All categories

3 years ago

the various geologic settings in which rocks can melt, metamorphose, or become sediment are ultimately generated by

1 answer:

3 0

Geological settings whereby there is melting of rock as well as formation of sediments can be attributed to Plate tectonics .

Plate tectonics can be regarded as theory which explained the Earth's outer shell and how its been divided into solid rocks as well as slabs and its often referred to as Plate.

Plate tectonics explained that there is a large-scale motion plates which Earth's lithosphere is made up.

Tectonic processes started about 3.3 and 3.5 billion years ago.

Through the generation of Tectonic plate, there are melting of rocks as well as generation of sediments

The types of boundaries that exist between those tectonic plates can be categorized into 4, base on the the movement of the plates, and these are;

divergent boundaries
convergent boundaries
transform fault boundaries
plate boundary zones

Therefore, plate tectonic gives explanation about Earth's outer shell.

brainly.com/question/20918142?referrer=searchResults

You might be interested in

4 and 5 pls and thank uuu

PtichkaEL [24]

Answer:

4. -ol

5. cyclic ketone

Explanation:

biology stuff

sorry it's hard to explain

5 0

2 years ago

A chemistry student weighs out of hypochlorous acid into a volumetric flask and dilutes to the mark with distilled water. He pla

riadik2000 [5.3K]

Answer:

Volume of NaOH, aka V2 = 6.32 mL to 3 sig. fig.

A chemistry student weighs out 0.0941 g of hypochlorous acid (HClo) into a 250. ml. volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.2000 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equivalence point. Round your answer to 3 significant digits mL.

Explanation:

1 mole HClO = 74.44g

0.0941g = $\frac{0.0941}{74.44}$ = 0.00126 moles

Concentration = no. of moles/volume in L

Hence, Concentration of HClO = 0.00126/ 0.250L

= 0.005M.

C1V1 =C2V2

0.005 × 250 mL = 0.2 × V2

Volume of NaOH, aka V2 = 6.32 mL to 3 sig. fig.

3 0

3 years ago

An unknown compound displays singlets at δ 2.1 ppm and 2.56 ppm in the ratio of 3:2. what is the structure of the compound?

olasank [31]

1) As can be seen from any 1H NMR chemical shift ppm tables, hydrogens which have δ values from 2ppm to 2.3ppm are hydrogens from carbon which is bonded to a carbonyl group. From this, we can conclude that our hydrogens belong to the type, but from 2 different alkyl groups because of 2 different signals.

2) So, one alkyl group is CH3 and second one can be CH or CH2.

3) If we know that ratio between two types of hydrogens is 3:2, it can be concluded that second alkyl group is CH2.

4) Finally, we don't have any other signals and it indicates that part of the compound which continues on CH2 is exactly the same as the first part.

The ratio remains the same, 3:2 ie 6:4

7 0

3 years ago

The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment

astraxan [27]

Answer:

The activation energy is $=8.1\,kcal\,mol^{-1}$

Explanation:

The gas phase reaction is as follows.

$A \rightarrow B+C$

The rate law of the reaction is as follows.

$-r_{A}=kC_{A}$

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = $10dm^{3}$

Temperature "T" = 300 K

Volumetric flow rate of the reaction $v_{o}=5dm^{3}s$

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

$V= \frac{v_{0}}{k}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]$

Rearrange the formula is as follows.

$k= \frac{v_{0}}{V}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]............(1)$

The feed has Pure A, mole fraction of A in feed $y_{A_{o}}$ is 1.

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacte.

$=1(2-1)=1$

Substitute the all given values in equation (1)

$k=\frac{5m^{3}/s}{10dm^{3}}[(1+1)ln \frac{1}{1-0.8}-1 \times 0.8] = 1.2s^{-1}$

Therefore, the rate constant in case of the plug flow reacor at 300K is $1.2s^{-1}$

The rate constant in case of the CSTR can be calculated by using the formula.

$\frac{V}{v_{0}}= \frac{X(1+\epsilon X)}{k(1-X)}.............(2)$

The feed has 50% A and 50% inerts.

Hence, the mole fraction of A in feed $y_{A_{o}}$ is 0.5

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacted.

$=0.5(2-1)=0.5$

Substitute the all values in formula (2)

$\frac{10dm^{3}}{5dm^{3}}=\frac{0.8(1+0.5(0.8))}{k(1-0.8)}=2.8s^{-1}$

Therefore, the rate constant in case of CSTR comes out to be $2.8s^{-1}$

The activation energy of the reaction can be calculated by using formula

$k(T_{2})=k(T_{1})exp[\frac{E}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})]$

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

$E= R \times(\frac{T_{1}T_{2}}{T_{1}-T_{2}})ln\frac{k(T_{2})}{k(T_{1})}$

Substitute the all values.

$=1.987cal/molK(\frac{300K \times320K}{320K \times300K})ln \frac{2.8}{1.2}=8.081 \times10^{3}cal\,mol^{-1}$

$=8.1\,kcal\,mol^{-1}$

Therefore, the activation energy is $=8.1\,kcal\,mol^{-1}$

8 0

3 years ago

Which electromagnet would have the most strength?

ycow [4]

The one with 20 coils as it can generate more electricity.

8 0

3 years ago

Read 2 more answers