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3 years ago

Explain how creativity can play a role in the construction of scientific questions and hypotheses.

1 answer:

4 0

Scientific questions and hypotheses come up frequently while one is engaged in investigating a scientific phenomenon such as natural geological phenomena as may occur in geological mapping in the field. For example, there may be a question does this canyon or deeply incised valley which is quite straight follow a weakness in the earth's crust like a major fault or the direction of bedding in well bedded sedimentary rocks. In a particular topographic area, some hypotheses which may be developed is that valleys follow geological structure whereas ridges follow resistant rocks like quartzites or quartz sandstones or in the ocean, points or capes may represent resistant quartz sandstones and bays may represent weak soft shales recessively weathering

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Which principal is exemplified by studying a modern river to learn about how ancient rock layers form?

faust18 [17]

A- unifromitariansim

7 0

3 years ago

A possible mechanism for the gas phase reaction of NO and H2 is as follows: Step 1 2NO N2O2 Step 2 N2O2 + H2 N2O + H2O Step 3 N2

mel-nik [20]

Answer: Option (d) is the correct answer.

Explanation:

Steps involved for the given reaction will be as follows.

Step 1: $2NO \Leftrightarrow N_{2}O_{2}$ (fast)

Rate expression for step 1 is as follows.

Rate = k $[NO]^{2}$

Step 2: $N_{2}O_{2} + H_{2} \rightarrow N_{2}O + H_{2}O$

This step 2 is a slow step. Hence, it is a rate determining step.

Step 3. $N_{2}O + H_{2} \rightarrow N_{2} + H_{2}O$ (fast)

Here, $N_{2}O_{2}$ is intermediate in nature.

All the steps are bimolecular and it is a second order reaction. Also, there is no catalyst present in this reaction.

Thus, we can conclude that the statement step 1 is the rate determining step, concerning this mechanism is not directly supported by the information provided.

4 0

3 years ago

Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find

FromTheMoon [43]

Answer: The concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

Explanation:

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

$H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)$

0.025 0.025 0.025

Equation for the second dissociation of sulfuric acid:

$HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)$

Initial: 0.025 0.025

At eqllm: 0.025-x 0.025+x x

The expression of second equilibrium constant equation follows:

$Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}$

We know that:

$Ka_2\text{ for }H_2SO_4=0.01$

Putting values in above equation, we get:

$0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

4 0

3 years ago

Calculate the theoretical yield of glycolysis and complete glucose breakdown

rusak2 [61]

Eukaryotic cells, the theoretical maximum yield of ATP generated per glucose is 36 to 38, depending on how the 2 NADH generated in the cytoplasm during glycolysis enter the mitochondria and whether the resulting yield is 2 or 3 ATP per NADH

6 0

3 years ago

Please help ASAP please and thanks!! I'll also mark you!

skelet666 [1.2K]

Substitution Reactions are those reactions in which one nucleophile replaces another nucleophile present on a substrate. These reactions can take place via two different mechanism i.e SN¹ or SN². In SN¹ substitution reactions the leaving group leaves first forming a carbocation and nucleophile attacks carbocation in the second step. While in SN² reactions the addition of Nucleophile and leaving of leaving group take place simultaneously.

Example:
OH⁻ + CH₃-Br → CH₃-OH + Br⁻

In above reaction,

OH⁻ = Incoming Nucleophile

CH₃-Br = Substrate

CH₃-OH = Product

Br⁻ = Leaving group

Organic reactions are typically slower than ionic reactions because in organic compounds the covalent bonds are first broken, this breaking of bonds is a slower step, while, in ionic compounds no bond breakage is required as it consists of ions, so only bond formation takes place which is a quicker and fast step.

7 0

3 years ago