In a reaction progress curve, each peak of the curve corresponds to _____ of the reaction.

Answer:

(Ca2 +, Mg2 +, Na+, and K+) and four major anions ( , , SO 4 2 - , and Cl-) with ionic forms of N, P, Fe, and other trace elements at lower concentrations (Livingstone 1963, Meybeck 1979) (Table 4.7).

The correct answer is C. The liver and gallbladder.

Explanation

The liver, is an organ of the digestive system, is the most voluminous in the human body, and performs three vital functions essential for the organism: Detoxification in which it acts as a filter that collects and eliminates numerous toxins; the liver metabolizes carbohydrates, fats, lipids and proteins, secreting bile, an essential element for our digestion. Also, it prevents bleeding through a coagulation process; finally, works as a container for vitamins and glycogen, carbohydrates, and energy is stored in the form of sugar. On the other hand, the gallbladder is an organ of the digestive system, whose function is to store and concentrate the bile secreted by the liver required by the digestion process. The gallbladder stores bile until a stimulus caused by food intake, especially meat or fat. According to the above, if the functions of these two organs are impaired, the most likely is that the person will have trouble digesting fats. So, the correct answer is C. The liver and gallbladder.

C

came from edginuity

Answer:

The percent yield of oxygen in this chemical reaction is 73.71 %.

Explanation:

Experimental yield of oxygen gas = 115.0 g

Theoretical yield:

Mass of potassium chlorate = 400.0 g

Moles of potassium chlorate =

According to reaction, 2 moles of potassium chlorate gives 3 moles of oxygen gas.

Then 3.2653 mol of potassium chlorate will give:

Mass of oxygen gas :

Percentage yield:

The percent yield of oxygen in this chemical reaction is 73.71 %.

Answer:

Explanation:

1. Gather all the information in one place

M_r:   22.99

           2Na + 2H₂O ⟶ 2NaOH + H₂

T = 25.0 °C

V = 246 mL

2. Moles of H₂

To find the moles of hydrogen, we can use the Ideal Gas Law:

pV = nRT

(a) Calculate the partial pressure of the hydrogen

(b) Convert the volume to litres

V = 246 mL = 0.246 L

(c) Convert the temperature to kelvins

T = (25.0 + 273.15) K = 298.15 K

(d) Calculate the moles of hydrogen

3. Moles of Na

The molar ratio is 2 mol Na: 1 mol H₂

4. Mass of Na

The number of grams of sodium used in the reaction will be approximately 0.1387 grams.

To calculate the number of grams of sodium used in the reaction, we need to use the ideal gas law and consider the effect of the vapor pressure of water on the pressure of the collected hydrogen gas.

Given data:

Volume of hydrogen gas (V) = 246 mL = 0.246 L

Pressure of hydrogen gas (P) = 1.00 atm

Vapor pressure of water (P_water vapor) = 0.0313 atm (subtracted from the total pressure)

Temperature (T) = 25.0°C = 298.15 K

The ideal gas law is given by the equation: PV = nRT, where n is the number of moles of the gas.

First, calculate the total pressure by subtracting the vapor pressure of water from the given pressure of the gas:

Total pressure (P_total) = P - P_water vapor

= 1.00 atm - 0.0313 atm = 0.9687 atm

Now, rearrange the ideal gas law equation to solve for n (moles of gas):

n = PV / RT

Plug in the values:

n = (0.9687 atm × 0.246 L) / (0.0821 L·atm/mol·K × 298.15 K)

n ≈ 0.01206 mol

According to the balanced chemical equation, 2 moles of sodium (Na) produce 1 mole of hydrogen gas (H₂). Therefore, the number of moles of sodium used in the reaction is half of the calculated moles of hydrogen gas:

Moles of sodium = 0.01206 mol / 2 = 0.00603 mol

Finally, calculate the mass of sodium (molar mass of sodium = 22.99 g/mol)

Mass of sodium = Moles of sodium × Molar mass of sodium

Mass of sodium = 0.00603 mol × 22.99 g/mol

≈ 0.1387 g

Therefore, the number of grams of sodium used in the reaction is approximately 0.1387 grams.

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The paleoatmosphere refers to the Earth’s atmosphere in the geological past. The paleoatmosphere was dominated by nitrogen, carbon dioxide and methane. These gases are generated by the radioactive decay of Potassium. During this pre-biological atmosphere, it is thought to have been reducing, to have highly contained virtually no free oxygen, and virtually no argon. Therefore the major gases on this period are the Nitrogen and Carbon Dioxide.

Answer:

A. No photoelectrons < C. KE = 51 kJ·mol⁻¹ <  B. KE = 155 kJ·mol⁻¹

              632 nm         <              455 nm      <           325 nm

Explanation:

The equation for the photoelectric effect is

E = hf = Φ + KE  

That is, the energy (hf) of the incident photon is used to eject an electron from the surface of the metal ( the work function, Φ), and anything left over goes into the kinetic energy (KE) of the electron.

1. Calculate the relative energies of the three wavelengths

E = hf

fλ = c or f = c/λ. So,  

E = (hc)/λ

The energy is inversely proportional to the wavelength. Thus, the order of increasing energies is

632 nm < 455 nm < 325 nm

2. Assign the radiation to the observations.  

Solve the photoelectric equation for KE.

E = Φ + KE

KE = E - Φ

If E < Φ, the light does not have enough energy to eject a photoelectron.

Once E > φ, the greater the value of E, the greater the value of  KE.

The order of energies is

A. No photoelectrons < C. KE = 51 kJ·mol⁻¹ < B. KE = 155 kJ·mol⁻¹

               632 nm        <            455 nm         <               325 nm

Final answer:

Observation A corresponds to the 325 nm wavelength, observation B corresponds to the 632 nm wavelength, and observation C corresponds to the 455 nm wavelength.

Explanation:

The observation labeled A corresponds to the 325 nm wavelength, observation B corresponds to the 632 nm wavelength, and observation C corresponds to the 455 nm wavelength. This is because the behavior of photoelectric effect can be explained by Einstein's equation, E = hf, where E is the energy of a photon, h is Planck's constant, and f is the frequency of the light. Since frequency and wavelength are inversely proportional, higher frequency light corresponds to shorter wavelength and vice versa.

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