According to the equation below, what is the enthalpy change when 400.0 g of propane is burned in excess oxygen? c3h8 (g) + 5o2 (g) ) ↔ 3co2 (g) + 4h2o (l) δh = -2221kj

A chemical change involves the conversion of substances into different substances, as in the separation of a compound into its elements which requires breaking chemical bonds, unlike a physical change.

An example of a chemical change is the separation of a compound into its elements. This process involves breaking chemical bonds and forming new substances with different properties than those of the original compound. For instance, table salt (a compound of sodium and chlorine) cannot be separated into sodium and chlorine by physical means like filtering or distillation; it requires a chemical process. In contrast, physical changes, such as the separation of a mixture or the condensation of steam, do not produce new substances and are reversible.

Burning of gasoline and the souring of milk are also examples of chemical changes, where original substances are transformed into new ones, such as carbon dioxide and water from gasoline, and denatured proteins and produced acid from milk.

A classic chemical change can be observed when natural gas is burned in a furnace, converting methane and oxygen into water and carbon dioxide, signifying not just a visual change but a molecular transformation as well.

When heat is applied during a phase change, there is zero change in temperature. This is because all the heat supplied is used to break the bonds of the solid molecules to form liquid molecules. So in essence, there the temperature remains constant until all solid molecules bond are broken down (until all ice melts into liquid).

So the answer is Yes or True.

Ok, lets see the definitions of percent yield, actual yield and theoretical yield.

Percent yield is the ratio of actual yield to theoretical yield.

Actual yield- amount of product produced in the EXPERIMENT.

Theoretical yield- max amount of product produced through CALCULATIONS

% yield= actual yield (from experiment)/theoretical yield (from calculation) *100

1st Step: Write the reaction 

H2 + Br2 --> 2 HBr

2nd Step: Get the mass ratio of H2 and HBr to find theoretical yield

1 mole H2 gives 2 moles HBr ( molar mass of H2= 2 g/mol, HBr= 81 g/mol)

2 g H2 gives 2*81= 162 g HBr

30 g H2 gives 162*30/2 = 2430 g HBr ( The equation is 2 g H2/ 30 gH2= 162 g HBr/ x g HBr)

So theoretically 2430 g HBr are produced by calculation ( THEORETICAL YIELD)

By experiment 85 g HBr are produced as it is given at the question ( ACTUAL YIELD)

% yield= actual yield / theoretical yield *100 = 85/2430 *100= 3.5 %

The percent yield is 3.5 %.

Answer:The answer is 6

Explanation:

Answer:

It shows the reduction-oxidation tendency of chemical species present in the chart.

Explanation:

The reduction potential chart also known as activity series or electrochemical series is a reference series or chart constructed on the basis of reduction potential of standard hydrogen electrode (SHE). The standard reduction potential of SHE is considered to be zero.

Any species which can reduce SHE has negative reduction potential in the chart.

Any species which can oxidize SHE has positive reduction potential in the chart.

Higher the reduction potential more the tendency to undergo reduction and to oxidize others.

Lower the reduction potential more the tendency to undergo oxidation and to reduce others.

Answer is: covalent bond.

Covalent bond is bond between nonmetals. Carbon (C) and oxygen (O) are nonmetals.

Carbon atom and oxygen atom are connected by a triple bond (six shared electrons in three bonding molecular orbitals) that is formed of two covalent bonds and one dative covalent bond.

Dative covalent bond is formed by two electrons derive from the oxygen atom.

Carbon monoxide is a colorless, odorless, toxic and tasteless gas.

.The temperature levels in a nuclear reactor are maintained primarily by the use of

1. shielding

2. coolants

3. moderators

4. control rods

Ideal gas law is valid only for ideal gas not for vanderwaal gas.  At very low temperatures and very high pressures, real gases deviate from the behavior predicted by gas laws.

Ideal gas equation is the mathematical expression that relates pressure volume and temperature.

Mathematically the relation between pressure, temperature and volume is given as

PV=nRT

where,

P = pressure of ideal gas

V= volume of gas

n =number of moles of gas

T =temperature

R = Gas constant = 8.314J/K/mo

This law is not valid at very low temperatures and very high pressures because of the interaction between the particles increases and the factor of reduced volume come into picture.

Therefore, at very low temperatures and very high pressures, real gases deviate from the behavior predicted by gas laws.

To learn more about ideal gas equation, here:

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"The correct option is A. [tex]\( n=5 \) to \( n=1 \).[/tex]

In atomic physics, the energy of the emitted photon when an electron transitions from one energy level to another is given by the Rydberg formula:

[tex]\[ E = h \cdot f = R \cdot \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \][/tex] where \( h \) is Planck's constant, \( f \) is the frequency of the emitted photon, \( R \) is the Rydberg constant for hydrogen, \( n_f \) is the final energy level, and \( n_i \) is the initial energy level.

The energy of the photon is directly proportional to the frequency of the light emitted, and the frequency is inversely proportional to the wavelength of the light. Therefore, the greater the energy difference between the two levels, the higher the frequency of the emitted light and the more energy the light carries.

 Let's calculate the energy released for each transition:

[tex]A. \( n=5 \) to \( n=1 \):[/tex]

[tex]\[ E_{5 \to 1} = R \cdot \left( \frac{1}{1^2} - \frac{1}{5^2} \right) = R \cdot \left( 1 - \frac{1}{25} \right) = R \cdot \frac{24}{25} \][/tex]

 B. [tex]\( n=4 \) to \( n=5 \):[/tex]

[tex]\[ E_{4 \to 5} = R \cdot \left( \frac{1}{5^2} - \frac{1}{4^2} \right) = R \cdot \left( \frac{1}{25} - \frac{1}{16} \right) = R \cdot \left( -\frac{9}{400} \right) \][/tex]

(Note: This transition actually absorbs energy, as the electron moves to a higher energy level.)

C.[tex]\( n=2 \) to \( n=5 \):[/tex]

[tex]\[ E_{2 \to 5} = R \cdot \left( \frac{1}{5^2} - \frac{1}{2^2} \right) = R \cdot \left( \frac{1}{25} - \frac{1}{4} \right) = R \cdot \left( -\frac{21}{100} \right) \][/tex]

(Again, this transition absorbs energy.)

 D. [tex]\( n=5 \) to \( n=4 \):[/tex]

[tex]\[ E_{5 \to 4} = R \cdot \left( \frac{1}{4^2} - \frac{1}{5^2} \right) = R \cdot \left( \frac{1}{16} - \frac{1}{25} \right) = R \cdot \frac{9}{400} \][/tex]

 Comparing the magnitudes of the energy changes, it is clear that the transition from \( n=5 \) to \( n=1 \) releases the most energy because it has the largest difference in energy levels. This corresponds to the largest absolute value of the energy change, which means it will give off the most light energy.

To convert 1.00 kg of 235U into energy, about 9.14 × 10^-4 kg of rest mass is converted.

The energy released in the fission of a 235U nucleus is about 200 MeV. To find out how much rest mass is converted to energy, we need to calculate the number of 235U atoms in 1.00 kg. One mole of 235U has a mass of 235.04 grams, so there are 4.25 moles of 235U in 1.00 kg. Using Avogadro's number (6.02 × 10^23 U/mol), we can calculate that there are 2.56 × 10^24 atoms of 235U in 1.00 kg. The total energy released is the number of atoms times the given energy per U fission, which is:

(2.56 × 10^24 atoms)(200 MeV/atom) = 5.12 × 10^26 MeV

To convert this to kilograms, we can use the conversion factor 1 MeV = 1.783 × 10^-30 kg, so:

(5.12 × 10^26 MeV)(1.783 × 10^-30 kg/MeV) = 9.14 × 10^-4 kg

Therefore, about 9.14 × 10^-4 kg of rest mass is converted to energy in this fission.

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The first level must be filled with two electrons before electrons can be added to the second level.

(Plato users)