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insens350 [35]
3 years ago
10

6. As you go down a family or group on the periodic table *

Chemistry
2 answers:
maks197457 [2]3 years ago
4 0

Answer:

the atoms have more valence electrons

Explanation:

icang [17]3 years ago
4 0
Idnshansbdnsjsjsjsjs
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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:

  • brainly.com/question/25475410
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6 0
2 years ago
What is diffusion?<br>Class 9th <br>Chemistry <br><br><br><br><br>​
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Diffusion is the movement of a substance from an area of high concentration to an area of low concentration . Diffusion happens in liquids and gases because their particles move randomly from place to place . Diffusion is an important process for living things ; it is how substances move in and out of cells .

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6 0
3 years ago
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What is the theoretical yield of aluminum oxide if 3.00 mol of aluminum metal is exposed to 2.55 mol of oxygen?
lisabon 2012 [21]

Theoretical yield of Al₂O₃: 1.50 mol.

<h3>Explanation</h3>

2 \; \text{Al} + \dfrac{3}{2} \; \text{O}_2 \to {\bf 1} \; \text{Al}_2\text{O}_3;

4 \; \text{Al} + 3 \; \text{O}_2 \to 2 \; \text{Al}_2\text{O}_3 \; \textit{Balanced}.

How many moles of aluminum oxide formula units will be produced <em>if</em> aluminum is the limiting reactant?

Aluminum reacts to aluminum oxide at a two-to-one ratio.

3.00 \times \dfrac{1}{2} = 1.50 \; \text{mol}.

As a result, 3.00 moles of aluminum will give rise to 1.50 moles of aluminum oxide.

How many moles of aluminum oxide formula units will be produced <em>if</em> oxygen is the limiting reactant?

Oxygen reacts to produce aluminum oxide at a three-to-two ratio.

2.55 \times \dfrac{2}{3} = 1.70 \; \text{mol}

As a result, 2.55 moles of oxygen will give rise to 1.70 moles of aluminum oxide.

How many moles of aluminum oxide formula units will be produced?

Aluminum is the limiting reactant. Only 1.50 moles of aluminum oxide formula units will be produced. 1.70 moles isn't feasible since aluminum would run out by the time 1.50 moles was produced.

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3 years ago
Select the situation below that would produce a total displacement of zero
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Select the correct answer. The wavelength of a given region of the electromagnetic spectrum ranges from 1 x 10-11 - 1 x 10-8 met
Makovka662 [10]

X-rays are found in the given region.

Answer: Option B

<u>Explanation:</u>

The wavelengths of different waves are studied carefully, and their range is determined according to the frequency they exhibited. And based upon the studies, the electromagnetic waves are classified into eight different categories such as gamma, X-, UV, visible, near IR, middle IR, far IR, micro and radio waves.

And each category has different wavelengths and the given radiation falls under X-rays.  The wavelength range of X-rays varies from 1 nm to 1 pm. Here, “nm” stands for nano meter and “pm” stands for “pico meter”.

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