Compound is a pure substance which is made from atoms of different elements combined together in a fixed ratio by mass.It can be decomposed into simpler constituents using chemical reactions. Example: $H_2O$

Heterogeneous Mixture : Mixture that has its different components mixed unevenly within the substance . Example: Air

Homogeneous Mixture : Mixture that has its different components mixed evenly within the substance. Example: sugar dissolved in water

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4 years ago

Which combination of an element and an ion will react? View Available Hint(s) Which combination of an element and an ion will re

Aleks04 [339]

Answer: The combination of element ad an ion that will react is $Ni(s)\text{ and }Pt^{2+}(aq.)$

Explanation:

Oxidation reaction is defined as the reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

$X\rightarrow X^{n+}+ne^-$

Reduction reaction is defined as the reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

$X^{n+}+ne^-\rightarrow X$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

For a reaction to be spontaneous, the standard electrode potential must be positive.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$ ......(1)

For the given options:

Option 1: $Sn(s)\text{ and }Mn^{2+}(aq.)$

Here, tin must undergo oxidation reaction and manganese undergo reduction reaction.

Oxidation half reaction: $Sn(s)\rightarrow Sn^{2+}(aq.)+2e^-;E^o_{Sn^{2+}/Sn}=-0.14V$

Reduction half reaction: $Mn^{2+}(aq.)+2e^-\rightarrow Mn(s);E^o_{Mn^{2+}/Mn}=-1.18V$

Putting values in equation 1, we get:

$E^o_{cell}=-1.18-(-0.14)=-1.04V$

As, the standard potential is coming out to be negative, the given reaction will not take place.

Option 2: $Fe(s)\text{ and }Ca^{2+}(aq.)$

Here, iron must undergo oxidation reaction and calcium undergo reduction reaction.

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-;E^o_{Fe^{2+}/Fe}=-0.44V$

Reduction half reaction: $Ca^{2+}(aq.)+2e^-\rightarrow Ca(s);E^o_{Ca^{2+}/Ca}=-2.87V$

Putting values in equation 1, we get:

$E^o_{cell}=-2.87-(-0.44)=-2.43V$

As, the standard potential is coming out to be negative, the given reaction will not take place.

Option 3: $Ni(s)\text{ and }Pt^{2+}(aq.)$

Here, nickel must undergo oxidation reaction and platinum undergo reduction reaction.

Oxidation half reaction: $Ni(s)\rightarrow Ni^{2+}(aq.)+2e^-;E^o_{Ni^{2+}/Ni}=-0.25V$

Reduction half reaction: $Pt^{2+}(aq.)+2e^-\rightarrow Pt(s);E^o_{Pt^{2+}/Pt}=1.2V$

Putting values in equation 1, we get:

$E^o_{cell}=1.2-(-0.25)=1.45V$

As, the standard potential is coming out to be positive, the given reaction will take place.

Option 4: $H_2(g)\text{ and }Na^{+}(aq.)$

Here, hydrogen must undergo oxidation reaction and sodium undergo reduction reaction.

Oxidation half reaction: $H_2(g)\rightarrow 2H^{+}(aq.)+2e^-;E^o_{2H^{+}/H_2}=0V$

Reduction half reaction: $Na^{+}(aq.)+e^-\rightarrow Na(s);E^o_{Na^{+}/Na}=-0.27V$

Putting values in equation 1, we get:

$E^o_{cell}=-0.27-(-0)=-0.27V$

As, the standard potential is coming out to be negative, the given reaction will not take place.

Hence, the combination of element ad an ion that will react is $Ni(s)\text{ and }Pt^{2+}(aq.)$

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3 years ago

Which statement is true for most chemical reactions?

andriy [413]

The statement that is true for most chemical reactions is that A. An energy change occurs during the breaking and forming of chemical bonds. As the reactants react ultimately producing products the atoms of the respective elements within the compounds if present will break and reform ultimately to produce the products, during this change energy transformation from chemical bond formation and breakage occurs, and allows one to determine whether or not a reaction releases energy or needs a substantial amount of energy based on the energy content of the reactants and products.

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4 years ago

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