Phosphorous can form an ion called phosphide, which has the formula P3−.
This ion ______.

A. contains 18 electrons
B. forms when a phosphorus atom loses three protons
C. has properties very similar of P
D. is called a cation

Answer : The one part nitrogen gas combines with one part oxygen gas to form one part of dinitrogen monoxide.

Explanation :

Balanced chemical reaction : It is defined as the reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.

When nitrogen gas combines with oxygen gas then it react to give dinitrogen monoxide or nitrous oxide.

The balance chemical reaction will be:

[tex]2N_2(g)+O_2(g)\rightarrow 2N_2O(g)[/tex]

By the stoichiometry we can say that, 2 parts of nitrogen gas combines with 1 part of oxygen gas to give 2 parts of dinitrogen monoxide or nitrous oxide.

First we have to determine the limiting reagent.

From the reaction we conclude that,

As, 2 moles of nitrogen gas combine with 1 mole of oxygen gas

So, 1 moles of nitrogen gas combine with 0.5 mole of oxygen gas

It means that, oxygen gas is an excess reagent because the given moles are greater than the required moles and nitrogen gas is a limiting reagent and it limits the formation of product.

Now we have to determine the moles of dinitrogen monoxide.

As, 2 moles of nitrogen gas combine to give 2 mole of dinitrogen monoxide

So, 1 mole of nitrogen gas combine to give 1 mole of dinitrogen monoxide

Thus, the one part nitrogen gas combines with one part oxygen gas to form one part of dinitrogen monoxide.

Final answer:

One part nitrogen gas (N₂) combines with one part oxygen gas (O₂) to produce two parts of nitric oxide (NO).

Explanation:

When nitrogen gas (N₂) combines with oxygen gas (O₂), they can form several different nitrogen oxides, depending on the conditions and proportions in which they react. For the specific formation of nitric oxide (NO), we need to understand that one volume of nitrogen gas will combine with one volume of oxygen gas to form two volumes of nitric oxide. This is evident from the balanced chemical equation for this reaction:

N₂(g) + O₂(g) → 2NO(g)

This means 1 part of nitrogen gas combines with 1 part of oxygen gas to form 2 parts of nitric oxide (NO). The balancing of the equation indicates that two molecules of NO are formed from one molecule of nitrogen and one molecule of oxygen, due to the conservation of mass and volume in chemical reactions according to the Law of Combining Volumes.

The question is incomplete, here is the complete question:

A chemist adds 345.0 mL of a 0.0013 mM (MIllimolar) copper(II) fluoride [tex]CuF_2[/tex] solution to a reaction flask.

Calculate the mass in micrograms of copper(II) fluoride the chemist has added to the flask. Be sure your answer has the correct number of significant digits.

Answer: The mass of copper (II) fluoride is 0.13 mg

Explanation:

We are given:

Millimolarity of copper (II) fluoride = 0.0013 mM

This means that 0.0013 millimoles of copper (II) fluoride is present in 1 L of solution

Converting millimoles into moles, we use the conversion factor:

1 moles = 1000 millimoles

So, [tex]0.0013mmol\times \frac{1mol}{1000mmol}=1.3\times 10^{-6}mol[/tex]

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

We are given:

Moles of copper (II) fluoride solution = [tex]1.3\times 10^{-6}mol[/tex]

Molar mass of copper (II) fluoride = 101.5 g/mol

Putting values in above equation, we get:

[tex]1.3\times 10^{-6}mol=\frac{\text{Mass of copper (II) fluoride}}{101.5g/mol}\\\\\text{Mass of copper (II) fluoride}=(1.3\times 10^{-6}mol\times 101.5g/mol)=1.32\times 10^{-4}g[/tex]

Converting this into milligrams, we use the conversion factor:

1 g = 1000 mg

So,

[tex]\Rightarrow 1.32\times 10^{-4}g\times (\frac{1000mg}{1g})=0.13mg[/tex]

Hence, the mass of copper (II) fluoride is 0.13 mg

Answer:

The mass of copper(II) fluoride is 45.54 micrograms

Explanation:

A chemist adds 345.0 mL  of a 0.0013 mM copper(II) fluoride solution to a reaction flask. Calculate the mass in micrograms of copper(II) fluoride the chemist has added to the flask. Round your answer to significant digits.

Step 1: Data given

Molarity of the copper(II) fluoride solution = 0.0013 mM =

Volume of the solution 345.0 mL = 0.345 L

Molar mass copper(II) fluoride = 101.54 g/mol

Step 2: Calculate moles of copper(II) fluoride

Moles CuF2 = molarity * volume

Moles CuF2 = 0.0000013 M * 0.345 L

Moles CuF2 = 0.0000004485 moles

Step 3: Calculate mass of  CuF2

Mass CuF2 = moles * Molar mass

Mass CuF2 = 0.0000004485 moles * 101.54 g/mol

Mass CuF2 = 0.00004554 grams = 0.04554 miligrams = 45.54 micrograms

The mass of copper(II) fluoride is 45.54 micrograms

Answer:

10.6%

Explanation:

The determined percent mass of water can be calculated from the formula of the hydrate by  

dividing the mass of water in one mole of the hydrate by the molar mass of the hydrate and  

multiplying this fraction by 100.

Manganese(ii) sulphate monohydrate is MnSO4 . H2O

1. Calculate the formula mass. When determining the formula mass for a hydrate, the waters of  

hydration must be included.

1 Manganes  52.94 g = 63.55 g  

1 Sulphur  32.07 g =  

32.07 g 2 Hydrogen is  = 2.02 g

4 Oygen       =  

64.00 g 1 Oxygen 16.00 = 16.00 g

151.01 g/mol  18.02 g/mol

Formula Mass = 151.01 + (18.02) = 169.03 g/mol

2. Divide the mass of water in one mole of the hydrate by the molar mass of the hydrate and  

multiply this fraction by 100.

Percent hydration = (18.02 g /169.03 g) x (100) = 10.6%

The final result is 10.6% after the two steps calculations

The mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.

The percentage mass is the ratio of the mass of the element or molecule in the given compound.

The percentage can be given as:

[tex]\text{Percent Mass} = \frac{\text{Mass of molecule}}{\text{total mass of compound}} \times 100 \%[/tex]

The mass of the water is 18.02 g/mol and the molar mass of hydrated magnesium sulfate (MnSO4 . H2O) is 169.03 g/mol.

Thus,

[tex]\text{Percent Mass} = \frac{\text{18.02}}{\text{169.03 }} \times 100 \%\\\\\text{Percent Mass} = 10.6 \%}[/tex]

Therefore, the mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.

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Answer: option B and option C

Explanation:

Explanation:

For the given reaction, the temperature of liquid will rise from 298 K to 339 K. Hence, heat energy required will be calculated as follows.

             [tex]Q_{1} = mC_{1} \Delta T_{1}[/tex]

Putting the given values into the above equation as follows.

           [tex]Q_{1} = mC_{1} \Delta T_{1}[/tex]

                      = [tex]27.3 g \times 1.70 J/g K \times 41[/tex]

                      = 1902.81 J

Now, conversion of liquid to vapor at the boiling point (339 K) is calculated as follows.

           [tex]Q_{2}[/tex] = energy required = [tex]mL_{v}[/tex]

    [tex]L_{v}[/tex] = latent heat of vaporization

Therefor, calculate the value of energy required as follows.

             [tex]Q_{2}[/tex] = [tex]mL_{v}[/tex]

                         = [tex]27.3 \times 444[/tex]

                         = 12121.2 J

Therefore, rise in temperature of vapor from 339 K to 373 K is calculated as follows.

            [tex]Q_{3} = mC_{2} \Delta T_{2}[/tex]

Value of [tex]C_{2}[/tex] = 1.06 J/g,    [tex]\Delta T_{2}[/tex] = (373 -339) K = 34 K

Hence, putting the given values into the above formula as follows.

             [tex]Q_{3} = mC_{2} \Delta T_{2}[/tex]

                       = [tex]27.3 g \times 1.06 J/g \times 34 K[/tex]

                       = 983.892 J

Therefore, net heat required will be calculated as follows.

            Q = [tex]Q_{1} + Q_{2} + Q_{3}[/tex]

                = 1902.81 J + 12121.2 J + 983.892 J

                = 15007.902 J

Thus, we can conclude that total energy (q) required to convert 27.3 g of THF at 298 K to a vapor at 373 K is 15007.902 J.

Explanation:

A)

We know, each mole contains [tex]N_A=[/tex] [tex]6.023 \times 10^{23}[/tex] atoms.

It is given that mass of one oxygen atom is m= [tex]2.66\times 10^{-23}\ g[/tex].

Therefore, mass of one mole of oxygen, [tex]M=m\times N_A[/tex].

Putting value of n and [tex]N_A[/tex],

[tex]M=2.66\times 10^{-23}\times 6.023\times 10^{23} \ gm\\M=16.0\ gm[/tex]

B)

Given,

Mass of water in glass=0.050 kg = 50 gm.

From above part mass of one mole of oxygen atoms = 16.0 gm.

Therefore, number of mole of oxygen equivalent to 50 gm oxygen[tex]=\dfrac{50}{16}=3.1 \ moles.[/tex]

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The mass of 1 mole of oxygen atoms is 16.00 grams. There are 3125 moles of oxygen atoms with a mass equal to that of a glass of water.

To find the mass of 1 mole of oxygen atoms, we need to use the molar mass of oxygen. The molar mass is the mass of one mole of a substance, and it is equal to the atomic mass in grams. The atomic mass of oxygen is 16.00 g/mol, so the mass of 1 mole of oxygen atoms is 16.00 grams.

To determine how many moles of oxygen atoms have a mass equal to the mass of a glass of water, we need to use the given mass of the water and the molar mass of oxygen. The molar mass of oxygen is 16.00 g/mol, and the mass of a glass of water is 0.050 kg (or 50,000 grams). Using the molar mass, we can set up a proportion to find the number of moles of oxygen atoms.

Moles of oxygen atoms = (Mass of water / Molar mass of oxygen) = (50,000 g / 16.00 g/mol) = 3125 moles of oxygen atoms.

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Answer:

Basic

Explanation:

[tex]Sodium\ Fluoride\ +\ Water = Hydrogen\ Fluoride +\ Sodium\ Hydroxide[/tex]  

[tex]NaF + H_{2}O\rightarrow HF + NaOH[/tex]

Sodium fluoride, NaF, is a soluble salt that dissociates completely in aqueous solution to give sodium cations, Na+, and fluoride anions, F-

[tex]NaF\rightarrow Na^{+} +F^{-}[/tex]

and when it dissolve in water the pH of the solution becomes greater than seven thereby becoming basic.

Final answer:

After sodium fluoride is dissolved in pure water, the solution becomes slightly basic due to the hydrolysis of fluoride ions which generates hydroxide ions.

Explanation:

When sodium fluoride is added to pure water and stirred until it dissolves, fluoride ions (F⁻) are released into the solution. These ions are capable of reacting, to a small extent, with water in a process known as hydrolysis. During this reaction, fluoride ions accept a proton from the water molecules, resulting in the formation of hydrofluoric acid (HF) and hydroxide ions (OH⁻). Since hydroxide ions increase the pH level of the solution, the new solution becomes slightly basic compared to pure water. Therefore, the corrected statement is: Sodium fluoride is added to pure water and stirred to dissolve. Compared to pure water, the new solution is slightly basic.

Answer: No hydrogen bond cannot occur in this alpha helix structure.

Explanation: For hydrogen bond to form, the electronegativity difference should be more than 1.7. carbon has an electronegativity of 2.5 whereas hydrogen has 2.1 so their electronegativity difference is 0.3. So in this alpha helix structure cannot occur.

The question is incomplete, the complete question is:

For a certain chemical reaction, the standard Gibbs free energy of reaction is 144. kJ. Calculate the temperature at which the equilibrium constant K = [tex]5.9\times 10^{-26}[/tex] .

Round your answer to the nearest degree.

Answer:

25°C is the temperature at which the equilibrium constant is [tex]5.9\times 10^{-26}[/tex].

Explanation:

[tex]\Delta G^o=-RT\ln K[/tex]

where,

[tex]\Delta G^o[/tex] = standard Gibbs free energy = 144.0 kJ=144,000 J  (Conversion factor: 1kJ = 1000J)

R = Gas constant = [tex]8.314 J/K mol[/tex]

T = temperature at which reaction is occurring = ?

K = Equilibrium constant of the reaction =[tex]5.9\times 10^{-26}[/tex]

Putting values in above equation, we get:

[tex]144,000 J/mol=-(8.3145J/Kmol)\times T\times \ln [5.9\times 10^{-26}][/tex]

[tex]T=\frac{144,000 J/mol}{-(8.314 J/Kmol)\times \ln [5.9\times 10^{-26}]}[/tex]

T = 298.15 K

T = 298.15 - 273 °C = 25°C

25°C is the temperature at which the equilibrium constant is [tex]5.9\times 10^{-26}[/tex].

Using the equation ΔG° = -RTlnK, which relates the Gibbs free energy change to the equilibrium constant, you can solve for temperature by rearranging the formula. Ensure that the units for Gibbs energy and the gas constant match. Plugging the given values into the rearranged formula will provide the temperature.

The student's question asked how to calculate the temperature at which the equilibrium constant K equals 5.9 x 10 with a standard Gibbs free energy of reaction of 144 kJ.

The relationship between the Gibbs free energy change and the equilibrium constant is defined by the equation ΔG° = -RTlnK, where R is the gas constant (8.314 J/K mol), T is the absolute temperature in Kelvin, and K is the equilibrium constant.

To solve for temperature, rearrange the equation as T = -ΔG / (R * lnK). But first, you must ensure that the units for Gibbs energy and the gas constant match. If ΔG is given in kJ, it should be converted to J (1 kJ = 1000 J).

So, T = -(144,000 J) / (8.314 J/K mol * ln(5.9 x 10)), which should give you the answer.

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Answer: the new volume will be 0.245L

Explanation:Please see attachment for explanation

Answer:

The volume will be 238 mL or 0.238 L

Explanation:

Step 1: Data given

Pressure = constant = 5.80 atm

The initial moles of helium = 0.04 moles

Temperature = 240.00 K

Volume = 0.14L

The number of moles increases to 0.07 moles

Step 2: Calculate the new volume

p*V = n*R*T

V = (n*R*T)/p

⇒ with n = the number of moles = 0.07 moles

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 240.00 K

⇒ with p = the pressure = 5.80 atm

V = (0.07 * 0.08206 * 240.00) / 5.80

V = 0.238 L = 238 mL

The volume will be 238 mL or 0.238 L

Molality  is the measure of concentration of solute in 1 kg of solution. The molality of the solution is 3.05 mol/kg.

10% of HCl (by mass) means 10 g of HCl and in 90 g of water.

Molar mass of HCl = 36.5 g/mol

Molality:

It is the measure of concentration of solute in 1 kg of solution. It can be calculated by the formula.

[tex]\bold {m = \dfrac {n }{w}\times 1000}[/tex]

Where,

m- molality

n - number of moles

w - weight of solvent in grams

Number of moles of HCl

[tex]\bold {n = \dfrac w{m} = \dfrac {10}{36.5} = 0.274 g}[/tex]

put the value in molality formula,

[tex]\bold {m = \dfrac {0.274 }{90}\times 1000}\\\\\bold {m = 3.05\ g/mol}[/tex]

Therefore, the molality of the solution is 3.05 mol/kg.

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Final answer:

To calculate the molarity of a 10.0% HCl solution, convert 10 grams of HCl to moles, and then divide by the volume of the solution in liters. This results in a molarity of 2.74 M.

Explanation:

To calculate the molarity of a 10.0% (by mass) aqueous solution of hydrochloric acid (HCl), start by understanding that a 10.0% solution means there are 10 grams of HCl in 100 grams of the solution. First, since the solution is aqueous, we can assume the density is close to that of water, which is approximately 1 g/mL, so 100 grams of the solution is roughly equivalent to 100 mL (0.1 L) of solution.

Next, convert the mass of HCl to moles. The molar mass of HCl is about 36.46 g/mol:

10 grams HCl × (1 mol HCl / 36.46 grams) = 0.274 moles HCl

Then, divide the moles of HCl by the volume of the solution in liters to find the molarity:

Molarity = Moles of solute / Volume of solution in liters

Molarity = 0.274 moles HCl / 0.1 L = 2.74 M

The molarity of the 10.0% HCl solution is therefore 2.74 M.

Answer: No crystals of potassium sulfate will be seen at 0°C for the given amount.

Explanation:

We are given:

Mass of potassium nitrate = 47.6 g

Mass of potassium sulfate = 8.4 g

Mass of water = 130. g

Solubility of potassium sulfate in water at 0°C = 7.4 g/100 g

This means that 7.4 grams of potassium sulfate is soluble in 100 grams of water

Applying unitary method:

In 100 grams of water, the amount of potassium sulfate dissolved is 7.4 grams

So, in 130 grams of water, the amount of potassium sulfate dissolved will be [tex]\frac{7.4}{100}\times 130=9.62g[/tex]

As, the soluble amount is greater than the given amount of potassium sulfate

This means that, all of potassium sulfate will be dissolved.

Hence, no crystals of potassium sulfate will be seen at 0°C for the given amount.