Consider a buffer solution that is 0. 50 M in NH3 and 0. 20 M in NH4Cl. For ammonia, pKb=4. 75. Calculate the pH of 1. 0 L of the solution upon addition of 30. 0 mL of 1. 0 M HCl to the original buffer solution.

Express your answer to two decimal places

Answers

Answer 1

The pH of 1. 0 L of the solution on addition of 30. 0 mL of 1. 0 M HCl to the original buffer solution will be 12.50.

The reaction that occurs when HCl is added to the buffer solution is:

HCl + NH₃ → NH₄⁺ + Cl⁻

The HCl reacts with NH₃ to form NH₄⁺ and Cl⁻. This will cause the concentration of NH₄⁺ in the buffer to increase and the concentration of NH₃ to decrease. However, since we started with a buffer solution, it will still be able to resist changes in pH.

To solve this problem, we will use the Henderson-Hasselbalch equation:

pH = pKb + log([NH₄⁺]/[NH₃])

where [NH₄⁺] is the concentration of the ammonium ion and [NH3] is the concentration of ammonia.

Calculate the moles of HCl added

The volume of HCl added is 30.0 mL = 0.0300 L. The concentration of HCl is 1.0 M, so the moles of HCl added are:

moles of HCl = concentration x volume = 1.0 M x 0.0300 L = 0.0300 moles

Calculate the new concentrations of NH₄⁺ and NH₃

The moles of NH₄⁺ and NH₃ in the original buffer solution can be calculated as:

moles of NH₄⁺ = 0.20 M x 1.0 L = 0.20 moles

moles of NH₃ = 0.50 M x 1.0 L = 0.50 moles

When HCl is added, it reacts with NH₃ to form NH₄⁺ and Cl⁻. The amount of NH₄⁺ produced is equal to the amount of HCl added, since the reaction is 1:1. Therefore, the new concentration of NH₄⁺ is:

[NH₄⁺] = moles of NH₄⁺ / (volume of buffer + volume of HCl added)

[NH₄⁺] = 0.20 moles / (1.0 L + 0.0300 L)

[NH₄⁺] = 0.196 M

The new concentration of NH₃ can be calculated using the buffer equation:

[NH₃] = Ka x [NH₄⁺] / [H⁺]

where Ka is the equilibrium constant for the reaction NH₄⁺ + H₂O → NH₃ + H₃O⁺, which is equal to the acid dissociation constant of NH₃, Kb. Since pKb is given as 4.75, we can calculate Kb:

Kb = 10^(-pKb) = [tex]10^{-4.75}[/tex]  = 1.78 x 10⁻⁵

Substituting the values we have:

[NH3] = Kb x [NH₄⁺] / [H⁺]

[NH3] = 1.78 x 10⁻⁵ x 0.196 M / [tex]10^{-pH}[/tex]

[NH3] = 3.49 x 10⁻⁶ / [tex]10^{-pH}[/tex]

Calculate the new pH of the buffer

Substituting the values we have into the Henderson-Hasselbalch equation:

pH = pKb + log([NH₄⁺]/[NH₃])

pH = 4.75 + log(0.196 M / (3.49 x 10⁻⁶ / [tex]10^{-pH}[/tex])))

Simplifying and solving for pH:

pH = 4.75 + log(5.61 x 10⁷) + log([tex]10^{pH}[/tex])

pH = 4.75 + 7.75 + pH

pH = 12.50

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Related Questions

What mass of methane (CH4) must be burned to give 1575 kJ of energy? (891 KJ/Mol)

Answers

The mass of methane (CH4) must be burned to give 1575 kJ of energy is 28.4 g.

The first step is to determine the number of moles of methane required to produce 1575 kJ of energy. We can do this using the conversion factor provided:

891 kJ of energy is produced by 1 mole of methane (CH4)

1575 kJ of energy is produced by how many moles of methane?

1575 kJ / 891 kJ/mol = 1.77 mol

Next, we can use the molar mass of methane to convert it from moles to grams:

Molar mass of methane = 12.01 g/mol (for carbon) + 4 x 1.01 g/mol (for hydrogen) = 16.05 g/mol

Mass of methane = number of moles x molar mass

Mass of methane = 1.77 mol x 16.05 g/mol = 28.4 g

Therefore, 28.4 g of methane must be burned to produce 1575 kJ of energy.

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chemistry need help please

Answers

The number of moles of hydrogen that can be made from 4.89 x 10-22 atoms of iron is 6.65 x 10-26 moles H2.

What is hydrogen ?

Hydrogen is the most abundant element found in the universe. It is a colorless, odorless gas that is the lightest of all elements. Hydrogen has the symbol H and the atomic number 1. It is the most basic building block of all matter. Hydrogen is an important part of many molecules, including water (H2O), proteins, and fats. It is a key component of many fuels, including gasoline, natural gas, and propane. Hydrogen is used in the production of ammonia, methanol, and other chemicals. It is also used in fuel cells.

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A solution is made using 145. 7 g of water (MM = 18. 02 g/mol) and 120. 0 g of ethanol (MM = 46. 07 g/mol). What is the mole fraction of the water in the solution?

Answers

Mole fraction of the water in the solution made using 145.7 g of water and 120.0 g of ethanol is 0.7564 mole.

Mass of water = 145.7 g

molar mass of water = 18.02 g/mole

Number of moles = mass ÷ molar mass

                             = 145.7 g ÷ 18.02 g/mole

                             = 8.085 mole

mass of ethanol = 120 g

molar mass of ethanol = 46.07 g/mole

Number of moles = mass ÷ molar mass

                              = 120 g ÷ 46.07 g/mole

                              = 2.6047 mole

Total number of moles = moles of water + moles of ethanol

                                      = 8.085 mole + 2.6047 mole

                                      = 10.6897 mole

The term mole fraction is also known as molar fraction. It is defined as unit of the amount of a constituent divided by the total amount of all constituents in a mixture.

Mole fraction of water = moles of water ÷ Total number of moles

                                     = 8.085 mole ÷  10.6897 mole

                                     = 0.756335

                                     = 0.7564 mole approximately.

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14) If 3 moles of methane reacts with oxygen to produce carbon dioxide and water,
what mass of water is produced?

Answers

Answer:

The balanced chemical equation for the combustion of methane (CH4) with oxygen (O2) is:

CH4 + 2O2 → CO2 + 2H2O

This equation tells us that for every mole of methane that reacts, we get 2 moles of water produced. So, for 3 moles of methane, we would get:

3 moles CH4 × (2 moles H2O / 1 mole CH4) = 6 moles H2O

Now we need to convert moles of water to mass. The molar mass of water (H2O) is 18.015 g/mol. So, the mass of water produced is:

6 moles H2O × 18.015 g/mol = 108.09 g

Therefore, if 3 moles of methane reacts with oxygen to produce carbon dioxide and water, 108.09 g of water is produced.

Explanation:

A 50.0-g ice cube at 0.0°C is added to a glass containing 400.0 g of water at 45.0°C. What is the final temperature of the system? Assume that no heat is lost to the surroundings.

Answers

At the final temperature, the ice has melted completely and the water has cooled down to the same temperature. Therefore, we can set the final temperature as T_f for both ice and water.

What is the principle of conservation of energy?

To determine the final temperature of the system, we need to use the principle of conservation of energy. The energy lost by the water when it cools down must be equal to the energy gained by the ice when it melts and warms up. We can express this as:

Q_water = Q_ice

where Q_water is the heat lost by the water and Q_ice is the heat gained by the ice.

The heat lost by the water can be calculated using the formula:

Q_water = m_water * C_water * ΔT

where m_water is the mass of water, C_water is the specific heat capacity of water, and ΔT is the change in temperature of water.

The heat gained by the ice can be calculated using the formula:

Q_ice = m_ice * L_f + m_ice * C_ice * ΔT

where m_ice is the mass of ice, L_f is the heat of fusion of ice, C_ice is the specific heat capacity of ice, and ΔT is the change in temperature of ice

Now we can set up the equation:

m_water * C_water * (T_i - T_f) = m_ice * L_f + m_ice * C_ice * (T_f - T_i)

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How is the acidic concentration of a solution determined?

How do you know when the endpoint is reached in a titration? When does this occur?


PLEASE BE ACCURATE! THANK YOU!!:)

Answers

Answer:

A. The acidic concentration of a solution can be determined by titrating with a strong base. You calculate the number of moles of strong base required to reach the equivalence point of the titration. Then, using the mole ratio from the balanced neutralization equation, convert from moles of strong base to moles of acid. [source, Khan Academy]

B. The endpoint is reached in a titration when the indicator used changes color (i.e, pink).

Explanation:

For B, if you do it correctly, the equivalence point is eventually reached and this is signified by the change in color of the indicator in the analyte solution.

Grain Alcohol (C2H5OH) can be produced by the fermentation of glucose. If 125g of alcohol is obtained from the fermentation of .511 kg of glucose, what is the percent yield?

Answers

i’m so glad you are able and enjoy yourself too .511 kg and i and

the answer is yes. .i92737.18 (C2) i die frage mich jetzt noch mal ob du das mit deinem mann mit dir passt

a solution is prepared by mixing 529.0 ml of ethanol with 594.0 ml of water. the molarity of ethanol in the resulting solution is 8.407 m. the density of ethanol at this temperature is 0.7893 g/ml. calculate the difference in volume between the total volume of water and ethanol that were mixed to prepare the solution and the actual volume of the solution.

Answers

To calculate the difference in volume between the total volume of water and ethanol that were mixed to prepare the solution and the actual volume of the solution, we must first calculate the mass of ethanol present in the solution. We can do this by multiplying the molarity of ethanol (8.407 M) by the volume of ethanol present (529.0 mL) and the molar mass of ethanol (46.07 g/mol).

Mass of ethanol = 8.407 M x 529.0 mL x 46.07 g/mol = 24617.3 g


We can then calculate the volume of ethanol in the solution by dividing the mass of ethanol (24617.3 g) by the density of ethanol (0.7893 g/mL):


Volume of ethanol = 24617.3 g/0.7893 g/mL = 31202.1 mL


Therefore, the difference in volume between the total volume of water and ethanol that were mixed to prepare the solution and the actual volume of the solution is:


Difference in volume = Volume of ethanol + Volume of water - Actual volume of solution

Difference in volume = 31202.1 mL + 594.0 mL - (529.0 mL + 594.0 mL) = 277.1 mL


The difference in volume between the total volume of water and ethanol that were mixed to prepare the solution and the actual volume of the solution is 277.1 mL.

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Calcium hydroxide reacts with hydrofloric acid according to the following reaction.
Ca(OH)2 + HF ===> CaF2+ HOH
If 0.5 moles of hydrofloric acid are reacted, what mass of calcium fluoride will be produced?

Answers

Answer:

To answer this question, one must first understand the stoichiometric equation for this particular reaction. The equation states that for each one mole of calcium hydroxide (Ca(OH)2) that reacts with one mole of hydrofloric acid (HF), one mole of calcium fluoride (CaF2) and one mole of water (HOH) is produced. Therefore, if 0.5 moles of hydrofloric acid are reacted with the calcium hydroxide, then 0.5 moles of calcium fluoride will be produced.

To calculate the mass of calcium fluoride that will be produced, one must first look up the atomic mass of both calcium and fluorine and multiply them by the number of moles of each that are present in the reaction. In this case, the atomic mass of calcium is 40.08 and the atomic mass of fluorine is 19. Therefore, the mass of calcium fluoride that will be produced is equal to (40.08 x 0.5) + (19 x 0.5) = 29.54 g.

In conclusion, if 0.5 moles of hydrofloric acid are reacted with calcium hydroxide, then a mass of 29.54 g of calcium fluoride will be produced.

Answer:

19.52 grams of calcium fluoride will be produced.

Step by step Explanation :
The balanced chemical equation for the reaction between calcium hydroxide and hydrofluoric acid is:

Ca(OH)2 + 2HF → CaF2 + 2H2O

According to the equation, 1 mole of calcium hydroxide reacts with 2 moles of hydrofluoric acid to produce 1 mole of calcium fluoride.

So, if 0.5 moles of hydrofluoric acid react, then the moles of calcium fluoride produced will be half of that, which is:

0.5 moles HF x (1 mole CaF2 / 2 moles HF) = 0.25 moles CaF2

To calculate the mass of calcium fluoride produced, we need to use the molar mass of CaF2, which is:

40.08 g/mol (for Ca) + 2(18.99 g/mol) = 78.06 g/mol

So, the mass of calcium fluoride produced will be:

0.25 moles CaF2 x 78.06 g/mol = 19.52 g

Therefore, if 0.5 moles of hydrofluoric acid are reacted, 19.52 grams of calcium fluoride will be produced.

how to calculate equilibrium concentration from initial concentration without kc with acids and bases

Answers

We can determine the equilibrium concentration when we are given the initial concentration and reaction of an acid or base. In the given reaction,

HA is an acid that dissociates in water to form its conjugate base A− and a hydronium ion H3O+.

HA(aq) + H2O(l) ⇌ A−(aq) + H3O+(aq)

For any weak acid, the dissociation constant is defined as follows:

Ka = [A−][H3O+]/[HA]

We can determine the equilibrium concentration from the above formula by using the initial concentration.

Let’s consider a weak acid HA with an initial concentration of [HA]0.

It will form [A−] and [H3O+] at equilibrium.

If the degree of dissociation of the acid is x, then the concentration of [H3O+] at equilibrium will be [H3O+] = x, and the concentration of [A−] will be [A−] = x. And the concentration of [HA] at equilibrium will be [HA] = [HA]0 − x.

So, we can write the equilibrium expression for the dissociation of HA as:

Ka = [A−][H3O+]/[HA]Ka = x * x/([HA]0 − x)

By solving the above equation, we can obtain the value of x, which is the degree of dissociation of the acid.

At equilibrium, the equilibrium concentration of HA is [HA] = [HA]0 − x. The equilibrium concentration of A− is [A−] = x. The equilibrium concentration of H3O+ is [H3O+] = x. Thus, we can determine the equilibrium concentration from the initial concentration of a weak acid or base without using the equilibrium constant Kc.

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Give an example of solid in solid solution.

Answers

Solid-solid solutions such as brass, bronze, and sterling silver are called alloys.

please make me brainalist and keep smiling dude

How many grams do 4.56 x 1024 formula units of lithium chloride weigh?

Answers

Answer:

321.27 grams.

Explanation:

To determine the weight of 4.56 x 10^24 formula units of lithium chloride, we first need to know the molar mass of lithium chloride.

Lithium chloride (LiCl) has a molar mass of approximately 42.39 g/mol. This means that one mole of lithium chloride weighs 42.39 grams.

To convert formula units to moles, we need to use Avogadro's number, which is approximately 6.022 x 10^23 particles per mole.

So, to find the number of moles in 4.56 x 10^24 formula units of lithium chloride, we can divide by Avogadro's number:

4.56 x 10^24 formula units / (6.022 x 10^23 formula units/mol) = 7.58 moles

Now that we know the number of moles, we can use the molar mass to find the weight:

7.58 moles x 42.39 g/mol = 321.27 grams

Therefore, 4.56 x 10^24 formula units of lithium chloride weigh approximately 321.27 grams.

When 23 grams of sodium react with 32 grams of sulfur according to the equation, how many total grams of sodium sulfide should be formed?

Answers

Answer:

78 grams of sodium sulfide should be formed

Explanation:

The balanced chemical equation for the reaction between sodium and sulfur is:

2 Na + S → Na2S

According to the equation, 2 moles of sodium react with 1 mole of sulfur to produce 1 mole of sodium sulfide. The molar mass of sodium is approximately 23 g/mol and the molar mass of sulfur is approximately 32 g/mol.

We need to determine which reactant is limiting and which is in excess in order to calculate the amount of sodium sulfide produced.

Using the given masses, we can calculate the number of moles of each reactant:

moles of sodium = 23 g / 23 g/mol = 1 mol

moles of sulfur = 32 g / 32 g/mol = 1 mol

From this calculation, we can see that both reactants are present in the stoichiometric ratio required by the balanced equation, so neither is limiting.

Therefore, the amount of sodium sulfide formed will be based on the amount of either reactant, which is 1 mole. Using the molar mass of sodium sulfide (78 g/mol), we can calculate the mass of sodium sulfide formed:

mass of Na2S = 1 mol x 78 g/mol = 78 g

Therefore, when 23 grams of sodium react with 32 grams of sulfur, a total of 78 grams of sodium sulfide should be formed.

what is the formula for caculating time

Answers

Answer:  Time = Distance ÷ Speed

A 750.0 mL metal bulb is filled with 42.1 g of CH4 and 3.23 g of NH3. If the amount of pressure contributed by CH4 is 111.2 atm, then what is pressure due to NH3?

Answers

If the pressure contributed by CH4 is 111.2 atm, the pressure due to NH3 would be 109.1 atm.

Pressure calculation

To solve the problem, we can use the ideal gas law equation:

PV = nRT

where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature.

We need to find the pressure due to NH3, so let's start by finding the number of moles of each gas:

moles of CH4 = 42.1 g / (16.04 g/mol) = 2.623 molmoles of NH3 = 3.23 g / (17.03 g/mol) = 0.190 mol

Now, we can use the pressure contributed by CH4 to find the total pressure:

Ptotal = PCH4 + PNH3

111.2 atm = PCH4 + PNH3

Finally, we can solve for the pressure due to NH3:

PNH3 = Ptotal - PCH4PNH3 = 111.2 atm - 0.190 mol x (0.0821 L·atm/mol·K) x (298 K) / 0.750 LPNH3 = 109.1 atm

Therefore, the pressure due to NH3 is 109.1 atm.

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PH= 4. 45, volume 40 mL, from an aqueous acid and solid conjugate. What mass of conjugate? Please explain with work

Answers

We are unable to give the mass of the conjugate a numerical value without knowing the precise conjugate that was employed.

We must apply the equation for estimating the pH of a buffer solution to estimate the mass of the solid conjugate:

pH equals pKa plus log([A-]/[HA])

Finding the [A-] and [HA] concentrations comes first. Since we are aware that the solution has a 40 mL volume, we can write:

[HA] = [A-] = conjugate moles / solution volume

The Henderson-Hasselbalch equation must then be used to determine how many moles of conjugate are present:

pH equals pKa plus log([A-]/[HA])

4.45 = log(1/1) + pKa

pKa = 4.45

We can now enter the predetermined values:

4.45 is equal to 4.45 plus log(moles of conjugate / 0.04 L).

Calculating the conjugate moles

moles of conjugate = 0.04 L x 1 M / log(moles of conjugate / 0.04 L) = 0 moles of conjugation

Using the molar mass of the conjugate, we can finally determine its mass:

mass is determined by multiplying the number of moles by the molar mass.

mass = molar mass x 0.04 moles.

We are unable to give the mass of the conjugate a numerical value without knowing the precise conjugate that was employed.

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Is this fission ir fusion

Answers

Answer:

fusion

Explanation: hope this helps

If ΔHfus of sodium is 2.60 kJ/mol, what is the Latent Heat of Fusion?

Answers

The Latent Heat of Fusion of sodium is 2.60 kJ/mol. Fusion is an endothermic process, which means that it requires energy to occur.

What is Fusion?

Fusion, also known as melting, is the physical process in which a substance changes from a solid state to a liquid state as a result of absorbing heat. When a substance is heated to its melting point, the energy absorbed causes the molecules or atoms of the substance to vibrate more rapidly, eventually breaking the bonds between them and allowing them to move more freely. As a result, the substance transitions from a solid state, where the molecules are tightly packed and fixed in position, to a liquid state, where the molecules are more spread out and can move around each other more freely.

The Latent Heat of Fusion (Lf) is defined as the amount of heat required to melt one mole of a substance at its melting point.

Lf = ΔHfus / n

where ΔHfus is the enthalpy of fusion and n is the number of moles of the substance.

In this case, ΔHfus of sodium is given as 2.60 kJ/mol. To calculate the Latent Heat of Fusion, we need to know the number of moles of sodium. Let's assume we have one mole of sodium.

Then,

Lf = ΔHfus / n

= 2.60 kJ/mol / 1 mol

= 2.60 kJ/mol

Therefore, the Latent Heat of Fusion of sodium is 2.60 kJ/mol.

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Chadwick worked to isolate the neutral particle Rutherford had proposed.

True
False

Answers

Answer: True

Explanation: Chadwick, a British physicist, worked to isolate the neutral particle that Rutherford had proposed.  This particle, known as the neutron, was hypothesized by Rutherford to exist within atomic nuclei. In 1932, Chadwick successfully conducted experiments that confirmed the existence of the neutron and provided evidence for its role in nuclear structure.

Rutherford's model of the atom suggested that the nucleus contains positively charged protons and that electrons orbit around it. However, this model could not explain why the positively charged protons did not repel each other and cause the nucleus to break apart. Rutherford proposed the existence of neutral particles, later identified as neutrons, to account for the stability of the atomic nucleus.

Chadwick's work involved bombarding various elements with alpha particles and observing the resulting radiation. Through careful experimentation, he was able to demonstrate that the radiation consisted of neutrons, which had the ability to penetrate and interact with atomic nuclei without being deflected by electromagnetic forces. This discovery revolutionized our understanding of atomic structure and paved the way for further advancements in nuclear physics.

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Answer:

True

Explanation:

I answered so the person who answered first can get branliest!

HELP HELP HELP!!! THIS IS DUE TOMORROW WILL MARK BRAINLIEST IF ALL QUESTIONS ARE SOLVED

8. How many molecules of aspartame (C₁4H₁N₂O5) are in 745 grams of C14H18N₂O5?

9. What is the mass of 5 x 10^25 molecules of Fe₂(SO3)3?

10. Calculate the mass of 6.3 moles of Ba3(PO4)2

Answers

Answer:

8. 1.52 x 10^24 molecules

9. 29206 g

10. 3791.97 g

Explanation:

8. Molecular mass of C14H18N2O5 = 294.3 g/mol

745 g / 294.3 g/mol = 2.53 moles

1 mole has 6.022 x 10^23 molecules

=> 2.53 x 6.022 x 10^23 = 1.52 x 10^24 molecules

9. 1 mole of Fe₂(SO3)3 has 6.022 x 10^23 molecules

so 5 x 10^25/6.022 x 10^23 = 83.0 moles

1 mole of Fe₂(SO3)3 is equal 351.88g

so 83.0 x 351.88 = 29206 g

10. Molar mass of Ba3(PO4)2 is 601.9 g

so 6.3 x 601.9 = 3791.97 g

What volume of a 3. 00 M glucose stock solution is necessary to prepare 2500 mL of 0. 400 M solution

Answers

Answer:333.3333 mL or 333mL

Explanation:

M1V1 = M2 V2

M1= 3.00 M

V2= 2500 ml

M2= 0.400M

V1= solving for

V1= M2V2/M1

0.400 X 2500 / 3.00  =

HELP ME ASAP

What is the frequency of an
orange light with a wavelength of
483 nm?
? ] × 10[²]
10
c = 3.0 x 108 m/s
Hint: Watch your units!
[?
Hz
Be sure to enter both the coefficient and the exponent.
Coefficient (green)
Exponent (yellow)

Answers

Answer:

Explanation:

λν = c

Wavelength, λ (lambda), times frequency, ν (nu), equals the speed of light.

c = 3.0 x 108 m/s

===

We are given a wavelength of 483 nanometers.  1 nanometer = 1x10^-9 meter.   483nm = 4.83x10^-7 meters

---

λν = c

ν = c/λ

ν = (3.0x10^8 m/s)/(4.83x10^-7 m)

ν = 6.21x10^14 Hz  (1/s)

Which process occurred when the water droplets formed on the grass? Choose the correct answer. Responses deposition deposition sublimation sublimation evaporation evaporation condensation condensation Skip to navigation

Answers

Answer:

Condensation is the process that took place when the water droplets formed on the grass.

Explanation:

Condensation is the process that took place when the water droplets formed on the grass. This occurs when airborne water vapour meets a colder surface, like the grass blades in the morning, and transforms into liquid form, generating tiny water droplets on the surface.

predict sodium nitrate and ammonium chloride are soluble in water. sodium chloride and ammonium nitrate are also soluble in water. on a molecular scale, describe the solution that results when solution of sodium nitrate and ammonium chloride are mxed

Answers

The resulting solution is a homogeneous mixture of Na+, NO3^-, NH4+, and Cl^- ions surrounded by water molecules.

When solutions of sodium nitrate (NaNO3) and ammonium chloride (NH4Cl) are mixed, the resulting solution contains the ions Na+, NO3^-, NH4+, and Cl^-. All of these ions are soluble in water, so the solution remains clear and homogeneous. On a molecular scale, the Na+ ions are surrounded by water molecules, as are the NO3^- ions, NH4+ ions, and Cl^- ions. This is known as hydration, and it is the reason why these compounds are soluble in water. The ions are free to move around in the solution, which allows them to conduct electricity.

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what are examples of monovalent atomic groups

Answers

Answer:

hydrogen,lithium, sodium

which factor is most sensitive to changes in temperature- x) the frequency of collisions, y) the orientation factor, or z) the fraction of molecules with energy greater than the activation energy?

Answers

The factor that is most sensitive to changes in temperature is the fraction of molecules with energy greater than the activation energy.

The statement that best represents the effect of temperature on reaction rates is called the Arrhenius equation. It is expressed as [tex]k = Ae^{-Ea/RT},[/tex]

where k is the rate constant,

A is the pre-exponential factor or frequency factor,

Ea is the activation energy,

R is the gas constant, and T is the absolute temperature.

At high temperatures, the reaction rate increases, and at low temperatures, the reaction rate decreases.

The Arrhenius equation shows that the rate constant depends on two factors:

the activation energy and the fraction of molecules with energy greater than the activation energy.

Thus, the most sensitive factor to changes in temperature is the fraction of molecules with energy greater than the activation energy.

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98 kJ +NaHCO, -> Na,CO, + CO2 + H20
How much heat will be absorbed when 3. 0 grams of NaHCO, decompose?

Answers

3.0 grams of NaHCO₃  will receive about 1.76 kJ of heat during the decomposition process.

To calculate the amount of heat absorbed when 3.0 grams of NaHCO₃ decompose, we need to first determine the limiting reactant and then use the balanced chemical equation and the enthalpy change to calculate the amount of heat absorbed.

The balanced chemical equation for the decomposition of NaHCO₃ is:

2 NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)

The enthalpy change for this reaction is not given, but assuming it is an endothermic reaction, the heat absorbed can be represented as a positive value.

First, we need to determine the limiting reactant. The molar mass of NaHCO₃ is:

NaHCO₃: 23.0 + 1.0 + 12.0 + 48.0 = 84.0 g/mol

Using the molar mass, we can convert 3.0 g of NaHCO₃ to moles:

3.0 g NaHCO₃ x (1 mol NaHCO₃/84.0 g NaHCO₃) = 0.0357 mol NaHCO₃

From the balanced equation, we know that 2 moles of NaHCO₃ produces 1 mole of CO₂. So, the moles of CO₂ produced from 0.0357 mol of NaHCO₃ is:

0.0357 mol NaHCO₃ x (1 mol CO₂/2 mol NaHCO₃) = 0.0179 mol CO₂

Next, we can use the enthalpy change for the reaction and the moles of CO₂ produced to calculate the heat absorbed:

0.0179 mol CO₂ x (98 kJ/1 mol) = 1.76 kJ

Therefore, the amount of heat absorbed when 3.0 grams of NaHCO₃ decompose is approximately 1.76 kJ.

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I need help with this<3 I'm so lost and need this assignment done

Answers

You must compare the total amount of oxygen atoms on the reactant and product sides of the chemical equations in order to determine whether the oxygen atoms are evenly distributed on both sides.

Why must each element have the same number of atoms on both sides of the equation?

Because matter cannot be generated or destroyed, there must be an equal amount of atoms of each element on both sides of the equation. Coefficients are the only variables that can be altered while balancing equations. An equation cannot be balanced by changing the subscripts in a chemical formula.

Reactant side:

B2Br has no oxygen atoms

6 HNO3 has 18 oxygen atoms (6 x 3)

Product side:

2 B(NO3)3 has 18 oxygen atoms (2 x 3 x 3)

6 HBr has no oxygen atoms

18 oxygen atoms total are present on the reactant side.

18 oxygen atoms total are on the product side.

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Macmill
20.8 g 0₂ x
grams of KCIO:
122.55 g KCIO,
32.00 g 0₂
74.55 g KCI
Answer Bank
1 mole KCIO,
1 mole O₂
1 mole KCI
2 moles KCIO,
3 moles 0₂
2 moles KCI
= g KCIO,
g KCIO,

Answers

Answer:

To solve this problem, we need to use the balanced chemical equation for the reaction between oxygen gas (O₂) and potassium chloride (KCI) to form potassium chlorate (KCIO₃):

2KCI + 3O₂ → 2KCIO₃

We can use the given mass of O₂ (20.8 g) and the molar mass of O₂ (32.00 g/mol) to find the moles of O₂:

20.8 g O₂ x (1 mol O₂ / 32.00 g O₂) = 0.65 mol O₂

According to the balanced equation, 3 moles of O₂ react with 2 moles of KCI to produce 2 moles of KCIO₃. Therefore, we can use the moles of O₂ to find the moles of KCIO₃:

0.65 mol O₂ x (2 mol KCIO₃ / 3 mol O₂) = 0.43 mol KCIO₃

Finally, we can use the molar mass of KCIO₃ (122.55 g/mol) to convert moles of KCIO₃ to grams of KCIO₃:

0.43 mol KCIO₃ x (122.55 g KCIO₃ / 1 mol KCIO₃) = 52.71 g KCIO₃

Therefore, the grams of KCIO₃ that can be produced from 20.8 g O₂ is 52.71 g KCIO₃. However, the problem does not ask for the grams of KCIO₃, but instead asks for the grams of KCIO, which is not a valid compound. It is possible that there is a typo in the problem and that it should have asked for the grams of KCIO₃ instead.

Explanation:

CHEMISTRY QUESTION, PLEASE HELP!!!


Reaction B (attached); The change in enthalpy for the forward reaction is -91kJ/mol. (Energy is a product, flowing from the chemical reaction to the surroundings. )


The forward reaction for Reaction B (attached) is.


- endothermic

- exothermic


If Reaction B (attached) was ta equilibrium and then was heated ______ CH3OH would be present after the reaction adjusts to the new temperature.


- more

- less

- the same amount of


If Reaction B (attached) was at equilibrium and then the pressure in its container was increased, ____ CH3OH would be present after the reaction adjusts to the new pressure.


- more

- less

- the same amount of


If Reaction B (attached) was at equilibrium and then H2 was added, _____ CH3OH would be present after the reaction adjusts.


- more

- less

- the same amount of


If Reaction B (attached) was at equilibrium and then H2 was added, ____ CO would be present after the reaction adjusts.


- more

- less

- the same amount

Answers

The forward reaction for Reaction B is endothermic, as indicated by the negative change in enthalpy (-91 kJ/mol) for the forward reaction.

If Reaction B was at equilibrium and then was heated, the amount of CH₃OH would be less after the reaction adjusts to the new temperature. This is because the forward reaction is endothermic, meaning that an increase in temperature would shift the equilibrium towards the reactants side to counteract the increase in temperature.

If the pressure in the container of Reaction B was increased, the amount of CH₃OH would be more after the reaction adjusts to the new pressure. This is because the forward reaction produces fewer moles of gas than the reverse reaction, so increasing the pressure would shift the equilibrium towards the side with fewer moles of gas (the products side) to counteract the increase in pressure.

If H₂ was added to Reaction B at equilibrium, the amount of CH₃OH would be more after the reaction adjusts. This is because H₂ is a reactant in the reverse reaction, so adding more H₂ would shift the equilibrium towards the products side to counteract the increase in H₂.

If H₂ was added to Reaction B at equilibrium, the amount of CO would be less after the reaction adjusts. This is because CO is a product in the forward reaction, so adding more H₂ would shift the equilibrium towards the products side to counteract the increase in H₂, resulting in a decrease in CO.

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