Nitrogen dioxide gas and liquid water react to form aqueous nitric acid and nitrogen monoxide gas. Suppose you have 5.0 mol of NO2 and 11.0 mol of H2O in a reactor.

Calculate the largest amount of HNO3 that could be produced. Round your answer to the nearest 0.1 mol

The chemical molecule CrO2 is also known as chromium(IV) oxide or chromic acid. It has the molecular formula CrO2 and is an inorganic substance.

In the solid state, CrO2 exists as a solid with a layered structure, and it is considered a cationic compound. The cation present in CrO2 is chromium(IV) ion, denoted as Cr4+.

On the other hand, the anion present in CrO2 is oxide ion, denoted as O2-. The oxidation state of oxygen in this compound is -2.

So, the cations present in CrO2 are Cr 4+ ions, and the anions present are O2 -2 ions.

In CrO2, the cation present is Chromium (Cr) with a charge of +4, and the anion present is Oxygen (O) with a charge of -2.

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Answer:

A solvent front is the point on a chromatography paper or plate where the solvent has reached the end of the stationary phase and has migrated as far as it can go. It is the farthest point reached by the solvent in the chromatography process.

Answer:

Explanation:

The change in entropy of a system can be determined by comparing the entropy of the reactants to the entropy of the products. The reaction that leads to an increase in the number of moles of gas or particles will generally have a positive change in entropy.

I. 2N2(g) + O2(g) → 2N2O(g)

The reactants have 3 moles of gas, while the product also has 3 moles of gas. Therefore, there is no change in the number of moles of gas, and the change in entropy is likely to be small.

II. CaCO3(s) → CaO(s) + CO2(g)

The reactant is a solid, while the products are a solid and a gas. The formation of a gas from a solid leads to an increase in the number of moles of particles, and therefore an increase in entropy.

III. Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

The reactants consist of a solid and a liquid, while the products consist of an aqueous solution and a gas. The formation of a gas leads to an increase in the number of moles of particles, and therefore an increase in entropy.

Therefore, reactions II and III have a positive change in entropyentropy

Answer:

Explanation:

Coal deposits have been found in Antarctica, particularly in the Transantarctic Mountains where they are interbedded in sedimentary rocks of the flat-lying Beacon Supergroup. The presence of coal in Antarctica supports Wegener's continental drift hypothesis because it suggests that Antarctica was once part of a larger landmass that had a warm climate suitable for the formation of coal. Coal is formed from ancient plant matter that has been compressed and heated over millions of years. The presence of coal in Antarctica suggests that the continent was once located closer to the equator and had a climate that supported lush vegetation.

The mass of sodium sulfate [tex](Na_2SO_4)[/tex] formed by the complete reaction of 137.3 grams of sodium hydroxide [tex](NaOH)[/tex] is 486.74 grams.

The balanced chemical equation for the reaction between sodium hydroxide [tex](NaOH)[/tex] and sulfuric acid [tex](H_2SO_4)[/tex] is:

[tex]NaOH[/tex] + [tex](H_2SO_4)[/tex] → [tex]Na_2SO_4[/tex] + [tex]2H_2O[/tex]

From the balanced equation, we can see that 1 mole of [tex]NaOH[/tex] reacts with 1 mole of [tex](H_2SO_4)[/tex] to produce 1 mole of [tex]Na_2SO_4[/tex] and 2 moles of water [tex](H_2O).[/tex]

The molar mass of [tex]NaOH[/tex] is 40.00 g/mol, which means that 137.3 grams of [tex]NaOH[/tex] is equal to 137.3 g / 40.00 g/mol = 3.4325 moles of [tex]NaOH[/tex].

Since 1 mole of [tex]NaOH[/tex] reacts with 1 mole of [tex](H_2SO_4)[/tex] to produce 1 mole of [tex](Na_2SO_4)[/tex] , we can say that 3.4325 moles of [tex]NaOH[/tex] will react with 3.4325 moles of [tex]H_2SO_4[/tex] to produce 3.4325 moles of [tex]Na_2SO_4[/tex].

The molar mass of [tex](Na_2SO_4)[/tex] is 142.04 g/mol, which means that 1 mole of [tex](Na_2SO_4)[/tex] has a mass of 142.04 g.

Therefore, 3.4325 moles of [tex](Na_2SO_4)[/tex] has a mass of 3.4325 moles x 142.04 g/mol = 486.74 grams.

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The information given can be used to construct the empirical formula for a tin oxide. We must first determine the mass of tin in the sample. This may be achieved by deducting the mass of the crucible, cover, and sample (21.76 g) from the mass of the empty crucible and cover (19.66 g).

This gives us a mass of 2.10 g of tin in the sample. The mass of oxygen in the sample must then be determined. To achieve this, we must deduct the mass of the crucible, cover, and sample (21.76 g) from the mass of the same components (22.29 g) prior to protracted heating. This provides us with an oxygen mass of 0.53 g.

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The theoretical yield of sodium sulfate, Na₂SO₄, formed from the reaction of 4.9 g of sulfuric acid, H₂SO₄ and 5.0 g of sodium hydroxide, NaOH is 7.1 g

First, we shall determine the limiting reactant. This is shown below:

H₂SO₄ + 2NaOH -> Na₂SO₄ + 2H₂O

From the balanced equation above,

98 g of H₂SO₄ reacted with 80 g of NaOH

Therefore,

4.9 g of H₂SO₄ will react with = (4.9 × 80) / 98 = 4 g of NaOH

From the above calculation, we can see that only 4 g of NaOH out of 5 g is needed to react with 4.9 g H₂SO₄.

Thus, the limiting reactant is H₂SO₄

Finally, we shall determine theoretical yield of sodium sulfate, Na₂SO₄ formed. Details below:

H₂SO₄ + 2NaOH -> Na₂SO₄ + 2H₂O

From the balanced equation above,

98 g of H₂SO₄ reacted to produce 142 g of Na₂SO₄

Therefore,

4.9 g of H₂SO₄ will react to produce = (4.9 × 142) / 98 = 7.1 g of Na₂SO₄

Thus, the theoretical yield of sodium sulfate, Na₂SO₄ formed is 7.1 g

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We would need about 16 hydrogen atoms so that we can convert the compound to a saturated fat.

In animal products like meat and dairy, saturated fat is a form of dietary fat that is normally solid at room temperature. It is known as being "saturated" because each molecule of fat has the most hydrogen atoms possible, giving it a stable structure.

We can see this by counting the number of double bonds in the fat and there are eight of them so sixteen hydrogen atoms are needed for saturation.

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A 6.50-g sample of copper metal at 25.0 °C is heated by the addition of 145 J of energy. The final temperature of the copper is 83.7 °C. The specific heat capacity of copper is 0.38 J/g-K.

The correct answer choice is "83.7"

To solve this problem, we can use the equation:

q = mcΔT

where q is the amount of energy absorbed by the copper, m is the mass of the copper, c is the specific heat capacity of copper, and ΔT is the change in temperature of the copper.

Rearranging this equation to solve for ΔT, we get:

ΔT = q / (mc)

Substituting the given values, we get:

ΔT = 145 J / (6.50 g x 0.38 J/g-K)

ΔT = 58.7 K

Therefore, the final temperature of the copper is:

25.0 °C + 58.7 °C = 83.7 °C

So the correct option is 83.7.

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The rate constant, k = 5.27 E-2 s⁻¹, determines the rate law for the reaction P → E + Z.

The rate of the reaction P → E + Z can be expressed as:

Rate = - d[P]/dt = d[E]/dt = d[Z]/dt

where d[P], d[E], and d[Z] = changes in the concentrations of P, E, and Z, respectively, over a small time interval dt.

Use the experimental data to determine the rate constant and the order of the reaction.

Calculate the initial rate of the reaction in each trial by dividing the change in concentration of P by the time interval:

rate1 = (d[P]/dt)1 = (0.30 M - 0 M)/(20 s) = 0.015 M/s

rate2 = (d[P]/dt)2 = (0.60 M - 0.30 M)/(20 s) = 0.015 M/s

rate3 = (d[P]/dt)3 = (0.90 M - 0.60 M)/(20 s) = 0.015 M/s

The initial rates are the same in all three trials, which suggests that the reaction is first-order with respect to P.

Now using any of the three trials to determine the value of the rate constant k, trial 1:

Rate1 = k[P]1

k = Rate1/[P]1 = (1.58 E-2 M/s)/(0.30 M) = 5.27 E-2 s⁻¹

Therefore, the rate law for the reaction P → E + Z is:

Rate = k[P]

where k = 5.27 E-2 s⁻¹ is the rate constant.

Use the rate law to calculate the expected rates of the reaction at different concentrations of P. For example:

Rate2 = k[P]2 = (5.27 E-2 s⁻¹)(0.60 M) = 3.16 E-2 M/s

Rate3 = k[P]3 = (5.27 E-2 s⁻¹)(0.90 M) = 4.74 E-2 M/s

These expected rates are close to the experimental rates, which suggests that the rate law is a good approximation for the reaction under these conditions.

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The reaction is second order with a rate constant of 0.043 mol/l min.

For CA = 0.1 mol/l, -rA = 0.1 mol/l min

For CA = 0.2 mol/l, -rA = 0.3 mol/l min

For CA = 0.3 mol/l, -rA = 0.5 mol/l min

For CA = 0.4 mol/l, -rA = 0.6 mol/l min

The slope of this line is equal to the order of the reaction (n), and the y-intercept is ln(k).

Slope = (0.6931 - (-2.3026)) / (0.3010 - (-0.9163)) = 1.929

ln(k) = -2.3026 + 1.929 * (-0.3010)

ln(k) = -3.1504

k = e^(-3.1504) = 0.043 mol/l min

The reaction is second order with a rate constant of 0.043 mol/l min.

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the oxidation number of copper in copper oxide is 2...

Answer: +2

Explanation: Copper has a +2 oxidation number in CuO.

This is due to the fact that oxygen has an oxidation number of 2, and the entire chemical has a neutral charge. Consequently, the following equation can be used to determine copper's oxidation number:

(+2) + (-2) = 0

In order to counteract the -2 oxidation number of oxygen in CuO, copper must have an oxidation number of +2.

Answer:

There are 1000 milliliters (ml) in one liter. Therefore, the teacher has a total of 6 x 1000 = 6000 ml of sodium nitrate solution.

Explanation:

To divide the solution evenly among the 24 students, we need to divide the total volume of the solution by the number of students:

6000 ml ÷ 24 students = 250 ml per student

Therefore, each student will receive 250 milliliters of sodium nitrate solution.

Answer:

Answer- 0.25ml

Explanation:

So there are 24 students and 6 liters of Solution.So to evenly distribute

Just divide 6 by 24(6÷24/)... So the answer will be 0.25

Answer:

The answer is 966.67 mm of Hg.

Explanation:

To solve this problem, we can use Boyle's Law which states that the pressure of a gas is inversely proportional to its volume when the temperature and the amount of gas are kept constant. The formula for Boyle's Law is:

P1V1 = P2V2

where P1 is the initial pressure, V1 is the initial volume, P2 is the final pressure, and V2 is the final volume.

Using the given values:

P1 = 580 mmHg

V1 = 500 cm³

V2 = 300 cm³

We can solve for P2:

P1V1 = P2V2

580 mmHg x 500 cm³ = P2 x 300 cm³

290,000 mmHg·cm³ = P2 x 300 cm³

P2 = 290,000 mmHg·cm³ / 300 cm³

P2 = 966.67 mmHg (rounded to the nearest hundredth)

Therefore, the pressure when the volume is reduced to 300 cm³ is approximately 966.67 mmHg.

The mass (in grams) of aluminum, Al that must react with iodine to form AlI₃, given that -836.0 KJ of heat is relaesd is 149.0 g

The mass of aluminum required to react with iodine to produce AlI₃ can be obtain as shown below:

2Al(s) + 3I₂(s) → 2AlI₃(s) ∆H = -302.9 KJ

From the balanced equation above,

When -302.9 KJ of heat energy is released, 54 g of aluminum, Al reacted.

Therefore,

When -836.0 KJ of heat energy will be release = (-836.0KJ × 54 g) / -302.9 KJ = 149.0 g of aluminum, Al will react.

Thus, from the above calculation, we can conclude that the mass of aluminum, Al required is 149.0 g

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The ΔH°vap of water at the critical point is approximately 0.04614 kJ/mol.

To calculate the ΔH°vap (enthalpy of vaporization) of water at the critical point, we can use the Clausius-Clapeyron equation;

ln(P₂/P₁) = ΔH°vap/R [1/T₁ - 1/T₂]

where P₁ and T₁ are the pressure and temperature at which the enthalpy of vaporization is known (usually at standard conditions of 1 atm and 100 °C), P₂ and T₂ are the pressure and temperature at the critical point, ΔH°vap is the enthalpy of vaporization, and R is the gas constant (8.314 J/mol∙K).

Using the given values, we can plug them into the equation and solve for ΔH°vap;

ln(3.2 atm / 1 atm) = ΔH°vap / R [1/373 K - 1/275 K]

Simplifying;

ln(3.2) = ΔH°vap / R [0.0026819]

ΔH°vap / R = ln(3.2) / 0.0026819

ΔH°vap / R = 5.552

Multiplying both sides by R:

ΔH°vap = 5.552 x R

ΔH°vap = 5.552 x 8.314 J/mol∙K

ΔH°vap = 46.14 J/mol

Converting to kJ/mol;

ΔH°vap = 0.04614 kJ/mol

Therefore, the ΔH°vap of water is 0.04614 kJ/mol.

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Charles's law of gases states that the density of an ideal gas is inversely proportional to its temperature at constant pressure.

The equation is as follows;

Va/Ta = Vb/Tb

Where;

0.67/362 = 1.12/Tb

0.00185Tb = 1.12

Tb = 605.41K

This temperature in °C is 605.41 - 273 = 332°C

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Answer:

zn + Hcl cual es su rreaccion

1. The star that has a temperature of approximately 9000 K and luminosity about 20 times greater than the Sun’s luminosity is Vega.

2. The type of star that is considered part of the main sequence is red dwarfs.

3. The star that is cooler than the Sun is Barnard’s Star.

4. The Sun is classified as a main sequence star.

5. The force most responsible for the formation of stars is gravity.

6. The star that has a higher luminosity and a lower temperature than the Sun is Aldebaran.

7. Compared to the temperature and luminosity of the star Polaris, the star Sirius is hotter and more luminous.

The equation that represents a balanced reaction involving 14 atoms of oxygen is:

2C2H6 + 7O2 --> 4CO2 + 6H2O

a. The relationship between variables in this equation is that 2 moles of C2H6 (ethane) react with 7 moles of O2 (oxygen) to produce 4 moles of CO2 (carbon dioxide) and 6 moles of H2O (water). This equation follows the law of conservation of mass, where the total number of atoms on both sides of the equation is the same, indicating a balanced reaction.

b. The graph is linear, as the coefficients of the reactants and products in the equation are whole numbers and form a consistent ratio. The coefficients of 2, 7, 4, and 6 represent the stoichiometry of the reaction, indicating a fixed relationship between the reactants and products.

c. An example of a situation where this balanced equation could be applicable is the combustion of ethane (C2H6) in the presence of excess oxygen (O2) to produce carbon dioxide (CO2) and water (H2O), which is a common reaction in the combustion of hydrocarbon fuels. The equation represents the balanced stoichiometry of this reaction, where 2 moles of ethane react with 7 moles of oxygen to produce 4 moles of carbon dioxide and 6 moles of water, involving a total of 14 atoms of oxygen in the reaction.

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Answer:

5. 0.566 g
6. A. 100 times more

Explanation:

5. The pH of a solution is defined as the negative logarithm (base 10) of the hydrogen ion concentration. For a solution with pH=2, the concentration of hydrogen ions is 10^-2 mol/L. Since HBr is a strong acid, it dissociates completely in water to produce H+ and Br- ions. Therefore, the concentration of HBr in the solution is also 10^-2 mol/L.

The molar mass of HBr is 80.91194 g/mol

So, in a 700 mL solution (0.7 L), there are

0.7 L * 10^-2 mol/L = 0.007 mol of HBr.

This corresponds to 0.007 mol * 80.91194 g/mol = 0.566 g of HBr dissolved in the solution.

6. The pH of a solution is defined as the negative logarithm (base 10) of the hydrogen ion concentration. This means that for each decrease in pH by 1 unit, the hydrogen ion concentration increases by a factor of 10. Since the difference in pH between the two solutions is 3 units (6-3=3), the hydrogen ion concentration in the solution with pH=3 is 10^3 = 100 times more than in the solution with pH=6.

The term that would best describe the solution formed if 80g KBr dissolved in 100g water is unsaturated solution.

A saturated solution is a solution with solute that dissolves until it is unable to dissolve anymore, leaving the undissolved substances at the bottom.

On the other hand, an unsaturated solution is that solution that is capable of dissolving more of a solute at the same temperature.

According to this question, 80 grams of KBr were dissolved in 100 grams of water at 35 degrees Celsius. This means that the solution is unsaturated because it can still dissolve more KBr.

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Answer:

Explanation:

Here are the labels for each situation:

1.Usually conducts electricity & heat well - Metal (1)

2.At room temperature these are gases or liquids - Nonmetal (2)

3.Will lose valance electrons to form compounds - Metal (1)

4.Can be used as semiconductors - Metalloid (3)

5.Will gain valance electrons to form compounds - Nonmetal (2)

Answer: They have eight outer electrons.

Explanation: They don't need any more electrons to be added to them and can't give out any electrons to other groups. They have a complete outer shell.

The molar volume of a gas can be calculated using the ideal gas law:

PV = nRT

where P is the pressure of the gas in atmospheres (atm), V is the volume of the gas in liters (L), n is the number of moles of gas, R is the ideal gas constant (0.08206 L·atm/mol·K), and T is the temperature of the gas in Kelvin (K).

To solve for the molar volume of CO2 at 39°C (312 K) and 652 torr (0.859 atm), we can rearrange the ideal gas law as follows:

V = (nRT) / P

First, we need to calculate the number of moles of CO2. We can use the following equation, which relates the pressure, volume, number of moles, and temperature of a gas:

PV = nRT

Solving for n, we get:

n = (PV) / (RT)

Substituting the given values, we get:

n = (0.859 atm * V) / (0.08206 L·atm/mol·K * 312 K)

Now we can substitute this expression for n into the equation for the molar volume:

V = (nRT) / P

V = [(0.859 atm * V) / (0.08206 L·atm/mol·K * 312 K)] * (0.08206 L·atm/mol·K * 312 K) / (0.859 atm)

Simplifying, we get:

V = 24.45 L/mol

Therefore, the molar volume of CO2 at 39°C and 652 torr is 24.45 L/mol.

The value of K at equilibrium, for the reaction is 0.0049

First, we shall list out the given parameters from the question. This is shown below:

Equilibrium constant is defined as:

Equilibrium constant = [Product]ᵐ / [Reactant]ⁿ

Where

With the above formula, we can obtain the equilibrium constant, K as follow:

Equilibrium constant, K = [B₂][AC] / [AB₂C]

K = (0.007 × 0.0118) / 0.0168

K = 0.0049

Thus, the equilibrium constant, K for the reaction is 0.0049

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The best solution to suppress the dissolution of MgCO3 is option D 0.200 M Na2CO3

We need to add an ion or compound that will react with MgCO3 and form a precipitate, thus removing Mg2+ and CO32- ions from the solution.

Therefore, Option D, 0.200 M Na2CO3, contains CO32- ions that can react with Mg2+ ions to form MgCO3 precipitate. This would effectively suppress the dissolution of MgCO3 by removing Mg2+ and CO32- ions from the solution.

Therefore, option D is the best solution to suppress the dissolution of MgCO3.

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If in a reaction between vinegar and antacid tablets, the antacid is the limiting reagent. cm of gas. At constant pressure and temperature, three tablets produce 600 cm³ . The  volume that four tablets will produce is: C. 800 cm³.

Since the antacid is the limiting reagent, the amount of gas produced will be directly proportional to the number of tablets used.

We know that three tablets produced 600 cm³ of gas. Therefore, we can set up a proportion:

3 tablets produce 600 cm³ of gas

4 tablets produce x cm³ of gas

To solve for x, we can use cross-multiplication:

3 tablets × x cm³ of gas = 4 tablets × 600 cm³ of gas

3x = 2400

x = 800 cm³

Therefore the answer is C. 800 cm³.

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