what is the mass or 1.53 x 10^23 molecules of H2SO4?

The two chemical compounds are not the same, because their ratio is not equal. In both samples the composition of lead and oxygen is different.

A chemical compound is a substance made of numerous similar molecules (or molecular entities) joined by chemical bonds and comprising atoms from various chemical elements. Therefore, a molecule made up of only one type of atom is not a compound. Chemical reactions, which may entail interactions with other molecules, can change a compound into a distinct substance. Atomic bonds may be broken or new ones created during this process.

sample 1 = mass of lead / mass of oxygen = 12.92g/2g = 6.46 .

sample 2 = mass of lead/ mass of oxygen = 34.27 - 14.39/14.39 = 1.38 .

so, the ratios are not the same.

Hence, the two chemical compounds are not the same, because their ratio is not equal. In both samples the composition of lead and oxygen is different.

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Answer:

im guessing its A run off

Explanation:

Answer: Run off

Explanation:

Wetlands have plants that clean water through. As water flows into a wetland it encounters the the plants growing there. This slows the water down making it less likely to cause erosion. The nutrient pollutants nitrogen and phosphorus are absorbed by the roots of the plants.

This happens trough the process of exfoliation. And the temperature changes are called thermal stress weathering.

Hope this helps.

Answer:

The human body needs both saturated fats and unsaturated fats to remain healthy. Most dietary recommendations suggest that, of the daily intake of fat, a higher proportion should be from unsaturated fats, as they are thought to promote good cholesterol and help prevent cardiovascular disease, whereas an overabundance of saturated fats is thought to promote bad cholesterol.

Explanation:

The ion with a +3 charge, 28 electrons, and an atomic mass of 71 is the aluminum ion (\(Al^{3+}\)).

Aluminum (Al) typically has an atomic number of 13, which means it has 13 protons and 13 electrons in its neutral state. However, in the given ion, \(Al^{3+}\), the ion has lost three electrons, resulting in a +3 charge. This means that the ion now has 13 protons and only 10 electrons remaining, giving it a net positive charge of +3.

The atomic mass of aluminum is 26.98 atomic mass units (amu). The given ion has an atomic mass of 71 amu, which suggests that the ion has gained additional particles. In this case, the ion has also gained three neutrons, resulting in a higher atomic mass.

The total number of particles (protons, neutrons, and electrons) in the ion can be calculated by adding the number of protons (13) and the number of neutrons (3), which equals 16. Since the ion has a net charge of +3, it only contains 10 electrons.

In summary, the ion with a +3 charge, 28 electrons, and an atomic mass of 71 is the aluminum ion (\(Al^{3+}\)), which has 13 protons, 10 electrons, and 3 neutrons.

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Answer:

CH₂

Explanation:

From the question given above, the following data were obtained:

Mass of Carbon (C) = 5.41 g

Mass of compound = 6.32 g

Empirical formula =?

Next, we shall determine the mass of Hydrogen (H) in the compound. This can be obtained as follow:

Mass of Carbon (C) = 5.41 g

Mass of compound = 6.32 g

Mass of Hydrogen (H) =?

Mass of H = mass of compound – mass of C

Mass of H = 6.32 – 5.41

Mass of H = 0.91 g

Finally, we shall determine the empirical formula of the compound. This can be obtained as follow:

Mass of C = 5.41 g

Mass of H = 0.91 g

Divide by their molar mass

C = 5.41 / 12 = 0.451

H = 0.91 / 1 = 0.91

Divide by the smallest

C = 0.451 / 0.451 = 1

H = 0.91 / 0.451 = 2

Thus, the empirical formula for the compound is CH₂

The volume of the tank is approximately 130.75 m³.

To solve this problem, we need to use the concept of air and water vapor mixture. The given data includes the mass of dry air and water vapor, temperature, and total pressure. We can calculate the specific humidity, relative humidity, and volume of the tank using the following steps:

(a) Specific humidity:

The specific humidity (ω) is defined as the ratio of the mass of water vapor (m_w) to the total mass of the air-water vapor mixture (m_t):

ω = m_w / m_t

Given that the mass of water vapor is 0.17 kg and the total mass of the mixture is 15 kg + 0.17 kg = 15.17 kg, we can calculate the specific humidity:

ω = 0.17 kg / 15.17 kg ≈ 0.0112

So, the specific humidity is approximately 0.0112.

(b) Relative humidity:

Relative humidity (RH) is the ratio of the partial pressure of water vapor (P_w) to the saturation vapor pressure of water (P_ws) at the given temperature, multiplied by 100:

RH = (P_w / P_ws) * 100

To find the relative humidity, we need to determine the saturation vapor pressure at 30°C. Using a vapor pressure table or equation, we can find that the saturation vapor pressure at 30°C is approximately 4.246 kPa.

Given that the total pressure is 100 kPa, the partial pressure of water vapor is 0.17 kg / 15.17 kg * 100 kPa = 1.119 kPa.

Now we can calculate the relative humidity:

RH = (1.119 kPa / 4.246 kPa) * 100 ≈ 26.34%

So, the relative humidity is approximately 26.34%.

(c) Volume of the tank:

To find the volume of the tank, we can use the ideal gas law equation:

PV = nRT

Where P is the total pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

First, we need to calculate the number of moles of dry air and water vapor in the tank. The number of moles (n) can be obtained using the equation:

n = m / M

Where m is the mass and M is the molar mass.

The molar mass of dry air is approximately 28.97 g/mol, and the molar mass of water vapor is approximately 18.015 g/mol.

For dry air:

n_air = 15 kg / 0.02897 kg/mol ≈ 517.82 mol

For water vapor:

n_water = 0.17 kg / 0.018015 kg/mol ≈ 9.43 mol

Now we can calculate the volume using the ideal gas law:

V = (n_air + n_water) * R * T / P

Given that R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (30°C + 273.15 = 303.15 K), and P is the total pressure (100 kPa), we can calculate the volume:

V = (517.82 mol + 9.43 Mol) * 8.314 J/(mol·K) * 303.15 K / (100,000 Pa) ≈ 130.75 m³

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Answer:

The rate constant of the reaction at 125˚ is \(0.3115 \ \text{sec}^{-1}\).

Explanation:

The Arrhenius equation is a simple equation that describes the dependent relationship between temperature and the rate constant of a chemical reaction. The Arrhenius equation is written mathematically as

                                                  \(k \ = \ Ae^{\displaystyle\frac{-E_{a}}{RT}}\)

                                               \(\ln k \ = \ \ln A \ - \ \displaystyle\frac{E_{a}}{RT}\)

where \(k\) is the rate constant, \(E_{a}\) represents the activation energy of the chemical reaction, \(R\) is the gas constant, \(T\) is the temperature, and \(A\) is the frequency factor.

The frequency factor, \(A\), is a constant that is derived experimentally and numerically that describes the frequency of molecular collisions and their orientation which varies slightly with temperature but this can be assumed to be constant across a small range of temperatures.

Consider that the rate constant be \(k_{1}\) at an initial temperature \(T_{1}\) and the rate constant \(k_{2}\) at a final temperature \(T_{2}\), thus

                         \(\ln k_{2} \ - \ \ln k_{1} = \ \ln A \ - \ \displaystyle\frac{E_{a}}{RT_{2}} \ - \ \left(\ln A \ - \ \displaystyle\frac{E_{a}}{RT_{1}}\right) \\ \\ \\ \rule{0.62cm}{0cm} \ln \left(\displaystyle\frac{k_{2}}{k_{1}}\right) \ = \ \displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)\)

                                         \(\rule{1.62cm}{0cm} \displaystyle\frac{k_{2}}{k_{1}} \ = \ e^{\displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)} \\ \\ \\ \rule{1.62cm}{0cm} k_{2} \ = \ k_{1}e^{\displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)}\)

Given that \(E_{a} \ = \ 26.5 \ \ \text{kJ/mol}\), \(R \ = \ 8.3145 \ \ \text{J mol}^{-1} \ \text{K}^{-1}\), \(T_{1} \ = \ \left(40 \ + \ 273\right) \ K\), \(T_{2} \ = \ \left(125 \ + \ 273\right) \ K\), and \(k_{1} \ = \ 0.0354 \ \ \text{sec}^{-1}\), therefore,

           \(k_{2} \ = \ \left(0.0354 \ \ \text{sec}^{-1}\right)e^{\displaystyle\frac{26500 \ \text{J mol}^{-1}}{8.3145 \ \text{J mol}^{-1} \ \text{K}^{-1}}\left(\displaystyle\frac{1}{313 \ \text{K}} \ - \ \displaystyle\frac{1}{398 \ \text{K}} \right)} \\ \\ \\ k_{2} \ = \ 0.3115 \ \ \text{sec}^{-1}\)                      

Answer:

0.686 M

Explanation:

To find the molarity of C₂H₆O, you need to (1) convert grams to moles (using the molar mass of C₂H₆O) and then (2) calculate the molarity (using the molarity ratio). The final answer should have 3 sig figs like the given values.

(Step 1)

Atomic Mass (C): 12.011 g/mol

Atomic Mass (H): 1.008 g/mol

Atomic Mass (O): 15.999 g/mol

Molar Mass (C₂H₆O): 2(12.011 g/mol) + 6(1.008 g/mol) + 15.999 g/mol

Molar Mass (C₂H₆O): 46.069 g/mol

  74.6 g C₂H₆O               1 mole
-------------------------  x  --------------------  =  1.62 moles C₂H₆O
                                      46.069 g

(Step 2)
Molarity = moles / volume (L)

Molarity = 1.62 moles C₂H₆O / 2.36 L

Molarity = 0.686 M

Answer:

1s22s22p6

Explanation:

Neon is an element in the periodic table and has an atomic number of 10, which means it has 10 protons in its nucleus and thus since the number of protons and electrons is the same then it has 10 electrons.

Therefore, it has 2 electrons in the first energy shell and 8 electrons in the second energy shell. To elaborate further, the first shell has a single s-sub shell that contains a single s-orbital that can hold two electrons. The second energy shell has a single s-sub-shell whose s-orbital will occupy 2 electrons, and also has a p-orbital which can hold 6 electrons, making the second shell to have 8 electrons.

Answer:

A. Chemical potential to Kinetic energy

Explanation:

The car engine burns gas to produce motion. Burning is a form of chemical reaction and the motion produced is the kinetic energy.

To draw the two circles, we would need to draw a smaller circle with a diameter of 2,532.5 miles (representing the moon) and a larger circle with a diameter of 75,974.4 miles (representing the Earth) that is 30 times larger than the smaller circle.

If we assume that the larger circle represents the Earth, then the diameter of the Earth would be 30 times the diameter of the smaller circle representing the moon. Let's say that the diameter of the smaller circle is x. Then the diameter of the larger circle (Earth) would be 30 times x or 30x.

To find the correct distance from Earth to the moon at this scale, we need to know the actual distance from Earth to the moon, which is approximately 238,855 miles or 384,400 kilometers. If we divide this distance by the scale factor of 30, we get:

238,855 miles / 30 = 7,961.8 miles

Therefore, the diameter of the smaller circle (moon) would be approximately 7,961.8 miles / π = 2,532.5 miles (rounded to one decimal place). And the diameter of the larger circle (Earth) would be 30 times that or 75,974.4 miles

So, to draw the two circles, we would need to draw a smaller circle with a diameter of 2,532.5 miles (representing the moon) and a larger circle with a diameter of 75,974.4 miles (representing the Earth) that is 30 times larger than the smaller circle.

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Answer:

When conducting a chemistry lab experiment using chalk (calcium carbonate) in vinegar (acetic acid), there are several limitations and precautions to be aware of:

Limitations of chalk in vinegar chemistry experiment:

Reaction rate: The reaction between chalk and vinegar is relatively slow, which may require a longer observation period or higher concentration of vinegar to observe significant changes within a reasonable time frame.

Solubility: Chalk may not dissolve completely in vinegar, resulting in incomplete reaction or difficulty in obtaining accurate results.

Product formation: The reaction between chalk and vinegar produces carbon dioxide gas, water, and calcium acetate. The carbon dioxide gas may escape into the atmosphere, leading to loss of product and inaccurate measurements.

pH: Chalk is a basic substance, and the reaction with vinegar, which is acidic, may result in neutralization, leading to a decrease in the overall acidity of the reaction mixture.

Precautions to take in chalk in vinegar chemistry experiment:

Ventilation: The reaction between chalk and vinegar produces carbon dioxide gas, which can displace air and potentially cause asphyxiation in a closed or poorly ventilated area. Conduct the experiment in a well-ventilated area or under a fume hood to ensure adequate air circulation.

Eye and skin protection: Vinegar is an acid and can cause skin and eye irritation. Wear appropriate personal protective equipment (PPE), such as gloves and goggles, to protect yourself from contact with vinegar or any other chemicals used in the experiment.

Chemical handling: Handle the chemicals, including chalk and vinegar, with care, following proper lab safety protocols. Avoid ingestion, inhalation, or direct contact with the chemicals, and dispose of them properly according to local regulations.

Accuracy in measurements: Use calibrated and accurate measuring tools, such as graduated cylinders or burettes, to measure the amount of chalk, vinegar, and other reagents accurately. This will ensure the reliability and accuracy of the experimental results.

Observations: Make careful and detailed observations during the experiment, noting any changes in appearance, gas evolution, or other relevant observations. Take measurements at appropriate intervals and record the data accurately for analysis and interpretation.

It is important to follow good laboratory practices, including proper chemical handling, accurate measurements, and cautious observations, to ensure safe and reliable results in a chalk in vinegar chemistry lab experiment. Consult with a qualified instructor or supervisor for specific guidelines and precautions related to your experiment.

Answer:

Explanation:

There are 5 carbon and one double hydrogen bond. The 1 indicates where the double bond goes. Each end of a line is a carbon.  

Answer: pure water is homogeneous and pure substance.

Explanation: However, when a homogeneous substance consists of two or more different types of molecules uniformly intermingled with one another, then it’s called a homogeneous mixture. A mixture’s composition can vary, but a pure substance does not.

Answer:

True

Explanation:

since the gases and minerals dissolved in water are in the same state as water and they do not form separate layers

The energy traveled from you to your friend through the medium of the air. When you moved your arm, the energy created a vibration in the air which then propagated in all directions. As the wave traveled, it passed through the air and eventually reached your friend, carrying energy with it.

Energy is the ability to do work or cause change. It is the capacity to cause motion, to create heat, to transform matter, or to generate electricity. It can come from a variety of sources, such as the sun, wind, water, fossil fuels, nuclear energy, and other renewable sources. Energy can also be stored in batteries or fuel cells. It is a fundamental component of the universe and is essential for life. All life forms require energy to survive and grow, and energy is also necessary for the development of technology. Without energy, the world would not function as it does today.

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Answer: 17.91 grams of CO2 will be produced

Explanation:

The balanced chemical equation for the combustion of pentane is:

C5H12(l) + 8 O2(g) → 5 CO2(g) + 6 H2O(g)

From the equation, we can see that for every 1 mole of C5H12 that reacts, 5 moles of CO2 are produced. Therefore, to determine the mass of CO2 produced, we need to first calculate the number of moles of CO2 produced by 0.08142 moles of C5H12:

0.08142 moles C5H12 x (5 moles CO2 / 1 mole C5H12) = 0.4071 moles CO2

Now we can use the molar mass of CO2 (44.01 g/mol) to calculate the mass of CO2 produced:

0.4071 moles CO2 x 44.01 g/mol = 17.91 g CO2

Therefore, the mass of CO2 produced by the complete combustion of 0.08142 moles of pentane is 17.91 g.

Answer: 17.912 gm

Explanation: C5H12 + 8O2 ------- 5CO2 +6H2O

IF ONE MOLE OF C5H12 produces 5 moles of CO2 then 0.08142 produces x mole

then x = 5 X 0.08142

X = 0.4071

NO. OF MOLES = given mass/ molecular mass of CO2

0.4071= mass/44

Mass= 0.4071 X 44

mass = 17.912

therefore mass of CO2 produced is 17.912

The volume of the sodium hydroxide solution required is 4.9 * 10^-5 L.

We know that the reaction between the sodium hydroxide in aqueous solution and para-nitrophenol is a 1:1 reaction. We now have to find the volume of the sodium hydroxide solution that must be added to the solution in order to ensure that formation of the yellow anion is complete.

We should know that the reaction is equimolar so we can write that;

0.105 M * V =  7.20 × 10−4 g/ 139 g/mol

V = 7.20 × 10−4 g/ 139 g/mol/0.105 M

V = 4.9 * 10^-5 L

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Answer:

480 mins

Explanation:

The temperature in Celsius and kelvin is 50°C and 323.15 K.

The three most used temperature scales for usage in business, science, and daily life are Kelvin, Celsius, and Fahrenheit.

A scale in Kelvin is absolute. There are no degree symbols after its values, which begin at absolute zero.

Both Celsius and Fahrenheit are relative scales. The degree symbol is used to indicate Fahrenheit and Celsius temperatures.

Given the temperature is 122°F

Temperature in Celsius is -

⇒ C = (F - 32)/1.80

⇒ C = (122 - 32) / 1.80

⇒ C = 90 / 1.80

⇒ C = 50°C

Temperature in Kelvin is -

K = 5/9(F - 32) + 273.15

⇒ K = 5/9 (122 - 32) + 273.15

⇒ K = 323.15 K

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Answer:

To find the pressure at which the solubility of the gas would be 0.2 g/L, we can use the concept of Henry's Law. Henry's Law states that the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid.

The equation for Henry's Law is:

S = k * P

Where:

S is the solubility of the gas in the liquid (in g/L)

k is the Henry's Law constant (which depends on the specific gas and liquid)

P is the partial pressure of the gas above the liquid (in atm)

We can set up a proportion to find the unknown pressure (P2) when the solubility (S2) is 0.2 g/L:

S1 / P1 = S2 / P2

Substituting the given values:

0.028 g/L / 3.5 atm = 0.2 g/L / P2

Now we can solve for P2:

P2 = (0.2 g/L * 3.5 atm) / 0.028 g/L

P2 = 24.5 atm

Therefore, at a pressure of 24.5 atm, the solubility of the gas would be 0.2 g/L.

Answer

The type of structural formula for each of the hydrocarbons given are shown below:

(a) Propyne (expanded):

(b) 2,2-dimethybutane (skeletal):

(c) Benzene (line-and-angle)

:

(d) Heptane (condensed)

CH₃CH₂CH₂CH₂CH₂CH₂CH₃

Answer:

biochemist is both life science and a chemical . it explores the chemistry of the living organisms and the molecular basis for the changes occurring in the living organisms. it uses the method of chemistry . BIOCHEMIST has become the foundation for understanding all biological processes

The Kc of the system is 1.29 * 10^22.

We know that the equilibrium constant has to do with the kind of constant that we can be able  to use to the conversion of the reactants to the products in the course of the reaction.

We know that;

Kp= Kc (RT)^Δn

Kp = Equilibrium constant in terms of pressure

Kc = Equilibrium constant in terms of concentration

R = gas constant

T = temperature

Δn = change in the number of moles

Then;

Kc = Kp/ (RT)^Δn

Kc = 7.7x10^24 /(0.082 * 298)^2

Kc = 1.29 * 10^22

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Answer:

a. Decrease

b. Increase

c. Increase

d. No effect

Explanation:

Glycolysis is present in muscle cells which converts glucose to pyruvate, water and NADH. It produces two molecules of ATP. Cellular respiration produces more molecules of ATP from pyruvate in mitochondria. Glycolysis increases in pyruvate kinase.

a. Loss of binding site for fructose 1,6-bisphosphate in pyruvate kinase: Decrease

b. Loss of allosteric binding site for ATP in pyruvate kinase: No effect

c. Loss of allosteric binding site for AMP in phosphofructokinase: Increase

d. Loss of regulatory binding site for ATP in phosphofructokinase: Increase

A. An important substrate in the glycolysis pathway is fructose 1,6-bisphosphate. It stimulates pyruvate kinase, an essential enzyme in glycolysis. The amount of pyruvate kinase that is activated will decrease if the fructose 1,6-bisphosphate binding site in pyruvate kinase is eliminated. As a result the rate of glycolysis in the muscle cells will probably decrease.

B. The allosteric ATP binding site of pyruvate kinase controls how active the enzyme is. However, pyruvate kinase is not significantly regulated by ATP in muscle cells. Therefore, it is unlikely that deletion of the ATP-binding allosteric site in pyruvate kinase would have no effect on the rate of glycolysis in muscle cells.

C. The rate-limiting enzyme in glycolysis, phosphofructokinase, is activated from all forms by AMP. It increases the rate of glycolysis by stimulating the activity of phosphofructokinase. If the allosteric binding site for AMP is eliminated, phosphofructokinase activation will be reduced. As a result, the rate of glycolysis in muscle cells will decrease.

D. Phosphofructokinase is inhibited allosterically by ATP. It regulates the rate of glycolysis by a feedback mechanism. High ATP concentrations cause phosphofructokinase to bind to its regulatory site, limiting its activity and delaying glycolysis. If the regulatory binding site for ATP is eliminated, the inhibitory action of ATP on phosphofructokinase would be lost. As a result, muscle cells will glycolysis at a faster rate.

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The ionization of the following acids can be represented as:

Tetraoxosulphate (VI) Acid (\(H_{2}SO_{4}\)) ionizes as H+ and SO4^2- ions.

Trioxonitrate (V) Acid (\(HNO_{3}\)) ionizes as H+ and \(NO_{3-}\) ions.

Ethanoic Acid (\(CH_{3}COOH\)) ionizes as H+ and \(CH_{3}COO^{-}\) ions.

Tetraoxosulphate (VI) Acid, also known as sulfuric acid (\(H_{2}SO_{4}\)), ionizes as follows:

\(H_{2}SO_{4}\) → \(H+\) + \(SO_{4}^{2-}\)

In this reaction, sulfuric acid donates two hydrogen ions (H+) to the solution, forming sulfate ions (\(SO_{4}^{2-}\)).

Trioxonitrate (V) Acid, commonly known as nitric acid (\(HNO_{3}\)), ionizes as follows:

\(HNO_{3}\) → \(H+_{}\) + \(NO_{3-}\)

Nitric acid dissociates to release one hydrogen ion (\(H+\)) and a nitrate ion  (\(NO_{3-}\)).

Ethanoic Acid, also known as acetic acid (\(CH_{3}COOH\)), ionizes as follows:

\(CH_{3}COOH\) → H+ + \(CH_{3}COO^{-}\)

Acetic acid donates a hydrogen ion (H+) to the solution, forming an acetate ion (\(CH_{3}COO^{-}\)).

In all cases, the acids dissociate in water, producing hydrogen ions (H+) as positively charged ions and their corresponding anions. The hydrogen ions are responsible for the acidic properties of these substances, while the anions contribute to the overall charge balance in the solution. The ionization of acids allows them to conduct electricity in aqueous solutions and react with other substances.

The question was incomplete. find the full content below:

Indicate the ionization of the following acids,

Tetraoxosulphate (VI) Acid

Trioxonitrate (V) Acid

Ethanoic Acid.​

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