How much time does it take light to travel 6.03 billion km? (billion=109)

Answer to 3 sig figs.

I think it is 2,4-dimethylpentane

To calculate the cell potential, Ecell, for the given reaction at 298K, we need to use the Nernst equation. The Nernst equation relates the cell potential to the standard cell potential, temperature, and the concentrations of the reactants and products. The Nernst equation is given as follows:

Ecell = E°cell - (RT/nF) ln(Q)

where,

Ecell = cell potential

E°cell = standard cell potential

R = gas constant (8.314 J/K.mol)

T = temperature (298 K)

n = number of electrons transferred in the balanced redox reaction

F = Faraday constant (96,485 C/mol)

Q = reaction quotient

The given reaction is a redox reaction, which involves the transfer of two electrons from Co to Ag+. The balanced half-reactions are as follows:

Co(s) → Co2+(aq) + 2 e-

Ag+(aq) + e- → Ag(s)

The standard reduction potentials for these half-reactions are:

Co2+(aq) + 2 e- → Co(s) E°red = -0.28 V

Ag+(aq) + e- → Ag(s) E°red = +0.80 V

The overall standard cell potential can be calculated by subtracting the standard reduction potential of the anode from that of the cathode:

E°cell = E°red,cathode - E°red,anode

= +0.80 V - (-0.28 V)

= +1.08 V

Now we need to calculate the reaction quotient Q using the concentrations of the reactants and products. According to the given information, [Ag+] = 0.010 M and [Co2+] = 0.015 M.

Q = ([Co2+][Ag+]^2)/([Ag+]^2)

= ([0.015][0.010]^2)/([0.010]^2)

= 0.015 M

Substituting the values in the Nernst equation, we get:

Ecell = E°cell - (RT/nF) ln(Q)

= 1.08 - (8.314 x 298 / (2 x 96485)) ln(0.015)

= 0.829 V

Therefore, the cell potential, Ecell, for the given reaction at 298K is 0.829 V.

The eutectic point is the lowest temperature at which the liquid phase is constant at a particular pressure.

A melting composition known as a eutectic consists of at least two components that melt and freeze at the same rates. The components combine during the crystallisation phase, operating as a single component as a result.

The eutectic is the system's lowest melting point under its own pressure; it has a matching temperature called the eutectic temperature and produces the eutectic liquid as a result. In terms of composition, eutectic liquids are located between the system's solid phases.

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Determine the amount of KOH present in the resulting solution. KOH was initially 0.00875 mol, then 0.00525 mol of it interacted with HCl. As a result, 0.00875 mole - 0.00525 mol (= 0.00350 mol of KOH is left. The resulting solution has a volume of 70.0 mL (35.0 mL plus 35.0 mL).

The titrant (NaOH), which is added gradually throughout the course of a titration, is added to the unknown substance. The equivalency point is the moment at which precisely the right quantity of titrant (NaOH) has indeed been added that react to the entire analyte (HCl).

What took place during titration: One mole of NaOH interacts with one mole of HCl inside the reaction between the two substances. NaOH with HCl equals NaCl plus H2O. (NaOH and HCl have a mole ratio of 1:1.) • The NaOH concentration is 0.1 M, or 0.1 molecules per litre.

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Answer:

colloid

Explanation:

there's no explanation

The state of matter that tends to have unique factors to consider when discussing solubility compared to the other two states (solid and gas) is the liquid state.

Solubility refers to the ability of a substance (solute) to dissolve in a particular solvent to form a homogeneous mixture (solution). Various factors can affect the solubility of a substance, including temperature, pressure, and the nature of the solute and solvent.

In the case of liquids, the unique factor to consider when discussing solubility is often temperature. The solubility of many solid solutes in liquids generally increases with increasing temperature. This is because higher temperatures provide more energy to break the intermolecular forces between solute particles, allowing them to disperse more evenly throughout the solvent.

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The pH of the solution can be calculated using the following steps:

Write the chemical equation for the dissociation of ethanoic acid:

CH3COOH + H2O ⇌ CH3COO- + H3O+

Write the equilibrium expression for the dissociation of ethanoic acid:

Ka = [CH3COO-][H3O+] / [CH3COOH]

Since the solution is equimolar in CH3COOH and CH3COO-, we can assume that the initial concentrations of CH3COOH and CH3COO- are equal. Let's use the variable x to represent the concentration of CH3COO- and CH3COOH in mol/L.

[CH3COOH] = x mol/L [CH3COO-] = x mol/L

Since CH3COOH is a weak acid, we can assume that only a small fraction of it dissociates in water. Let's use the variable y to represent the concentration of H3O+ ions in mol/L that are produced from the dissociation of CH3COOH. From the dissociation of ethanoic acid, we know that [CH3COO-] = [H3O+].

[CH3COO-] = y mol/L [H3O+] = y mol/L

Use the equilibrium expression to solve for the concentration of H3O+ ions:

Ka = [CH3COO-][H3O+] / [CH3COOH] 1.79 x 10^-5 = y^2 / x

Solving for y in terms of x, we get:

y = sqrt(Ka * x)

Calculate the pH of the solution using the equation:

pH = -log[H3O+]

pH = -log(y)

Substituting in the value of y from Step 5, we get:

pH = -log(sqrt(Ka * x))

Simplifying, we get:

pH = -0.5 * log(Ka * x)

Substituting in the value of Ka, we get:

pH = -0.5 * log(1.79 x 10^-5 * x)

Now we can calculate the pH for the solution by substituting the value of x as it is equimolar.

pH = -0.5 * log(1.79 x 10^-5 * x)

pH = -0.5 * log(1.79 x 10^-5 * 1)

pH = -0.5 * log(1.79 x 10^-5)

pH = 4.74

Therefore, the pH of an equimolar solution of ethanoic acid and Na+CH3COO- is 4.74.

Around 80.5 KJ

Multiply Heat of Fusion and Mass to get the q value.

3.21 grams of Aluminum Sulfate are got when 4 g of Aluminum Nitrate reacts chemcially with 3 g of Sodium Sulfate.

The equation given is not balanced. Thus,  when balanced the equation becomes:

2 Al(NO₃)₃ + 3 Na₂SO₄ → Al₂(SO₄)₃ + 6 NaNO₃

The molar mass of Al(NO₃)₃ is:

Al(NO₃)₃ = 1(Al) + 3(N) + 9(O) = 213 g/mol

The molar mass of Na₂SO₄ is:

Na₂SO₄ = 2(Na) + 1(S) + 4(O) = 142 g/mol

From the balanced equation, we can see that 2 moles of Al(NO₃)₃ react with 3 moles of Na2SO4 to produce 1 mole of Al₂(SO₄)₃. Therefore, we can calculate the number of moles of Al(NO₃)₃ and Na₂SO₄ that react:

Number of moles of Al(NO₃)₃ = 4 g / 213 g/mol = 0.0188 mol

Number of moles of Na₂SO₄ = 3 g / 142 g/mol = 0.0211 mol

From the balanced equation, we can see that 2 moles of Al(NO₃)₃ produce 1 mole of Al₂(SO₄)₃. Therefore, the number of moles of Al₂(SO₄)₃ produced is:

Number of moles of Al₂(SO₄)₃ = 0.0188 mol / 2 * 1 = 0.0094 mol

The molar mass of Aluminum Sulfate (Al₂(SO₄)₃) is:

Al₂(SO₄)₃ = 2(Al) + 3(S) + 12(O) = 342 g/mol

Therefore, the mass of Aluminum Sulfate produced is:

Mass of Al₂(SO₄)₃ = Number of moles of Al₂(SO₄)₃ * Molar mass of Al₂(SO₄)₃

= 0.0094 mol * 342 g/mol

= 3.21 g

Hence, 3.21 grams of Aluminum Sulfate are liberated when 4 g of Aluminum Nitrate change state with 3 g of Sodium Sulfate.

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Answer: C)Deposition (gas to solid)

Explanation: According to the phase diagram, at room temperature (25°C) and 1 atm, the sample is in the gas phase.  As the temperature decreases to 80 K, it falls below the sublimation curve. T he sublimation curve represents the conditions at which a substance can change directly from a solid to a gas or from a gas to a solid without passing through the liquid phase.

Since the sample is in the gas phase at room temperature, cooling it to 80 K would cause it to go through the process of deposition, where the gas particles directly transform into a solid without first becoming a liquid.  This is indicated by the section of the phase diagram below the sublimation curve.

The order for reactant A is 2 and the order for reactant B is 1. For the first reaction, the overall order of the reaction is 3 and for the second reaction, the overall order of the reaction is 5.

The order of a reaction is the sum of the exponents in the rate law expression that relates the rate of a chemical reaction to the concentrations of the reactants.

To determine the order of each reactant, we need to compare the initial rates of reaction at different concentrations while keeping the concentration of the other reactant constant.

For reactant A:

Exp#1 (0.100 M A, 0.100 M B): initial rate = k(0.100)^2(0.100) = 0.001 k

Exp#2 (0.200 M A, 0.100 M B): initial rate = k(0.200)^2(0.100) = 0.004 k

Exp#3 (0.100 M A, 0.200 M B): initial rate = k(0.100)^2(0.200) = 0.002 k

We can see that when the concentration of A doubles (Exp#1 to Exp#2), the initial rate quadruples, which indicates that A is second order. When the concentration of B doubles (Exp#1 to Exp#3), the initial rate doubles, which indicates that B is first order.

Therefore, the order for reactant A is 2 and the order for reactant B is 1.

To determine the overall order of the reaction, we add the orders of the reactants:

Overall order = 2 (order of A) + 1 (order of B) = 3

Therefore, the overall order of the reaction is 3.

For the second reaction, we can see that the rate depends on the concentration of both reactants, and we cannot determine their individual orders without further information or experiments. However, we can determine the overall order of the reaction by adding the exponents of the concentration terms in the rate law:

Overall order = 1 + 4 = 5

Therefore, the overall order of the reaction is 5.

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Answer:

Neutrons and Protons

Explanation:

Different elements can have subatomic particles of varying sizes. The size of an atom is defined by the size of its electron cloud, which is composed of electrons, and the size of its nucleus, which is composed of protons and neutrons. The atomic number and subsequently the identity of an element are determined by the number of protons in the nucleus. The quantity of protons and neutrons in the nucleus determines its size. The quantity of electrons in the electron cloud and the energy levels they are located at define its size. The size of atoms can differ depending on the element due to differences in the amount of protons, neutrons, and electrons.

1.8 moles of water are formed when 1.20 moles of ammonia reacts

How is ammonia used?

Ammonia produced by industry is used as fertilizer in agriculture to the tune of 80%. In addition to these uses, ammonia is used to make polymers, explosives, textiles, insecticides, dyes, and other compounds. It is also used to purify water sources.

Ammonia is a colorless, intensely unpleasant gas with a pungent, choke-inducing smell. It readily dissolves in water to produce an ammonium hydroxide solution that can irritate the skin and burn. Ammonia gas is easily compressed and, when put under pressure, turns into a clear, colorless liquid.

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

4 moles of ammonia gives 6 moles of water

Moles of H₂O = 1.2 moles of NH₃ x 6 moles of H₂O/4 moles of NH₃

Moles of H₂O = 1.8moles

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The Ammonium Chloride solution at 0.25 M has a pH of 2.67.

As a result, the weak basic (Chlorine) in the solution is overpowered by the conjugate acid (Ammonium cation), making the solution mildly acidic. According to the equation pH =log[Hydrogen ion], an acidic solution has a pH lower than 7. Aqueous ammonium chloride solution has a pH that is less than 7.

Ammonium cation + Water ⇌ Nitrogen trihydride + Hydronium ion

Kb = [Nitrogen trihydride][Hydronium ion] / [Ammonium cation]

[Nitrogen trihydride] = [Hydronium ion] = x

[Ammonium cation] = 0.25 - x

Kb = [Nitrogen trihydride][Hydronium ion] / [Ammonium cation]

1.8 × 10–5 = x² / (0.25 - x)

1.8 × 10–5 = x² / 0.25

x² = 4.5 × 10–6

x = 2.12 × 10–3

pH = -log[Hydronium ion] = -log(2.12 × 10–3) = 2.67

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Here's a more detailed step-by-step calculation for the theoretical yield and percent yield of chromium (Cr) in the given reaction:

Given: Mass of chromium(III) oxide (Cr2O3) = 80.0 g Mass of carbon (C) = 8.00 g Actual yield of Cr = 21.7 g

Step 1: Calculate the molar mass of Cr2O3 and C. Molar mass of Cr2O3 = 2 x (51.996 g/mol) + 3 x (15.999 g/mol) = 151.996 g/mol Molar mass of C = 12.011 g/mol

Step 2: Convert the masses of Cr2O3 and C to moles. Moles of Cr2O3 = Mass of Cr2O3 / Molar mass of Cr2O3 = 80.0 g / 151.996 g/mol = 0.527 mol (rounded to three decimal places)

Moles of C = Mass of C / Molar mass of C = 8.00 g / 12.011 g/mol = 0.666 mol (rounded to three decimal places)

Step 3: Determine the limiting reactant. The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed. In this case, we compare the moles of Cr2O3 and C to see which one is limiting.

From the balanced equation: Cr2O3 + 3C -> 2Cr + 3CO

We can see that 1 mol of Cr2O3 requires 3 moles of C to react completely and produce 2 moles of Cr. Therefore, the limiting reactant is C, as we have 0.666 mol of C, which is less than the 0.527 mol of Cr2O3.

Step 4: Calculate the theoretical yield of Cr. The theoretical yield of Cr is the maximum amount of Cr that can be obtained based on the limiting reactant.

Moles of limiting reactant (C) = 0.666 mol Molar mass of Cr = 51.996 g/mol

Theoretical yield of Cr = Moles of limiting reactant (C) x Molar mass of Cr = 0.666 mol x 51.996 g/mol = 34.65 g (rounded to two decimal places)

Step 5: Calculate the percent yield of Cr. The percent yield is a measure of how much of the theoretical yield was actually obtained.

Actual yield of Cr = 21.7 g Theoretical yield of Cr = 34.65 g

Percent yield = (Actual yield / Theoretical yield) x 100% = (21.7 g / 34.65 g) x 100% = 62.7% (rounded to three significant figures)

Therefore, the percent yield for the reaction is approximately 62.7%.

K and Br since an halogen and a metal make a salt

The new volume of the helium sample would be 2.4 L.

According to the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in kelvins.

At STP (standard temperature and pressure), which is defined as 0°C (273.15 K) and 101.325 kPa, the volume of 2.0 liters of helium gas contains one mole of helium atoms.

To find the new volume of the helium sample when the temperature is increased to 27°C (300.15 K) and the pressure is decreased to 80 kPa, we can use the following equation:

(P1V1)/T1 = (P2V2)/T2

where P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 are the final pressure, volume, and temperature, respectively.

Plugging in the values, we get:

(101.325 kPa)(2.0 L)/(273.15 K) = (80 kPa)(V2)/(300.15 K)

Solving for V2, we get:

V2 = (101.325 kPa)(2.0 L)/(273.15 K) * (300.15 K)/(80 kPa) = 2.36 L

Therefore, the new volume of the helium sample is approximately 2.4 L (rounded to the nearest tenth).

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The addition of concentrated HCl to the equilibrium mixture will result in the precipitation of more NH₄Cl(s) as the equilibrium shifts towards the left. This can be observed as cloudiness or precipitation forming in the solution.

When concentrated hydrochloric acid (HCl) solution is added to the equilibrium mixture of NH₄Cl(s) + heat <=> NH₄+(aq) + Cl-(aq), the equilibrium will shift towards the left, meaning more solid NH₄Cl will be formed.

This is because HCl is a strong acid that will react with NH₄+ ion to form NH₄Cl(s) and H+ ion:

NH₄+(aq) + Cl-(aq) + HCl(aq) → NH₄Cl(s) + H₂O(l)

The increase in H+ ion concentration due to the addition of HCl will result in the shift of the equilibrium to the left to reduce the excess H+ ion concentration. This will favor the formation of more solid NH₄Cl.

Therefore, the addition of concentrated HCl to the equilibrium mixture will result in the precipitation of more NH₄Cl(s) as the equilibrium shifts towards the left. This can be observed as cloudiness or precipitation forming in the solution.

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The percent yield of copper(I) sulfide in this experiment is 31.83%.

What is percent yield?

To calculate the percent yield, we need to compare the actual yield (the amount of product that was obtained in the experiment) with the theoretical yield (the amount of product that should have been obtained if the reaction had gone to completion).

The balanced chemical equation for the reaction between copper and sulfur to form copper(I) sulfide is:

Cu + S →  [tex]Cu_{2}S[/tex]

The molar mass of Cu is 63.55 g/mol, and the molar mass of S is 32.06 g/mol. The molar mass of  [tex]Cu_{2}S[/tex]  is 159.17 g/mol.

First, we need to calculate the theoretical yield of copper(I) sulfide using the amount of copper used in the experiment:

5 g Cu × (1 mol Cu / 63.55 g Cu) × (1 mol [tex]Cu_{2}S[/tex] / 1 mol Cu) × (159.17 g  [tex]Cu_{2}S[/tex] / 1 mol [tex]Cu_{2}S[/tex] ) = 12.57 g  [tex]Cu_{2}S[/tex]

So the theoretical yield of copper(I) sulfide is 12.57 g.

The actual yield obtained in the experiment is 4 g.

The percent yield is then:

percent yield = (actual yield / theoretical yield) × 100%

percent yield = (4 g / 12.57 g) × 100%

percent yield = 31.83%

Therefore, the percent yield of copper(I) sulfide in this experiment is 31.83%.

What is theoretical yield ?

The theoretical yield is the amount of product that would be obtained in a chemical reaction if it went to completion, meaning that all the limiting reactant was used up and no product was lost. It is calculated using stoichiometry, which involves balancing the chemical equation for the reaction and using the coefficients to determine the mole ratio between the reactants and products.

Theoretical yield is often used as a reference value to compare with the actual yield obtained in an experiment, which is the amount of product actually obtained from the reaction. The percent yield can then be calculated by dividing the actual yield by the theoretical yield and multiplying by 100%.

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Answer:

140.3 *C

Explanation:

(P1 * V1) / T1 = (P2 * V2) / T2

where P1 = 3.000 atm, V1 = 200.0 ml, T1 = 37.0°C + 273.15 = 310.15 K, P2 = 2.000 atm, V2 = 400.0 ml.

Substituting these values into the formula gives:

(3.000 atm * 200.0 ml) / 310.15 K = (2.000 atm * 400.0 ml) / T2

Solving for T2 gives:

T2 = (2.000 atm * 400.0 ml * 310.15 K) / (3.000 atm * 200.0 ml)

T2 ≈ 413 K or 140°C.

Answer:

M=0.06998 mol/L

Explanation:

[tex] \ddots[/tex] The heat energy can be deduced as -

[tex] \odot\sf \footnotesize{Heat \:energy = Mass\: of\: substance\times Specific \:heat\times Change\: in \:temperature}\\[/tex]

[tex] \qquad :\implies\sf \boxed{\sf Q = mS\Delta T}\\[/tex]

Where-

In this instant, we are given -

[tex] \ddots[/tex] Now that we have all the required values except ∆T,so we can plug the rest of the known values into the formula and solve for ∆T -

[tex] \qquad :\implies\sf \underline{Q = mS\Delta T}\\[/tex]

[tex] \qquad :\implies\sf 14500 = 485 \times 4.18 \times \Delta T\\[/tex]

[tex] \qquad :\implies\sf 14500 = 2027.3\times \Delta T\\[/tex]

[tex] \qquad :\implies\sf \Delta T = \dfrac{14500}{2027.3}\\[/tex]

[tex] \qquad :\implies\sf \Delta T = 7.152370........\:°C\\[/tex]

[tex] \qquad :\implies\sf \underline{\boxed{\sf \Delta T=7.15\:°C}}\\[/tex]

[tex] \ddots[/tex]Correct answer - [tex]\boxed{\sf \Delta T=7.15\:°C}.[/tex]

This is an exercise in specific heat and thermal conductivity which are two important physical properties that describe how materials interact with heat. Specific heat refers to the amount of energy required to raise the temperature of a material by a given amount, while thermal conductivity refers to a material's ability to transfer heat through itself.

The formula for specific heat is Q = mcΔT, where Q is the amount of heat transferred, m is the mass of the material, c is the specific heat, and ΔT is the change in temperature. The unit of measure for specific heat is J/(g*°C).

On the other hand, thermal conductivity is measured in terms of the amount of heat that is transferred through a material per unit time and area, given a temperature difference. It is expressed as the amount of heat transferred per second, per square meter, per meter of material thickness, when the temperature difference between the extremes is one Kelvin. Its formula is Q/t = -kA(∆T/∆x), where Q/t is the heat transfer rate, k is the thermal conductivity, A is the cross-sectional area, ∆T is the temperature difference, and ∆ x is the thickness of the material.

These properties are useful for understanding how materials interact with heat in a variety of situations, from building design to heating and cooling equipment manufacturing.

Now to calculate the temperature rise of 485 g of liquid water when 14.5 kJ of heat is added to it, we can use the formula:

Q = mcΔT

We must know that it has a quantity of heat of 14.5 Kj, with a mass of 485 g. The specific heat capacity of water is 4.18 J/(g °C).

First, we need to convert the heat added to joules:

Q = 14.5 KJ × (1000 J/1 KJ)

Q = 14500 J

We can then solve for ΔT. We clear the formula.

ΔT = Q / (m × c)

We substitute our data in the formula and solve the temperature change:

ΔT = Q / (m × c)

ΔT = (14500 J)/(485 g × 4.18 J/(g·°C))

ΔT ≈ 7.15 °C

The water would increase its temperature by approximately 7.15°C if 14.5 kJ of heat were added.

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The compound NaOH as shown is a strong base.

NaOH (sodium hydroxide) is considered a strong base. A strong base is a base that dissociates completely in water to form hydroxide ions (OH-) and cations. NaOH is highly soluble in water and, when added to water, it completely dissociates into Na+ and OH- ions, which makes it a strong base.

The strength of a base depends on the extent of its dissociation in water. Strong bases dissociate completely in water, while weak bases dissociate only partially. The dissociation of a base is usually represented by its base dissociation constant (Kb), which is the equilibrium constant for the reaction of the base with water to form hydroxide ions.

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The concentrations of all species in a 0.510 M NaCH₃COO (sodium acetate) solution: [Na+]= 0.510 M , [OH-]= 1.8x10⁻⁵ M , [H₃O+]= 1.8x10⁻⁵ M , [CH₃COO-]= 0.510 M and [CH₃COOH]= 0.510 - (1.8x10⁻⁵) = 0.50982 M.

Concentration is the ability to focus your attention on a single task or thought for a prolonged period of time. It involves being able to ignore distractions and to be able to work through any difficulties or obstacles that may arise. Concentration is an important skill to master in order to achieve success in any endeavor, whether it be academic, professional, or personal. Good concentration can help you to stay focused, organized, and productive. When you are able to concentrate, you can take in the information needed to make better decisions and solve problems. Concentration is a skill that can be developed with practice, such as by setting goals, breaking down tasks into smaller, manageable pieces, and avoiding distractions.

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Answer:

-191 kJ

Explanation:

The given reaction is:

N₂(g) + 3H₂(g) → 2NH₃(g) ΔH = -76.4 kJ/mol

From the balanced equation, we can see that the stoichiometric ratio between hydrogen (H₂) and ammonia (NH₃) is 3:2. This means that 3 moles of hydrogen react to produce 2 moles of ammonia.

To determine the heat energy when 5.0 g of hydrogen (H₂) burns, we need to follow these steps:

Step 1: Calculate the moles of hydrogen (H₂)

Using the molar mass of hydrogen (H₂), which is 2 g/mol, we can calculate the moles of hydrogen (H₂) in 5.0 g of hydrogen:

Moles of H₂ = Mass of H₂ / Molar mass of H₂

Moles of H₂ = 5.0 g / 2 g/mol

Moles of H₂ = 2.5 mol

Step 2: Use the stoichiometry of the reaction

Based on the stoichiometry of the reaction, we know that 3 moles of hydrogen (H₂) react to produce 2 moles of ammonia (NH₃), and the enthalpy change (ΔH) is -76.4 kJ/mol.

Step 3: Calculate the heat energy

The heat energy for 2.5 moles of hydrogen (H₂) can be calculated using the given enthalpy change (ΔH) and the stoichiometry of the reaction:

Heat energy = Moles of H₂ x ΔH

Heat energy = 2.5 mol x -76.4 kJ/mol

Heat energy = -191 kJ (rounded to three significant figures)

So, the heat energy when 5.0 g of hydrogen (H₂) burns is -191 kJ (rounded to three significant figures), and the negative sign indicates that the reaction is exothermic, releasing heat.

It requires 10.15 kilojoules of energy.

The term "vaporisation" (or "evaporation") often refers to the transformation of a liquid's condition into a vapour phase below its boiling point. The phrase, however, can also refer to the process of removing a solvent, independent of the temperature used.

When a body moves to exert force, it is said to be exerting work. Energy is the capacity to accomplish work. Energy is something we always need, and it can take many different forms.

If the gold is present in the liquid state, you only have to determine the latent heat of vaporization, or lvap. The empirical data for gold is 330 kJ/mol.

Q = mlvap

Q = (2 kg)(1 kmol/197 kg)(1,000 mol/1 kmol)

Q = 10.15 kJ

It needs an energy of 10.15 kilojoules

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0.1 mol of sodium hydroxide (NaOH) would react exactly with 7.5 g of the diprotic acid [tex]H_{2}[/tex]A.

What is Molar Mass?

Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). It is calculated by adding up the atomic masses of all the atoms in a molecule or the formula mass of all the ions in an ionic compound.

The balanced chemical equation for the reaction between diprotic acid, [tex]H_{2}[/tex]A, and sodium hydroxide, NaOH, can be represented as follows:

2[tex]H_{2}[/tex]A + 2 NaOH -> [tex]Na_{2}[/tex]A + 2 [tex]H_{2}[/tex]O

From the balanced equation, we can see that 2 moles of [tex]H_{2}[/tex]A react with 2 moles of NaOH to produce 1 mole of [tex]Na_{2}[/tex]A and 2 moles of water ([tex]H_{2}[/tex]O).

First, we need to calculate the number of moles of [tex]H_{2}[/tex]A in 7.5g using the formula:

moles = mass / molar mass

moles of [tex]H_{2}[/tex]A = 7.5g / 150 g/mol = 0.05 mol

Since diprotic acid, [tex]H_{2}[/tex]A, reacts in a 1:2 ratio with NaOH, we need to multiply the moles of [tex]H_{2}[/tex]A by 2 to determine the moles of NaOH required for complete reaction:

Moles of NaOH = 2 * Moles of [tex]H_{2}[/tex]A

Moles of NaOH = 2 * 0.05 mol

Moles of NaOH = 0.1 mol

0.1 mol of sodium hydroxide (NaOH) would react exactly with 7.5 g of the diprotic acid [tex]H_{2}[/tex]A.

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252,212 Joules of energy are required to heat 100g of water from 35°C to 115°C water vapor.

To calculate the amount of energy required to heat water from 35°C to 100°C, we use the specific heat capacity of water, which is 4.18 J/(g°C). This means that it takes 4.18 Joules of energy to heat one gram of water by one degree Celsius.

So, the energy required to heat 100 g of water from 35°C to 100°C can be calculated as follows:

Q1 = m × c × ΔT

Q1 = 100 g × 4.18 J/(g°C) × (100°C - 35°C)

Q1 = 26,212 Joules

Next, we need to calculate the amount of energy required to vaporize the water at 100°C. This is done using the heat of vaporization of water, which is 2260 J/g.

So, the energy required to vaporize 100 g of water at 100°C is:

Q2 = m × Lv

Q2 = 100 g × 2260 J/g

Q2 = 226,000 Joules

Therefore, the total energy required to heat 100 g of water from 35°C to 115°C water vapor is:

Q = Q1 + Q2

Q = 26,212 Joules + 226,000 Joules

Q = 252,212 Joules

Thus, 252,212 Joules of energy are required to heat 100g of water from 35°C to 115°C water vapor.

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