"Henry mixed salt and water together in a cup until he observed a clear solution. He measured the mass of the solution. Then he placed the cup outside for several sunny days during the summer. After a week, he observed that only solid salt remained in the cup and the mass had decreased. Henry concluded that a physical and chemical change occurred in this investigation."

Which statements correctly defend or dispute his conclusion?

A.)He is correct. Dissolving salt in water is a physical change, but evaporating the water is a chemical change. Formation of a solid is evidence that a chemical change occurred.

B.)He is correct. Evaporation is a physical change, but dissolving salt in water is a chemical change. The change in mass is evidence that a chemical change occurred.

C.)He is incorrect. Dissolving salt in water and evaporation of the water are both physical changes. The reappearance of salt is evidence that the change was reversible by a physical change, so it could not be a chemical change.

D.)He is incorrect. Dissolving salt in water and evaporation of the water are both chemical changes. The reappearance of salt is evidence that the change was reversible by a chemical change, so it could not be a physical change.

Nuclear processes alter the sorts of atoms present, but chemical reactions do not. The electrons in the atom play a significant role in chemical reactions when nucleus reactions occur in atom nuclei.

The energy found in an atom's nucleus, or core, is known as nuclear energy. Energy maintains the nucleus of atoms together, the minuscule units that make up all matter in the universe. The dense nucleus of an atom has an enormous quantity of energy.

The most lethal weapons on earth are nuclear weapons. One can wipe out an entire metropolis, perhaps killing millions of people, endangering the ecosystem and the lives of future generations due to its long-term devastating impacts.

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Answer:

\(\boxed {\boxed {\sf V_2=75 \ mL}}\)

Explanation:

Since the pressure is constant, the only variables we need to work with are temperature and volume. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula is:

\(\frac{V_1}{T_1}=\frac{V_2}{T_2}\)

Originally, the gas was 50 milliliters at 20 degrees celsius. Substitute these values into the left side of the equation.

\(\frac{50 \ mL}{20 \textdegree C}=\frac{ V_2}{T_2}\)

We don't know the volume of the new gas, but we know the temperature was changed to 30 degrees celsius.

\(\frac{50 \ mL}{20 \textdegree C}=\frac{ V_2}{30 \textdegree C}\)

Since we are solving for the new volume, we must isolate the variable. It is being divided by 30 °Cand the inverse of division is muliplication. Multiply both sides by 30 °C.

\(30 \textdegree C*\frac{50 \ mL}{20 \textdegree C}=\frac{ V_2}{30 \textdegree C}* 30 \textdegree C\)

\(30 \textdegree C*\frac{50 \ mL}{20 \textdegree C}= V_2\)

The units of degrees celsius cancel, so we are left with milliliters as the units.

\(30*\frac{50 \ mL}{20}= V_2\)

\(\frac{1500 \ mL}{20}= V_2\)

\(75 \ mL=V_2\)

The new volume of the gas is 75 milliliters.

An atom of C-14 has the same number of protons as an atom of C-12, but a greater mass number due to the presence of more neutrons.

An atom of carbon has six protons and its atomic number is six. An atom of C-12, a stable isotope of carbon, has six neutrons while an atom of C-14, an unstable isotope of carbon, has eight neutrons. Thus, an atom of C-14 has a greater mass number than an atom of C-12 due to the presence of more neutrons, even though they have the same number of protons and atomic number.

An atom's atomic number is the number of protons present in its nucleus, whereas its mass number is the number of protons plus the number of neutrons. Since C-14 has the same number of protons as C-12 but more neutrons, it has a greater mass number than C-12, while the number of electrons in both atoms is the same. As a result, the correct answer is that compared to an atom of C-12, an atom of C-14 has a greater mass number.

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Complete question is:

Compared to an atom of c-12 an atom of c-14 has a greater

number of electrons

number of protons

atomic number

mass number​

The pH after 30.0 ml of NaOH has been added is 2.18.

To find the pH after 30.0 ml of 0.175 M NaOH has been added to a 30.00 ml sample of 0.125 M HCOOH, you need to use the balanced chemical equation for the reaction between HCOOH and NaOH:

HCOOH + NaOH → NaCOOH + H2O

This equation shows that 1 mole of HCOOH reacts with 1 mole of NaOH, so the number of moles of NaOH added to the HCOOH solution is:

n(NaOH) = C(NaOH) x V(NaOH) = 0.175 mol/L x 0.0300 L = 0.00525 mol

Since the stoichiometry of the reaction is 1:1, this means that 0.00525 mol of HCOOH has been neutralized by the NaOH. The remaining amount of HCOOH is:

n(HCOOH) = C(HCOOH) x V(HCOOH) - n(NaOH) = 0.125 mol/L x 0.0300 L - 0.00525 mol = 0.00225 mol

Now you can use the Ka expression for HCOOH to find the concentration of H+ ions:

Ka = [H+][COO-]/[HCOOH] = 1.8 x 10^-4

[H+][0.00225]/[0.1225 - 0.00525] = 1.8 x 10^-4

[H+] = 0.00659 M

pH = -log[H+] = 2.18

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The density of the unknown sample is 1.025 g / mL and its salt composition is 3.55 %.

PART A: Density of a regular shaped object:

Trial 1: mass of the object = 162.20 g

volume of object = L x H x W = 4.90 cm x 3.90 cm x 2.90 cm

= 55.419 cm^3

Therefore density of the object = mass / volume = 162.20 g / 55.419 cm^3

= 2.9268 g/cm^3

trial 2: mass of the object = 162.18 g

volume of object = L x H x W = 4.89 cm x 3.90 cm x 2.88 cm

= 54.92448 cm^3

Therefore density of the object = mass / volume = 162.18 g / 54.92448 cm^3

= 2.9528 g/cm^3

Average = [ 2.9268 + 2.9528 ] /2 = 5.8796 / 2 = 2.9398 g / cm^3 = 1.94 g / cm^3.

The accepted value is 2.73 g / cm^3 for aluminium. The difference is 0.21

% error = 100 x difference / accepted value = 100 x 0.21/2.73 = 7.7 %.

---------------------------------------------------------------------------------------------------

Part B: Determination of density of an irregular shaped object:

Trial 1:

mass of the marble chips = 10.25 g

Volume of the marble chip = final volume of water - initial volume of water

= 53.8 - 50 = 3.8 mL

Therefore density of marble chip = mass / volume = 10.25 g / 3.8 mL

= 2.697 g / mL

Trial 2:

mass of the marble chips = 10.32 g

Volume of the marble chip = final volume of water - initial volume of water

= 53.9 - 50.1 = 3.8 mL

Therefore density of marble chip = mass / volume = 10.32 g / 3.8 mL

= 2.716 g / mL

Average = [2.697 + 2.716] / 2 = 5.413 / 2 = 2.71 g / mL

The accepted density of marble chip = 2.70 g / mL The difference is 0.01

% error = 100 x difference / accepted value = 100 x 0.01/ 2.70 = 0.37 %.

--------------------------------------------------------------------------------------------------------------------

PART C: Determination of density of saline solution:

Trial 1:

Volume of the saline solution = 10 mL

mass of the saline solution = finall mass - initial mass

= 35.66 - 25.36 = 10.3 g

Density of the saline solution = mass / volume = 10.3 g / 10 mL = 1.03 g / mL

Trial 2:

Volume of the saline solution = 10 mL

mass of the saline solution = finall mass - initial mass

= 35.55 - 25.35 = 10.2 g

Density of the saline solution = mass / volume = 10.2 g / 10 mL = 1.02 g / mL

Average =[ 1.03 + 1.02 ] / 2 = 1.025 g / mL

Thus the unknown sample B has the density of 1.025 g / mL.

The composition of salt in this solution can be determined by interpolation.

salt % = 0 + 5 x [ 1.025-0.998] / [1.036 - 0.998] ( using the values given in the table )

= 0 + 5 x 0.027 / 0.038

= 3.55 %.

Thus the density of the unknown sample is 1.025 g / mL and its salt composition is 3.55 %.

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As the term "peroxide" is used in Chapter 10 it can refer to RCOOCR structure.

Hydrogen peroxide solutions up to 9% in concentration are often safe when consumed; nonetheless, even a 3% solution is slightly irritating to nasal mucosa and might result in vomiting and diarrhea. Industrial-strength solutions (10%) when consumed induce systemic toxicity and have been linked to deaths.

The fact that oxygen is an anion and hydrochloric acid is a chemical substance is the main distinction between the two peroxides. The specific class of oxygen compounds known as peroxides has distinctive characteristics. Thenard, a French chemist, made the discovery of hydrogen peroxide in 1818.

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Answer:

A. 32 g

Explanation:

Density = 2.7 g/cm3

Volume = 2 x 2 x 3 = 12 cm3

Mass = dV = 2.7 x 12 = 32.4 g or 32 g

Explanation:

A because A galvanic cell consists of two half-cells, such that the electrode of one half-cell is composed of metal A, and the electrode of the other half-cell is composed of metal B; the redox reactions for the two separate half-cells are thus: An+ + ne− ⇌ A.

Answer:

A

Explanation:

Answer:

The answer is 100 m/p/h because you divide the total 200 miles and the 50 speed and you get about 100 m/p/h

Explanation:

Answer:

The mass of one mole of a substance is equal to that substance's molecular weight. ... water is 18.015 atomic mass units (amu), so one mole of water weight 18.015 grams. ... Avogadro's number is a proportion that relates molar mass on an atomic ... one molecule of water (H2O), one mole of oxygen (6.022×1023 of O atoms)

Gasoline blending in petroleum refineries involves analyzing the properties of different components and determining the optimal mixing ratios to produce gasoline that meets specific octane rating and quality requirements.

Gasoline blending is a critical process in petroleum refineries where different components are combined to produce the desired gasoline product. In this example, the challenge is to blend gasoline from three different components.

To solve the gasoline blending problem, various factors need to be considered such as the desired octane rating, volatility, and environmental regulations. The first step is to determine the optimal proportion of each component based on their individual characteristics. This involves analyzing the properties of each component, such as its research octane number (RON), motor octane number (MON), and vapor pressure.

The second step is to develop a blending strategy that achieves the desired gasoline specifications. This involves determining the appropriate mixing ratios of the three components to meet the target octane rating and other quality requirements. The blending process requires precise calculations and adjustments to ensure the final gasoline product meets the desired specifications.

Additionally, economic considerations play a role in gasoline blending. The cost of each component and the market demand for specific gasoline grades can influence the blending decisions. Refineries aim to optimize the blend to minimize costs while meeting quality standards.

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The molar mass of the compound is 128.4 g/mol.

The empirical formula is an empirical formula that represents the lowest whole-number ratio of the atoms present in a compound. The empirical formula for the molecular compound is calculated using the percentage composition of the elements present in the compound. The steps used to find the empirical formula are as follows:

Find the mass of each element present in the compound.Convert each mass to moles.Divide each mole value by the smallest number of moles.Round to the nearest whole number and write the subscripts.The molar mass is the mass of a substance that contains 6.02 × 10²³ atoms or molecules. To calculate the molar mass of a compound, add the masses of all the atoms present in the compound.

C=0.707g,12.01 g/mol=0.0588 molCnH=0.2372 =1.01g/m=0.235 mol H

nH=4nC

The empirical formula of the compound is CH4. The molar mass of the compound can be calculated using the empirical formula.

M=12.01 g/mol+4(1.01 g/mol)=16.05 g/mol

The molar mass of the compound is 8 times the empirical mass, so the actual molar mass is;

M=8(16.05{g/mol})=128.4g/mol. The molar mass of the compound is 128.4 g/mol.

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Answer:

Examples of EM waves are radio waves, microwaves, infrared waves, X-rays, gamma rays, etc.

Answer:

Electromagnetic waves can be split into a range of frequencies. This is known as the electromagnetic spectrum. Examples of EM waves are radio waves, microwaves, infrared waves, X-rays, gamma rays, etc.

Explanation:

Answer:

Explanation:

a substance's enthalpy of fusion tells you how much heat is needed in order to convert

1 g

of said substance from solid at its melting point to liquid at its melting point.

In water's case, an enthalpy of fusion equal to

333.55 J g

−

1

tells you that

1 g

of ice at

0

∘

C

can be converted to

1 g

of liquid water at

0

∘

C

by supplying

333.55 J

of heat.

Your ice cube has a mass of

55.0 g

, which means that it will require

55.0

g

⋅

=

Δ

H

fus



333.55 J

1

g

=

18,345.25 J

Rounded to three sig figs, the number of sig figs you have for the mass of the ice cube, the answer will be

heat needed

=

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

18,300 J

a

a

∣

∣

−−−−−−−−−−

The kinetic-molecular principle of gases assumes that perfect gasoline molecules are constantly moving.

The gasoline debris has a negligible extent. The gas debris is similarly sized and no longer has intermolecular forces (enchantment or repulsion) with other fuel particles. The gasoline debris flows randomly in settlement with Newton's legal guidelines of motion. The gasoline debris has perfect elastic collisions without energy loss.

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Here are the structures of the three primary amines with molecular formula C5H13N that contain five carbon atoms in a continuous chain:

Structure 1: 1-Aminopentane

Structure 2: 2-Aminopentane

Structure 3: 3-Aminopentane

To draw the structures of the three primary amines with molecular formula C5H13N that contain five carbon atoms in a continuous chain, we first need to determine the possible ways of arranging the functional group NH2 on a 5-carbon chain.

Aliphatic amines with one amino group and one hydrocarbon group less than the corresponding alcohol are called primary amines. We can arrange the functional group NH2 in three ways on a 5-carbon chain:

On carbon 1

On carbon 2

On carbon 3

The three primary amines with the molecular formula C5H13N are as follows:

Structure 1: N attached to carbon 1 (1-aminopentane)

Structure 2: N attached to carbon 2 (2-aminopentane)

Structure 3: N attached to carbon 3 (3-aminopentane)

Here are the structures of the three primary amines with molecular formula C5H13N that contain five carbon atoms in a continuous chain:

Structure 1: 1-Aminopentane

Structure 2: 2-Aminopentane

Structure 3: 3-Aminopentane

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The balanced equation of the reaction is:

The precipitate formed is magnesium hydroxide.

The mass of precipitate produced is 0.58 g.

The concentration of the ions remaining in the solution are:

[Mg²⁺] = 0.05 M

[NO₃⁻] = 0.1 M

The balanced equation of the reaction of aqueous potassium hydroxide and aqueous magnesium nitrate is given below:

2 KOH (aq) + Mg(NO₃)₂ ---> Mg(OH)₂ (s) + 2 KNO₃ (aq)

Magnesium hydroxide is obtained as a precipitate in the reaction above.

The mass of precipitate produced is calculated as follows:

moles of KOH (aq) reacting = 0.200 * 100 / 1000 = 0.02 moles

moles of Mg(NO₃)₂ reacting = 0.2 * 100 / 1000 = 0.02 moles

From the equation of reaction, KOH is the limiting reagent.

Moles of precipitate formed = 0.02 / 2 = 0.01 moles

molar mass of Mg(OH)₂ (s) = 58 g/mol

Mass of precipitate formed = 0.01 * 58

Mass of precipitate formed = 0.58 g

Moles of Mg(NO₃)₂ remaining = 0.02 - 0.01 = 0.01 moles

total volume of solution = 200 mL or 0.2 L

The concentration of each ion remaining in the solution is as follows:

[Mg²⁺] = 0.01 / 0.2 = 0.05 M

[NO₃⁻] = 0.01 * 2 / 0.2 = 0.1 M

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Answer:

D

Explanation:

Luck

In an ecosystem, the difference in eye color among the same species is called variation.

Ecosystem is defined as a system which consists of all living organisms and the physical components with which the living beings interact. The abiotic and biotic components are linked to each other through nutrient cycles and flow of energy.

Energy enters the system through the process of photosynthesis .Animals play an important role in transfer of energy as they feed on each other.As a result of this transfer of matter and energy takes place through the system .Living organisms also influence the quantity of biomass present.By decomposition of dead plants and animals by microbes nutrients are released back in to the soil.There are different  ecosystems present in the environment.

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Compact Bone is composed of mineralized matrix and bone cells. whereas, Spongy Bone consists of a network of trabeculae, which are thin bony spicules or plates interconnected to create a porous framework.

Compact Bone:

Compact bone, also known as cortical bone, forms the outer layer of bone and provides strength and support. It is composed of mineralized matrix and bone cells.The main minerals found in compact bone include hydroxyapatite, which is a crystalline form of calcium phosphate, and calcium carbonate. These minerals contribute to the hardness and rigidity of the bone tissue.

In addition to minerals, compact bone contains several proteins that contribute to its structure and function. Collagen is the predominant protein found in the bone matrix. It provides flexibility and tensile strength to the bone, allowing it to resist breaking under stress. Other proteins, such as osteocalcin and osteopontin, are involved in mineralization processes and regulate bone remodeling.

Spongy Bone:

Spongy bone, also called trabecular or cancellous bone, is found at the inner layer of bone and forms a lattice-like structure. It consists of a network of trabeculae, which are thin bony spicules or plates interconnected to create a porous framework. This arrangement provides strength to the bone while keeping it lightweight.

Similar to compact bone, spongy bone contains mineralized matrix and bone cells. The minerals present in spongy bone are also hydroxyapatite and calcium carbonate. However, spongy bone has a higher proportion of spaces within its structure compared to compact bone.

Proteins found in spongy bone include collagen, which provides structural support, and other non-collagenous proteins involved in bone development, remodeling, and mineralization.

Overall, both compact and spongy bone consist of mineralized matrix containing hydroxyapatite and calcium carbonate, along with collagen and other proteins that contribute to the structure, strength, and function of the bone tissue. The specific arrangement and density of these components differ between compact and spongy bone, allowing them to fulfill different roles within the skeletal system.

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Bones are made up of two types of tissue: compact and spongy bone. Compact bone (cortical) forms the hard external layer of all bones while spongy (cancellous) bone forms the inner layer of all bones. Both types of bones are composed of specialized cells, mineral salts, and collagen fibers.

Our bones are made up of two types of tissue: compact bone and spongy bone. Compact bone, also known as cortical bone, forms the hard external layer of all bones and surrounds the medullary cavity, or bone marrow. This bone tissue consists of units called osteons or Haversian systems, featuring mineral matrix and living osteocytes connected by canaliculi, which transport blood.

Spongy bone, on the other hand, or cancellous bone, forms the inner layer of all bones. Unlike compact bone tissue, spongy bone tissue does not contain osteons. It consists of trabeculae: lamellae that are arranged like rods or plates. In between these trabeculae, we'll find the red bone marrow that is responsible for forming blood cells.

Both types of bone tissues contain specialized cells, mineral salts (mainly calcium and phosphorus), and collagen fibers. The integration of these minerals and proteins is critical for the robust structure and resilience of the skeletal system.

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As a result, the aluminum emits -180 J of heat when it is reduced from 70 to 50 degrees Celsius.

The mobility of atoms and molecules is accelerated when energy is added (heating), increasing the temperature. By withdrawing energy (cooling), atoms and molecules slow down and the temp drops as a result. Conduction is a technique that allows energy to be added to or taken away from a medium.

The following formula is used to calculate how much heat is emitted when aluminum is reduced from 70°C to 50°C:

Q = mcΔθ

where;

m is the aluminum's mass.

Specific heat capacity is c.

Δθ is a temperature change.

Q = (10)(0.9)(50 - 70) (50 - 70)

Q = -180 J

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The amount of heat released by the production of 1.0 mole of SO3(g) is 196 kJ.

To determine the amount of heat released by the production of 1.0 mole of SO3(g), we need to first balance the chemical equation:

2SO2(g) + O2(g) = 2SO3(g) + 392 kJ

Now, we can see that 2 moles of SO3 are produced by releasing 392 kJ of heat. To find the heat released for 1 mole of SO3, we can set up a proportion:

(392 kJ) / (2 moles of SO3) = x kJ / (1 mole of SO3)

Solving for x:

x = (1 mole of SO3) * (392 kJ) / (2 moles of SO3)
x = 196 kJ

So, the amount of heat released by the production of 1.0 mole of SO3(g) is 196 kJ.

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Answer:

While the weight of an object varies in proportion to the strength of the gravitational field, its mass is constant, as long as no energy or matter is added to the object. Therefore, false.

Explanation:

Answer:

The correct answer should be false

Explanation:

Answer:

Explanation:

To find the value of Kc for the reaction HBr(g) ⇌ 1/2H2(g) + 1/2Br2(g), we can use the relationship between the equilibrium constants of reverse reactions.

For the given reaction: H2(g) + Br2(g) ⇌ 2HBr(g)

We are given the value of Kc as 2.0 × 10^19 at 298 K.

To find the equilibrium constant for the reverse reaction, we take the inverse of Kc.

Kc(reverse) = 1/Kc(forward)

Kc(reverse) = 1/(2.0 × 10^19)

Kc(reverse) = 5.0 × 10^(-20)

Now, we can use the equilibrium constant expression for the reverse reaction to determine the equilibrium constant for the reaction HBr(g) ⇌ 1/2H2(g) + 1/2Br2(g).

Kc = ([1/2H2][1/2Br2]) / [HBr]

Since the coefficients are halved in the reverse reaction, the equilibrium constant is also squared:

Kc = ([H2]^0.5[Br2]^0.5) / [HBr]

Kc = (√[H2]√[Br2]) / [HBr]

Therefore, the equilibrium constant Kc for the reaction HBr(g) ⇌ 1/2H2(g) + 1/2Br2(g) is approximately 5.0 × 10^(-20).

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Explanation:

We are given different isotopes and we have to identify the number of protons and neutrons that are in the nuclueus of each atom.

a) Carbon-14:

By definition two isotopes are atoms that have the same atomic number but different mass number. The atomic number of an atom is equal to the number of protons of that atom, and the mass number is equal to the number of protons plus the number of neutrons.

atomic number = n° of protons

mass number = n° of protons + n° of neutrons

n° of protons = atomic number

n° of neutrons = mass number - n° of protons

n° of neutrons = mass number - atomic number

If two isotopes have the same atomic number but different mass number we can say that two isotopes have the same number of protons but different number of neutrons.

In we pay attention to carbon-14 we can look for its atomic number in the period table: 6. And its mass number is the one that we are given after the name of the element: 14.

n° of protons = atomic number = 6

n° of protons = 6

n° of neutrons = mass number - atomic number = 14 - 6

n° of neutrons = 8

b) Cobalt-60:

atomic number = 27 (from the periodic table)

mass number = 60

n° of protons = atomic number = 27

n° of protons = 27

n° of neutrons = mass number - atomic number = 60 - 27

n° of neutrons = 33

c) Gold-197:

atomic number = 79 (from the periodic table)

mass number = 197

n° of protons = atomic number = 79

n° of protons = 79

n° of neutrons = mass number - atomic number = 197 - 79

n° of neutrons = 118

d) Uranium-235:

atomic number = 92 (from the periodic table)

mass number = 235

n° of protons = atomic number = 92

n° of protons = 92

n° of neutrons = mass number - atomic number = 235 - 92

n° of neutrons = 143

Answer:

a) Carbon-14: n° of protons = 6 n° of neutrons = 8

b) Cobalt-60: n° of protons = 27 n° of neutrons = 33

c) Gold-197: n° of protons = 79 n° of neutrons = 118

d) Uranium-235: n° of protons = 92 n° of neutrons = 143

12.33 liters of oxygen are produced when 45.0 g of KClO3 decomposes.

The mass in grams of one mole of a material is its molar mass. The molar mass of a material may be found by adding the molar masses of its constituent atoms, as demonstrated in this video. The computed molar mass may then be used to convert between mass and the quantity of moles in the material.

KClO3 ------>  kcl+ 3/2 O2

KClO3 122.55 gm in terms of molar mass.

Given mass = 459 moles of KCl03= 45 10.367 moles 122.55,

we know that the reaction produces 1 mole of kuo and 3 moles of O2.

I mole + 3/2 mule 02.. 0.367 mole X mole 02

X= 0.367X3 =0.5505 mol using the straightforward unitary technique.

Oxygen capacity at STP = 0.5505 x 22.41

12.33 liters

12.33 liters of oxygen are produced when 45.0 g of KClO3 decomposes.

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Answer:

19.5

Please tell me if wrong.

When a car uses a gallon of gas, 22.0 miles per gallon of carbon dioxide is emitted into the environment.

After being released into the atmosphere, carbon dioxide (CO₂) is initially quickly dispersed among the atmosphere, the upper ocean, and plants.

The sources of carbon dioxide emissions include both natural and human activities. Decomposition, oceanic discharge, and respiration are examples of natural sources.

The annual carbon dioxide emissions from a typical passenger automobile are around 4.6 metric tonnes. This estimates that the typical gasoline car on the road today gets roughly 22.0 miles per gallon of fuel and travels about 11,500 miles annually.

Thus, 22.0 miles per gallon of carbon dioxide is emitted into the environment.

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Answer:

Prokaryotic cells contain DNA, contain ribosomes, and lack a nucleus.

Answer:

Contains DNA

Contains Ribosomes

lacks a nucleus

Explanation: I took the quiz on edge