Phosphorus-32 has a half-life of 14. 0 days. Starting with 8. 00 g of 32P , how many grams will remain after 98. 0 days ?

The final volume of the gas is 31.4 L when cooled from 45.0°C to 12.0°C.

The Charles's law states the relationship between the volume and the temperature of a gas when the pressure is constant. We can use the formula for the relationship between volume and temperature of a gas: [tex]\frac{V_{1} }{T_{1} }[/tex] = [tex]\frac{V_{2} }{T_{2} }[/tex]

where [tex]V_{1}[/tex] and [tex]T_{1}[/tex] are the initial volume and temperature, and [tex]V_{2}[/tex] and [tex]T_{2}[/tex] are the final volume and temperature.

We are given [tex]V_{1}[/tex] = 35.0 L and [tex]T_{1}[/tex] = 45.0°C = 45.0°C + 273.15 = 318.15 K,

and we need to find [tex]V_{2}[/tex] when [tex]T_{2}[/tex] = 12.0°C = 12.0°C + 273.15 = 285.15 K .  

Now by using the formula:

35.0 L / 318.15 K = [tex]V_{2}[/tex] / 285.15 K

[tex]V_{2}[/tex] = (35.0 L / 318.15 K) × 285.15 K

[tex]V_{2}[/tex] = 31.4 L

Therefore, the final volume of the gas is 31.4 L when cooled from 45.0°C to 12.0°C.

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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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The molarity of 0.50 moles of CaCl₂ in 3500 mL of solution is approximately 0.143 M.

To calculate the molarity of 0.50 moles of CaCl₂ in 3500 mL of solution, follow these steps:

1. Convert the volume of the solution from milliliters (mL) to liters (L). There is 1000 mL in 1 L, so divide the given volume by 1000:
  3500 mL ÷ 1000 = 3.5 L

2. Use the formula for molarity (M), which is the number of moles of solute (in this case, CaCl₂) divided by the volume of the solution in liters (L):
  M = moles of solute/volume of solution in L

3. Plug in the values given in the problem: 0.50 moles of CaCl₂ and 3.5 L of solution:
  M = 0.50 moles / 3.5 L

4. Calculate the molarity:
  M ≈ 0.143 M

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Answer:

cellular respiration

Explanation:

Living animals release carbon back into the carbon cycle through the process of respiration. During respiration, animals take in oxygen and release carbon dioxide as a waste product. This carbon dioxide can be taken up by plants during photosynthesis and used to build organic compounds, which can then be consumed by other animals, continuing the carbon cycle. Additionally, when animals defecate or when their bodies naturally decompose after death, the organic matter can be broken down by decomposers, such as bacteria and fungi, which release carbon back into the cycle as well.

I just finished my biology class in high school with an A. Trust me lol

Hope you have a nice day

Answer:

One process that returns carbon from living animals to the cycle is cellular respiration. Cellular respiration converts the organic carbon in the food molecules into carbon dioxide gas, which is released into the atmosphere or water. Another process that returns carbon from living animals to the cycle is excretion1. Excretion removes waste products that contain carbon, such as urea and uric acid, from the body of animals. These waste products can be decomposed by bacteria and fungi, releasing carbon dioxide back into the environment.

Explanation:

After 100 years, a person who had 100 g of pure radioactive nuclei with a half-life of 100 years would have 50 g of radioactive nuclei left.

The half-life of a radioactive substance is the time it takes for half of the substance's original amount to decay. In this case, since the half-life is 100 years, after 100 years, half of the original amount of radioactive nuclei would have decayed.

After the first 100 years, 50 g of radioactive nuclei would remain, and the other 50 g would have decayed. If we wait for another 100 years, half of the remaining 50 g, which is 25 g, would decay, leaving only 25 g of the original amount. This process will continue until all the radioactive nuclei have decayed.

It's worth noting that the rate of decay is exponential, which means that the amount of radioactive substance remaining decreases at a constant rate over time. Knowing the half-life of a radioactive substance is important in determining the amount of time it takes for the substance to decay to a safe level.

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The excess reactant will be oxygen.

To determine the excess reactant, we need to compare the amount of moles of each reactant to the stoichiometry of the balanced equation. The stoichiometric ratio between magnesium and oxygen is 2:1, which means that for every 2 moles of magnesium, 1 mole of oxygen is required for complete reaction.

In this case, we have 7.8 moles of magnesium and 4.7 moles of oxygen. Based on the stoichiometric ratio, we can see that 7.8 moles of magnesium require 3.9 moles of oxygen (2 moles of oxygen for every 1 mole of magnesium). Since we only have 4.7 moles of oxygen, it is the limiting reactant, and magnesium will be in excess.

Therefore, after the reaction is complete, all of the magnesium will be consumed, and some oxygen will be left over. The product of the reaction will be 7.8 moles of magnesium oxide.

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The most likely source of a substance containing carbon, hydrogen, oxygen, and nitrogen is a living organism, such as a plant or an animal.

This is because these four elements are the main components of organic matter, which is found in living things. Carbon is the backbone of organic molecules, while hydrogen and oxygen are also found in many organic compounds, including carbohydrates and lipids.

Nitrogen is an essential component of amino acids, which are the building blocks of proteins. Therefore, if a substance contains all four of these elements, it is likely that it was produced by a living organism or is a byproduct of a living organism's metabolism.

However, this is not always the case, as there are other sources of these elements, such as fossil fuels, which contain carbon and hydrogen, and water, which contains hydrogen and oxygen.

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Answer:

15717.124

Explanation:

124 moles of FeCl2.

The molar mass of FeCl2 is 126.751 g/mol.

To find grams of FeCl2, multiply the number of moles by its molar mass.

124 moles * 126.751 g/mol  = 15717.124 grams.

You can check the ending unit. moles * grams / moles leaves just grams, which is the answer you're looking for.

To determine the mass of copper (II) sulfate in the hydrate, we need to understand the concept of a hydrate. A hydrate is a compound that has water molecules bound to it. Copper (II) sulfate is a hydrate, meaning it has water molecules attached to it. To find the mass of copper (II) sulfate in the hydrate, we need to remove the water molecules from the compound and calculate the remaining mass of the anhydrous salt.

To do this, we need to use the molar mass of the hydrate and the molar mass of the anhydrous salt. The molar mass of copper (II) sulfate pentahydrate is 249.68 g/mol, and the molar mass of anhydrous copper (II) sulfate is 159.61 g/mol. This means that the water molecules in the hydrate account for 90.07 g/mol of the total mass.

Now, let's assume we have 5 grams of the hydrate. We can use this information to calculate the mass of copper (II) sulfate in the hydrate. First, we need to calculate the number of moles of the hydrate by dividing the mass by the molar mass:

5 g / 249.68 g/mol = 0.02002 mol
Next, we need to calculate the number of moles of water in the hydrate by multiplying the total number of moles by the molar mass of water:

0.02002 mol x 18.015 g/mol = 0.3609 g
Finally, we can calculate the mass of anhydrous copper (II) sulfate by subtracting the mass of water from the total mass of the hydrate:
5 g - 0.3609 g = 4.6391 g

Therefore, the mass of copper (II) sulfate in the hydrate is:

4.6391 g * (159.61 g/mol / 249.68 g/mol) = 2.9647 g

In conclusion, to find the mass of copper (II) sulfate in the hydrate, we need to subtract the mass of water from the total mass of the hydrate and then convert the remaining mass to the mass of anhydrous copper (II) sulfate.

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The balance and net ionic equation are;

1. Ca (s) + Br2 (l) → CaBr2 (s)

2. HNO3 (aq) + BaCl2 (aq) → Ba(NO3)2 (aq) + 2HCl (aq)

3. C7H16 (l) + 11O2 (g) → 7CO2 (g) + 8H2O (l)

4. balanced equation:Cl2 (g) + 2KI (aq) → 2KCl (aq) + I2 (s),

Net ionic equation:

Cl2 (g) + 2I- (aq) → 2Cl- (aq) + I2 (s)

5. Zn (s) + Ca(NO3)2 (aq) → No reaction (since Ca is less reactive than Zn)

6. 2Na3PO4 (aq) + 3Mg(NO3)2 (aq) → Mg3(PO4)2 (s) + 6NaNO3 (aq)

Net ionic equation: 2PO4^3- (aq) + 3Mg^2+ (aq) → Mg3(PO4)2 (s)

7. 2Al (s) + 3ZnCl2 (aq) → 2AlCl3 (aq) + 3Zn (s)

8. 2Fe2O3 (s) → 4Fe (s) + 3O2 (g)

9. Balanced equation: LiOH (aq) + HCl (aq) → LiCl (aq) + H2O (l)

Net ionic equation: OH- (aq) + H+ (aq) → H2O (l)

10A. Solid sodium in water.

2Na (s) + 2H2O (l) → 2NaOH (aq) + H2 (g)

10B. What would happen if you bring a burning splint to the previous reaction?

10 C - The burning splint would go out (since the H2 produced in the reaction may ignite and cause a "pop" sound, but the burning splint itself would go out).

A balanced equation is a chemical equation with equal numbers of atoms for each element on both sides, following the law of conservation of mass.

A net ionic equation is a simplified version of a balanced equation that only shows species participating in the reaction as ions, excluding spectator ions that remain unchanged throughout the reaction. This highlights the actual chemical changes occurring in the reaction.

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- Dialcohol: used in polyester synthesis
- Diester: used in polyester synthesis
- Dinitrodiacid: neither polyester nor nylon synthesis
- Diamine: used in nylon synthesis
- Diether: neither polyester nor nylon synthesis

1. Dialcohol: This type of bifunctional molecule is used in the synthesis of polyesters. Polyesters are formed through the condensation reaction between a dialcohol and a diacid or diester.

2. Diester: Diesters are also used in the synthesis of polyesters. They react with dialcohols to form polyester chains.

3. Dinitrodiacid: Dinitrodiacids are not commonly used in the synthesis of either polyesters or nylons. Their nitro functional groups make them less reactive for the condensation reactions required for these polymer types.

4. Diamine: Diamines are used in the synthesis of nylons. Nylons are formed through the condensation reaction between a diamine and a diacid or a diester with a specific type of functional groups, such as adipoyl chloride.

5. Diether: Diethers are not used in the synthesis of polyesters or nylons. They lack the necessary functional groups (alcohol, ester, or amine) for the condensation reactions needed to form these polymers.

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The number of moles of nitrogen gas that will occupy a volume of 5L at 3.85 atm and 27°C is determined using the ideal gas law equation. After calculations, it is found to be approximately 0.7919 moles. Thus, 0.7919 moles of nitrogen gas occupy 5L at 3.85 atm and 27°C.

To calculate the number of moles of nitrogen gas that will occupy a volume of 5L at 3.85 atm and 27°C, we can use the ideal gas law equation:

PV = nRT

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

First, we need to convert the temperature from Celsius to Kelvin:

T = 27°C + 273.15 = 300.15 K

Next, we can rearrange the ideal gas law equation to solve for the number of moles:

n = PV / RT

Plugging in the values, we get:

[tex]n = \frac{{(3.85 \, \text{atm}) \cdot (5 \, \text{L})}}{{(0.08206 \, \text{L} \cdot \text{atm/K} \cdot \text{mol}) \cdot (300.15 \, \text{K})}}[/tex]

Simplifying the expression, we get:

n = 0.7919 moles

Therefore, 0.7919 moles of nitrogen gas will occupy a volume of 5L at 3.85 atm and 27°C.

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The temperature of the gas when the volume of the container increases to 1074 mL is 358.15 K or 85.0°C

The behavior of gases is affected by several factors including temperature, pressure, and volume. One important principle that applies to gases is that they tend to occupy the entire volume of their container. Therefore, if the volume of the container increases, the gas will occupy more space.

In this particular scenario, the gas initially occupies 900 mL at a temperature of 27.0°C. When the volume of the container increases to 1074 mL, we need to determine the corresponding temperature of the gas. To do this, we can use the formula:

(V1/T1) = (V2/T2)

Where V1 and T1 represent the initial volume and temperature of the gas, respectively, and V2 and T2 represent the final volume and temperature of the gas, respectively.

Substituting the given values into the formula, we get:

(900/300.15) = (1074/T2)

Simplifying the equation, we can cross-multiply and solve for T2:

900T2 = 1074 x 300.15

T2 = 1074 x 300.15 / 900

T2 = 358.15 K

Therefore, the temperature of the gas when the volume of the container increases to 1074 mL is 358.15 K or 85.0°C (rounded to one decimal place).

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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 changes that occur in the parent nucleus during gamma decay are limited to the emission of a gamma ray and the associated decrease in energy. The mass and atomic number of the nucleus remain unchanged.

In gamma decay, the parent nucleus does not undergo any changes in terms of its mass or atomic number. Instead, the nucleus emits a gamma ray, which is a high-energy photon. This gamma ray is released as the nucleus transitions from an excited state to a lower energy state.

The emission of a gamma ray does not affect the number of protons or neutrons in the nucleus. This means that the atomic number and mass number of the nucleus remain the same before and after gamma decay.

However, the emission of a gamma ray does result in a decrease in the energy of the nucleus. This is because gamma rays have a very high frequency and carry a lot of energy. By releasing a gamma ray, the nucleus is able to shed some of this excess energy and move to a lower energy state.

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The number of moles in the room depends on the temperature.

Assuming that the temperature and volume in the room are the same as those outside, we can use the ideal gas law to calculate the number of moles of gas in the room.

Ideal gas law is given by:

PV = nRT

Number of moles:

n = PV/RT

Since the temperature and pressure are the same in both cases, we can write:

n(room) = (P × V(room)) / RT

n(bulb) = (P × V(bulb)) / RT

We are given that the bulb contains 9.00 moles of gas at the same temperature and pressure as the room. Therefore, we can use the number of moles in the bulb to find the pressure and temperature:

n(bulb) = (P × V(bulb)) / RT

9.00 mol = (P × 61.0 L) / (R × T)

Similarly, for the room, we can write:

n(room) = (P × V(room)) / RT

n(room) = (P × 85.0 L) / (R × T)

P = (n × RT) / V

P = (PV / RT) × RT / V

P = nRT / V

We can use the value of n from the bulb to find the pressure and temperature:

9.00 mol × R × T / 61.0 L = P

P = 3.17 atm

Now we can use this value of pressure to find the number of moles in the room:

n(room) = (P × V(room)) / RT

n(room) = (3.17 atm × 85.0 L) / (R × T)

n(room) = (3.17 atm × 85.0 L) / (0.08206 L atm/mol K × T)

n(room) = 129.3 L atm / (R × T)

Therefore, the number of moles in the room depends on the temperature.

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Turtle refuges are usually created on beaches where turtles lay their eggs, hatch, and return to the sea. Therefore, beaches that are known as nesting grounds for sea turtles may be suitable for creating a turtle refuge.

In general, turtle nesting sites are characterized by sandy beaches, dunes, and undisturbed vegetation. Female sea turtles come ashore to lay their eggs on sandy beaches, and the hatchlings make their way to the ocean once they emerge from the nest.

Turtle refuges provide protection for these nesting sites, allowing the turtles to lay their eggs and for the hatchlings to safely make their way to the ocean.

It is important to note that the location of a turtle refuge should be based on careful research and consideration of a variety of factors, such as the species of turtles that inhabit the area, the presence of human and natural threats to the nesting sites, and the availability of resources and support for the conservation efforts.

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152 grams of [tex]C2H6[/tex]would release 152 kJ of energy when it condenses at its boiling point.

Assuming you meant "[tex]C2H6[/tex]" instead of "[tex]CH Ch[/tex]", the heat of vaporization of [tex]C2H6[/tex]is 30.1 kJ/mol. The molar mass of [tex]C2H6[/tex] is 30.07 g/mol.

To calculate the heat of vaporization for 152 g of [tex]C2H6[/tex], we need to first calculate the number of moles of [tex]C2H6[/tex]:

152 g / 30.07 g/mol = 5.05 mol

Then, we can calculate the energy released using the heat of vaporization:

5.05 mol x 30.1 kJ/mol = 152 kJ

Therefore, 152 grams of [tex]C2H6[/tex]would release 152 kJ of energy when it condenses at its boiling point.

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The pressure inside a container that contains 25 grams of argon is 0.34 atm.

The pressure inside a container can be calculated using the following expression;

PV = nRT

Where;

According to this question, 25 grams of argon in a container has a volume of 60 liters and at a temperature of 400 K.

P × 60 = 0.625 × 0.0821 × 400

60P = 20.525

P = 0.34 atm

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We can deduce from the computations that the mass of the acetic acid produced is 28.2 g.

The reactant that is totally consumed during a chemical reaction involving two or more reactants is known as the limiting reactant. This limits the amount of product that can be generated. Excess reactants are the additional reactant(s) that are still present after the limiting reactant has been completely consumed.

CH3CHO's molecular weight is 20.8 g/44 g/mol.

= 0.47 moles

O2 molecular weight is 14.5 g/32 g/mol.

= 0.45 moles

If 1 mole of O2 interacts with 2 moles of CH3CHO

CH3CHO containing 0.47 moles would react with 0.47 * 1/2.

= 0.24 moles

Thus, the limiting reactant is CH3CHO.

Acetic acid mass produced is 0.47 moles * 60 g/mol.

= 28.2 g

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The chemical reaction is

2 K + Na2O -> K2O + 2 Na

The given chemical equation represents a reaction between potassium (K) and sodium oxide (Na2O). The products formed in this reaction are potassium oxide (K2O) and sodium (Na).

On the reactant side, we have two atoms of potassium and two atoms of sodium, while on the product side, we have two atoms of potassium and two atoms of sodium as well.

Therefore, the equation is already balanced with respect to the number of potassium and sodium atoms.

However, we need to balance the oxygen atoms. On the reactant side, we have one molecule of Na2O, which contains two atoms of oxygen. On the product side, we have one molecule of K2O, which also contains two atoms of oxygen. Thus, the equation is balanced.

Finally, we can write the balanced equation as:

2 K + Na2O → K2O + 2 Na

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Answer:

-2.6 mL.

Explanation:

To solve this question, we need to use the formula:

V1/T1 = V2/T2

where V1 and T1 are the initial volume and temperature of the gas, and V2 and T2 are the final volume and temperature of the gas. We also need to convert the temperatures from degrees Celsius to kelvins by adding 273.15. Plugging in the given values, we get:

50.0 mL / (48.0 + 273.15) K = V2 / (3 + 273.15) K

Solving for V2, we get:

V2 = 50.0 mL x (3 + 273.15) K / (48.0 + 273.15) K V2 = 47.4 mL

Therefore, the change in volume is:

ΔV = V2 - V1 ΔV = 47.4 mL - 50.0 mL ΔV = -2.6 mL

The negative sign indicates that the volume decreases when the gas is cooled.

The answer is -2.6 mL.

480 mL of the HCl solution with a pH of 1.3 can be neutralized by one dose of milk of magnesia assuming the concentration of magnesium hydroxide is 0.2 M.

To determine the volume of [tex]HCl[/tex] solution that can be neutralized by milk of magnesia, we need to know the concentration of the milk of magnesia.

Assuming milk of magnesia is a suspension of solid magnesium hydroxide in water, we need to know the concentration of magnesium hydroxide [tex](Mg(OH)2)[/tex] in the suspension.

Let's assume that the concentration of magnesium hydroxide in milk of magnesia is 0.2 M.

The balanced chemical equation for the neutralization reaction between [tex]HCl[/tex] and[tex]Mg(OH)2[/tex]is:

[tex]2HCl + Mg(OH)2 - > MgCl2 + 2H2O[/tex]

From the equation, we can see that two moles of [tex]HCl[/tex] react with one mole of [tex]Mg(OH)2[/tex].

To determine the volume of [tex]HCl[/tex] solution, we need to calculate the number of moles of [tex]Mg(OH)2[/tex] in one dose of milk of magnesia:

0.2 M = 0.2 moles / liter

Let's assume one dose of milk of magnesia is 30 mL, or 0.03 L. Then the number of moles of [tex]Mg(OH)2[/tex] in one dose is:

0.2 moles / L x 0.03 L = 0.006 moles Mg(OH)2

Therefore, this amount of [tex]Mg(OH)2[/tex] would require:

2 x 0.006 = 0.012 moles of [tex]HCl[/tex] for complete neutralization

Now, let's calculate the volume of [tex]HCl[/tex] solution needed to provide 0.012 moles of [tex]HCl[/tex].

The volume of [tex]HCl[/tex] solution can be calculated using the balanced chemical equation and the molarity of the [tex]HCl[/tex] solution:

2 moles HCl / 1 mole [tex]Mg(OH)2[/tex] x 0.012 moles [tex]Mg(OH)2[/tex] / 1 = 0.024 moles HCl

[tex]pH = -log[H+]1.3 = -log[H+]\\[H+] = 5 x 10^-2 M[/tex]

Now we can calculate the volume of the HCl solution using the equation:

moles = concentration x volume

0.024 moles = [tex]5 x 10^-2 M x volume[/tex]

volume = 0.48 L or 480 mL

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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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Total, 1.77 g of silver sulfide are formed, when 1. 90 g of silver reacts with 0.

Balanced chemical equation for the reaction is;

4Ag(s) + 2H₂S(g) + O₂(g) → 2Ag₂S(s) + 2H₂O(g)

To determine the limiting reactant, we need to compare the number of moles of each reactant to their stoichiometric ratio in the balanced equation.

First, we need to convert the given masses of silver, hydrogen sulfide, and oxygen to moles;

molar mass of Ag = 107.87 g/mol

moles of Ag = 1.90 g / 107.87 g/mol

= 0.0176 mol

molar mass of H₂S = 2(1.01 g/mol) + 32.06 g/mol = 34.08 g/mol

moles of H₂S = 0.280 g / 34.08 g/mol = 0.00821 mol

molar mass of O₂ = 2(16.00 g/mol) = 32.00 g/mol

moles of O₂ = 0.160 g / 32.00 g/mol = 0.00500 mol

Next, we need to compare the number of moles of each reactant to their stoichiometric ratio in the balanced equation;

Ag ; H₂S ; O₂ = 4 : 2 : 1

The stoichiometric ratio tells us that we need 2 moles of H2S and 0.5 moles of O₂ for every 4 moles of Ag.

Let's calculate the number of moles of each reactant we actually have, starting with H₂S;

H₂S is the limiting reactant if it produces fewer moles of Ag₂S than either of the other reactants. We can calculate the number of moles of Ag₂S that each reactant would produce, assuming that it is the limiting reactant;

If H₂S is the limiting reactant;

moles of Ag₂S = (0.00821 mol H₂S) x (2 mol Ag₂S / 2 mol H₂S)

= 0.00821 mol

If O₂ is the limiting reactant;

moles of Ag₂S = (0.00500 mol O₂) x (2 mol Ag2S / 1 mol O₂)

= 0.0100 mol

If Ag is the limiting reactant;

moles of Ag₂S = (0.0176 mol Ag) x (0.5 mol Ag₂S / 4 mol Ag)

= 0.00220 mol

Since H₂S produces the fewest moles of Ag₂S, it is the limiting reactant.

To calculate the mass of Ag₂S produced, we can use the number of moles of Ag₂S produced by the limiting reactant:

mass of Ag₂S = (0.00821 mol Ag₂S) x (2 x 107.87 g/mol)

= 1.77 g

Therefore, 1.77 g of silver sulfide are formed.

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One way to reduce the volume of hydrogen gas needed to power a car for 400 km is to use a technology called on-board hydrogen storage.

This involves compressing the hydrogen gas to very high pressures, typically between 5,000 and 10,000 psi, which significantly reduces its volume.

Another method is to use liquid hydrogen storage, which involves cooling hydrogen gas to its boiling point (-423.17°F or -252.87°C) and storing it in a cryogenic tank. At this temperature, hydrogen gas is in its liquid state and takes up much less space than when it is in its gaseous state.

Both of these methods of hydrogen storage can greatly reduce the volume of hydrogen needed to power a car for 400 km, making hydrogen fuel cell cars more practical and feasible for everyday use.

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The types of chemical bonds present in a material determine the arrangement of atoms or molecules at the microscopic level, which in turn determines the properties of the material at the macroscopic level.

For example, materials with ionic bonds tend to have high melting and boiling points due to the strong electrostatic attraction between positively and negatively charged ions. Covalently bonded materials tend to have lower melting and boiling points due to the weaker intermolecular forces between molecules.

Metallic bonding leads to high electrical and thermal conductivity due to the delocalization of electrons within the metal lattice. These different bond types and resulting material properties are important in understanding the behavior and applications of different materials.

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When zinc reacts with hydrochloric acid, the response bubbles vigorously as hydrogen fueloline is produced.

The manufacturing of a fueloline is likewise an illustration that a chemical response is occurring. When dilute hydrochloric acid is introduced to granulated zinc positioned in a take a look at tube, zinc metallic is transformed to zinc chloride and hydrogen fueloline is developed withinside the response. In the response we will see that a zinc chloride salt is fashioned and hydrogen fueloline is developed. The developed hydrogen fueloline is colourless and odourless. When Zinc granules reacts with Hydrochloric acid ,it'll produces hydrogen fueloline and zinc chloride.

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Answer:

Carbon and silicon are both in Group 14 of the periodic table, which means they have similar electronic configurations and therefore similar bonding properties. However, the difference in melting and sublimation temperatures of their oxides, silicon oxide and solid carbon dioxide, respectively, can be attributed to differences in their structure and bonding.

Silicon oxide (SiO2) has a giant covalent structure, in which each silicon atom is covalently bonded to four oxygen atoms and each oxygen atom is covalently bonded to two silicon atoms. This gives rise to a three-dimensional network of strong covalent bonds, which requires a large amount of energy to be broken. Therefore, silicon oxide has a high melting point of 2440°C because a lot of energy is required to overcome the strong covalent bonds and melt the solid.

On the other hand, solid carbon dioxide (CO2) has a molecular structure, in which each carbon atom is double bonded to two oxygen atoms. The carbon dioxide molecules are held together by weak intermolecular forces, such as Van der Waals forces, which are much weaker than the strong covalent bonds present in silicon oxide. As a result, solid carbon dioxide can sublime at -70°C, without melting into a liquid, because the intermolecular forces can be overcome by relatively low energy input.

In summary, the difference in melting and sublimation temperatures of silicon oxide and solid carbon dioxide can be explained by the difference in their bonding types and structures. Silicon oxide has a giant covalent structure with strong covalent bonds that require a large amount of energy to break, resulting in a high melting point. Solid carbon dioxide has a molecular structure held together by weak intermolecular forces, which can be overcome by relatively low energy input, resulting in a low sublimation point.