According to the nemst equation, when ecell = 0 then: _______

(A) Q=K (B) K=1 (C) AG=0 (D) Ecell = 0

Answer: d=2.70 g/mL

I think

Substance C is a solid solute according to the solubility curve. So option B is correct.

Solubility is the maximum solubility that a solute can have in a 100 g solvent at a specific temperature. Solubility curves are plots of the temperature and the solubility value of a specific solute.

The curve of solubility is a curved line on a graph that indicates the relationship between temperature and solubility for a given substance at different temperatures. The graph of the relationship of solubility to temperature is called the Solubility curve. Most solubility curves are sigmoidal, meaning that the peak solubility occurs at the inflection point.

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Answer:

Hypothesis is a statement of expectation or prediction that will be tested by research

Answer:

Explanation:

Here, we want to know what happens to a solution that becomes more acidic

A lesser ph (1-7) indicates acidity with the acidity being higher as the number becomes smaller

What this means is that a solution with a pH of 3 is more acidic than a solution with a pH of 5

Thus, when the acidity increases, it is expected that the pH of the solution decreases (it becomes smaller in number)

The diagrams show the arrangements of carbon atoms in diamond and in graphite. The diamond is sp³ hybridized and have tetrahedral shape while graphite is sp² hybridized and is hexagonal ring shaped structure.

Diamond containing carbon atoms are strongly bonded chemically with other four carbons. The carbon atoms are sp³ hybridized and make percept tetrahedral shape. Diamond used in jewelry makin g. they are used by surgeons in the cataract surgery. diamond is hard, used for cutting or drilling rock.

Graphite containing carbon atom are form the weak bond with weak forces. the carbon are sp² hybridized and make a hexagonal ring shape structure. The force in graphite in weak that makes it useful material.  it is used as lubricant. it is used in pencil. used in making electrodes.

The diagrams show the arrangements of carbon atoms in diamond and in graphite. The diamond is sp³ hybridized and have tetrahedral shape while graphite is sp² hybridized and is hexagonal ring shaped structure.

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The marginal density function of X and Y is 1 and thus they are probability density functions and the probability of the conditions in (c) and (d) is 0.2 and 0.8 respectively.

A probability is a number that expresses the likelihood or chance that a given event will occur. Probabilities can be stated as proportions ranging from 0 to 1, as well as percentages ranging from 0% to 100%.

A probability density function, also known as the density of a continuous random variable in probability theory, is a function whose value at each given sample in the sample space may be read as offering a relative likelihood that the random variable's value will be near to that sample. Probability density functions are statistical measures that are used to predict the outcome of a discrete value. The probability density function is nonnegative all the way through, and the area under the entire curve is equal to one.

a) The marginal density of X is

\(f_X (x) = \int\limits^x_0 {f_{X,Y} (x,y)} \, dx = \int\limits^x_0 {2} \, dx = 2x\)

If it is a pdf, then

\(\int\limits^1_0 {f_X(x)} \, dx = 1\)

\(\int\limits^1_0 {2x} \, dx = x^2 |_0^1 = 1\)

Since integral is 1, marginal density function of X is pdf.

b) The marginal density of Y is

\(f_Y (y) = \int\limits^y0 {f_{X,Y} (x,y)} \, dy = \int\limits^y_0 {2} \, dy = 2y\)

If it is a pdf, then

\(\int\limits^1_0 {f_Y(y)} \, dy = 1\)

\(\int\limits^1_0 {2y} \, dy = y^2 |_0^1 = 1\)

Since integral is 1, marginal density function of Y is pdf.

c)

       \(P (x < 0.2,y > 0.5) = \int\limits^1_{0.5}\int\limits^{0.2}_0 {f_{X,Y} (x,y) \, dx \, dy = \int\limits^1_{0.5}\int\limits^{0.2}_0 2dxdy\)

       \(= \int\limits^1_{0.5} [2x]_0^{0.2}dy = 0.4[y]_{0.5}^1 = 0.4 (1 - 0.5) = 0.2\)

d)

      \(P (x < 0.8,y > 0.5) = \int\limits^1_{0.5}\int\limits^{0.8}_0 {f_{X,Y} (x,y) \, dx \, dy = \int\limits^1_{0.5}\int\limits^{0.8}_0 2dxdy\)

      \(= \int\limits^1_{0.5} [2x]_0^{0.8}dy = 1.6[y]_{0.5}^1 = 1.6 (1 - 0.5) = 0.8\)

The complete question is:

A chemical system that results from a chemical reaction has two important components in a blend among others. The joint distribution that describes the proportion X and Y of these two components is given by f(x,y) = 2 for 0 < x < y < 1 and f(x,y) = 0 otherwise. (a) Find the marginal density function of X and verify if it is a pdf. (b) Find the marginal density function of Y and verify if it is a pdf. (c) Find the probability that the component proportions produce results X < 0.2, Y > 0.5. (d) Find the probability that the component proportions produce results X < 0.8, Y > 0.5.

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The final concentration of the new copper (II) solution (solution B) is approximately 1.198 M. The absorbance of the new copper (II) solution would be approximately 1.198.

To determine the absorbance of the new copper (II) solution, we need to apply the dilution equation:

C₁V₁ = C₂V₂

Where;

C₁ = initial concentration of solution A

V₁ = initial volume of solution A

C₂ = final concentration of new solution

V₂ = final volume of the new solution

Given;

Initial volume of solution A (V₁) = 22.5 mL

Absorbance of solution A = 1.065

Volume of solution A used (V₁) = 8 mL

Final volume of the new solution (V₂) = 20.0 mL

To find the final concentration of the new solution (C₂), we rearrange the dilution equation:

C₂ = (C₁V₁) / V₂

Now we can calculate C₂:

C₂ = (C₁  ×  V₁) / V₂

C₂ = (1.065  ×  22.5 mL) / 20.0 mL

Calculating the values:

C₂ = 23.9625 / 20.0

C₂ = 1.198125

Therefore, the final concentration of the new copper (II) solution (solution B) is approximately 1.198 M.

To determine the absorbance of the new solution, we can use the Beer-Lambert Law:

A = ε × c × l

Where:

A = absorbance

ε = molar absorptivity (a constant specific to the compound and the wavelength of light used)

c = concentration of the solution

l = path length (the width of the cuvette or cell used for measurement)

Since the path length remains constant, we can assume it is the same for both solution A and solution B. The molar absorptivity (ε) for copper (II) solution also remains the same.

Now we can calculate the absorbance of the new solution (A₂);

A₂ = ε × c₂ × l

A₂ = ε × (1.198) × l

Since the molar absorptivity (ε) and path length (l) are constant, the absorbance of the new copper (II) solution (solution B) would be 1.198 times the absorbance of solution A.

Therefore, the absorbance of the new copper (II) solution would be approximately 1.198.

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The time required to coat one side of the 10x10 cm square plate with Ag using the process of electrolysis is approximately 65.7 minutes.

The amount of silver (Ag) required to form a coating of 0.84 mm thickness on a 10x10 cm square plate can be calculated as follows:

Area of the plate = length x width = 10 cm x 10 cm = 100 cm²

Volume of the coating = area x thickness = 100 cm² x 0.084 cm = 8.4 cm³

Mass of Ag required = density x volume = 10.5 g/cm³ x 8.4 cm³ = 88.2 g

Now, let's calculate the amount of charge required to deposit 88.2 g of Ag:

1 mole of Ag has a mass of 107.87 g, which contains 6.022 x 10²³ atoms.

Hence, 88.2 g of Ag contains (88.2/107.87) x 6.022 x 10²³ atoms = 4.92 x 10²³ atoms

Each Ag atom requires one electron to be deposited on the plate during electrolysis.

Therefore, the total amount of charge required to deposit 4.92 x 10²³ Ag atoms is 4.92 x 10²³ x 1.602 x 10⁻¹⁹ C/atom = 7.88 x 10⁴ C

Now, let's use the formula Q = I x t (where Q is the amount of charge, I is the current, and t is the time) to calculate the time required to deposit this amount of charge:

t = Q/I = 7.88 x 10⁴ C/20 A = 3940 s = 65.7 minutes

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Answer:

Shift it to the right toward the products,

Explanation:

Adding more reactant will increase the concentration of the reactants, to offset this increased concentration, more products will be formed.

A. build proteins

Explanation:

The endoplasmic reticulum can either be smooth or rough, and in general its function is to produce proteins for the rest of the cell to function. The rough endoplasmic reticulum has on it ribosomes, which are small, round organelles whose function it is to make those proteins

Substance A is likely a transition metal such as silver (Ag) with a higher atomic number, and substance B is aluminum (Al), a widely distributed metal occurring only in a bound state

a)The substance A and B are not defined in the question. However, from the information given in the question, we can deduce the following:

Elements E1 and E2 have the same number of electrons in their outer fourth electron shell. The total number of electrons in the electron shells of their atoms differs by 10. The element E2 is among the top ten in distribution in the earth's crust and occurs in nature only in a bound state. Their simple substances A and B are silver-white, electrically and thermally conductive.

This indicates that elements E1 and E2 have the same number of valence electrons but different total numbers of electrons in the shells. Since element E2 is among the top ten in distribution in the earth's crust and occurs only in a bound state, it is likely to be a metal. As a result, the simple substances A and B, which are silver-white and electrically and thermally conductive, are also metals.

b) Metal atoms, according to the metallic bond theory, are held together by the sharing of valence electrons. These electrons are responsible for the excellent electrical and thermal conductivity of metals. Silver-white is a typical color for many metals. As a result, substances A and B may be any metal that fits the criteria described above.To sum up, substance A and B can be any metal that has silver-white color, and is electrically and thermally conductive.

Substance A: Based on the higher number of electrons and similar outer electron configuration, substance A can be identified as a transition metal. Transition metals typically have high atomic numbers, exhibit metallic properties, and are located in the d-block of the periodic table. Examples of transition metals include copper (Cu), silver (Ag), and gold (Au).

Substance B: Given that substance B occurs only in a bound state and is among the top ten in distribution in the earth's crust, it can be identified as a highly abundant and widely distributed metal. One such metal that meets these criteria is aluminum (Al). Aluminum is a silver-white metal that is highly abundant in the earth's crust and is commonly found in minerals and compounds.

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Chemicals like bacterial toxins that poison cells are commonly described as being toxic or poisonous. These substances have the ability to disrupt normal cellular functions and processes, leading to harmful effects on the cells and organisms they come into contact with.

Toxins can have various mechanisms of action. Some toxins interfere with essential biochemical pathways, disrupt cellular membranes, or inhibit vital enzymes, while others may directly damage DNA or disrupt cellular signaling. Regardless of their specific mode of action, toxins are designed to have a detrimental impact on cellular function and can cause a wide range of adverse effects, from mild symptoms to severe illness or even death.

The toxicity of a substance is often determined by its concentration, exposure duration, and the specific vulnerability of the target cells or organisms. Toxins produced by bacteria can be classified into exotoxins, which are secreted by bacteria and released into the surrounding environment, or endotoxins, which are part of the bacterial cell wall and are released upon cell death or lysis.

Chemicals like bacterial toxins are referred to as toxic or poisonous due to their ability to disrupt cellular function and cause harm to cells and organisms.

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Citric acid and water are both polar molecules and dissolve polar molecules in orange juice, whereas grease in French fries is made up of nonpolar molecules that cannot be dissolved in water. Soap contains both polar and nonpolar molecules, allowing it to interact with both water and grease to cleanse hands.

Citric acid, which is found in oranges, is a polar molecule due to its asymmetric molecular structure that contains polar functional groups such as carboxylic acid and hydroxyl groups. Similarly, water molecules are polar due to their bent shape and the polarity of the oxygen-hydrogen bond. When combined, the polar molecules of citric acid and water are able to dissolve other polar molecules in the orange juice, such as sugar.

In contrast, grease is made up of nonpolar molecules, such as long-chain hydrocarbons, which cannot be dissolved in water due to the polar nature of water molecules. This is why washing hands with plain water is not effective in removing grease from the skin. However, soap contains both polar and nonpolar molecules, allowing it to interact with both water and grease. The polar end of the soap molecule is attracted to the water molecules, while the nonpolar end is attracted to the grease, allowing it to remove grease from the skin when washing hands.

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Answer:

The specific heat of the metal is 0.212 J/(g°C).

Explanation:

We can calculate the specific heat of the metal by the following equilibrium:

\( q_{a} = -q_{b} \)                          

\( m_{a}C_{a}\Delta T_{a} = -m_{b}C_{b}\Delta T_{b} \)

\(m_{a}C_{a}(T_{f_{a}} - T_{i_{a}}) = -m_{b}C_{b}(T_{f_{b}} - T_{i_{b}})\)

In the above equation, we have that the heat loses by the metal (b) is gained by the water (a).

\(m_{a}\): is the water's mass = 72.0 g

\(C_{a}\): is the specific heat of water = 4.184 J/(g°C)            

\(T_{i_{a}}\): is the initial temperature of the water = 19.2 °C

\(T_{f_{a}}\): is the final temperature of the water = 25.5 °C

\(m_{b}\): is the metal's mass = 141 g

\(C_{b}\): is the specific heat of metal =?

\(T_{i_{b}}\): is the initial temperature of the metal = 89.0 °C

\(T_{f_{b}}\): is the final temperature of the water = 25.5 °C

\(m_{a}C_{a}(T_{f_{a}} - T_{i_{a}}) = -m_{b}C_{b}(T_{f_{b}} - T_{i_{b}})\)

\(72.0 g*4.184 J/(g^{\circ} C)*(25.5 ^{\circ} C - 19.2 ^{\circ} C) = -141 g*C_{b}*(25.5 ^{\circ} C - 89.0 ^{\circ} C)\)            

\( C_{b} = -\frac{72.0 g*4.184 J/(g^{\circ} C)(25.5 ^{\circ} C - 19.2 ^{\circ} C)}{141 g(25.5 ^{\circ} C - 89.0 ^{\circ} C)} = 0.212 J/(g^{\circ} C) \)

Therefore, the specific heat of the metal is 0.212 J/(g°C).

I hope it helps you!

The one that is not a name given for the four outer planets terrestrial planets. The correct option is 3.

The planets Mercury, Venus, Earth, and Mars are referred to as terrestrial because their surfaces are compact and rocky, similar to Earth's terra firma. The terrestrial planets are the solar system's four innermost planets.

A terrestrial planet, telluric planet, or rocky planet is one that is primarily composed of silicate rocks or metals.

Mercury, Venus, Earth, and Mars are the inner planets in the Solar System that are closest to the Sun and are classified as terrestrial planets by the International Astronomical Union.

Astronomers who use the geophysical definition of a planet may consider two or three planetary-mass satellites, including the Earth's Moon, Io, and occasionally Europa, as terrestrial planets, along with the rocky protoplanet-asteroids Pallas and Vesta.

Thus, the correct option is 3.

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Answer: volume of gas at 175°C = 18.10mL.

Given :

• Temperature (T1 = 175°C

• Volume (V1 ) = 132 mL

• Temperature (T2) = 24 °C

• Volume 2 = ....mL?

We will consider Charles law "volume of an ideal gas at constant pressure is directly proportional to the absolute temperature"

• Represented as :

Replacing the given parameters into chales law equation , we get that V2 =

This means that the volume of gas at 175 = 18.10mL.

Answer:

Freezing T° of solution = - 4.52°C

Explanation:

ΔT = Kf . m . i

That's the formula for colligative property about freezing point depression.

Li₂O is an oxide that can not be dissociated but, if we see it's a ionic compound.

Li₂O →  2Li⁺  +  O⁻²

3 moles of ions have been formed. Ions dissolved in solution are i, what we call Van't Hoff factor.

m is molality → 0.811 m, this is data

Kf →Cryoscopic constant, for water is 1.86 °C/m

and ΔT = Freezing T° of pure solvent - Freezing T° of solution

We replace: 0°C - Freezing T° of solution = 1.86°C/m . 0.811 m . 3

Freezing T° of solution = - 4.52°C

The concentration of the Nitric acid solution : 0.114 M

Titration is a procedure for determining the concentration of a solution (analyte) by reacting with another solution whose known concentration (usually a standard solution) is called the titrant. Determination of the endpoint/equivalence point of the reaction can use indicators according to the appropriate pH range  

Titrations can be acid-base titration, depositional titration, and redox titration. An acid-base titration is the principle of neutralization of acids and bases  

Reaction

HNO₃ + NaOH → NaNO₃ + H₂O​

Concentration a standard solution  of sodium hydroxide :  0.0998 mol/dm³ , and the volume = 25 cm³

moles NaOH=

\(\tt mol=M\times V\\\\mol=0.0998\times 25\\\\mol=2.495~mlmoles\)

From the equation, mol ratio HNO₃ : NaOH = 1 : 1, so mol HNO₃ = mol NaOH=2.495 mlmoles

The volume of HNO₃ = 21.8 cm³, so the concentration :

\(\tt M=\dfrac{n}{V}\\\\M=\dfrac{2.495}{21.8}\\\\M=0.114\)

The concentration of the nitric acid solution is 0.11445 mol/dm³

From the question,

We are to calculate the concentration of the nitric acid solution

The given balanced chemical equation for the reaction is

HNO₃ + NaOH → NaNO₃ + H₂O​

This means 1 mole of HNO₃ is needed to completely neutralize 1 mole of NaOH

Using the formula

\(\frac{C_{A}V_{A} }{C_{B}V_{B}} = \frac{n_{A}}{n_{B}}\)

Where \(C_{A}\) is the concentration of acid

\(C_{B}\) is the concentration of base

\(V_{A}\) is the volume of acid

\(V_{B}\) is the volume of base

\(n_{A}\) is the mole ratio of acid

\(n_{B}\) is the mole ratio of base

From the given information

\(C_{B}= 0.0998\ mol/dm^{3}\)

\(V_{A} = 21.80 \ cm^{3}\)

\(V_{B} = 25.0 \ cm^{3}\)

From the balanced chemical equation

\(n_{A} = 1\)

\(n_{B} =1\)

Putting the parameters into the formula, we get

\(\frac{C_{A} \times 21.80 }{0.0998 \times 25.0} = \frac{1}{1}\)

Then,

\(C_{A} \times 21.80=0.0998 \times 25.0\)

∴ \(C_{A}=\frac{0.0998 \times 25.0}{21.80}\)

\(C_{A} =\frac{2.495}{21.80}\)

\(C_{A} = 0.11445 \ mol/dm^{3}\)

Hence, the concentration of the nitric acid solution is 0.11445 mol/dm³

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The strengths of intermolecular forces vary from CO₂ to CS₂ to CSe₂ as as it increase from CO₂ to CS₂ to CSe₂. CO₂ has the weakest intermolecular forces mainly London dispersion forces. CS₂ exhibits stronger London dispersion forces due to the larger size of sulfur atoms. CSe₂ has the strongest intermolecular forces among the three compounds, due to highest boiling point.

In CO₂, the intermolecular forces are weak London dispersion forces. CO₂ is a nonpolar molecule, and the only intermolecular forces present are the temporary fluctuations in electron density that induce weak attractions between molecules.

In CS₂, the intermolecular forces are stronger than in CO₂ due to the presence of polarizable sulfur atoms. CS₂ molecules experience London dispersion forces, but in addition, they also exhibit dipole-dipole interactions. The sulfur atoms have a larger electron cloud and create temporary dipoles, leading to stronger intermolecular attractions.

In CSe₂, the intermolecular forces are the strongest among the three compounds. Similar to CS₂, CSe₂ has London dispersion forces and dipole-dipole interactions. However, the larger and more polarizable selenium atoms in CSe₂ result in even stronger intermolecular forces compared to CS₂.

The increased electron cloud size and polarity of the selenium atoms contribute to the higher boiling and melting points of CSe₂ compared to CS₂.

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The strong acid in aqueous solution among the given options is HClO4 (option a).

HClO4, also known as perchloric acid, is a strong acid because it completely dissociates in water, releasing H+ ions. Strong acids are acids that ionize completely in water, resulting in a high concentration of H+ ions.

On the other hand, the other options listed are not strong acids:

b. HOCH2CH2OH is ethylene glycol, which is a non-acidic compound and does not dissociate into H+ ions in water.

c. NH3 is ammonia, which is a weak base, not a strong acid.

d. Ca(OH)2 is calcium hydroxide, which is a strong base, not a strong acid.

e. H3PO4 is phosphoric acid, which is a weak acid but not a strong acid like HClO4.

Therefore, the correct answer is option a. HClO4.

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Answer:

B. NH4Cl

Explanation:

When NH4Cl is dissolved in water, it hydrolyzes to give HCl and NH4OH

NH4Cl + H2O = HCl + NH4OH

NH4OH is a weak base.

HCl is a strong acid which changes blue litmus to red.

We know the standard reduction potentials of the half-cells involved, so we can find the cell voltage and the spontaneous reaction. Thus;

The anode reaction is:

Pb(s) → Pb2+(aq) + 2e-

This is the oxidation half-reaction that occurs in the Pb half-cell.

The cathode reaction is:F2(g) + 2e- → 2F-(aq).

This is the reduction half-reaction that occurs in the F2 half-cell.

The spontaneous cell reaction is

:Pb(s) + F2(g) → Pb2+(aq) + 2F-(aq).

This is the combination of the oxidation and reduction half-reactions, with the electrons canceled out from both sides.

The cell voltage is 2.996 V The standard cell potential is calculated as follows:

standard cell potential = E°(reduction) - E°(oxidation)standard cell potential = 2.870 V - (-0.126 V)standard cell potential = 2.996 V, The cell voltage is positive, indicating that the reaction is spontaneous.

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The charge on ammonium is +1, indicating that it has one less electron than protons. In ammonium, nitrogen forms four covalent bonds with four hydrogen atoms, using four of its valence electrons. Each hydrogen atom contributes one valence electron to the bonding process.

As a result, the total number of valence electrons in ammonium is 5 (from nitrogen) + 4 (from four hydrogen atoms) = 9. Since nitrogen's valence electron count is less than its usual five, it has an electron deficiency. This deficiency is balanced by a positive charge on the ammonium ion.

In summary, the charge on ammonium is +1 because nitrogen forms four covalent bonds with hydrogen atoms, resulting in an electron deficiency in the nitrogen atom.

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Answer: 

hope it helps!! ..............

The mass of ammonia in the container is  B. 0.180 g.

The ideal gas law is defined as PV = nRT

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

According to the given data; the pressure of the gas (P) is 833 torr, the volume of the gas (V) is 250.0 mL, temperature of the gas (T) is 42.4 °C. Ammonia, NH₃ is the gas that we are dealing with. The molecular weight of NH₃ is 17.03 g/mol.

To calculate the mass of ammonia, we must convert the given volume to liters and the given temperature to Kelvin.

Let's solve this by using the ideal gas law equation as follows;

PV = nRT

n = PV/RT

Where P = 833 torr

V = 250.0 mL = 0.25 L (1 L = 1000 mL)

R = 0.08206 Latm/molK (or) 62.3637 mmHg/L molK (I'm using torr so I will use mmHg/L molK)

T = 42.4 + 273 = 315.4 K

Substituting the values into the above formula,

n = (833 mmHg/L × 0.25 L) / (62.3637 mmHg/L molK × 315.4 K)

n = 0.0104 moles

From here we can convert moles of NH₃ to grams. The molecular weight of NH₃ is 17.03 g/mol.

Therefore, mass = moles × molar mass

mass = 0.0104 mol × 17.03 g/mol

mass = 0.177 g (approx)

Therefore, the correct answer is option B. 0.180 g.

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The measurement of expired CO₂ that typically provides a numeric display of the partial pressure of CO₂ in torr or mmHg is the capnometry.

The capnometry measurement of the  CO₂ , carbon dioxide is the measurement in the expired air and it provides the clinician with the noninvasive measurement of systemic metabolism and the ventilation. it is used to determine the correlation in between the tidal CO₂  and the partial CO₂  in the non intubated breathless patents.

Thus, the measurement if the expired CO₂  that shows the partial pressure of the CO₂  is the capnometry.

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Answer:

Explanation:

1) Organic compounds always contain only p-block elements (Groups III-VII), at least one of which must be carbon. 2) Organic compounds almost always contain one or more C-H bonds. ... Thus, all bonds are typically covalent in organic compounds. Methane (CH4) is the prototypical organic molecule.

Methane contains carbon, forming covalent bonds, and is found in living organisms, making it an organic compound.

Methane (CH₄) is considered an organic compound due to its molecular structure and occurrence in living organisms. Organic compounds are primarily composed of carbon atoms covalently bonded to hydrogen and often other elements like oxygen, nitrogen, sulfur, and more.

Methane consists of one carbon atom bonded to four hydrogen atoms through covalent bonds. Carbon's unique ability to form stable covalent bonds with other elements, including itself, leads to the vast diversity of organic molecules found in living organisms.

Methane is a crucial component of natural gas and is produced by various biological and geological processes. It is present in the digestive systems of animals, formed during decomposition, and plays a role in carbon and energy cycles.

Its prevalence in living systems and its molecular structure classify methane as an organic compound, reflecting the foundational principles of organic chemistry.

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The complete question is :

Methane is called an organic compound why​?