Write the Lewis structures of HNNH and H2NNH2. Predict which molecule has the greater N–N bond energy.

a) Compound C contains only covalent bonds.

Covalent bonds are formed when atoms share electrons. Compound C, which represents carbon (C), consists only of covalent bonds. Carbon is a nonmetal and typically forms covalent compounds with other nonmetals.

In contrast, compounds such as H (hydrogen), Mg (magnesium), and Zn (zinc) can form both ionic and covalent bonds. Hydrogen can exist as H2, a diatomic molecule held together by a covalent bond.

Magnesium (Mg) and zinc (Zn) are metals that predominantly form ionic compounds, where electrons are transferred from the metal to a nonmetal.

A molecule containing a triple covalent bond is represented by the formula C2H2, which corresponds to ethyne (also known as acetylene).

Ethyne consists of two carbon atoms bonded by a triple covalent bond and two hydrogen atoms bonded to each carbon atom.

A formula representing a molecular substance is represented by the compound XCl4, where X can be any nonmetal element.

This formula signifies a molecular compound consisting of covalent bonds between X and four chlorine (Cl) atoms.

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Answer: 10y/s

Explanation:

The rate at which N\(_{2}\) disappears is 0.016 M/s, while the rate at which H\(_{2}\) disappears is 0.0213 M/s.

In the balanced chemical equation N\(_{2}\)(g) + 3 H\(_{2}\) (g) → 2NH\(_{3}\)(g), the stoichiometric coefficients represent the mole ratios between the reactants and products.

Since the reaction rate is given for NH\(_{3}\), we can determine the rates of disappearance of  N\(_{2}\) and H\(_{2}\) by comparing their stoichiometric ratios in the reaction.

The stoichiometric ratio between  N\(_{2}\) and NH\(_{3}\) is 1:2, meaning for every mole of  N\(_{2}\) consumed, 2 moles of NH\(_{3}\) are produced. Therefore, the rate of disappearance of  N\(_{2}\) is half of the rate of formation of NH\(_{3}\).

Similarly, the stoichiometric ratio between H\(_{2}\) and NH\(_{3}\) is 3:2. This means that for every 3 moles of H\(_{2}\) consumed, 2 moles of NH\(_{3}\) are produced. Therefore, the rate of disappearance of  H\(_{2}\) is (2/3) times the rate of formation of NH\(_{3}\).

Given the rate of formation of NH\(_{3}\) as 0.032 M/s, the rate of disappearance of  N\(_{2}\) would be 0.016 M/s (0.032 M/s ÷ 2), and the rate of disappearance of  H\(_{2}\) would be approximately 0.0213 M/s (0.032 M/s × 2/3).

Therefore, the rate of disappearance of  N\(_{2}\) is 0.016 M/s, and the rate of disappearance of  H\(_{2}\) is 0.0213 M/s.

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Answer:

1 mole of Sulfuric Acid

Explanation:

Given

\(Compound = H_2SO_4\)

\(Grams = 98.078\)

Required

Determine the amount of moles

First, we need to determine the atomic mass of the acid.

\(H = 1.008g\) -- Hydrogen

\(S = 32.065\) --- Sulfur

\(O =15.999\) -- Oxygen

So:

\(H_2SO_4 = H * 2 + S + O * 4\)

\(H_2SO_4 = 1.008 * 2 + 32.065 + 15.999 * 4\)

\(H_2SO_4 = 2.016 + 32.065 + 63.996\)

\(H_2SO_4 = 98.077g\)

Number of moles (n) is then calculated as thus:

\(n = \frac{Grams}{Atomic\ Mass}\)

\(n = \frac{98.078g}{98.077g}\)

\(n = \frac{98.078}{98.077}\)

\(n = 1.00001019607\)

\(n =1\) (approximated).

Hence, there is 1 mole of Sulfuric Acid

Answer:

Explanation:

Here, we want to get the name for the given chemical formula

Looking at the formula, we look at the names of the element

The names are sodium and fluorine

We write the name of the metal element first, then the non-metal atom

Thus, we have it as

Since they are the main producers of the organic compounds needed for the development and survival of other organisms, phototrophs are essential for the flow of energy in the biosphere.

The flow of energy in the biosphere begins primarily with phototrophs. Phototrophs are organisms that use energy from the sun to convert carbon dioxide and water into organic compounds, such as sugars and carbohydrates, through the process of photosynthesis. These organic compounds provide a source of energy and nutrients for other organisms in the biosphere, which ultimately fuels the flow of energy through the food chain. Other organisms, such as chemotrophs, can also obtain energy from oxidized chemicals or inorganic compounds, but the primary source of energy in most ecosystems is derived from photosynthesis.

Therefore, phototrophs play a crucial role in the flow of energy in the biosphere, as they are the primary producers of organic compounds that support the growth and survival of other organisms.

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Answer:

The protons determine the name of an element and also atomic number because its the same as protons.

Answer:

C.

Explanation:

pH > 7, basic, so pH = 9  is basic.

When solution is basic, it will turn phenolphthalein pink.

A piece of lithium metal is added to a beaker that contains water and phenolphthalein. Using what you know about the properties of bases, choose the best explanation for what you observe.

Answer:

The phenolphthalein is turning pink because a base is forming.

i believe the answer is 2, good luck

The number of moles of NaF produced in the reaction between sodium bromide and calcium fluoride is 5.35 moles (approx 5 moles).

To determine the number of moles of NaF produced in the reaction between sodium bromide and calcium fluoride when 550 grams of sodium bromide are used, we first need to write the balanced chemical equation for the reaction:

2 NaBr + \(CaF_2\) → 2 NaF + \(CaBr_2\)

We can then use the molar mass of sodium bromide and the given mass of sodium bromide to determine the number of moles of sodium bromide:

Number of moles of NaBr = mass of NaBr / molar mass of NaBr

= 550 grams / 102.89 grams/mol

= 5.35 moles

Since the balanced chemical equation tells us that 2 moles of NaBr are needed to produce 2 moles of NaF, the number of moles of NaF produced when 550 grams of NaBr are used is 5.35 moles of NaF.

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As the grain size decreases, the grain size-number (g) in the Hall-Petch equation increases.

Equation 4.17, commonly known as the Hall-Petch equation, relates the yield strength of a metal to its grain size:

\(σy = σ0 + Kd^(-1/2)\)

where σy is the yield strength, σ0 is the frictional stress, K is the Hall-Petch constant, and d is the average grain size.

According to the Hall-Petch equation, the yield strength of a metal increases with decreasing grain size. This means that the value of the Hall-Petch constant K is positive, indicating that the yield strength increases as the grain size decreases.

In other words, as the grain size decreases, the grain size-number (g) in the Hall-Petch equation increases.

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Answer:

an acid is a substance which produces hydrogen ions as the only as the only positive ion when dissolved in water.

The oxidation number of an element in a compound is equal to the charge of the atoms if the compound was an ionic compound.

The types of compounds are:

Ionic compounds:

Covalent compound:

Molecular compounds:

Therefore, the oxidation number of an element in a compound is equal to the charge of the ion that would have if the compound was an ionic compound.

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This chemical is known as lead (II) nitrate. It is an ionic assembly (salt compound) comprised of lead cations in the +2 oxidation state. With regard to the naming convention, each lead (II) cation is paired with two nitrate anions, each having a charge of -1.

Chemical nomenclature is a set of principles for naming chemical substances in a systematic manner. The International Union of Pure and Applied Chemistry designed and developed the most widely used nomenclature in the world (IUPAC).

The basic goal of chemical nomenclature is to guarantee that no ambiguity exists between a spoken or written chemical name and the chemical compound to which the name refers. Each chemical name should only relate to one substance.

It is required to indicate the charge of these cations or compounds containing these cations when identifying them. Ionic compounds are formed when cations and anions interact. The cation of an ionic compound is named first, followed by the anion. When writing their chemical formulae, they use the same format.

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Answer:

How much energy does it take to melt 50g of ice at 0°C?

The amount of energy required to melt 50g of ice at 0°C is approximately 334 Joules.

The concentration of A after 30 seconds when the given reaction obeys the rate law rate = k[A]²[B].

We use the initial concentration of A and B and the rate constant of the reaction to find the rates at these concentrations. Using the integrated rate law for a second-order reaction, we find the concentration of A after 30 seconds to be 0.0934 M.

Given reaction obeys the rate law, rate=k[A]²[B].

Here, the initial concentration of A= 0.10 M,

initial concentration of B = 0.05 M, and

rate constant, k = 2.0 × 10⁻⁴ M⁻¹s⁻¹

We have to find the concentration of A, after 30 seconds.

To find the concentration of A, we need to know the rate at 0.10 M and 0.05 M. Therefore, we have to calculate the rates at these concentrations.

rate1 = k[A]²[B]

= (2.0 × 10⁻⁴ M⁻¹s⁻¹)(0.10 M)²(0.05 M)

= 1.0 × 10⁻⁷ M/srate2

= k[A]²[B] = (2.0 × 10⁻⁴ M⁻¹s⁻¹)(0.09 M)²(0.04 M)

= 6.48 × 10⁻⁸ M/s

Using the integrated rate law for a second-order reaction: [A] = [A]₀ - kt where [A]₀ = initial concentration of A, k = rate constant, and t = time in seconds.

We know [A]₀ = 0.10 M and k = 2.0 × 10⁻⁴ M⁻¹s⁻¹.

Substituting the values in the above equation, we get: [A] = [A]₀ - kt= 0.10 M - (2.0 × 10⁻⁴ M⁻¹s⁻¹)(30 s)≈ 0.0934 M

Therefore, the concentration of A in the vessel after 30 seconds is 0.0934 M.

This question requires us to calculate the concentration of A after 30 seconds when the given reaction obeys the rate law rate = k[A]²[B].

We are given the initial concentration of A and B and the rate constant of the reaction. To find the concentration of A after 30 seconds, we need to calculate the rates at the initial concentrations of A and B.

Using the integrated rate law for a second-order reaction, we can find the concentration of A at any given time. We substitute the given values in the formula and solve for [A]. We get the concentration of A as 0.0934 M after 30 seconds. This calculation is based on the assumption that no other reaction is important.

The concentration of A after 30 seconds when the given reaction obeys the rate law rate = k[A]²[B]. We use the initial concentration of A and B and the rate constant of the reaction to find the rates at these concentrations. Using the integrated rate law for a second-order reaction, we find the concentration of A after 30 seconds to be 0.0934 M. This calculation assumes that no other reaction is important.

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To prepare 1.00 L solution of 0.1 M NaOH, weight 4 grams of NaOH and place in a 1 L volumetric flack and make it up with water to the mark.

Step 1: Determine the mole of NaOH in the 0.1 M NaOH solution. Details below:

Mole of NaOH = molarity × volume

= 0.1 × 1

= 0.1 mole

Step 2: Determine the mass of NaOH in the solution. Details below:

Mass of NaOH = Mole × molar mass

= 0.1 × 40

= 4 grams

Step 3: Weight 4 grams of NaOH and place in a 1 L volumetric flack and make it up with water to the mark.

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The power output of the turbine is 45.7 kW (approx).

The adiabatic turbine is operating with steam that has a flow rate of 1.2 kg/s, inlet pressure (Pi) of 3 MPa and inlet temperature (Ti) of 400°C.

After going through the adiabatic turbine, the pressure (P2) drops to 30 kPa.

The isentropic efficiency (ηi) of the turbine is 0.92.

We have to calculate the power output by the turbine.

Power output of the turbine can be calculated using the formula,

Power Output (W) = (m * h1 - m * h2) / ηi

Where, m = Mass flow rate of the steam

h1 = Enthalpy of steam at the inlet to the turbine

h2 = Enthalpy of steam at the outlet from the turbine

h1 is determined from the steam table at 3 MPa and 400°C.

Similarly, h2 is determined from the steam table at 30 kPa (P2).

h1 = 3249.9 kJ/kg (from steam table)

h2 = 2445.3 kJ/kg (from steam table)

Substitute these values in the formula,

Power Output (W) = (1.2 * 3249.9 - 1.2 * 2445.3) / 0.92

Power Output (W) = 45684.35 watts ≈ 45.7 kW

The power output of the turbine is 45.7 kW (approx).

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The power output of the turbine is `W = -127.92 kW` (negative sign indicates that work is done by the system).

Given,

Mass flow rate of water, `m = 1.2 kg/s`

Inlet conditions of water,

`P1 = 3 MPa` and

`T1 = 400°C`Exit pressure of water,

`P2 = 30 kPa`

Isentropic efficiency of the turbine,

`η = 0.92`

First, we need to determine the exit state of water using the given inlet and exit conditions.

The inlet state of water can be determined using the steam tables.

Using steam tables,

h1 = 3496.6 kJ/kg` (Enthalpy of saturated water at 3 MPa)

Using the same steam tables, the entropy of water at 3 MPa and 400°C is

`s1 = 6.8466 kJ/kgK`.At 30 kPa,

the entropy of water is

`s2s = 7.0305 kJ/kgK`.

The isentropic enthalpy of exit state can be determined using the inlet entropy and exit pressure.

`s2s = s1 = 6.8466 kJ/kgK`

`h2s = 249.5 kJ/kg` (enthalpy of water at 30 kPa and 400°C using steam tables).

Now, we can determine the actual exit state of water using the given isentropic efficiency.

`η = (h1 - h2)/(h1 - h2s)`

where,

`h2` = actual exit enthalpy of water.

Substituting the values,`0.92 = (3496.6 - h2)/(3496.6 - 249.5)

`Solving for `h2`, we get `h2 = 3600.8 kJ/kg

`The power output of the turbine is given by the expression:

Power output = Mass flow rate * (Enthalpy drop across the turbine)

W = m * (h1 - h2)`W = 1.2 kg/s * (3496.6 - 3600.8) kJ/kg

`Therefore, the power output of the turbine is `W = -127.92 kW` (negative sign indicates that work is done by the system).

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An insulated cup contains 75 g of water at 24 °C. A 26 g sample of metal at 82.25°C is added. The final temperature of the water and metal is 28.34°C. The specific heat of the metal is 0.972 J/ g°C.

Given that :

Mass of the metal = 26 g

The specific heat of the metal = ?

Initial temperature of the metal = 82.25°C

Equilibrium temperature = 28.34°C

Mass of the water = 75 g

Specific heat of water = 4.18 J/ g°C

Initial temperature of the water = 24 °C

The specific heat capacity is given as :

26 × c × ( 82.25 - 28.34) = 75 × 4.18 ( 28.34 - 24)

26 × c × 53.91 = 313.5  × 4.34

c = 0.972 J/ g°C

Thus, the specific heat capacity of metal is 0.972 J/ g°C.

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Answer:

independent variable is how it changes how it affects something else dependent variable is something being measured

Answer: An independent variable is the variable which is changed by the scientist also called the manipulated variable. The dependent variable is the result of the action of the independent variable. it is the factor that changes due to the independent variable. It is also known as the responding variable.

Explanation

To transport even more calcium to the inside of the cell, the cell can use active transport mechanisms such as calcium pumps or exchangers. These proteins can actively transport calcium ions against their concentration gradient, allowing the cell to accumulate more calcium from the extracellular environment.

Additionally, the cell can also increase the expression or activity of these transporters to facilitate more calcium uptake.Dynamic vehicle is a vehicle component of particles across the layer from a district of lower fixation to one of higher focus (against the fixation inclination) by using energy (frequently ATP). The transport is sometimes driven by an electrochemical gradient.

The human body absorbs glucose in the intestine and plants absorb minerals or ions into their root hair cells are two examples of active transport.

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Each species contains a large number of valence electrons. N2 = 16; NO = 13; HBr = 10; NO+ = 13.

a. N2: Nitrogen is in group 15 of the periodic table, so it has 5 valence electrons. N2 has a triple bond between the two nitrogen atoms, so there are 6 electrons shared between them. Therefore, the total number of valence electrons in N2 is 5 x 2 + 6 = 16.

b. NO: Nitrogen is in group 15 of the periodic table, so it has 5 valence electrons. Oxygen is in group 16 of the periodic table, so it has 6 valence electrons. NO has a single bond between nitrogen and oxygen, so there is one electron pair shared between them. Therefore, the total number of valence electrons in NO is 5 + 6 + 2 = 13.

c. HBr: Hydrogen is in group 1 of the periodic table, so it has 1 valence electron. Bromine is in group 17 of the periodic table, so it has 7 valence electrons. HBr has a single covalent bond between hydrogen and bromine, so there is one electron pair shared between them. Therefore, the total number of valence electrons in HBr is 1 + 7 + 2 = 10.

d. NO+: Nitrogen is in group 15 of the periodic table, so it has 5 valence electrons. Oxygen is in group 16 of the periodic table, so it has 6 valence electrons. NO+ has a single bond between nitrogen and oxygen, and a positive charge on the nitrogen atom. Therefore, the nitrogen atom has lost one electron and has 4 valence electrons. The oxygen atom has gained one electron and has 7 valence electrons. There is one electron pair shared between nitrogen and oxygen. Therefore, the total number of valence electrons in NO+ is 4 + 7 + 2 = 13.

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Two electrons are lost by Mg and gained by N

Electrons can be lost or gained in a chemical reaction. In fact, the process of gaining or losing electrons is central to many types of chemical reactions.

Chemical reactions involve the transfer of electrons between atoms or molecules, which can result in the formation of new compounds or the breakdown of existing ones. When an atom or molecule gains one or more electrons, it becomes negatively charged and is said to be reduced. Conversely, when an atom or molecule loses one or more electrons, it becomes positively charged and is said to be oxidized.

We can see that we can write this reaction as;

3Mg^0 + N2^0  ---->3Mg^+ + N^2-

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The charge concentration due to the positive charge carriers (Na+ ions) in the saline solution is 14,840 coulombs per cubic meter.

A saline solution for medical use contains 9.0 g of sodium chloride (NaCl) dissolved in 1.0 L of water. To determine the charge concentration due to the positive charge carriers (Na+ ions) in the solution, follow these steps:
1. Determine the number of moles of Na+ ions in the solution:
Moles = mass / molar mass = 9.0 g / (58.44 g/mol) = 0.154 moles
2. Calculate the number of Na+ ions:
Number of ions = moles × Avogadro's number = 0.154 moles × (6.022 × 10^23 ions/mol) = 9.27 × 10^22 ions
3. Calculate the total positive charge in the solution:
Total charge = number of ions × charge per ion = 9.27 × 10^22 ions × (1.602 × 10^-19 C/ion) = 14.84 C
4. Convert the volume of the solution to cubic meters:
1.0 L = 0.001 m^3
5. Determine the charge concentration in C/m^3:
Charge concentration = total charge / volume = 14.84 C / 0.001 m^3 = 14,840 C/m^3
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The molecular formula of the Allicin is, C₁₈H₃₀S₆O₃ and the empirical formula is C₆H₁₀S₂O

the molecular formula can be calculated as follows:

If a percentage is provided, we will assume that the entire mass is 100 grams.

As a result, each element's mass is equal to the percentage indicated.

C weighs 44.4 g.

H weighs 6.21 g.

S weighs 39.5 g.

O's mass is 9.86 g.

C has a molar mass of 12 g/mole.

H has a molar mass of 1 g/mole.

S has a molar mass of 32 g/mole.

O has a molar mass of 16 g/mole.

first, convert given masses into moles.

Moles of C =      mass C       =    44.4 g      = 3.7 moles

                    molar mass C       12 g/mole.

Moles of H =      mass H       =    6.21 g.      = 6.21 moles

                      molar mass           1 g/mole.

Moles of S =      mass S       =    39.5 g.      = 1.23 moles

                      molar mass          32 g/mole.

Moles of O =        mass O       =    9.86 g.      = 0.62 moles

                      molar mass          16 g/mole.

then we find the mole ratio by divide each value of moles by the smallest number of moles calculated.

For C = 3.7 moles = 5.96 ≈ 6

           0.62 moles

For H = 6.21  moles = 10.01≈ 10

           0.62 moles

For S = 1.23 moles = 1.98 ≈ 2

           0.62 moles

For O = 0.62 moles = 1

             0.62 moles

The ratio of C : H : S : O = 6 : 10 : 2 : 1

The Empirical formula is C₆H₁₀S₂O

The empirical formula weight = 6(12) + 10(1) + 2(32) + 1(16) = 162 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

n= molecular formula = 486 =3

    empirical formula     162

Molecular formula = (C₆H₁₀S₂O)₃ =C₁₈H₃₀S₆O₃

Therefore, the molecular of the compound is, C₁₈H₃₀S₆O₃

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