In what direction does thermal energy transfer?

The carbon atoms in graphite and the carbon atoms in diamond have different

kicyunya [14]

Answer:

<u>structural arrangements</u>

_______________________________________

<h2>properties of daimond: </h2><h3>appearance: transparent</h3><h3>hardness: very hard</h3><h3>thermal conductivity :very poor</h3><h3>electric conductivity: poor</h3><h3>density:</h3>

$3510 {kgm}^{3}$

<h3>uses: jewellery and drilling</h3>

_______________________________________

<h2>properties of graphite:</h2>

<h3>appearance: black shiny</h3><h3>hardness: soft ,slippery to touch</h3><h3>thermal conductivity : moderate</h3><h3>electric conductivity: good</h3><h3>density:</h3>

$2250 {kgm}^{3}$

<h3>uses:dry cell, electric arc, pencil lead, lubricant</h3>

_______________________________________

<h2>How Diamond and Graphite are chemically identical?</h2>

On heating diamond or graphite in the air, they burn completely to form carbon dioxide.
- Equal quantities of diamond and graphite when burned, produce exactly the same amount of carbon dioxide.

_______________________________________

<h2>Why the physical properties of diamond and graphite are so different?</h2>

Due to the difference in the arrangement of carbon atoms in diamond and graphite

_______________________________________

<h2><em><u>hope</u></em><em><u> it</u></em><em><u> helps</u></em><em><u> you</u></em><em><u><</u></em><em><u>3</u></em></h2>

7 0

3 years ago

Abbreviated electron configuration for tellurium

Zinaida [17]

Answer:

The ground state electron configuration of ground state gaseous neutral tellurium is [Kr]. 4d10.

4 0

3 years ago

A sample of neon has a volume of 40.81 m3 at 23.5C. At what temperature, in Kelvins, would the gas occupy 50.00 cubic meters? As

mezya [45]

At $\fbox{\begin \\363 K \end{minispace}}$ temperature, a sample of neon gas will occupy $50.00 \text{ m}^{3}$ volume.

Further Explanation:

The given problem is based on the concept of Charles’ law. Charles’ law states that “at constant pressure and fixed mass the volume occupied an ideal gas is directly proportional to the Kelvin temperature.”

Mathematically the law can be expressed as,

$\fbox{ \begin \\ V \propto T \end{minispace}}$

Or,

$\frac{V}{T}=k$

Here, <em>V</em> is the volume of the gas, <em>T</em> is Kelvin temperature, and <em>k</em> is proportionality constant.

Given information:

The initial volume of neon gas is $40.81 \text{ m}^{3}$ .

The final volume of neon gas is $50.00 \text{ m}^{3}$ .

The initial temperature value is $23.5 \text{ } ^{\circ} \text{C}$ .

To calculate:

The final temperature

Given Condition:

The pressure is constant.

Mass of gas is fixed.

Solution:

Step 1: Modify the mathematical expression for Charles’ law for two different temperature and volume values as follows:

$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$

Here,

$V_{1}$ is the initial volume of the gas.

$V_{2}$ is the final volume of the gas.

$T_{1}$ is the initial temperature of the gas.

$T_{2}$ is the final temperature of the gas.

Step 2: Rearrange equation (2) for .

$\fbox {\begin \\T_{2}=\frac{(V_{2}) \times (T_{1})}{V_{1}}\\\end{minispace}}$ …… (2)

Step 3: Convert the given temperature from degree Celsius to Kelvin.

The conversion factor to convert degree Celsius to Kelvin is,

$T(\text{K}) = T(^{\circ}\text{C}) + 273.15$ …… (3)

Substitute $23.5\text{ }^{\circ} \text{C}$ for $T(^{\circ}\text{C})$ in equation (3) to convert temperature from degree Celsius to Kelvin.

$T(\text{K}) = 23.5 \text{ } ^{\circ} \text{C} + 273.15\\T(\text{K})= 296.65 \text{ K}$

Step 4: Substitute $40.81 \text{ m}^{3}$ for $V_{1}$ , $50.00 \text{ m}^{3}$ for $V_{2}$ and $296.65 \text{ K}$ for $T_{1}$ in equation (2) and calculate the value of $T_{2}$ .

$T_{2}=\frac{(50.00 \text{ m}^{3}) \times (296.65 \text{ K})}{40.81 \text{ m}^{3}}\\T_{2}=363.45 \text{ K}\\T_{2} \approx 363 \text{ K}$

Important note:

The temperature must be in Kelvin.

The condition of fixed mass and fixed pressure must be fulfilled in order to apply Charles’ law.

Learn More:

1. Gas laws brainly.com/question/1403211

2. Application of Charles’ law brainly.com/question/7434588

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: States of matter

Keywords: neon, volume, occupies, temperature, Kelvin, degree Celsius, Charle’s law, constant pressure, fixed mass, 40.81 m^3 , 50.00 m^3 , 23.5 degree C , celsius , 363 K , sates of matter, initial volume, final volume, initial temperature, final temperature, V1 , V2 , T1 , T2 .

5 0

3 years ago

Read 2 more answers

A 0.0000792 M sample of Compound X in a solvent has an absorbance of 0.341 at 528 nm in a 1.000-cm cuvet. The solvent alone has

natima [27]

Answer:

a) a = 3485 M⁻¹cm⁻¹

b) C = 0,000127 M

Explanation:

Lambert-Beer law says that there is a linear relationship between absorbance and concentration of a chemical substance. The formula is:

A = a×b×C

Where A is absorbance, a is molar absorptivity, b is path length and C is concentration.

a) In the problem Concentration is 0.0000792 M, b is 1,000cm and Absorbance is absorbance of sample-absorbance of blank: 0,341-0,065 = 0,276

Replacing:

0,276 = a×1,000cm×0,0000792M

b) As the experiment consist in the same compound in the same solvent, the molar absorptivity will be the same, a = 3485 M⁻¹cm⁻¹, path length will be 1,000cm and absorbance: 0,508-0,065 = 0,443

Replacing:

0,443 = 3485 M⁻¹cm⁻¹×1,000cm×C

I hope it helps!

8 0

3 years ago

Do free electrons appear anywhere in the balanced equation for a redox reaction?

Andru [333]

Yes, free electrons appear in balanced redox reaction equations. However, this is only true for half-reactions. This is because redox reactions primarily involve the transfer of electrons, which are better visualized if explicitly shown in the balanced reactions. In reduction reactions, electrons are placed on the left side of the equation. Oxidation reactions show electrons on the right side of the equation.

Explanation:

A half reaction is either the chemical reaction or reduction reaction part of an oxidoreduction reaction. A half reaction is obtained by considering the amendment in chemical reaction states of individual substances concerned within the oxidoreduction reaction. Half-reactions are usually used as a way of leveling oxidoreduction reactions.The half-reaction on the anode, wherever chemical reaction happens, is Zn(s) = Zn2+ (aq) + (2e-).

The metal loses 2 electrons to create Zn2+. The half-reaction on the cathode wherever reduction happens is Cu2+ (aq) + 2e- = Cu(s).

Here, the copper ions gain electrons and become solid copper.

4 0

3 years ago

Read 2 more answers