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 1

Answer:

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.

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How many grams of sucrose must be added to 375 mL of watertoprepare a 2.75m/m percent solution of sucrose?

Answers

Answer : The mass of sucrose added to 375 mL of water must be, 10.6 grams.

Explanation :

As we are given that 2.75 m/m percent solution of sucrose. That means, 2.75 grams of sucrose present in 100 grams of solution.

Mass of solution = 100 g

Mass of sucrose = 2.75 g

Mass of water = Mass of solution - Mass of sucrose

Mass of water = 100 g - 2.75 g

Mass of water = 97.25 g

First we have to calculate the mass of water.

$\text{Mass of water}=\text{Density of water}* \text{Volume of water}$

Density of water = 1.00 g/mL

Volume of water = 375 mL

$\text{Mass of water}=1.00g/mL* 375mL=375g$

Now we have to calculate the mass of sucrose in 375 g of water.

As, 97.25 grams of water contain 2.75 grams of sucrose

So, 375 grams of water contain $(375)/(97.25)* 2.75=10.6$ grams of sucrose

Therefore, the mass of sucrose added to 375 mL of water must be, 10.6 grams.

To make a 2.75% m/m sucrose solution, you need to add approximately 1062 grams of sucrose to 375 mL of water, considering the density of water as 1 g/mL.

To prepare a mass/mass (m/m) percent solution of sucrose, you need to calculate the mass of sucrose (in grams) that needs to be added to 375 mL of water to achieve a 2.75% concentration.

Here's how you can calculate it:

1. Convert the volume of water to grams, considering the density of water:

Density of water ≈ 1 g/mL

Mass of water = Volume of water × Density of water

Mass of water = 375 mL × 1 g/mL = 375 g

2. Determine the desired mass of sucrose as a percentage of the total mass:

Desired m/m percent = 2.75%

3. Calculate the mass of sucrose needed:

Mass of sucrose = (Desired m/m percent / 100) × Total mass

Mass of sucrose = (2.75 / 100) × (375 g + Mass of sucrose)

4. Rearrange the equation to solve for the mass of sucrose:

Mass of sucrose = (2.75 / 100) × (375 g) / (1 - (2.75 / 100))

Now, calculate:

Mass of sucrose = (2.75 / 100) × (375 g) / (1 - 0.0275)

Mass of sucrose ≈ (2.75 / 100) × (375 g) / 0.9725

Mass of sucrose ≈ (2.75 × 375 g) / 100 / 0.9725

Mass of sucrose ≈ (1031.25 g) / 0.9725

Mass of sucrose ≈ 1061.98 g

So, approximately 1062 grams of sucrose must be added to 375 mL of water to prepare a 2.75 m/m percent solution of sucrose.

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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 in a CSTR. When pure A is fed to a 10 dm 3 PFR at 300 K and a volumetric flow rate of 5 dm 3 /s, the conversion is 80%. When a mixture of 50% A and 50% inert (I) is fed to a 10 dm 3 CSTR at 320 K and a volumetric flow rate of 5 dm 3 /s, the conversion is also 80%. What is the activation energy in cal/mol

Answers

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.

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

Rearrange the formula is as follows.

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

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)

$k=(5m^(3)/s)/(10dm^(3))[(1+1)ln (1)/(1-0.8)-1 * 0.8] = 1.2s^(-1)$

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.

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

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)

$(10dm^(3))/(5dm^(3))=(0.8(1+0.5(0.8)))/(k(1-0.8))=2.8s^(-1)$

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

$k(T_(2))=k(T_(1))exp[(E)/(R)((1)/(T_(1))-(1)/(T_(2)))]$

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

Rearrange the above formula is as follows.

$E= R *((T_(1)T_(2))/(T_(1)-T_(2)))ln(k(T_(2)))/(k(T_(1)))$

Substitute the all values.

$=1.987cal/molK((300K *320K)/(320K *300K))ln (2.8)/(1.2)=8.081 *10^(3)cal\,mol^(-1)$

$=8.1\,kcal\,mol^(-1)$

Therefore, the activation energy is $=8.1\,kcal\,mol^(-1)$

Express the equilibrium constant for the combustion of propane in the balanced chemical equation c3h8(g)+5o2(g)???3co2(g)+4h2o(g)

Answers

according to the balanced equation:
C3H8(g) + 5O2(g) ↔ 3CO2(g) + 4 H2O (g)
aA + bB ↔ cC + dD

according to K formula when:
K = concentration of the products / concentration of the reactants

K = [C]^c[D]^d / [A]^a[B]^b
when [A],[B],[C]and[D] is the concentrations
and a,b,c and d is the no of moles
∴ K = [CO2]^3[H2O]^4 / [C3H8] [ O2]^5

Answer:

combustion reaction

Explanation:

there is oxides in the equation c3h8(g)+5o2(g)3co2(g)+4h2o(g)

A spontaneous reaction occurs

Answers

Answer:

A spontaneous reaction is a reaction that occurs in a given set of conditions without intervention. A spontaneous reactions are accompanied by an increase in overall entropy, or disorder

Explanation:

If a cell is 80% water and the outside environment is 90% water. What is likely to happen? A. water will rush into the cell
B. net movement of water will be equal
C. water will not move into or out of the cell
D. water will rush out of the cell

Answers

Water will rush into the cell, because water likes everything to be equal. This process is called diffusion (osmosis).

A lab group was calculating the speed of a radio car. They measured the distance traveled to be 6 meters and the time to be 3.5 seconds. Then they divided the distance by the time to find the speed. The actual speed was 2.2 m/s. What was their percent error?

Answers

Percent error is a measure of the difference between an observed value and a true value.

Actual Speed (True Value) = 2.2 m/s

Experimental Speed (Calculated Value) = Distance / Time = 6 m / 3.5 s = 1.714 m/s

The formula for calculating percent error is:

Percent Error = ((|Actual Value - Experimental Value|) / |Actual Value|) * 100%

Calculate the absolute difference between the actual speed and the experimental speed:

|2.2 - 1.714| = 0.486

Calculate the absolute value of the actual speed:

|2.2| = 2.2

Percent Error = (0.486 / 2.2) * 100%

= 0.221 * 100%

= 22.1%

The calculated percent error is approximately 22.1%. This means that the lab group's calculated speed of 1.714 m/s is about 22.1% lower than the true speed of 2.2 m/s.

Percent error is a way to quantify the accuracy of experimental measurements. A positive percent error indicates that the experimental value is higher than the true value, while a negative percent error indicates that the experimental value is lower. In this case, since the calculated speed is lower than the true speed, we have a positive percent error.

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Answer:456

Explanation:

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