Which statement best describes why specific heat capacity is often more useful than heat capacity for scientists when comparing two materials?Specific heat capacity is an intensive property and does not depend on sample size.
Specific heat capacity is an extensive property and does not depend on sample size.
Specific heat capacity is an intensive property and depends on the sample size.
Specific heat capacity is an extensive property and depends on the sample size.

Answers

Answer 1

Answer:

The correct option is this: SPECIFIC HEAT CAPACITY IS AN INTENSIVE PROPERTY AND DOES NOT DEPEND ON SAMPLE SIZE.

Generally, all the properties of matters can be divided into two classes, these are intensive and extensive properties. Intensive properties are those properties that are not determined by the quantity of the material that is present or available. Examples of intensive properties are colour, density and specific heat capacity. For instance, whether you have a bucket of water or a cup of water, the quantity does not matter, the colour of water will always remain the same. Extensive properties in contrast, are those properties that depend on the quantity of material that is available. Examples are mass, heat capacity and volume.

Answer 2

Answer:

Specific heat capacity is defined as the amount of heat required to increase the temperature of 1 gram of a substance by 1-degree Celsius.

The specific heat capacity is more often used for comparing materials, as it is an intensive property. Thus, statement A is correct.

What is the use of Specific Heat Capacity?

Specific heat capacity, which is also referred to as massic heat capacity, is the amount of heat added to 1 gram of mass to raise the temperature by 1-degree Celsius.

Specific heat is an intensive property and is not dependent on the size of the sample. This property of the specific heat is used to compare two materials.

Intensive properties are those properties that are not affected by the quantity of the material. Thus, statement A is correct, Specific heat capacity is an intensive property and does not depend on sample size.

Learn more about specific heat capacity, here: brainly.com/question/13369050

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Which of the following is an acid-base neutralization reaction? (1 point) Sn + 2HBr yields SnBr2 + H2 HCl + KOH yields KCl + H2O 2AlCl3 + 3Ca(OH)2 yields 2Al(OH)3 + 3CaCl2 2C2H6 +7O2 yields 4CO2 + 6H2O

If 6.49 mol of ethane (C2H6) undergo combustion according to the unbalanced equationC2H6 + O2 −→ CO2 + H2O ,
how many moles of O2 is required?
1. 27.79
2. 7.63
3. 8.4
4. 25.48
5. 12.635
6. 21.035
7. 23.8
8. 19.04
9. 22.715
10. 26.775
Answer in units of mol.

Answers

Answer:

22.715 moles of oxygen are used

Explanation:

Given data:

Number of moles of ethane = 6.49 mol

Number of moles of O₂ required = ?

Solution:

Chemical equation:

2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

Now we will compare the moles of oxygen with ethane.

C₂H₆ : O₂

2 : 7

6.49 : 7/2×6.49 = 22.715 mol

Thus, 22.715 moles of oxygen are used.

A chemical reaction between iron atoms and oxygen molecules can only occur if(1) the particles are heated
(2) the atmospheric pressure decreases
(3) there is a catalyst present
(4) there are effective collisions between the particles

Answers

Answer: option (4) there are effective collisions between the particles

Justification:

The collision theory explains the chemical reactions in terms of collisions. A chemical reaction takes place if two molecules collide effectively.

So, the first conditon is that the collision occur.

The second condition, that the collision be effective requires two conditions:

1) that the collision occurs with the proper orientation of the molecules, and

2) that the energy of collision be high enough to start the break of the original bonds ant the creation of the new bonds in the activated complex.

The energy required to form this activated complex is the activation energy.

So, an effective collision requires the proper orientation of the molecules when they collide and that the energy be equal or greater than the activation energy.

Temperature and catalyst are factors that affect the occurrence of those collisions, so they affect the rate of the reaction, by modifiying the number of collisions.

the answer is D because in order for a reaction to take place there must be collisions between the iron and oxygen atoms

when copper sulphate + carbon = carbon sulphate + copper , does copper turn into a solid or still stays a liquid

Answers

Solid, copper becomes insoluble when it is alone.

1. What is the volume in liters of 8.20 moles of CO₂ at standard temperature and pressure (STP)?2. A sample of 3.05 mol of gas in a 10.00 L container is at 45.0 °C. What is the pressure (in atm) of the gas?
3. What temperature (in °C) did an ideal gas shift to if it was initially at -17.00 °C at 4.620 atm and 35.00 L and the pressure was changed to 8.710 atm and the volume changed to 15.00 L?
4. A mixture of two gases with a total pressure of 1.98 atm contains 0.70 atm of Gas A. What is the partial pressure of Gas B in atm?
5. A chamber contains equal molar amounts of He, Ne, Ar, and Kr. If the total chamber pressure is 1 atm, then the partial pressure (in atm) of Kr is:

Answers

Answer:

1. To find the volume of 8.20 moles of CO₂ at standard temperature and pressure (STP), we can use the ideal gas law. At STP, the temperature is 0 °C or 273.15 K, and the pressure is 1 atm. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

We can rearrange the equation to solve for V:

V = (nRT) / P

Substituting the values:

V = (8.20 mol * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm

Calculating the numerical value:

V ≈ 181.3 L

Therefore, the volume of 8.20 moles of CO₂ at STP is approximately 181.3 liters.

2. To find the pressure of a gas sample with 3.05 moles in a 10.00 L container at 45.0 °C, we can still use the ideal gas law. However, we need to convert the temperature to Kelvin by adding 273.15 to it.

The ideal gas law equation can be rearranged to solve for pressure:

P = (nRT) / V

Substituting the values:

P = (3.05 mol * 0.0821 L·atm/mol·K * (45.0 + 273.15) K) / 10.00 L

Calculating the numerical value:

P ≈ 4.083 atm

Therefore, the pressure of the gas sample is approximately 4.083 atm.

3. To find the final temperature in °C when the initial temperature was -17.00 °C, and the pressure changed from 4.620 atm to 8.710 atm, and the volume changed from 35.00 L to 15.00 L, we can use the combined gas law.

The combined gas law states that (P₁ V₁) / T₁ = (P₂ V₂) / T₂, where P is pressure, V is volume, and T is temperature.

Rearranging the equation to solve for T₂:

T₂ = (P₂ V₂ T₁) / (P₁ * V

Final answer:

The volume of 8.20 moles of CO₂ at standard temperature and pressure (STP) is approximately 180.4 liters.

Explanation:

Gas Laws

Gas laws describe the behavior of gases under different conditions. One of the fundamental gas laws is the ideal gas law, which relates the pressure, volume, temperature, and number of moles of a gas. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

Question 1: Volume of CO₂ at STP

To find the volume of 8.20 moles of CO₂ at standard temperature and pressure (STP), we can use the ideal gas law equation. At STP, the temperature is 0 degrees Celsius (273.15 Kelvin) and the pressure is 1 atmosphere (atm).

Given:

Number of moles (n) = 8.20 moles
Temperature (T) = 0 degrees Celsius (273.15 Kelvin)
Pressure (P) = 1 atmosphere (atm)

Using the ideal gas law equation, we can rearrange it to solve for the volume (V):

V = (nRT) / P

Substituting the given values:

V = (8.20 moles * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm

V ≈ 180.4 liters

Question 2: Pressure of Gas

To find the pressure of a gas in a given container, we can use the ideal gas law equation. The temperature must be converted to Kelvin before using the equation.

Given:

Number of moles (n) = 3.05 mol
Volume (V) = 10.00 L
Temperature (T) = 45.0 °C (318.15 Kelvin)

Using the ideal gas law equation, we can rearrange it to solve for the pressure (P):

P = (nRT) / V

Substituting the given values:

P = (3.05 mol * 0.0821 L·atm/mol·K * 318.15 K) / 10.00 L

P ≈ 7.79 atm

Question 3: Temperature Change

To find the temperature change of an ideal gas, we can use the ideal gas law equation. The initial and final conditions of the gas must be known.

Given:

Initial temperature (T1) = -17.00 °C (256.15 Kelvin)
Initial pressure (P1) = 4.620 atm
Initial volume (V1) = 35.00 L
Final pressure (P2) = 8.710 atm
Final volume (V2) = 15.00 L

Using the ideal gas law equation, we can rearrange it to solve for the final temperature (T2):

T2 = (P2 * V2 * T1) / (P1 * V1)

Substituting the given values:

T2 = (8.710 atm * 15.00 L * 256.15 K) / (4.620 atm * 35.00 L)

T2 ≈ 303.6 °C

Question 4: Partial Pressure of Gas B

To find the partial pressure of Gas B in a mixture of gases, we need to know the total pressure and the partial pressure of Gas A.

Given:

Total pressure = 1.98 atm
Partial pressure of Gas A = 0.70 atm

The partial pressure of Gas B can be calculated by subtracting the partial pressure of Gas A from the total pressure:

Partial pressure of Gas B = Total pressure - Partial pressure of Gas A

Partial pressure of Gas B = 1.98 atm - 0.70 atm

Partial pressure of Gas B ≈ 1.28 atm

Question 5: Partial Pressure of Kr

To find the partial pressure of Kr in a chamber containing equal molar amounts of He, Ne, Ar, and Kr, we need to know the total chamber pressure.

Given:

Total chamber pressure = 1 atm

Since the chamber contains equal molar amounts of gases, the partial pressure of Kr is equal to the total chamber pressure divided by the number of gases:

Partial pressure of Kr = Total chamber pressure / Number of gases

Partial pressure of Kr = 1 atm / 4

Partial pressure of Kr = 0.25 atm

Learn more about gas laws here:

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What does the symbol 2CO represent in the model below? 2CO + o^2 -> 2CO2

Answers

Answer: a reactant

Explanation: online physics class

Answer:2CO (g) + O2 (g) 2CO2 (g)This Equation Means That No Matter How Much Carbon Monoxide And Oxygen ... Question: 2CO (g) + O2 (g) 2CO2 (g)This Equation Means That No Matter How Much Carbon Monoxide And Oxygen Gas Is Added To The Flask, Only 2 Moles Of CO And 1 Mole Of O2 Will React To Make Two Moles Of Co2.What Is Wrong With This Statement?

Explanation:

After 32 days, 5 milligrams of an 80-milligram sample of a radioactive isotope remains unchanged. What is the half-life of this element?(1) 8 days (3) 16 days
(2) 2 days (4) 4 days

Answers

Unstable heavy atoms will undergo radioactive decay to produce stable species. The half life time of the isotope which undergone a decay of 75 mg in 32 days is 18 days.

What is half life time?

The half life time of a radioactive sample is the time taken to reduce it to half of the initial amount by decay.

The heavy unstable material have very short half life and they will easily undergoes radioactive decay by emitting certain radiation.

Radioactive decay is a firs order reaction and have the equation to find the radioactive constant as follows:

$\lambda = (1)/(t) log([Ni])/([Nt])$

Where, t is the time of decay and Ni and Nt be the initial and final amount respectively.

It is given that 5 mg is remaining out of 80 mg after 32 days. Thus the radioactive constant is calculated as follows:

$\lambda = (1)/(32 days ) log (80)/(5) \n \n = 0.0376. days ^(-1)$

Now the half life time of the decay is calculated as below:

t(1/2) = 0.693 /decay constant

= 0.693/0.0376

= 18 days

Therefore, the half life time of the isotope which undergone a decay of 75 mg in 32 days is 18 days.

To find more about radioactive decay, refer the link below:

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$m=m_0 * ((1)/(2))^(t)/(t_(1/2))$
m - the mass that remains unchanged, m₀ - the inital mass, t - the time of decay, t1⁄2 - the half-life

$t=32 \ days \n m=5 \ mg \n m_0 = 80 \ mg \n \n 5 = 80 * ((1)/(2))^(32)/(t_(1/2)) \ \ \ \ \ \ \ |/ 80 \n (5)/(80)= ((1)/(2))^(32)/(t_(1/2)) \n (1)/(16)=((1)/(2))^(32)/(t_(1/2)) \n ((1)/(2))^4=((1)/(2))^(32)/(t_(1/2)) \n 4=(32)/(t_(1/2)) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |* t_(1/2) \n 4t_(1/2)=32 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |/ 4 \n t_(1/2)=8$

The half-life is (1) 8 days.

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