What statement correctly describes what is occurring in this chemical reaction?
Note: Zinc (Zn), Sulfur (S)
Zn + S→ ZnS
ZnS is breaking apart into Zn and S
ZnS reacts with S to form Zn
Zn and S combine to form ZnS
Zn combusts with S to form ZnS

Answer:  **

H-N-H

   |

  H

Explanation:

Look at a periodic table to determine how many electrons you need to account for. Hydrogen (H) only has 1 electron, while Nitrogen (N) has 5. We have three Hydrogen atoms and one Nitrogen atom, so the total number of electrons will be 3 * 1 + 5 = 8 e-.

Now, place the center atom, which will be Nitrogen and place the three Hydrogens on three sides of it as above in the answer. You should use single bonds for this. Each single bond is a pair of electrons, so since we have three single bonds so far, we have accounted for 2 * 3 = 6 electrons. However, we need 2 more electrons for the total of 8. We put these electrons in as a lone pair above Nitrogen.

We check to see if everything follows the octet rule: Nitrogen has three single bonds, so that's 6 e-, as well as one lone pair, so that's another 2 e- for a total of 8 e-. Check. Now look at Hydrogen: H is the only element whose full orbital is 2 e-. Each H has a single bond with Nitrogen, so each does have 2 e-.

Thus, we know this is the correct diagram, and we are done.

Explanation:

A bond line structure of a compound has H N H in linear plane and a hydrogen is branching upward, and the compound is H N (H) H. The nitrogen has two dots at its bottom represents a lone pair of electrons. So ,the correct answer is option C.

The correct Lewis structure for ammonia ([tex]NH_3[/tex]) is option C. It shows a bond line structure with three hydrogen atoms (H) bonded to a central nitrogen atom (N) in a linear plane.

One hydrogen atom branches upward from the plane. Additionally, the nitrogen atom in this structure has two dots at its bottom, indicating a lone pair of electrons. This arrangement follows the octet rule, as nitrogen has formed three covalent bonds with hydrogen, completing its valence shell. The lone pair on nitrogen gives ammonia its characteristic properties.

Thus, option C accurately represents the Lewis structure of ammonia, showing the bonding and lone pair arrangement of its atoms.

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The molarity of the solution containing 0.05 g of KCl in a total volume of 0.310 L is approximately 0.0022 M.

To determine the molarity of a solution, we need to know the moles of solute and the volume of the solution. First, we need to convert the mass of KCl to moles using its molar mass. The molar mass of KCl is approximately 74.55 g/mol (39.10 g/mol for potassium + 35.45 g/mol for chlorine).

Using the formula:

moles = mass / molar mass

moles = 0.05 g / 74.55 g/mol = 0.00067 mol

Next, we convert the total volume of the solution to liters:

0.310 L

Finally, we calculate the molarity (M) using the formula:

Molarity = moles of solute / volume of solution (in liters)

Molarity = 0.00067 mol / 0.310 L ≈ 0.0022 M

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The highest boiling point based on the data is option 4

Compared to alcohols of comparable molecular weight, carboxylic acids often have higher boiling temperatures. Between the hydrogen atoms of adjacent molecules and the oxygen in the carboxyl group of carboxylic acids, strong intermolecular hydrogen bonds can develop. Because it takes more energy to break the intermolecular interactions and change the substance from a liquid to a gas during boiling, these hydrogen bonds help materials have higher boiling temperatures.

Although carboxylic acids and alcohols are both capable of forming hydrogen bonds, carboxylic acids have higher boiling temperatures due to the extra carboxyl group that they contain.

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An example of a potassium compound containing both ionic and covalent bonds is potassium chloride and an example containing only ionic bonds is potassium sulfite.

A potassium ion (K+) can combine with various monatomic and polyatomic ions to create compounds through different types of chemical bonds. In the context of potassium compounds, both ionic and covalent bonds can be observed.

An example of a potassium compound that exhibits both ionic and covalent bonds is potassium chloride (KCl). In this compound, potassium donates its valence electron to chlorine, forming an ionic bond between the positively charged potassium ion and the negatively charged chloride ion. However, there is also some degree of covalent character present in the compound, as the electron is not fully transferred from potassium to chlorine. This partial sharing of electrons results in a degree of covalent bonding.

On the other hand, an example of a potassium compound containing only ionic bonds is potassium sulfate (K2SO4). In this compound, two potassium ions (K+) combine with one sulfate ion (SO4^2-) through ionic bonds. The transfer of electrons is complete, resulting in a strong electrostatic attraction between the oppositely charged ions, forming an entirely ionic compound.

In summary, potassium chloride exemplifies a potassium compound containing both ionic and covalent bonds, whereas potassium sulfate represents a potassium compound consisting solely of ionic bonds.

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A potassium ion can combine with several monatomic and polyatomic ions to form compounds.

An example of a potassium compound containing both ionic and covalent bonds is ___(potassium chloride, potassium hydride, potassium nitrate, potassium oxide), and an example containing only ionic bonds is _____(potassium chloride, potassium hypochlorite, potassium sulfate, potassium sulfite).

The United States experienced a decrease in the real GDP, high inflation, and arising unemployment rate.

The United States appeared to be entering a recession.A recession is a decline in economic activity, characterized by declining GDP, high unemployment rates, and increased unemployment benefits. Economic analysts and the media commonly use a two-quarter consecutive decline in real GDP as a definition of a recession.

The United States is considered to have entered a recession in the 1970s, which was characterized by an energy crisis, inflation, and recession. However, by the end of the decade, the economy had improved, and it entered into the 1980s with a strong economic performance.

The 1970s were a period of high inflation, low growth, and an oil crisis, which had a significant impact on the United States economy. Therefore, it can be concluded that The United States was in the middle of an economic boom before the 1970s recession and entered a recession in the 1970s due to a decrease in the real GDP, high inflation, and arising unemployment rate.

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It would require approximately 600 kilojoules of energy to vaporize 2 kg of copper.

To calculate the energy required to vaporize a substance, we need to consider the heat of vaporization, which is the amount of energy required to convert a given amount of substance from its liquid state to its gaseous state at a constant temperature.

The heat of vaporization for copper is approximately 300 kJ/kg (kilojoules per kilogram) at its boiling point, which is around 2567 degrees Celsius (4649 degrees Fahrenheit). This means that for every kilogram of copper, 300 kJ of energy is needed to vaporize it.

Given that you have 2 kg of copper, we can calculate the total energy required as follows:

Energy = Heat of Vaporization × Mass

Energy = 300 kJ/kg × 2 kg

Energy = 600 kJ

Therefore, it would require approximately 600 kilojoules of energy to vaporize 2 kg of copper.

It's worth noting that the heat of vaporization can vary slightly depending on the purity of the copper and the specific conditions, such as temperature and pressure. The value provided here is an approximation. Additionally, it's important to handle copper and any high-temperature processes with caution, as they can pose safety hazards.

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The amount of energy needed to raise the temperature of the liquid benzene sample from 42.40 °C to 95.40 °C is approximately 1.865 kJ.

To calculate the amount of energy needed to raise the temperature of a sample, we can use the equation:

q = m × C × ΔT

Where:

q = heat energy (in joules)

m = mass of the sample (in grams)

C = specific heat capacity of the substance (in J/g°C)

ΔT = change in temperature (in °C)

First, we need to determine the specific heat capacity of liquid benzene. The specific heat capacity of benzene is approximately 1.74 J/g°C.

Using the given values:

m = 20.60 g

C = 1.74 J/g°C

ΔT = (95.40 °C - 42.40 °C) = 53 °C

Plugging these values into the equation:

q = 20.60 g × 1.74 J/g°C × 53 °C = 1865.044 J

To convert the energy to kilojoules, divide by 1000:

q = 1865.044 J ÷ 1000 = 1.865 kJ

Therefore, the amount of energy needed to raise the temperature of the liquid benzene sample from 42.40 °C to 95.40 °C is approximately 1.865 kJ.

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Answer:

Explanation:

1.1 we have to find mass of Na3PO4;

for that we have to Calculate the moles of Na3PO4 needed:

volume is 100mL = 0.1L

Molar concentration = Moles of solute / Volume of solution in L

0.1 mol/L = Moles of Na3PO4 / 0.1 L

Moles of Na3PO4 = 0.1 mol/L * 0.1 L

Moles of Na3PO4 = 0.01 mol

Now, Calculate the mass of Na3PO4 needed:

so, Mass = Moles of Na3PO4 * Molar mass of Na3PO4

Mass = 0.01 mol * 164 g/mol

Mass = 1.64 g of Na3PO4.

1.2 materials and glassware needed:

1.64 g Sodium phosphate (Na3PO4)

100 mL volumetric flask

weighing balance

Distilled water

Glass rod

Pipette and burette

Answer:

The balanced chemical equation for the reaction is:

2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

From the equation, 2 moles of ethane react with 7 moles of oxygen gas to produce 6 moles of water.

First, we need to determine which reactant is limiting. To do this, we can use the given masses and convert them to moles using the molar masses of the compounds:

n(C2H6) = 3.01 g / 30.07 g/mol = 0.100 mol

n(O2) = 19.3 g / 32.00 g/mol = 0.603 mol

The mole ratio of C2H6 : O2 is 2 : 7, so to react completely with 0.100 mol of C2H6, we need:

n(O2) = (7/2) x 0.100 mol = 0.350 mol

Since we have more than 0.350 mol of O2, O2 is not the limiting reactant. Therefore, the limiting reactant is C2H6.

The theoretical yield of H2O can be calculated from the number of moles of C2H6:

n(H2O) = (6/2) x 0.100 mol = 0.300 mol

Finally, we can convert the number of moles of H2O to grams:

m(H2O) = n(H2O) x M(H2O)

m(H2O) = 0.300 mol x 18.02 g/mol

m(H2O) = 5.41 g

Therefore, the theoretical yield of H2O formed is 5.41 g. Rounded to three significant figures, the answer is 5.41 g.

Explanation:

Ratio of moles of NH₃ produced to moles of N₂ used: 2 moles of NH₃ / 1 mole of N₂

Ratio of moles of NH₃ produced to moles of H₂ used: 2 moles of NH₃ / 3 moles of H₂

From the balanced chemical equation:

N₂ + 3 H₂ ⟶ 2 NH₃

We can determine the ratio of moles of products to the moles of each reactant.

Ratio of moles of NH₃ produced to moles of N₂ used:

From the balanced equation, we can see that 1 mole of N₂ reacts to produce 2 moles of NH₃. Therefore, the ratio is:

2 moles of NH₃ / 1 mole of N₂

Ratio of moles of NH₃ produced to moles of H₂ used:

From the balanced equation, we can see that 3 moles of H₂ react to produce 2 moles of NH₃. Therefore, the ratio is:

2 moles of NH₃ / 3 moles of H₂

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Given the equation of reaction;

N₂ + 3 H₂ ---> 2 NH₃

Calculate the ratio of the moles of produced to the moles of each of the reactants used. (Write two separate ratios.)