A 0. 3 gram piece of copper is heated and fasioned into a bracelet. The amount of energy transferred by heat to the copper is 66,300 Joules. If the specific heat of copper is 390J/gxC, what is the change of the copper's temperature? (4 sig figs)

The number of moles of hydrogen gas comes out to be 5.700 that can be calculated using the ideal gas equation.

Using ideal gas equation,

PV = nRT ......(1)

It is given that,

T = 150.0 °C

P = 23.3 atm

V = 8.50 L

To calculate the number of moles, substitute the known values in equation (1).

PV = nRT

23.3 atm x 8.50 L  =  n x 0.0821 L atm/mol/K x 423.15 K

n = (23.3 atm x 8.50 L) / (0.0821 L atm/mol/K x 423.15 K)

  =  198.05 / 34.74 mole

  = 5.700 moles

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The percent by mass of carbon in the hydrocarbon is 85.7%.

To find the percent by mass of carbon in the hydrocarbon, follow these steps:

1. Determine the mass of hydrogen in the hydrocarbon: 2.86g.
2. Calculate the mass of carbon in the hydrocarbon by subtracting the mass of hydrogen from the total mass: 20.0g (total mass) - 2.86g (mass of hydrogen) = 17.14g (mass of carbon).
3. Calculate the percent by mass of carbon by dividing the mass of carbon by the total mass of the hydrocarbon, and then multiply by 100: (17.14g carbon / 20.0g total mass) * 100 = 85.7%.

So, the percent by mass of carbon in the 20.0g hydrocarbon sample is 85.7%.

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The name of the branched alkene given in the question is:

6-ethyl-8-methyl-5-propyl-2-nonene or 6-ethyl-8-methyl-5-propylnon-2-ene

The naming of compound is obtained by the of IUPAC standard. This is illustrated below:

Thus, the IUPAC name for the branched alkene is:

6-ethyl-8-methyl-5-propyl-2-nonene or 6-ethyl-8-methyl-5-propylnon-2-ene

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The percent composition of the sample containing 1.29 grams of carbon and 1.71 grams of oxygen is 43% carbon and 57% oxygen.

The percent composition of a sample can be calculated by dividing the mass of each element in the sample by the total mass of the sample and then multiplying by 100%.

To find the percent composition of a sample containing 1.29 grams of carbon and 1.71 grams of oxygen, we need to calculate the total mass of the sample first.

Total mass of the sample = mass of carbon + mass of oxygen
= 1.29 grams + 1.71 grams
= 3 grams

Now, we can calculate the percent composition of carbon and oxygen in the sample:

Percent composition of oxygen = (mass of oxygen / total mass of the sample) x 100%
= (1.71 grams / 3 grams) x 100%
= 57%

Percent composition of carbon =  (mass of carbon / total mass of the sample) x 100%

=(1.29 grams / 3 grams) x 100%

= 43%

Therefore, the sample contains 43% carbon and 57% oxygen by mass.

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A total of 28,000 grams of calcium oxide will be produced.

To find out how many grams of calcium oxide will be produced in a closed vessel containing 20.0 kg of calcium and 20.0 kg of oxygen gas, follow these steps:

1. Convert the given masses into moles using the molar mass of each element:
  - For calcium (Ca): 20,000 g / 40.08 g/mol ≈ 499 moles
  - For oxygen (O2): 20,000 g / 32 g/mol ≈ 625 moles

2. Determine the limiting reactant using the stoichiometry of the balanced equation:
  - The stoichiometric ratio of Ca to O2 is 2:1, so 625 moles of O2 would require 1,250 moles of Ca, but there are only 499 moles of Ca available. Therefore, calcium is the limiting reactant.

3. Calculate the moles of calcium oxide (CaO) produced using the stoichiometry of the balanced equation:
  - The ratio of Ca to CaO is 1:1, so 499 moles of Ca will produce 499 moles of CaO.

4. Convert the moles of calcium oxide back to grams using the molar mass:
  - The molar mass of CaO is 56.08 g/mol (40.08 g/mol for Ca + 16 g/mol for O). Therefore, 499 moles of CaO * 56.08 g/mol ≈ 28,000 grams of calcium oxide will be produced.

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Using this volume ratio, we can determine that (b) 2.4 liters of ammonia are produced when 3.6 liters of hydrogen reacts with an excess of nitrogen.

The given chemical equation represents the reaction between nitrogen and hydrogen to produce ammonia. The balanced equation shows that for every 3 volumes of hydrogen, 2 volumes of ammonia are produced.

According to the balanced chemical equation N₂ + 3H₂ → 2NH₃, for every 3 volumes of H₂, 2 volumes of NH₃ are produced.

Therefore, if 3.6 liters of H₂ reacts, the amount of NH₃ produced can be calculated as follows:

3.6 L H₂ * (2 L NH₃ / 3 L H₂) = 2.4 L NH₃

Therefore, 2.4 liters of NH₃ would be produced if 3.6 liters of H₂ reacts with an excess of N₂. The correct answer is option (b) 2.4 liters.

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The gas would occupy 90 L at a final pressure of 200 mmHg if the amount of gas does not change.

To determine how many liters this gas would occupy at a final pressure of 200 mmHg, we can use Boyle's Law. Boyle's Law states that the product of the initial pressure and volume of a gas is equal to the product of the final pressure and volume if the temperature and amount of gas remain constant. The formula for Boyle's Law is:

P₁ × V₁ = P₂ × V₂

Where P₁ is the initial pressure (1800 mmHg), V₁ is the initial volume (10 L), P₂ is the final pressure (200 mmHg), and V₂ is the final volume we need to find.

Rearranging the formula to find V₂:
V₂ = (P₁ × V₁) / P₂

Substituting the values:

V₂ = (1800 mmHg × 10 L) / 200 mmHg
V₂ = 18000 L·mmHg / 200 mmHg
V₂ = 90 L

So, this gas would occupy 90 L at a final pressure of 200 mmHg if the amount of gas does not change.

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The solubility of CO₂ in water at 263.4 kPa is 1.064 g/100 ml water.

We can use Henry's law to calculate the solubility of CO₂ in water at a different pressure. Henry's law states that the amount of gas dissolved in a liquid is directly proportional to the partial pressure of the gas above the liquid. Thus, we can set up the following equation:

We are given the solubility of CO₂ at a reference pressure of 101.3 kPa, which is 0.348 g/100 ml water. We want to find the solubility at a new pressure of 263.4 kPa. We can rearrange the equation above to solve for the solubility at the new pressure:

We know that the temperature is constant at 0°C, so we can assume that the solubility is directly proportional to the partial pressure. Thus, we can set up a ratio:

Plugging in the given values, we get:

Solving for the solubility at the new pressure, we get:

Therefore, the solubility of CO₂ in water at a pressure of 263.4 kPa is 1.064 g/100 ml water.

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The amount heat needed to evaporate 500 g of ice that starts at 0 °C is 1504800 J

First, we shall determine the heat needed to melt the ice. Details below:

H₁ = m × ΔHf

H₁ = 500 × 334

H₁ = 167000 J

Next, we shall determine the heat required to change the water from 0 °C to 100°C. Details below:

H₂ = MCΔT

H₂ = 500 × 4.186 × 100

H₂ = 209300 J

Next, we shall determine the heat required to vaporize the water. Details below:

H₃ = m × ΔHv

H₃ = 500 × 2257

H₃ = 1128500 J

Finally, we shall determine the heat required to evaporate the ice. Details below:

Q = H₁ + H₂ + H₃

Q = 167000 + 209300 + 1128500

Total heat required = 1504800 J

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A 31.4 g gold wafer initially at 69.7°C is submerged into 64.1 g of water at 26.8°C. The final temperature at which both substances reach thermal equilibrium is 31.9°C.

The final temperature of both substances at thermal equilibrium needs to be determined.

We can use the principle of conservation of energy. Since the system is insulated, the heat lost by the gold will be equal to the heat gained by the water.

The heat lost by the gold can be calculated using:

Q = mcΔT

where Q is the heat lost, m is the mass of the gold, c is its specific heat capacity, and ΔT is the change in temperature.

Similarly, the heat gained by the water can be calculated using:

Q = mcΔT

where Q is the heat gained, m is the mass of the water, c is its specific heat capacity, and ΔT is the change in temperature.

Setting these two equations equal to each other and solving for the final temperature, we get:

[tex]m_{\text{gold}} \cdot c_{\text{gold}} \cdot (T_{\text{final}} - T_{\text{initial\_gold}}) = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_{\text{final}} - T_{\text{initial\_water}})[/tex]

where [tex]$m_{\text{gold}}$[/tex] is the mass of the gold, [tex]c_{\text{gold}}[/tex] is its specific heat capacity, [tex]T_{\text{initial\_gold}}[/tex] is its initial temperature, [tex]m_{\text{water}}[/tex] is the mass of the water, [tex]$c_{\text{water}}$[/tex] is its specific heat capacity, and [tex]T_{\text{initial\_water}}[/tex] is its initial temperature.

Plugging in the values we get:

[tex]31.4 \, \text{g} \times 0.128 \, \text{J/(g} \cdot \text{°C)} \times (T_{\text{final}} - 69.7^\circ\text{C}) = 64.1 \, \text{g} \times 4.18 \, \text{J/(g} \cdot \text{°C)} \times (T_{\text{final}} - 26.8^\circ\text{C})[/tex]

Solving for [tex]$T_{\text{final}}$[/tex], we get:

[tex]T_{\text{final}} = \frac{(31.4 \, \text{g} \times 0.128 \, \text{J/(g} \cdot \text{°C)} \times 69.7^\circ\text{C}) + (64.1 \, \text{g} \times 4.18 \, \text{J/(g} \cdot \text{°C)} \times 26.8^\circ\text{C})}{(31.4 \, \text{g} \times 0.128 \, \text{J/(g} \cdot \text{°C)}) + (64.1 \, \text{g} \times 4.18 \, \text{J/(g} \cdot \text{°C)})}[/tex]

[tex]$T_{\text{final}}$[/tex] = 31.9°C

Therefore, the final temperature of both substances at thermal equilibrium is 31.9°C.

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A total of 96,320 kJ / 17 kJ per gram of protein = 5,670 grams of protein.

To determine the grams of protein needed to produce 23,000 calories of energy, we need to convert the calories to kilojoules (kJ) and then divide by the energy produced by each gram of protein.

23,000 calories = 96,320 kJ (1 calorie = 4.184 kJ)
Each gram of protein produces 17 kJ of energy.


Protein is an important nutrient for our bodies, as it provides the building blocks for our muscles, bones, and other tissues. It also plays a role in many cellular functions and processes. One of the functions of protein is to provide energy for our bodies, although this is not its primary role.

When we eat protein, our bodies break it down into amino acids, which can then be used for various purposes. One of these purposes is to produce energy.

Every gram of protein contains 4 calories, or 17 kilojoules, of energy. This is less than the amount of energy provided by a gram of fat (9 calories or 37 kilojoules) or a gram of carbohydrate (4 calories or 17 kilojoules), but it is still significant.

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The final volume of the gas is 3.75 L. This result can be explained by the fact that as the temperature of the gas increased, the kinetic energy of its particles also increased, causing them to move faster and occupy a larger volume.

According to the ideal gas law, PV = nRT, the volume of a gas is directly proportional to its temperature.

Therefore, if the temperature of a gas is increased while its pressure and amount remain constant, its volume will also increase.

In this case, the initial volume of the gas is 2.5 L and its temperature is increased from 200 K to 300 K. To find the final volume of the gas, we can use the following equation:

V2 = (T2/T1) x V1

where V1 is the initial volume of the gas, T1 is the initial temperature of the gas, T2 is the final temperature of the gas, and V2 is the final volume of the gas. Plugging in the values, we get:

V2 = (300 K/200 K) x 2.5 L

V2 = 3.75 L

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The percent yield of Ca₃(PO₄)2 is 100.6%. This means that the actual yield is slightly higher than the theoretical yield, which could be due to experimental error or incomplete reaction.

To calculate the theoretical yield of Ca₃(PO₄)2, we need to determine the limiting reagent in the reaction. We can do this by calculating the amount of Ca₃(PO₄)2 that can be produced from each reactant, using stoichiometry and the molar masses of the compounds.

The balanced chemical equation for the reaction is:

3 CaCl₂ + 2 Na₃PO₄ → Ca₃(PO₄)2 + 6 NaCl

The molar mass of CaCl₂ is 110.98 g/mol, and the molar mass of Ca₃(PO₄)2 is 310.18 g/mol.

First, we convert the mass of CaCl₂ to moles:

0.515 g CaCl₂ / 110.98 g/mol = 0.00464 mol CaCl₂

Next, we use stoichiometry to calculate the moles of Ca₃(PO₄)2 that can be produced from the CaCl₂:

0.00464 mol CaCl₂ × (1 mol Ca₃(PO₄)2 / 3 mol CaCl₂) = 0.00155 mol Ca₃(PO₄)2

Finally, we convert the moles of Ca₃(PO₄)2 to grams:

0.00155 mol Ca₃(PO₄)2 × 310.18 g/mol = 0.481 g Ca₃(PO₄)2 (theoretical yield)

Therefore, the theoretical yield of Ca₃(PO₄)2 is 0.481 g.

To calculate the percent yield, we use the formula:

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

The actual yield is given as 0.484 g. Plugging in the values, we get:

percent yield = (0.484 g / 0.481 g) × 100% = 100.6%

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To write the name "Sodium Chloride" from a chemical formula, follow these steps:

1. Identify the elements present in the formula: In this case, the formula is "NaCl," which contains the elements Sodium (Na) and Chlorine (Cl).
2. Write the name of the metal (cation) first: In this case, the metal is Sodium (Na). So, the first part of the name is "Sodium."
3. Write the name of the non-metal (anion) with the suffix "-ide": The non-metal is Chlorine (Cl), so the name changes to "Chloride."
4. Combine the names of the metal and non-metal: The final name is "Sodium Chloride."

To write the chemical formula "NaCl" from the name "Sodium Chloride," follow these steps:

1. Identify the elements from the name: In this case, the name is "Sodium Chloride," which contains the elements Sodium (Na) and Chlorine (Cl).
2. Determine the charges of the elements: Sodium has a +1 charge as a cation, and Chlorine has a -1 charge as an anion.
3. Balance the charges to form a neutral compound: Since the charges are +1 and -1, they balance out, and you don't need to add any subscripts.
4. Write the chemical formula using the element symbols: Combine the symbols to form the formula "NaCl."

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Answer: 4.7g NH4Cl

Explanation:

First we need to determine how many moles of NH4Cl we have:

0.250Lx0.35M= 0.0875moles

now we can multiply the molar mass of NH4Cl by how many moles we have

NH4Cl has a molar mass of 53.49g/mol

53.49 x 0.0875= 4.68g NH4Cl or 4.7g NH4Cl using 2 sig figs.

Baking soda is actually a compound known as sodium bicarbonate, which is water-soluble. Sugar, on the other hand, is also soluble in water and other liquids that contain water.

This means that it dissolves in water and can also dissolve in other liquids that contain water, such as soda. Therefore, baking soda is indeed soluble in soda.

Sugar, on the other hand, is also soluble in water and other liquids that contain water. This includes soda, which is a carbonated beverage that typically contains a high amount of dissolved sugar.

However, the solubility of sugar in soda can depend on various factors such as the temperature of the soda, the amount of sugar present, and the type of sugar used.

In general, both baking soda and sugar are soluble in soda and can dissolve to some extent. However, the exact degree of solubility can vary depending on various factors. It is worth noting that excessive consumption of sugary soda can have negative impacts on health, so it is important to consume such beverages in moderation.

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To solve this problem, we need to use the combined gas law, which relates the pressure, volume, and temperature of a gas.

The combined gas law is expressed as:

(P1V1)/T1 = (P2V2)/T2

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

In this problem, we are given the initial volume V1 as 2.32 liters and the initial temperature T1 as 24.0°C. We need to find the final volume V2 when the temperature is raised to 48.0°C.

We can set up the equation as:

(P1V1)/T1 = (P2V2)/T2

Since the pressure remains constant, we can cancel it out:

V1/T1 = V2/T2

We can rearrange this equation to solve for V2:

V2 = (V1/T1) x T2

Substituting the given values, we get:

V2 = (2.32 L/297.15 K) x 321.15 K

V2 = 2.86 L

Therefore, the new volume of the balloon is 2.86 liters when heated to 48.0°C.

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The percent dissociation of HNO₂ comes out to be 5.2% which is shown in the below secion.

The calculations of pKa is done as follows-

pKa = - log Ka

      = - log (4.0 x 10⁻⁴)

     = 3.398

Mole of NaNO₂ = mass / molar mass

                          = 0.058 g / 68.9953 g/mole

                          = 8.406 x 10⁻⁴ mole

Mole of HNO₂ = 0.110 L * 0.060 mole / L = 6.6 x10⁻³ mole.

Resulting solution is buffer solution.

pH = pKa + log [salt] / [acid]

Substituting the known values in the above formula.

pH = 3.398 + log ( 8.406 x 10⁻⁴ / 6.6 x 10⁻³ )

pH = 2.503

The pH can also be evaluated using the below expression.

pH = -log[H⁺]

-log[H] = 2.503

[H⁺]= 3.14 x 10⁻³ M

Thus

Percent of ionization = 3.14 x 10⁻³ M x 100 / 0.060 = 5.2 %

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The transition within an atom that is not possible is an electron begins in an excited state and then gains enough energy to become ionized. Option D is correct.

An excited electron already has excess energy above its ground state energy level. If it gains more energy, it can transition to a higher energy level or even become ionized by being ejected from the atom. However, an electron that has already been excited and has reached its highest energy level cannot gain any more energy from the atom and therefore cannot be ionized further.

Once an electron is in its highest energy level, it is said to be in the ionization continuum and cannot be further excited or ionized by the atom. Therefore, the transition of an electron beginning in an excited state and then gaining enough energy to become ionized is not possible. On the other hand, the other three transitions listed are possible and occur naturally in many physical and chemical processes, such as atomic emission and absorption spectra. Option D is correct.

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3MnO₂ + 4Al → 2Al₂O₃ + 3Mn

A chemical equation is said to be balanced when the number of atoms of each element on both sides of the equation are the same.

According to this question, manganese oxide reacts with aluminum to produce aluminum oxide and manganese. The balanced equation is given above.

49.7 grams of MnO₂ is equivalent to 0.57 moles

If 3 moles of MnO₂ reacts with 4moles of Al, then 0.57 moles of MnO₂ will react with 0.76 moles of Al.

0.76 moles of Al is equivalent to 20.52 grams of Al.

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The enthalpy of the reaction can be obtained as 118 kJ/mol.

Reaction enthalpy, also known as heat of reaction or ΔHrxn, is the change in enthalpy that occurs during a chemical reaction. It is defined as the difference between the enthalpy of the products and the enthalpy of the reactants.

We have;

Enthalpy of reaction = Bonds broken - Bonds formed

Enthalpy of reaction = [4(413) + 2(495) - [2(799) + 2(463)

= [1652 + 990] - [1598 + 926]

=2642 - 2524

= 118 kJ/mol

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There are 0.0492 moles of the compound C7H14O7 when given 10.34 grams.

To determine how many moles of the compound C7H14O7 you have when given 10.34 grams, you need to follow these steps:

1. Calculate the molar mass of the compound C7H14O7:
- For carbon (C), there are 7 atoms, each with a molar mass of 12.01 g/mol.
- For hydrogen (H), there are 14 atoms, each with a molar mass of 1.01 g/mol.
- For oxygen (O), there are 7 atoms, each with a molar mass of 16.00 g/mol.

2. Add up the molar masses:
- Molar mass of C7H14O7 = (7 * 12.01) + (14 * 1.01) + (7 * 16.00) = 84.07 + 14.14 + 112.00 = 210.21 g/mol.

3. Use the formula to convert grams to moles:
- Moles = mass (grams) / molar mass (g/mol)

4. Plug in the values and solve for moles:
- Moles of C7H14O7 = 10.34 grams / 210.21 g/mol = 0.0492 moles.

So, you have 0.0492 moles of the compound C7H14O7 when given 10.34 grams.

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The sample of crushed rock containing 4.81 x 10²¹ atoms of gold corresponds to 0.008 moles of gold.

The number of moles of gold in the sample of crushed rock can be calculated by dividing the total number of atoms of gold by Avogadro's number, which represents the number of particles in one mole of a substance.

First, we convert the given value of atoms of gold to moles by dividing by Avogadro's number (6.022 x 10²³ atoms per mole).

Number of moles of gold = 4.81 x 10²¹ atoms / 6.022 x 10²³ atoms/mol

Simplifying the calculation, we get:

Number of moles of gold = 0.00799 moles

Therefore, there are approximately 0.008 moles of gold present in the sample of crushed rock.

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The balanced chemical equation is:

2 MnI2 + 13 F2 → 2 MnF3 + 4 IF5

According to the stoichiometry of the reaction, for every 13 moles of F2 that react, 2 moles of MnI2 are consumed. Therefore, the conversion factor to use when converting moles of MnI2 to moles of F2 is:

13 moles F2 / 2 moles MnI2

This conversion factor can be used to convert moles of MnI2 to moles of F2 or vice versa, by multiplying the number of moles of the starting substance by the conversion factor.

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The reaction of 4.30 mol of magnesium with an excess of oxygen produces 6.40 mol of magnesium oxide (MgO).

Magnesium oxide is a white, odorless inorganic compound composed of magnesium and oxygen atoms. It is a strong basic oxide and an important mineral component of many rocks and soils. It has a wide range of industrial uses, such as in the production of cement, ceramics, and glass. It is also used as an antacid and laxative, and as a supplement to increase dietary magnesium intake.

The energy released in this reaction can be determined using the following equation:E = ΔHf (MgO) x (6.40 mol MgO)

In this equation, ΔHf (MgO) is the molar enthalpy of formation of magnesium oxide. The molar enthalpy of formation of magnesium oxide is -601.8 kJ/mol. Therefore, the total energy released in this reaction is:

E = -601.8 kJ/mol x (6.40 mol MgO)

E = -3,854.7 kJ.To determine the number of grams of MgO produced, we can use the following equation: Mass (MgO) = (6.40 mol MgO) x (Molar mass MgO) .

The molar mass of MgO is 40.3 g/mol. Therefore, the mass of MgO produced is: Mass (MgO) = (6.40 mol MgO.

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According to the question the q(gas) in J for this compression process is 0J.

Gas is a state of matter in which particles are spread out and have enough energy to move around freely. Gas is composed of molecules in constant motion and takes the shape and volume of its container. Gas can be either naturally occurring or man-made and is found in the atmosphere. Examples of naturally occurring gases include oxygen, nitrogen, and carbon dioxide. Man-made gases include helium, chlorine, and hydrogen. Gas is often used as a source of energy and is burned to produce heat, which can be used to power machines and vehicles. Gas is also used in many industries, such as in the production of chemicals and plastics.

In this case, n = 0.100 moles,
[tex]C_v[/tex] = (3/2)R = (3/2)(8.314 J/mol K) = 12.471 J/mol K, and
T₁ = 298.0 K,
T2 = 298.0 K.
Therefore, q(gas)
= nCv (T₂- T₁)
= 0.100 mol × 12.471 J/mol K × (298.0 K - 298.0 K)
= 0 J.

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The molality of the solution is 2.57 mol/kg.

We need to determine the number of moles of solute (in this case, KOH) dissolved in a specified mass of the solvent (in this case, water).

First, let's convert the given mass of KOH to moles:

molar mass of KOH = 56.11 g/mol

moles of KOH = mass of KOH / molar mass of KOH

moles of KOH = 85 g / 56.11 g/mol

moles of KOH = 1.515 mol

Next, we need to calculate the mass of the solvent (water) in kilograms:

mass of solvent = 590. g

mass of solvent in kg = mass of solvent / 1000

mass of solvent in kg = 590. g / 1000

mass of solvent in kg = 0.590 kg

Now we can use these values to calculate the molality of the solution:

molality = moles of solute / mass of solvent in kg

molality = 1.515 mol / 0.590 kg

molality = 2.57 mol/kg

Therefore, the molality of the solution is 2.57 mol/kg.

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The student collected 2.63 x 10¹⁰ grams of ammonium phosphate.

To determine the mass of the sample collected, we need to know the molar mass of ammonium phosphate, which is (NH₄)₃PO₄. The molar mass of (NH₄)₃PO₄ can be calculated by adding the atomic masses of the constituent atoms:

Molar mass of (NH₄)₃PO₄ = (3 x molar mass of NH₄) + (1 x molar mass of PO₄)

= (3 x 18.04 g/mol) + (1 x 94.97 g/mol)

= 149.99 g/mol

The number of moles of (NH₄)₃PO₄ in the sample can be calculated by dividing the number of molecules by Avogadro's number (6.022 x 10²³):

Number of moles of (NH₄)₃PO₄ = 9.52 x 10²⁵ molecules / 6.022 x 10²³ molecules/mol

= 15.8 mol

Finally, we can calculate the mass of the sample using the formula:

Mass = Number of moles x Molar mass

= 15.8 mol x 149.99 g/mol

= 2.63 x 10¹⁰ g

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When Lee mixed sodium and hydrogen chloride, a chemical reaction occurred. Sodium has a single valence electron, which it donates to hydrogen chloride, forming Na⁺ and Cl⁻ ions.

These ions then combine to form solid sodium chloride (NaCl) and hydrogen gas (H₂). This reaction is an example of a redox reaction, where the sodium undergoes oxidation and the hydrogen chloride undergoes reduction.

In the simulation or token activity, the reaction can be represented by the following equation:

2 Na + 2 HCl → 2 NaCl + H₂

Thus, the sodium and hydrogen chloride changed into two different substances, solid sodium chloride and gaseous hydrogen, as a result of a chemical reaction involving the transfer of electrons.

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