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3 years ago

Please answer the question circled in red

Chemistry

2 answers:

ki77a [65]3 years ago

5 0

Answer:

i need more info bro

Explanation:

Tresset [83]3 years ago

5 0

I need more information in order to help you solve the question bro.. lol otherwise I can’t be of much help.

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The actual volume of the object was 34.1 mL. What is the percent error of his average result?

zysi [14]

It’s .03 % error marist bro

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3 years ago

I have an unknown volume of gas at a pressure of 0.5 atm and a temperature of 325k if I raise the pressure to 1.2 atm, decrease

Nimfa-mama [501]

Use the Equation of Clapeyron:

$\frac{P_1.V_1}{T_1}=\frac{P_2.V_2}{T_2}\\
\\
\frac{0.5V_1}{235}=\frac{1.2*48}{320}\\
\\
320*0.5V_1=235*1.2*48\\
\\
V_1=\frac{235*1.2*48}{320*0.5}=84.5 liters$

7 0

3 years ago

a wooden block with a mass of 2.0kg starts from at the top of an inclined plane and ends with a force of 25N at the bottom what

Papessa [141]

Answer:

F= m*a

25= 2*a

25/2=a

a=12.5

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3 years ago

The a-helices and b- sheets in protein make up its

Marta_Voda [28]

Alpha helix and beta sheets are the secondary structure of protein

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3 years ago

A gas stream containing n-hexane in nitrogen with a relative saturation of 0.58 (as a fraction, multiply by 100% if you prefer %

kondor19780726 [428]

This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.

In this case, it is recommended to write the enthalpy for each substance as follows:

$H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)$

Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and $y_{C-6}$ and $y_{N_2}$ are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

$H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol$

It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:

$Q=-40036+(-672.21)=-40708.21J$

Finally we convert this result to kJ:

$Q=-40708.21J*\frac{1kJ}{1000J}\\\\Q=-40.7kJ$

Learn more:

6 0

2 years ago