How many molecules are in 2.50 moles of co2

Answers

Answer 1

Answer:

Explanation:

There are 1.51 x 1024 molecules of carbon dioxide in 2.50 moles of carbon dioxide.

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Suppose you have been given the task of distilling a mixture of hexane + toluene. Pure hexane has a refractive index of 1.375 and pure toluene has a refractive index of 1.497. You collect a distillate sample which has a refractive index of 1.441. Assuming that the refractive index of the hexane + toluene mixture varies linearly with mole fraction, what is the mole fraction of hexane in your sample?

Answers

Answer:

0.4590

Explanation:

How the refractive index of the hexane + toluene mixture varies linearly with mole fraction, it means that the mole fraction is the fraction that each pure index contribute for the mixture index, so, calling xh the mole fraction of hexane and xt the mole fraction of toluene:

1.375xh + 1.497xt = 1.441

And, xh + xt = 1 (because there are only hexane and toluene in the mixture), so xt = 1- xh

1.375xh + 1.497(1-xh) = 1.441

1.375xh + 1.497 - 1.497xh = 1.441

-0.122xh = -0.056

xh = -0.056/(-0.122)

xh = 0.4590

Perform the calculation and report the answer using the proper number of significant figures. Make sure the answer is rounded correctly. 1.012×10^-3 J/(0.015456 g)(298.3682−298.3567)K

Answers

Answer:

$=5.694(J)/(g*K)$

Explanation:

Hello,

In this case, since the result of the operation between two magnitudes is shown with the same significant figures of the shortest number, we obtain:

$1.012x10^(-3) J/[(0.015456 g)(298.3682-298.3567)]K$

Next, we proceed as follows:

$=0.065476J/[(g)(20.0115K)]\n\n=5.693582(J)/(g*K)$

Nevertheless, since 1.012 is the shortest number and has four significant figures, the result is rounded to four significant figures, that is until the three but it rounded due to the fact that the next digit is five:

$=5.694(J)/(g*K)$

Regards.

What is the maximum mass in grams of NH3 that can be produced by the reaction of of 2.5 g N2 with 2.5 g of H2 via the equation below?N2 (g) + 3 H2 (g) → 2 NH3 (g)

Answers

Answer: The mass of $NH_3$ produced is, 3.03 grams.

Explanation : Given,

Mass of $N_2$ = 2.5 g

Mass of $H_2$ = 2.5 g

Molar mass of $N_2$ = 28 g/mol

Molar mass of $H_2$ = 2 g/mol

First we have to calculate the moles of $N_2$ and $H_2$ .

$\text{Moles of }N_2=\frac{\text{Given mass }N_2}{\text{Molar mass }N_2}=(2.5g)/(28g/mol)=0.089mol$

and,

$\text{Moles of }H_2=\frac{\text{Given mass }H_2}{\text{Molar mass }H_2}=(2.5g)/(2g/mol)=1.25mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

From the balanced reaction we conclude that

As, 1 mole of $N_2$ react with 3 mole of $H_2$

So, 0.089 moles of $N_2$ react with $0.089* 3=0.267$ moles of $H_2$

From this we conclude that, $H_2$ is an excess reagent because the given moles are greater than the required moles and $N_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NH_3$

From the reaction, we conclude that

As, 1 mole of $N_2$ react to give 2 mole of $NH_3$

So, 0.089 mole of $N_2$ react to give $0.089* 2=0.178$ mole of $NH_3$

Now we have to calculate the mass of $NH_3$

$\text{ Mass of }NH_3=\text{ Moles of }NH_3* \text{ Molar mass of }NH_3$

Molar mass of $NH_3$ = 17 g/mole

$\text{ Mass of }NH_3=(0.178moles)* (17g/mole)=3.03g$

Therefore, the mass of $NH_3$ produced is, 3.03 grams.

Consider the following reaction at equilibrium: NO2(g) + CO(g) = NO(g) + CO2(g) Suppose the volume of the system is decreased at constant temperature, what change will this cause in the system? A shift to produce more NO A shift to produce more CO A shift to produce more NO2 No shift will occur

Answers

Answer: Option (d) is the correct answer.

Explanation:

According to Le Chaltelier's principle, when there occurs any change in an equilibrium reaction then the equilibrium will shift in a direction that will oppose the change.

This means that when pressure is applied on reactant side with more number of moles then the equilibrium will shift on product side that has less number of moles.

For example, $NO_(2)(g) + CO(g) \rightleftharpoons NO(g) + CO_(2)(g)$

Since here, there are same number of moles on both reactant and product side. So, when volume is decreased at a constant temperature in this system then there will occur no change in the equilibrium state.

Thus, we can conclude that in the given when volume of the system is decreased at constant temperature, then no shift will occur.

“the density of a subtance generally decreases as the temperature increases”

Answers

Answer:

The correct answer is D.

Explanation:

Water can evaporate, and if it does, the density decreases

Sodium tert-butoxide (NaOC(CH3)3) is classified as bulky and acts as Bronsted Lowry base in the reaction. It is reacted with 2-chlorobutane. Based on this information, the major organic product(s) of this reaction are expected to be _____.

Answers

Answer:but-1-ene

Explanation:This is an E2 elimination reaction .

Kindly refer the attachment for complete reaction and products.

Sodium tert-butoxide is a bulky base and hence cannot approach the substrate 2-chlorobutane from the more substituted end and hence major product formed here would not be following zaitsev rule of elimination reaction.

Sodium tert-butoxide would approach from the less hindered side that is through the primary centre and hence would lead to the formation of 1-butene .The major product formed in this reaction would be 1-butene .

As the mechanism of the reaction is E-2 so it will be a concerted mechanism and as sodium tert-butoxide will start abstracting the primary hydrogen through the less hindered side simultaneously chlorine will start leaving. As the steric repulsion in this case is less hence the transition state is relatively stabilised and leads to the formation of a kinetic product 1-butene.

Kinetic product are formed when reactions are dependent upon rate and not on thermodynamical stability.

2-butene is more thermodynamically6 stable as compared to 1-butene

The major product formed does not follow the zaitsev rule of forming a more substituted alkene as sodium tert-butoxide cannot approach to abstract the secondary proton due to steric hindrance.

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