The answer to this question would be: 2.36 mol

To answer this question, you need to know the molecular weight of copper. Molecular weight determines how much the weight of 1 mol of a molecule has. Copper molecular weight about 63.5g/mol. Then, the amount of mol in 150g copper should be: 150g / (63.5g/mol)= 2.36 mol

3 0

3 years ago

I need help pls Where does it belong to?

Daniel [21]

The answer is B. Fungi,Protists

6 0

2 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:

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

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6 0

2 years ago

Why LED is better option for testing the conduction of electricity?

Anettt [7]

Answer:

Led is a light-emitting diode, a semiconductor diode that glows when a voltage is applied. So, there is always a sure result of whether a substance conducts electricity or not as the light glows. That is why it is a better option for testing conduction of electricity.

Explanation:

5 0

2 years ago