If 8.00 moles of NH₃ of and 10.00 moles of O₂ react in the following reaction, how many moles of which reactant will be left over? 4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g)
A) 2.00 mol O₂
B) 2.00 mol NH₃
C) 8.00 mol NO
D) no moles will be left over of either reactant because both are limiting.

Answer:

please I don't understand the question, rewrite it

For the equilibrium Fe³⁺ (aq) + SCN⁻(aq) → FeSCN²⁺ (aq) + heat equations for the reactions is describe as below:

Since, it is an equilibrium if you apply any stress (concentration of product or reactant, heat and pressure) on the reaction. it will adjust itself in order to reduce the applied stress.

a. FeCl solution, 0.10 M

FeCl → 3 Fe³⁺ +3Cl³⁻

it increases the concentration of Fe³⁺ hence, equilibrium is shifted towards right.

b. KSCN solid

KSCN → K⁺ +SCN⁻

it increases the concentration of SCN (reactant) hence, equilibrium is shifted towards right.

c. Fe(NO₃) ₃ solution, 0.10 M

Fe(NO₃)₃  → Fe³⁺ + 3NO₃⁻

it increases the concentration of Fe³⁺, hence, equilibrium is shifted towards right

d. Na₂HPO₄, solution, 0.10 M

Na₂(HPO₄) → 2Na⁺ + HPO₄²⁻

HPO₄²⁻ +Fe³⁺ → Fe₂ (HPO₄)₃

since, it reduces the concentration of Fe³⁺ equilibrium will shifted towards left side

e. Na₂HPO₄ solid

since, it reduces the concentration of Fe³⁺ equilibrium will shifted towards left side.

f. Deionized water

It will reduce the formed heat, and the heat required for the backward reaction is not available. equilibrium will stay on right.

(will not affect)

g. Ice

It will reduce the formed heat and the heat required for the backward reaction is not available, equilibrium will stay on right.

So, this is the explanations for the given reaction.

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Approximately 166.67 mL of water needs to be evaporated from 250 mL of 1 M Ca(OH)2 to make it 3 M.

The relationship between the initial and final concentrations and volumes must be taken into account.

Given: Initial concentration \((C^1) = 1 M Initial volume (V^1) = 250 mL\)

\((C^2) = 3 M final concentration\)

We can use the equation:

\(C^1 * V^1 = C^2 * V^2\)

Where:

\(V^2\)is the final volume of the solution

Rearranging the equation to solve for V2:

\(V^2 = (C^1 * V^1) / C^2\)

Substituting the given values:

\(V^2 = (1 M * 250 mL) / 3 M\)

\(V^2 = 250 mL / 3\)

\(V^2\) ≈ \(83.33 mL\)

To find the amount of water that needs to be evaporated, we subtract the final volume from the initial volume:

Amount of water to be evaporated = \(V^1 - V^2\)

Amount of water to be evaporated = 250 mL - 83.33 mL

Amount of water to be evaporated ≈ 166.67 mL

Therefore, approximately 166.67 mL of water needs to be evaporated from 250 mL of 1 M Ca(OH)2 to make it 3 M.

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we would need 2625.13 grams (or 2.62513 kilograms) of NaCl.

To calculate the mass of NaCl required to produce a 26.4 mol/L solution with a 1.7 L volume, we need to use the formula that relates the mass of solute, moles of solute, and molarity:Molarity (M) = moles of solute / liters of solution Rearranging this formula, we get:moles of solute = Molarity (M) x liters of solutionWe can use this formula to find the moles of NaCl needed:moles of NaCl = 26.4 mol/L x 1.7 L = 44.88 molNow, we can use the molar mass of NaCl to convert from moles to grams. The molar mass of NaCl is 58.44 g/mol:mass of NaCl = moles of NaCl x molar mass of NaClmass of NaCl = 44.88 mol x 58.44 g/mol = 2625.13 gTo produce a 26.4 mol/L solution with a 1.7 L volume.

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Answer:

7.0

Explanation:

the maximum amount of product that can be formed is 124.39 g SO₃, and there will be 36.8 g of excess O₂ left over.

To find the amount of excess reagent, you need to first determine which reactant is limiting and which is in excess.

Determine the limiting reagent:

Use stoichiometry to determine how much product can be formed from each reactant:

mol SO2:

2 SO₂ + O₂ -> 2 SO₃

2 mol SO₃/2 mol SO₂ = 1 mol SO₃/mol SO₂

1 mol SO₃ = 80.06 g SO₍₃₎

From 2.7 mol O₂

2 SO₂ + O₂ -> 2 SO₃

1 mol SO₃/1 mol O₂ = 1 mol SO₃/mol O₂

1 mol SO₃ = 80.06 g SO₃

2.7 mol O₂ x (1 mol SO₂/1 mol O₂) x (80.06 g SO₂/mol SO₂) = 216.45 g SO₂

Since the amount of SO₂ produced from 3.1 mol of SO₂ is less than the amount produced from 2.7 mol of O₂, SO₂ is the limiting reagent.

Calculate the amount of excess reagent:

To find the amount of excess O₂, use the balanced equation to determine how much O₂ is required to react with all of the SO₂:

2 SO₂ + O₂ -> 2 SO

3.1 mol SO2 x (1 mol O₂/2 mol SO2) = 1.55 mol O₂

Subtract the amount of O₂ used from the initial amount of O₂:

2.7 mol O₂ - 1.55 mol O2 = 1.15 mol O₂

Finally, convert the excess O₂ to mass:

1.15 mol O₂ x 32.00 g/mol = 36.8 g O₂

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Answer:

they are known as halogens

Explanation:

everything else is correct

by fractional distillation of liquefied air or the use of zeolites to remove carbon dioxide and nitrogen from air or the electrolysis of water

Answer:

9991 J

9.991 kJ

2.78 × 10⁻³ kWh

Explanation:

Step 1: Given data

Energy consumed per day by a person (E): 2388 cal

Step 2: Convert "E" to J

We will use the conversion factor 1 cal = 4.184 J.

2388 cal × 4.184 J/1 cal = 9991 J

Step 3: Convert "E" to kJ

We will use the conversion factor 1 kJ = 1000 J.

9991 J × 1 kJ/1000 J = 9.991 kJ

Step 4: Convert "E" to kWh

We will use the conversion factor 1 kWh = 3600 kJ.

9.991 kJ × 1 kWh/3600 kJ = 2.78 × 10⁻³ kWh

Answer:

Explanation:

1. Cornstarch and water - decantation

2. Iron and sulfur

3. Rice grains and stones - sieving

4. Pebbles and soil - decantation

5. Muddy water - decantation

6. Ground pepper and water - decantation

7. Flour and coarsely ground corn - sieving

8. Needles in a box of beads - using a magnet

9. Sesame seeds in soy sauce -

10. Pebbles and iron nails - using a magnet

11. Fine sand from rocks - sieving

12. Sand from water - filtration

13. Gravel and sand - sieving

14. Salt and iron filings - sieving

15. Pins from assorted buttons - using a magnet

16. Coconut oil from coconut milk - filtration

17. Sand and pebbles - sieving

18. Pasta from water - sieving

19. Sand and thumb tacks - using a magnet

20. Salt solution - filtration

Answer:A cell is the smallest unit of life, also called the 'building blocks of life' because cells multiply and differentiate to form a multicellular organism as well as give rise to new organism by forming gametes or reproductive spores.

Explanation:

Cations are formed through lose of electrons and thus, possess positive charge. Cations has smaller ionic radius than the neutral atom, because, they have are fewer  electrons and thus experience more nuclear attractive pull.

Cations are charged particles or ions formed by the lose of electrons from the neutral atom. Atoms acquires positive charge when they lose electrons and acquire negative charge by gaining electrons  and the negative ions are called anions.

For example, Na metal loses one electrons to form the cation Na+. Similarly Mg loses two electrons, to form Mg²⁺  ion. When these atoms loses electrons, the outermost shell can be emptied and the atomic radius shrinks to the penultimate shell.

Similarly, as the number of electrons reduces, the screening of electron from neighboring electrons reduces results in greater nuclear attractive pull and thereby the atomic radius shrinks to smaller than the neutral atom.

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Answer:

Color.

Explanation:

Hello,

In this case, since matter has a lot of properties regarding its physical condition and chemical composition, those related to the appearance of matter are physical. In such a way, since the student observed to different substances with also different colors, we can infer that the property of matter that was observed in the scenario was color, which accounts for the graphical perception we have from them.

Best regards.

Answer:

c. a proton

Explanation:

A neutron is most equal in mass to a proton.

A neutron is a subatomic particle without any charges on them.

A proton is a subatomic particle with a positive charge.

Taking into account the Charles's law, the gas occupies a volume of 0.0182 m³ when the temperature is raised to 95.0 °C. ​

Charles's law shows the relationship between the volume and temperature of a gas sample at constant pressure.

This law states that the volume is directly proportional to the temperature of the gas: if the temperature increases, the volume of the gas increases, while if the temperature of the gas decreases, the volume decreases.

Mathematically, Charles' law is a law that says that when the amount of gas and pressure remain constant, the ratio between volume and temperature will always have the same value:

\(\frac{V}{T}= k\)

Considering an initial state 1 and an initial state 2:

\(\frac{V1}{T1}= \frac{V2}{T2}\)

In  this case, you know:

Replacing in Charles's Law:

\(\frac{0.0141 m^{3} }{284.5 K}=\frac{V2}{368 K}\)

Solving:

\(\frac{0.0141 m^{3} }{284.5 K}x368 K=V2\)

0.0182 m³= V2

Finally, the gas occupies a volume of 0.0182 m³ when the temperature is raised to 95.0 °C. ​

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A 50.0 mL solution of 0.250 M HC₂H₃O₂ before and after adding 50 mL of 0.250 M NaOH has a pH of 2.87, the pH changes due to the formation of a buffer solution, which can be calculated using the Henderson-Hasselbalch equation.

This problem requires us to calculate the pH of a buffer solution after the addition of a strong base. A buffer solution is one that resists pH changes when modest amounts of acid or base are added. The buffer system in this case is the weak acid, acetic acid (HC₂H₃O₂) and its conjugate base, acetate ion (C₂H₃O₂-).

Before the addition of NaOH, we have a solution of 0.250 M HC₂H₃O₂, which we can assume to be completely dissociated in water:

HC₂H₃O₂ + H₂O ⇌ H₃O+ + C₂H₃O₂-

HC₂H₃O₂ has a Ka value of 1.8 x 10-5. We can use this Ka value to calculate the equilibrium concentration of H3O+ and C₂H₃O₂- in the solution.

First, we need to calculate the initial concentrations of HC₂H₃O₂ and C₂H₃O₂-.

moles of HC₂H₃O₂ = 0.250 M × 0.0500 L = 0.0125 mol

moles of C₂H₃O₂- = 0 mol (since there is no NaOH added yet)

Since HC₂H₃O₂ is a weak acid, it only partially dissociates in water. The equilibrium concentrations of H₃O+ and C₂H₃O₂- can be calculated using the Ka expression:

Ka = [H₃O+][C₂H₃O₂-]/[HC₂H₃O₂]

We can assume that the initial concentration of H3O+ is negligible compared to the amount that will be produced by the dissociation of HC₂H₃O₂, so we can simplify the expression to:

Ka = [H₃O+]²/[HC₂H₃O₂]

[H₃O+]² = Ka × [HC₂H₃O₂]

[H₃O+]² = 1.8 × 10⁻⁵ × 0.0125

[H₃O+] = 1.34 × 10⁻³ M

Now that we know the equilibrium concentration of H₃O+, we can use the pH formula to calculate the pH:

pH = -log[H₃O+]

pH = -log(1.34 × 10⁻³)

pH = 2.87

This is the pH of the buffer solution before the addition of NaOH.

Next, we add 50 mL of 0.250 M NaOH to the solution. NaOH is a strong base, so it completely dissociates in water to produce OH- ions:

NaOH → Na+ + OH-

The OH- ions will react with the acetate ions in the buffer solution to form water and acetate ions:

OH- + C₂H₃O₂- → H₂O + C₂H₃O₂-

This reaction will consume some of the acetate ions in the buffer solution, causing the equilibrium to shift to the left to produce more acetate ions.

To calculate the new concentrations of H₃O+ and C₂H₃O₂-, we need to use the Henderson-Hasselbalch equation:

pH = pKa + log([C₂H₃O₂-]/[HC₂H₃O₂])

where pKa = -log(Ka), [C₂H₃O₂-] is the equilibrium concentration of acetate ions, and [HC₂H₃O₂] is the equilibrium concentration of acetic acid.

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The molarity of the ammonium nitrate solution is 2281 M.

Molarity is defined as the number of moles of solute per liter of solution. It is also known as molar concentration of the solution and is used to calculate amount of substances in the solution.

Molarity = n/M

n = m / MW

m = 2 lbs = 1000/2.2 = 909 g

V = 53 x 3.79/1 = 200.9

MW = 80.04 g

M = m/Mw / V

M = 909/80.04 /200.9

M = 2281 M

Thus, the molarity of the ammonium nitrate solution is 2281 M.

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To convert moles of copper to grams, we need to use the molar mass of copper, which is 63.55 g/mol.

Grams of copper = moles of copper x molar mass of copper

Grams of copper = 5.699 mol x 63.55 g/mol

Grams of copper = 362.071 g

Therefore, there are 362.071 grams of copper in 5.699 moles of copper.

So there are 361.28 grams in 5.699 moles of copper.

The amount of heat energy produced when 1 g of hydrogen and 2 g of nitrogen are reacted, is -6.61 KJ

First, we shall obtain the limiting reactant. Details below:

3H₂ + N₂ -> 2NH₃

From the balanced equation above,

28 g of N₂ reacted with 6 g of H₂

Therefore,

2 g of N₂ will react with = (2 × 6) / 28 = 0.43 g of H₂

We can see that only 0.43 g of H₂ is needed in the reaction.

Thus, the limiting reactant is N₂

Finally, we the amount of heat energy produced. Details below:

3H₂ + N₂ -> 2NH₃ ΔH = -92.6 KJ

From the balanced equation above,

When 28 grams of N₂ reacted, -92.6 KJ of heat energy were produced.

Therefore,

When 2 grams of N₂ will react to produce = (2 × -92.6) / 28 = -6.61 KJ

Thus the heat energy produced from the reaction is -6.61 KJ

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According to the question the reaction enthalpy is thus 10610 J / 0.278 moles = 38.3 kJ/mol.

Enthalpy is a thermodynamic property of a system that measures the total energy content of a system. It is a state function that is expressed in terms of internal energy, pressure, and volume of a system. Enthalpy represents the amount of energy that is associated with a chemical reaction or physical change.

The reaction is exothermic, meaning that heat is released during the reaction. The amount of heat released can be calculated with the equation q = mcΔT, where q is the heat released, m is the mass of the solution, c is the specific heat capacity of the solution, and ΔT is the change in temperature of the solution. Using the given data, the amount of heat released by the reaction can be calculated as q = (250 g)(4.184 J/g-K)(11.1 K) = 10610 J. The enthalpy change for the reaction can then be calculated by dividing the heat released by the number of moles of NaOH, which is 11.1 g / 40.00 g/mol = 0.278 moles. The reaction enthalpy is thus 10610 J / 0.278 moles = 38.3 kJ/mol.

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The concentration of the HCl (aq) solution was at least 1.0 M.

The abundance of a constituent divided by the sum of the mixture's volumes is the definition of concentration in chemistry. There are several different categories of mathematical description: mass concentration, molar concentration, number concentration, and volume concentration

P = 0.095atm(corresponds to 0.20g of CaCO3

so,moles of CaCO3 =0.20g/100gmol⁻¹=0.002moles

moles of HCl =(molesₓCaCO3ₓ2)=0.002ₓ2

M=0.004/4ₓ100

Concentration of HCl=1M

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Answer:

chlorous acid

Explanation:

The acidic equilibrium of weak acids, HX, occurs as follows:

HX ⇄ H+ + X-

Where Ka is written as:

Ka = [H+] [X-] / [HX]

The strongest acid is the acid that produce more H+. The acid with the higher Ka is the acid that produce more [H+] and is, thus, the strongest acid.

The higher Ka is the Ka of chlorus acid = 1.2x10-2

Right answer is:

Answer:

chlorous acid :)

Explanation:

Considering the definition of molar mass, 5 grams of hydrogen is contained in methane, which contains 15 grams of carbon.

Definition of molar mass

The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.

The molar mass of a compound (also called Mass or Molecular Weight) is the sum of the molar mass of the elements that form it (whose value is found in the periodic table) multiplied by the number of times they appear in the compound.

In this case, you know the molar mass of the elements is:

So, the molar mass of the compound CH₄ is calculated as:

CH₄= 12 g/mole + 4× 1 g/mole

Solving:

CH₄= 16 g/mole

In this case, from the above you can see that 1 mole of methane contains 4 grams of hydrogen.

In first place, you must know the moles de methane that contain 15 grams of carbon. For that, you can apply the following rule of three: If from the above you can see that 12 grams of carbon are contained in 1 mole of methane, 15 grams of carbon are contained in how many moles of methane?

\(moles of methane=\frac{15 grams of carbonx1 mole of methane}{12 grams of carbon}\)

moles of methane= 1.25 moles

Next, you can apply the following rule of three: If 1 mole of methane contains 4 grams of hydrogen, 1.25 moles of methane contains how many moles of hydrogen?

\(mass of hydrogen= \frac{1.25 moles of methanex4 grams of hydrogen}{1 moles of methane}\)

mass of hydrogen= 5 grams

Finally, 5 grams of hydrogen is contained in methane, which contains 15 grams of carbon.

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Answer:

27)

a. Silver nitrate

b. Lead chromate

c. Anhydrous magnesium acetate

d. Copper nitrate

28) a. KNo3

Answer:

math si hard

Explanation:

The mole fration of ethanol and water in a sample of rectified spirit which contains 95% of ethanol by mass is 0.8 and 0.11.

Mole fraction of any substance will be calculated by dividing the moles of that substance from the total moles of the solution.

Moles (n) will be calculated as:

n = W/M, where

W = given mass

M = molar mass

Given that 95% of ethanol by mass is present in the sample, so 95g of ethanol is present in 100g of solution.

Mass of solvent (water) = 100 - 95 = 5g

Moles of water = 5g / 18g/mol = 0.27mol

Moles of ethanol = 95g / 46g/mol = 2.06mol

Mole fraction of water  = 0.27 / (0.27+2.06) = 0.11

Mole fraction of ethanol = 2.06 / (0.27+2.06) = 0.8

Hence required mole fraction of water & ethanol is 00.11 and 0.8 respectively.

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The statement "All organic compounds contain the element carbon, but not all compounds containing the element 'carbon' are organic" can be justified based on the definition and characteristics of organic compounds.

Organic compounds are compounds primarily composed of carbon and hydrogen atoms, often with other elements like oxygen, nitrogen, sulfur, and phosphorus. These compounds are typically associated with living organisms and are known for their unique properties and behavior, including the ability to form complex structures, exhibit covalent bonding, and undergo organic reactions.

On the other hand, there are compounds that contain carbon but are not classified as organic. One notable example is carbon dioxide (\(CO_{2}\)), which is a simple inorganic compound composed of carbon and oxygen. Carbon dioxide does not possess the characteristic properties of organic compounds, such as the ability to form long chains or undergo organic reactions.

Additionally, there are inorganic compounds like carbonates (such as calcium carbonate) and carbides (such as calcium carbide) that contain carbon but are not considered organic. These compounds have distinct chemical and physical properties different from those of organic compounds.

In summary, while all organic compounds contain carbon, not all compounds containing carbon are organic. The classification of a compound as organic or inorganic depends on its overall molecular structure, bonding, and characteristic properties.

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The vapor pressure of methanol at 40°C is 0.234 atm.

Only two types of alcohol are methanol and ethanol. Ethanol, sometimes referred to as ethyl alcohol, has a chemical composition of two carbon atoms. Methanol, sometimes referred to as methyl alcohol, is made up of just one carbon atom.

ln(P2/P1) = (ΔHvap/R) x (1/T1 - 1/T2)

ΔGvap = -RTln(Pvap/P) = ΔHvap - TΔSvap

ΔGvap = -RTln(Pvap/P) = -166.6 kJ/mol

ΔSvap = S(g) - S(l) = 239.9 J/mol-K - 126.8 J/mol-K = 113.1 J/mol-K

ΔHvap = ΔGvap + TΔSvap = -166.6 kJ/mol + (298.15 K)(113.1 J/mol-K) = -134.6 kJ/mol

Now we can use the Clausius-Clapeyron equation to find the vapor pressure of methanol at -30°C and 40°C.

At -30°C, we have:

T1 = 25°C + 273.15 = 298.15 K

T2 = -30°C + 273.15 = 243.15 K

ΔHvap = -134.6 kJ/mol

R = 8.314 J/mol-K

ln(P2/5 atm) = (-134.6 kJ/mol / 8.314 J/mol-K) x (1/298.15 K - 1/243.15 K)

P2 = 0.0038 atm

Therefore, the vapor pressure of methanol at -30°C is 0.0038 atm.

At 40°C, we have:

T1 = 25°C + 273.15 = 298.15 K

T2 = 40°C + 273.15 = 313.15 K

ΔHvap = -134.6 kJ/mol

R = 8.314 J/mol-K

ln(P2/5 atm) = (-134.6 kJ/mol / 8.314 J/mol-K) x (1/298.15 K - 1/313.15 K)

P2 = 0.234 atm

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Answer:

x/2

Explanation:

X = 2 molecules of H2

For 2 molecules of H2, there's only 1 molecule of O2. Meaning, there's twice the amount of H2, so O2 = x/2 molecules.

I hope I'm understanding this question right.

An average brightness was calculated for each length of graphite tested to get a better understanding of the relationship between the length of the graphite and the brightness of the line it produced.

Graphite has many uses in various industries. Some of its uses include: Pencils, Lubricants, Refractories, Batteries, Electrodes, Nuclear reactors, Aerospace industry.

By calculating the average, it is possible to see if there is a trend in the data and if longer or shorter lengths of graphite produce brighter or duller lines. This information can be useful in determining the best length of graphite to use for a particular task or project.

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