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2 years ago

In the diagram, what is happening to the temperature at Point B?

Chemistry

1 answer:

taurus [48]2 years ago

8 0

In the diagram, an activity which is happening to the temperature at Point B is that: C. The temperature is not rising because the heat is being used to break the connections between the molecules.

<h3>What is matter?</h3>

Matter can be defined as any physical object or body that has both mass and takes up space. This ultimately implies that, the mass of a physical object measures the amount of matter the object contains.

In Science, heat is sometimes referred to as thermal energy and it can be defined as a form of energy that is transferred from a physical object (body) to another due to a difference in temperature.

By critically observing the diagram, we can reasonably infer and logically deduce that matter has dynamic states depending on the composition of its atoms and at point B the temperature remained constant (not rising) because the quantity of heat generated is used for breaking any form of connection between the molecules.

Read more on matter here: brainly.com/question/24783543

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The combustion of 0.25 mol of an unknown organic compound results in the release of 320 kJ of energy. Which of the compounds in

vichka [17]

Answer: C. ethanol

The enthalpy of combustion is the amount of heat produced when one mole of ethanol undergoes complete combustion at 25 ° C and 1 atmosphere pressure, yielding products also at 25 ° C and 1 atm.

<u>The enthalpy of combustion of the unknown compound is</u>

ΔH = - 320 kJ / 0.25 mol = - 1280 kJ / mol

<u>To choose a probable compound according to this combustion enthalpy, we must evaluate the deviation in relation to the values reported in the literature for the three probable compounds</u> (methane, ethylene and ethanol). The deviation (e%) will be calculated according to the following equation,

e% = ( | ΔHx - ΔH | / ΔHx ) x 100%

where ΔHx is the enthalpy of combustion of the probable compound.

The following table shows the combustion enthalpies of the probable compounds and their deviation in relation to the enthalpy of ΔH = - 1280 kJ / mol

Compound Enthalpy of combustion (kJ/mol) Deviation

Methane - 890.7 43.8%

Ehylene -1411.2 9.3%

Ethanol -1368.6 6.5%

According to the previous table, we can say that the most probable compound is ethanol, since it has the smallest deviation in relation to the experimental enthalpy value of combustion.

5 0

4 years ago

Read 2 more answers

1.Which form of energy is due to the motion of an object's particles .

tamaranim1 [39]

Answer:

thermal

Explanation:

4 0

3 years ago

The total pressure of all the gases inside a metal container is 9.8 atm. If the partial pressure of the oxygen in the container

mina [271]

5.6 atm

In a container, the sum of partial pressures equals the total pressure. If the sun is 9.8 atm and the partial pressure of oxygen is 4.2 atm, the rest of the gases must account for the other 5.6 atm.

7 0

3 years ago

In two or more complete sentences explain how to balance the chemical equation and classify its reaction type.

Ierofanga [76]

This is a double replacement reaction. To balance the equation, you can not change the subscript because it will fully change the equation. Instead you change the coefficient infront of the element.

4 NaI + 1 Pb(SO4)2 —> 1 PbI4 +2 Na2SO4

8 0

3 years ago

The freezing point of pure benzene (C₆H₆) is 5. 49 °C. The freezing point of a solution made using toluene (C₇H₈) in benzene is

Irina18 [472]

The molality of the toluene is 1.46 mol/kg.

<h3>What is molality?</h3>

The molality of a solution is its molal concentration. Molal concentration is denoted by m. It is the mol of solute dissolves in 1 Kg solvent.

The depression at freezing point is directly proportional to the molality of the solution. The equation we use for this type of problem is:

ΔT=iKm $f_m$

Where, ΔT $f_m$ is depression at freezing point, i is Van't hoff factor, m is molality and $K_f$ is the freezing point depression constant.

Toluene is a non-electrolyte and so the value of i for this is 1.

Depression at a freezing point is given as -13. $0^0$ and the $K_f$ is given as 5.12. So, we could calculate the molality of the solution using the equation written above.

Let's plug in the values in the equation:

ΔT=iKm $f_m$

-7.51 =1 x5. 12 xm

m = $\frac{7.51}{5. 12}$

m = 1.46 m

It means the molality of the solution is 1.46 mol/kg.

Hence, the molality of the solution is 1.46 mol/kg.

Learn more about the molality of the solution here:

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3 0

2 years ago