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2 years ago

An object which allows light to show through it unaltered is called?

1 answer:

6 0

Transparency (also called pellucidity or diaphaneity) in the field of optics is the physical property of allowing light to pass through the material without significant light dispersion.

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How much water should be added to 4.3 moles of LiBr to prepare a 2.05 m solution?

arsen [322]

Answer:

2.1 kg of water

Explanation:

Step 1: Given data

Moles of lithium bromide (solute): 4.3 moles

Molality of the solution (m): 2.05 m (2.05 mol/kg)

Mass of water (solvent): ?

Step 2: Calculate the mass of water required

Molality is equal to the moles of solute divided by the kilograms of solvent.

m = moles of solute/kilograms of solvent

kilograms of solvent = moles of solute/m

kilograms of solvent = 4.3 mol /(2.05 mol/kg) = 2.1 kg

8 0

2 years ago

Each of the following names is wrong. Draw structures based on them, and correct the names:(b) 1,1,1-trimethylheptane

devlian [24]

The correct IUPAC name of 1,1,1-trimethylheptane is 2,2-dimethyloctane.

<h3>IUPAC NAME:</h3>

It is systematic way of nomenclature of organic compounds.

It is based on the position of functional groups, preference of the functional group, long or short chain of carbon, preference of double bond, single and triple bond, branching of carbon chain, etc.

The given compound is 1,1,1-trimethylheptane.

The given name is wrong according to IUPAC name because the numbering of carbon atom should be done in that way in which the carbon atom chain is largest.

Here, in this case the numbering is done from right side. Thus the largest carbon chain have 8 carbon atom.

If the numbering is done according to the question, number of carbon atom in straight chain is 7.

Thus, we concluded that the Correct IUPAC name of 1,1,1-trimethylheptane is 2,2-dimethyloctane.

learn more about Nomenclature:

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5 0

1 year ago

Which statement is correct regarding the reaction below? 3A + 2B yields C + 2D The rate of formation of D is twice the rate of d

Vedmedyk [2.9K]

Answer:

The correct statements are:

The rate of disappearance of B is twice the rate of appearance of C.

Explanation:

Rate of the reaction is a change in the concentration of any one of the reactant or product per unit time.

3A + 2B → C + 2D

Rate of the reaction:

$R=-\frac{1}{3}\times \frac{d[A]}{dt}=-\frac{1}{2}\times \frac{d[B]}{dt}$

$-\frac{1}{3}\times \frac{d[A]}{dt}=\frac{1}{1}\times \frac{d[C]}{dt}$

$-\frac{1}{3}\times \frac{d[A]}{dt}=\frac{1}{2}\times \frac{d[D]}{dt}$

The rate of disappearance of B is twice the rate of appearance of C.

$\frac{1}{1}\times \frac{d[C]}{dt}=-\frac{1}{2}\times \frac{d[B]}{dt}$

$2\times \frac{1}{1}\times \frac{d[C]}{dt}=-\frac{1}{1}\times \frac{d[B]}{dt}$

8 0

3 years ago

Read 2 more answers

Plants and animals use different energy storage molecules, yet they both use the same mechanism to "burn" their stored energy. H

jek_recluse [69]

Answer:

The breaking of the chemical bonds of a storage molecule transfer energy, no what molecule is stored.

Explanation:

Being successful of plants and animals does not necessary depend on the stored molecule but on the energy being transferred during their breaking.

8 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

2 years ago

An object which allows light to show through it unaltered is called?​

An object which allows light to show through it unaltered is called?