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2 years ago

Is caffeine optically active

1 answer:

7 0

Caffeine is not optically active because it is a planar heterocycle. When you say optically active, it means that compounds have molecules which are chiral. Chirality is the property of molecules as a result of its structure. And when you say planar heterocycle, molecules are flat and the ring should contain atoms located in a similar plane.

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4. How does the kinetic theory of gases explain the weather changes happening in the troposphere? You can also research on the I

uranmaximum [27]

According to the kmt pressure is directly proportional to the number of collision between particles

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2 years ago

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On a cool morning, Uyen’s breath can form a cloud when she breathes out. Which changes of state are most responsible for Uyen se

marissa [1.9K]

The correct option is: VAPORIZATION AND CONDENSATION.

Matters have the ability to change from one state to another state, this is called state transition. In the question given above, the carbon dioxide, which Uyen's was breathing out came in form of vapors and forms a cloud by condensing. Condensation is the process by which water droplets are formed when a vapor from comes in contact with cold surfaces. In the question given above, the vaporized gas condenses when it comes in contact with the humid air.

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2 years ago

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What is the density (in g/L) of a gas with a molar mass of 146.06 g/mol at 1.03 atm and 297k

marshall27 [118]

Lets assume the gas is acting Ideally, then According to Ideal Gas Equation the density is given as,

d = P M / R T ------- (1)
Where;
P = Pressure = 1.03 atm

M = Molar Mass = 146.06 g/mol

R = Gas Constant = 0.08206 atm.L.mol⁻¹.K⁻¹

T = Temperature = 297 K

Putting Values in eq. 1,

d = (1.03 atm × 146.06 g/mol) ÷ (0.08206 atm.L.mol⁻¹.K⁻¹ × 297 K)

density = 6.17 g/L

6 0

2 years ago

balance the following reaction using LCM method by showing each steps Pb (N3)2 + Cr(MnO4)2  Cr2O3 + MnO2 + Pb3O4+ NO

blsea [12.9K]

here's the answer to your question

5 0

1 year ago

Diborane, B2H6 a possible rocket propellant, can be made by using lithium hydride (LiH): 6 LiH+ 2 BCl2àB2H6+ 6 LiCl . If you mix

Genrish500 [490]

Answer :

(a) Limiting reactant = $LiH$

(b) The excess reactant = $BCl_3$

(c) The percent of excess reactant is, 50.87 %

(d) The percent yield of $B_2H_6$ or percent conversion of $LiH$ to $B_2H_6$ is, 38.80 %

(e) The mass of $LiCl$ produced is, 1066.42 lb

Explanation : Given,

Mass of $LiH$ = 200 lb = 90718.5 g

conversion used : (1 lb = 453.592 g)

Mass of $BCl_3$ = 1000 lb = 453592 g

Molar mass of $LiH$ = 7.95 g/mole

Molar mass of $BCl_3$ = 117.17 g/mole

Molar mass of $B_2H_6$ = 27.66 g/mole

Molar mass of $LiCl$ = 42.39 g/mole

First we have to calculate the moles of $LiH$ and $BCl_3$ .

$\text{Moles of }LiH=\frac{\text{Mass of }LiH}{\text{Molar mass of }LiH}=\frac{90718.5g}{7.95g/mole}=11411.13moles$

$\text{Moles of }BCl_3=\frac{\text{Mass of }BCl_3}{\text{Molar mass of }BCl_3}=\frac{453592g}{117.17g/mole}=3871.23moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$6LiH+2BCl_3\rightarrow B_2H_6+6LiCl$

From the balanced reaction we conclude that

As, 6 moles of $LiH$ react with 1 mole of $BCl_3$

So, 11411.13 moles of $LiH$ react with $\frac{11411.13}{6}=1901.855$ moles of $BCl_3$

From this we conclude that, $BCl_3$ is an excess reagent because the given moles are greater than the required moles and $LiH$ is a limiting reagent and it limits the formation of product.

Moles of remaining excess reactant = 3871.23 - 1901.855 = 1969.375 moles

Total excess reactant = 3871.23 moles

Now we have to determine the percent of excess reactant $(BCl_3)$ .

$\% \text{ excess reactant}=\frac{\text{Moles of remaining excess reactant}}{\text{Moles of total excess reagent}}\times 100$

$\% \text{ excess reactant}=\frac{1969.375}{3871.23}\times 100=50.87\%$

The percent of excess reactant is, 50.87 %

Now we have to calculate the moles of $B_2H_6$ .

As, 6 moles of $LiH$ react to give 1 mole of $B_2H_6$

So, 11411.13 moles of $LiH$ react to give $\frac{11411.13}{6}=1901.855$ moles of $B_2H_6$

Now we have to calculate the mass of $B_2H_6$ .

$\text{Mass of }B_2H_6=\text{Moles of }B_2H_6\times \text{Molar mass of }B_2H_6$

$\text{Mass of }B_2H_6=(1901.855mole)\times (27.66g/mole)=52605.3093g$

Now we have to calculate the percent yield of $B_2H_6$ .

$\%\text{ yield of }B_2H_6=\frac{\text{Actual yield of }B_2H_6}{\text{Theoretical yield of }B_2H_6}\times 100=\frac{20411.7g}{52605.3093g}\times 100=38.80\%$

The percent yield of $B_2H_6$ or percent conversion of $LiH$ to $B_2H_6$ is, 38.80 %

Now we have to calculate the moles of $LiCl$ .

As, 6 moles of $LiH$ react to give 6 mole of $LiCl$

So, 11411.13 moles of $LiH$ react to give 11411.13 moles of $LiCl$

Now we have to calculate the mass of $LiCl$ .

$\text{Mass of }LiCl=\text{Moles of }LiCl\times \text{Molar mass of }LiCl$

$\text{Mass of }LiCl=(11411.13mole)\times (42.39g/mole)=483717.8007g=1066.42lb$

The mass of $LiCl$ produced is, 1066.42 lb

8 0

2 years ago