A reaction requires 22.4l of at stp. You have 32.0l of gas at 398k and 105.6 kpa. Will you have enough gas to carry out the reaction? True or false

The ion-molecular attraction between NaCl and water has been Na attracted with the oxygen of water, while Cl has been attracted with the hydrogen of water.

The ion has been the charged molecules that has been formed with the loss or gain of the electrons, while the molecules has been the molecules with the bonded anions and cations.

The water has been the polar molecule, with the development of partial positive charge over hydrogen atoms, and partial negative charge over the oxygen atom.

The like charges repel each other, while unlike charges attract each other.

The addition of NaCl to water results in the formation of positive Na ions, and negative Cl ions.

The interaction is demonstrated in the image attached.

Thus, the molecular attraction between NaCl and water has been Na has been attracted with the oxygen of water while Cl has been attracted with the hydrogen of water.

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Answer:

4.245 × 10²¹ cubic centimetre of Fe

Explanation:

First step is to find the number of atoms that is present in Fe(Iron)

The atomic mass of Fe(iron) is  55.845

We  have to convert  the atomic weight into grams

1 atomic mass = 1.66 x 10⁻²⁴grams

55.845(atomic mass of Fe) =

55.845 x 1.66 x 10 ⁻²⁴ grams = 9.27 x 10²² grams.

Therefore, the number of atoms in a cubic centimetre of Fe (Iron) =

Density of Fe(solid) ÷ number of grams of Fe

Density of Fe (solid) is known as = 7.874g/cm³

The number of atoms in a cubic centimetre of Fe (Iron =

7.874g/cm³ ÷9.27 x 10²²grams

= 8.494 × 10²¹atoms.

In the question we are told Fe adopted a body centered cubic unit cell

Hence , in Body centered cubic unit cell, we have:

We have one atom at the 8 corners of a cube

We also have one body atom the cube's center

8 corners × 1/8 per corner atom = 8 × 1/8 = 1 atom

(8 corners × 1/8 per corner atom) + (1× 1) = 2 atoms

Therefore, the total number of featoms present per unit cell = 2 atoms.

The number of unit cells are present per cubic centimeter of Fe =

Number of Fe atoms per cubic centimeter ÷ Number of Fe per unit cell

= 8.494 × 10²¹ atoms ÷ 2 atoms.

= 4.247 × 10²¹ cubic centimetre of Fe

Hence the number of unit cells that are present per cubic centimeter of Fe is

4.247 × 10²¹ cubic centimetre of Fe.

Answer: Option (a) is the correct answer.

Explanation:

An equation can only be balanced when number of reactants equal the number of products.

The given equation is as follows.

   [tex]Ca(OH)_{2} + H_{3}PO_{4} \rightarrow Ca_{3}(PO_{4})_{2} + H_{2}O[/tex]

The number of reactant atoms are as follows.

The number of product atoms are as follows.

To balance the equation, multiply [tex]Ca(OH)_{2}[/tex] by 3, and [tex]H_{3}PO_{4}[/tex] by 2 on the reactant side. Multiply [tex]H_{2}O[/tex] by 6 on the product side. Therefore, the equation will be as follows.

[tex]3Ca(OH)_{2} + 2H_{3}PO_{4} \rightarrow Ca_{3}(PO_{4})_{2} + 6H_{2}O[/tex]

Thus, we can conclude that the coefficient in front of [tex]Ca_{3}(PO_{4})_{2}[/tex] is 1.

Answer:

3.179 atm

Explanation:

Osmotic pressure is expressed as:

π = iMRT , Where

π = osmotic pressure

i = van't Hoff factor  

M = molar concentration

R = universal gas constant (0.08206 L·atm/mol·K )

T = absolute temperature in K

In this case, i = 2, M = 0.065, R = 0.08206 and T = 298. Substitute into the formula:

π = 2 x 0.065 x 0.08206 x 298

   = 3.179 atm

The osmotic pressure is 3.179 atm.

To produce a solution that is 35.0 wt% SrCl2 with 36.0 grams of SrCl2, one would need to add 66.9 grams of water. This is because the total weight of the solution is the weight of SrCl2 plus the weight of the water needed.

In order to determine how much water must be added to 36.0 g of SrCl2 to produce a solution that is 35.0 wt% SrCl2, we must understand that the 35% by weight represents that 35 g of SrCl2 is in 100 g of the solution. Thus, to find the total weight of the water and the SrCl2, we should set up the equation 36g (weight of SrCl2) / X g (total weight of solution) = 35%, which would give that X = 102.9 g. The total solution weight is the weight of the SrCl2 plus the weight of the water needed, so to find the weight of the water we subtract the weight of SrCl2 (36 g) from the total weight of solution (102.9 g), which gives us 66.9 g as the amount of water to be added to the SrCl2.

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number of moles of SCN⁻ = 0 .001 X 3 / 1000 = 3 X 10⁻⁶ moles    in the 25 ml of solution

2) SCN⁻ + Fe³⁺ ===> FeNCS²⁺

1 mole SCN⁻ produces 1 mole of FeNCS²⁺

Therefore moles of FeNCS²⁺ in 25 ml = 3 X 10⁻⁶

Molar Concentration of FeNCS²⁺ = (3 X10⁻⁶ )X 1000 / 25 = 1.2 X 10⁻⁴ Moles / liter

The molar concentration of Fe(SCN)²+ is 3.52 × 10⁻² M, assuming all SCN⁻ is complexed with Fe³+. Thus, the concentration of Fe(SCN)²+ is equal to 3.52 × 10⁻² M.

The molar concentration of Fe(SCN)²+, assume that all of the SCN⁻ is complexed with Fe³+ to form Fe(SCN)²+. The concentration of Fe(SCN)²+ after 10 s is 3.52 × 10⁻² M under conditions of excess Fe³+.

Therefore, the molar concentration of Fe(SCN)²+ is 3.52 × 10⁻² M. The reaction is stoichiometric, meaning the initial concentration of SCN⁻ is converted entirely to Fe(SCN)²+.

The energy required to heat the silver cube from 15°C to 32°C can be calculated via the formula for heat transfer. The mass of the silver cube is calculated to be 210g using the given volume and density.

The energy required to heat a substance is calculated using the formula q = mcΔT, where 'q' is the heat energy, 'm' is the mass, 'c' is the specific heat capacity and 'ΔT' is the change in temperature.

The volume of the silver cube is given as 20.0 cm3. The density of silver is given as 10.5 g/cm3. Therefore, you can calculate the mass of the silver cube by multiplying the volume with the density, that is m = volume x density = 20.0 cm3 x 10.5 g/cm3 = 210 g.

The specific heat capacity for silver, c, is given to be 0.235 J/g°C. The change in temperature, ΔT, is final temperature - initial temperature = 32°C - 15°C = 17°C.

Substituting the calculated and given values into the formula, we get q = (210 g)(0.235 J/g°C)(17°C) = 826.95 joules. Therefore, the energy required to heat the cube of silver from 15°C to 32°C is about 828 Joules.

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The quantum numbers (4, 2, 1, +½) and (4, 1, 1, +½) represent two different electrons in an atom.

The first quantum number, n, represents the principal energy level, which is 4 for both electrons.

The second quantum number, l, refers to the type of subshell, where l = 2 corresponds to the d subshell and l = 1 corresponds to the p subshell.

The third quantum number, m₁, specifies the specific orbital within the subshell. For the electron (4, 2, 1, +½), m₁ = 1, indicating a specific orbital within the d subshell. For the electron (4, 1, 1, +½), m₁ = 1, indicating a specific orbital within the p subshell.

The fourth quantum number, m², represents the spin of the electron. In both cases, it is +½, indicating a spin up orientation for both electrons.

The quantum numbers (4, 2, 1, +½) and (4, 1, 1, +½) represent two different electrons in an atom. The first quantum number, n, represents the principal energy level, which is 4 for both electrons. The second quantum number, l, refers to the type of subshell, where l = 2 corresponds to the d subshell and l = 1 corresponds to the p subshell.

The third quantum number, m₁, specifies the specific orbital within the subshell. For the electron (4, 2, 1, +½), m₁ = 1, indicating a specific orbital within the d subshell. For the electron (4, 1, 1, +½), m₁ = 1, indicating a specific orbital within the p subshell.

The fourth quantum number, m², represents the spin of the electron. In both cases, it is +½, indicating a spin up orientation for both electrons.

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At the final equilibrium temperature, the specific heat capacity of the metal is 5.999 J/g°C.

Given the following data:

To find the specific heat capacity of the metal:

Mathematically, quantity of heat is given by the formula;

[tex]Q = mc\theta[/tex]

Where:

The quantity of heat lost by the water = The quantity of heat gained by the metal.

[tex]Q_{lost} = Q_{gained}\\\\mc\theta = mc\theta\\\\20(4.18)(29.2 - 22.1) = 5.10c(48.6 - 29.2)\\\\83.6(7.1) = 5.10c(19.4)\\\\593.56 = 98.94c\\\\c = \frac{593.56}{98.94}[/tex]

Specific heat capacity of metal, c = 5.999 J/g°C

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You need to add 4.043 grams of anhydrous calcium bromide (CaBr2) to make a 0.202 M aqueous solution in a 125 mL volumetric flask.

To calculate the mass of CaBr2 needed, use the formula molarity  imes volume (in liters)  imes molar mass. The molarity is given as 0.202 M, and the volume needs to be converted to liters (0.125 L). The molar mass of CaBr2 is approximately 199.9 g/mol. Multiplying these values gives us the mass in grams:

0.202 mol/L  imes 0.125 L  imes 199.9 g/mol = 5.0475 g

However, since mass must be measured accurately in the laboratory, this final result must be weighed correctly using a balance and all the necessary safety precautions must be observed when handling calcium bromide.

Answer: The pressure under the water is 7.29 atm

Explanation:

We are given:

Pressure under the water = 739 kPa

To convert into atm, we use the conversion factor:

1 atm = 101.325 kPa

Converting the pressure from kilo pascals to atmospheres, we get:

[tex]\Rightarrow 739kPa\times \frac{1atm}{101.325kPa}=7.29atm[/tex]

Hence, the pressure under the water is 7.29 atm

Answer: the ionic compound calcium fluoride [tex](CaF_2)[/tex]

Explanation:

[tex]\Delta T_f=i\times k_f\times m[/tex]

[tex]T_f[/tex] = change in freezing point

i = Van'T Hoff factor

[tex]k_f[/tex] = freezing point constant

m = molality

1. For [tex]C_{12}H_{22}O_{11}[/tex] , i= 1 as it is a non electrolyte and does not dissociate.

2. For [tex]MgSO_{4}[/tex] , i= 2 as it is a electrolyte and dissociate to give 2 ions.

[tex]MgSO_4\rightarrow Mg^{2+}+SO_4^{2-}[/tex]

3. For [tex]LiCl[/tex], i= 2 as it is a electrolyte and dissociate to give 2 ions.

[tex]LiCl\rightarrow Li^{+}+Cl^{-}[/tex]

4. For [tex]CaF_{2}[/tex], i= 3 as it is a electrolyte and dissociate to give 3 ions.

[tex]CaF_2\rightarrow Ca^{2+}+2F^{-}[/tex]

Thus as vant hoff factor is highest factor for [tex]CaF_{2}[/tex] and the freezing point will be lowest.

Answer: The correct answer is Option D.

Explanation:

A balanced equation follows Law of conservation of mass. This law states that mass can neither be created nor be destroyed but it can only be transformed from one form to another. In a chemical reaction, mass is always conserved.

This also states that the total number of individual atoms on the reactants side must be equal to the total number of individual atoms on the product side.

For the given chemical reaction:

[tex]2CH_4+4O_2\rightarrow CO_2+4H_2O[/tex]

On the reactant side:

Number of Carbon atoms = 2

Number of Hydrogen atoms = 8

Number of Oxygen atoms = 8

On the product side:

Number of Carbon atoms = 1

Number of Hydrogen atoms = 8

Number of Oxygen atoms = 4

In order to balance the number of atoms, 2 must be added infront of [tex]CO_2[/tex] molecule, in order to balance the equation.

Hence, the correct answer is Option D.

The concentration of hydrofluoric acid in the solution, calculated using the given pH and the acid dissociation constant Ka, is about 0.069 M.

The problem is asking us to find the concentration of hydrofluoric acid in a solution where pH and Ka are given. We can use the following formula that relates pH, Ka, and the concentration of the acid [Ha]:
pH = -log([H+]), and since for weak acids [H+] ~= sqrt(Ka × [Ha]), we can substitute and solve for [Ha].
Thus, [Ha] = ((10^-pH)²) / Ka = ((10⁻³)²) / 6.8×10⁻⁴ = 0.069 M. So, the concentration of hydrofluoric acid in the solution is approximately 0.069 M.

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To find the total concentration of the hydrofluoric acid, we use the dissociation constant and the pH to determine the hydronium ion concentration. We then assume ionic product equals hydronium ion concentration. This allows us to calculate the initial concentration of the hydrofluoric acid as approximately 3.16 M.

The question deals with hydrofluoric acid (HF) whose dissociation constant, Ka is given to be 6.8 x 10^-4. The pH of the solution is given to be 3.67. First, we need to find the concentration of hydronium ions: [H3O+] from the given pH using the relationship pH = -log[H3O+], which gives us [H3O+] = 10^-3.67 = 2.15 x 10^-4 M.

Next, we use the formula for the dissociation constant Ka of an acid, which is [H3O+][F-]/[HF]. Here [H3O+] = 2.15 x 10^-4 and [F-] = x, where x is the concentration of the anion produced. [HF] is the initial concentration of HF which we are trying to find. By assuming that x is much smaller than [HF] and hence can be neglected in the denominator, we set [H3O+] = [F-] = x = 2.15 x 10^-4. Rearranging the equation, we find that [HF] = Ka/x = (6.8 x 10^-4) / (2.15 x 10^-4) = 3.16 M.

Thus, the total concentration of hydrofluoric acid in the solution is approximately 3.16 M.

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Answer:

After the well-known solids, liquids, and gases, plasmas are considered as the fourth state of matter. They are found rarely on Earth, however, they are found in enormous quantity all through the universe. As they comprise free-flowing charged particles, plasmas exhibit many specific features.  

In the majority of the plasmas, the electrons and protons take place in equal numbers, forming it electrically neutral. As they flow liberally, they are influenced by magnetic and electric fields in the manner not witnessed in the other forms of matter. These fields can affect plasmas over higher distances, warping, pinching, and modeling them, like the twisting flares observed on the Sun's surface.  

The oxidation numbers of nitrogen in NH₄⁺, NO₂, and NaNO₃  are -3, +4, and +5, as determined by balancing the charges of other atoms in the compounds.

The oxidation numbers of nitrogen in NH₄⁺, NO₂, and NaNO₃ can be calculated using some simple rules. In NH₄⁺, nitrogen must balance the charge from the four hydrogen atoms, each having an oxidation number of +1, leading to the oxidation number of nitrogen being -3. For NO₂, oxygen typically has an oxidation number of -2, with two oxygen atoms that would contribute an overall charge of -4, thus nitrogen must have an oxidation number of +4 to balance out and make the molecule neutral. Finally, in NaNO₃, with sodium having an oxidation number of +1, and three oxygen atoms giving a total of -6, nitrogen must have an oxidation number of +5 to balance the overall charge to zero due to the monatomic sodium ion.

Therefore, the respective oxidation numbers of nitrogen in NH₄⁺, NO₂, and NaNO₃ are -3, +4, and +5.