Precipitation reactions and ionic equations
(look at attachment)

The resultant temperature on the beach is 294.39 K.

Relation between the temperature and pressure of gas will be explained by using the ideal gas equation PV = nRT.

And for this question, required equation is:

P₁/T₁ = P₂/T₂, where

On putting values from question, we get

T₂ = (4.78)(20) / (4.50) = 21.24 degrees Celsius

T₂ = 21.24 degrees Celsius = 294.39 K

Hence required temperature is 294.39 K.

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Final answer:

To determine the new temperature on the beach causing an increase in pressure within an aerosol can from 4.50 atm to 4.78 atm, the Gas Law is utilized, yielding a final temperature of 35.1 degrees Celsius.

Explanation:

The question involves finding the new temperature at which the pressure of gases in an aerosol can increases from 4.50 atm to 4.78 atm, initially at 20.0 degrees Celsius. We will use the Gas Law, which states that for a constant volume and amount of gas, the pressure of the gas is directly proportional to its temperature. This can be mathematically represented as P1/T1 = P2/T2, where P is the pressure, T is the temperature in Kelvin, and subscripts 1 and 2 refer to the initial and final states, respectively.

First, convert the initial temperature from Celsius to Kelvin: T1 = 20.0 + 273.15 = 293.15 K. Then, solve for T2: T2 = (P2 × T1) / P1 = (4.78 atm × 293.15 K) / 4.50 atm. T2 = 308.25 K, which converts back to 35.1 degrees Celsius when subtracted by 273.15.

Therefore, the Celsius temperature on the beach where the pressure of the gases in the aerosol can increases to 4.78 atm is 35.1 degrees Celsius.

Answer:

pH= 0.369

Explanation:

The formic acid reacts with NaOH as

HCOOH + NaOH= HCOONa + H2O

Apply CaVa/ CbVb = Na/Nb

Ca×25/(29.8×0.3587) = 1/1

Ca= (29.8×0.3587)/25= 0.428M

pH = - log(H+)

Since only 0ne H+ is in the stoichiometric equation, it means H+ = 0.428M

pH = -log(0.428) =0.369

The approximate pH at the equivalence point of a weak acid-strong base titration is 8.35.

To find the pH at the equivalence point, we first need to determine the concentration of the formic acid solution. The balanced equation for the reaction between formic acid (HCOOH) and sodium hydroxide (NaOH) is:

[tex]\[ \text{HCOOH} + \text{NaOH} \rightarrow \text{NaCOO} + \text{H}_2\text{O} \][/tex]

Given that 25 mL of formic acid requires 29.80 mL of 0.3567 M NaOH to reach the equivalence point, we can calculate the initial moles of NaOH used:

[tex]\[ \text{moles of NaOH} = \text{volume of NaOH} \times \text{concentration of NaOH} \] \[ \text{moles of NaOH} = 0.02980 \text{ L} \times 0.3567 \text{ M} \] \[ \text{moles of NaOH} = 0.01064 \text{ mol} \][/tex]

Since the stoichiometry of the reaction is 1:1, the initial moles of formic acid are the same as the moles of NaOH used:

[tex]\[ \text{moles of HCOOH} = 0.01064 \text{ mol} \] Now, we can find the concentration of the formic acid solution: \[ \text{concentration of HCOOH} = \frac{\text{moles of HCOOH}}{\text{volume of HCOOH}} \] \[ \text{concentration of HCOOH} = \frac{0.01064 \text{ mol}}{0.025 \text{ L}} \] \[ \text{concentration of HCOOH} = 0.4256 \text{ M} \][/tex]

At the equivalence point, all of the formic acid has been neutralized, and we are left with a solution of sodium formate (NaCOO), which is the salt of a weak acid and a strong base. The concentration of the sodium formate solution at the equivalence point is the same as the initial concentration of formic acid:

[tex]\[ [\text{NaCOO}] = 0.4256 \text{ M} \][/tex]

The pH at the equivalence point is determined by the hydrolysis of the sodium formate salt. The hydrolysis reaction is:

[tex]\[ \text{NaCOO} \rightleftharpoons \text{Na}^+ + \text{HCOO}^- \] \[ \text{HCOO}^- + \text{H}_2\text{O} \rightleftharpoons \text{HCOOH} + \text{OH}^- \][/tex]

The equilibrium constant expression for the hydrolysis is:

[tex]\[ K_b = \frac{[\text{HCOOH}][\text{OH}^-]}{[\text{HCOO}^-]} \][/tex]

Since [tex]\( K_b = \frac{K_w}{K_a} \), where \( K_w \)[/tex] is the ionization constant of water, we can find [tex]\( K_b \)[/tex]:

[tex]\[ K_b = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-4}} \] \[ K_b = 5.56 \times 10^{-11} \][/tex]

Assuming that the hydrolysis of sodium formate is small, we can approximate the concentration of [tex]HCOO^-[/tex] to be equal to the concentration of NaCOO:

[tex]\[ [\text{HCOO}^-] \approx 0.4256 \text{ M} \][/tex]

The concentration of HCOOH produced by hydrolysis is negligible compared to the initial concentration of [tex]HCOO^-[/tex], so we can say:

[tex]\[ [\text{HCOOH}] \approx 0 \][/tex]

The concentration of [tex]OH^-[/tex] can be found using the equilibrium constant expression:

[tex]\[ K_b = \frac{[\text{HCOOH}][\text{OH}^-]}{[\text{HCOO}^-]} \] \[ 5.56 \times 10^{-11} = \frac{(0)[\text{OH}^-]}{0.4256} \][/tex]

[tex]Since \( [\text{HCOOH}] \) is approximately zero, we can rearrange the equation to solve for \( [\text{OH}^-] \):[/tex]

[tex]\[ [\text{OH}^-] = K_b \times \frac{[\text{HCOO}^-]}{[\text{HCOOH}]} \] \[ [\text{OH}^-] = 5.56 \times 10^{-11} \times \frac{0.4256}{0} \][/tex]

However, since [tex]\( [\text{HCOOH}] \)[/tex] cannot be zero, we can instead use the initial concentration of [tex]HCOO^-[/tex] to find an approximate value for [tex]\( [\text{OH}^-] \)[/tex]:

[tex]\[ [\text{OH}^-] \approx \sqrt{K_b \times [\text{HCOO}^-]} \] \[ [\text{OH}^-] \approx \sqrt{5.56 \times 10^{-11} \times 0.4256} \] \[ [\text{OH}^-] \approx \sqrt{2.36 \times 10^{-11}} \] \[ [\text{OH}^-] \approx 1.54 \times 10^{-6} \text{ M} \][/tex]

Now we can calculate the pOH and then the pH:

[tex]\[ \text{pOH} = -\log[\text{OH}^-] \] \[ \text{pOH} = -\log(1.54 \times 10^{-6}) \] \[ \text{pOH} \approx 5.82 \][/tex]

Since pH + pOH = 14:

[tex]\[ \text{pH} = 14 - \text{pOH} \] \[ \text{pH} = 14 - 5.82 \] \[ \text{pH} \approx 8.18 \][/tex]

However, we must consider that the assumption that [tex]\( [\text{HCOOH}] \)[/tex] is negligible may not be entirely accurate.

The actual pH at the equivalence point will be slightly higher due to the common ion effect, which suppresses the ionization of the weak acid (HCOOH) in the presence of its conjugate base [tex](HCOO^-)[/tex].

Taking this into account, the pH at the equivalence point is approximately 8.35.

In an HCl molecule, the chlorine atom is more electronegative than hydrogen, resulting in chlorine pulling the bonded electrons towards itself more and creating a dipole with a partial negative charge. This makes chlorine the atom that has a higher electronegativity in comparison to hydrogen.

In the molecule HCl, the chlorine atom pulls more on the bonded pair of electrons, creating a dipole. Thus, the correct answer is (b) chlorine for both parts of the question. Chlorine has a higher electronegativity compared to hydrogen, which is why the electron density in the HCl molecule is uneven and is greater around the chlorine nucleus. This difference in electronegativity between hydrogen (XH = 2.20) and chlorine (XCl = 3.16) results in a polar covalent bond with a dipole moment. Consequently, chlorine bears a partial negative charge (designated as δ−), and hydrogen bears a partial positive charge (designated as δ+), leading to dipole-dipole attractions between HCl molecules.

Answer:

The correct option is;

A) 1 to 1.

Explanation:

A stab;e nuclei requires the presence of a neutron to accommodate the the protons repulsion forces within the nucleus. An increase in the number of protons should be accompanied by an even more instantaneous increase in the number of neutrons to balance the forces in the nucleus. If there is an excess of neutrons or a deficit in protons a state of unbalance exists in the nucleus, which results to nuclear instability.

Therefore, the ratio of neutrons to protons is an appropriate way in foretelling nuclear stability and a stable nuclei is known to have a proton to neutron ratio of 1:1 and the number of protons and neutrons in the stable nuclei are usually even numbers.

Answer:

0.2, relatively inelastic

Explanation:

Price elasticity of demand is the degree of responsiveness of demand for a product/service to a unit change in the price of that product/service. Mathematically:

Price elasticity of demand = change in demand/change in price.

Price elasticity of demand can be perfectly elastic if elasticity is infinity, perfectly inelastic if elasticity is 0, relatively elastic if elasticity is greater than 1, unitary elastic if it is equal to 1 and relatively inelastic if it is less than 1.

For gasoline,

Change in price = 5%

Change in quantity demanded = 1%

Hence,

Price elasticity of demand for gasoline = 1/5 = 0.2

The price elasticity of demand for gasoline is 0.2 and can be described as being relatively inelastic.

When a titration is performed using an indicator that changes color slightly past the equivalence point, the calculated acid concentration will be marginally lower due to the slight excess of titrant added.

During a titration, the equivalence point is where the amount of titrant added neutralizes the analyte, resulting in a solution where the concentration of hydrogen ions ([tex][H^+][/tex]) equals the concentration of hydroxide ions ([tex][OH^-][/tex]). An indicator is used to visually signal this point through a color change.

When using an indicator that changes color slightly past the equivalence point, the student will observe a color change indicating a slightly larger volume of titrant has been added than necessary. Consequently, the calculated concentration of the acid will be slightly lower than its actual concentration, because the calculation will be based on a presumed complete neutralization that requires a bit more of the titrant.

It is vital to choose an indicator with a color change interval that brackets the pH at the equivalence point for an accurate titration. Methyl orange, for instance, changes color in the acidic range and is suitable for the titration of a weak base with a strong acid.

Phenolphthalein would be another choice as an indicator, changing from colorless to pink as the pH rises above 8.3, which can accurately signal the equivalence point in many titrations.

Answer : The age of the rock is, 2.60 billion years

Explanation :

Half-life = 1.3 billion years

First we have to calculate the rate constant, we use the formula :

[tex]k=\frac{0.693}{t_{1/2}}[/tex]

[tex]k=\frac{0.693}{1.3\text{ billion years}}[/tex]

[tex]k=0.533\text{ billion years}^{-1}[/tex]

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]

where,

k = rate constant  = [tex]0.533\text{ billion years}^{-1}[/tex]

t = time passed by the sample  = ?

a = initial amount of the reactant  = 12 mg

a - x = amount left after decay process = 3 mg

Now put all the given values in above equation, we get

[tex]t=\frac{2.303}{0.533}\log\frac{12}{3}[/tex]

[tex]t=2.60\text{ billion years}[/tex]

Therefore, the age of the rock is, 2.60 billion years

Answer:

The correct answer is 4. gives the relative number of atoms of each element per formula unit.

Explanation:

Empirical formula shows the ratio of elements of which a compound is composed. It is a minimal formula: it does not show the actual number of atoms of each element in the compound, but it shows the relative number of each element in the molecule. For example: butane has the molecular formula C₄H₁₀. It means that it is composed by 4 atoms of C and 10 atoms of hydrogen (actual number of atoms in the molecule). The ratio C:H is 4:10, so the simplest ratio is 2:5 (if we divide by 2). Thus, the empirical formula is C₂H₅ (there are 2 atoms of C for every 5 atoms of H).

Answer:

4

Explanation:

It gives the lowest ratio of the number of atoms of each element per formula unit. This is usually derived by dividing common multiples of the molecular formula thereby reducing the value.

Answer:

The new temperature is 829.5 K

Explanation:

Step 1: Data given

The initial pressure of neon = 786 mmHg

The final pressure of neon = 1811 mmHg

The initial temperature = 87 °C

Step 2: Calculate the new temperature

P1/T1 = P2/T2

⇒P1 = the initial pressure = 786 mmHg  = 1.03421 atm

⇒T1 = the initial temperature = 87 °C = 360 K

⇒P2 = the final pressure = 1811 mmHg = 2.382895 atm

⇒T2 = the final temperature = ?

1.03421 atm / 360 K = 2.382895 atm / T2

T2 = 829.5 K

786/360 = 1811 / T2

The new temperature is 829.5 K

Answer : The weight of first bar is, 1 kg

Explanation :

Let the weight of first bar and second bar be, x and y.

The ratio of gold and silver in first bar is, 2 : 3

The ratio of gold and silver in second bar is, 3 : 7

The final ratio of gold and silver in first and second bar is, 5 : 11

Total weight of bar = 8 kg

The equations will be:

[tex]\frac{2}{5}x+\frac{3}{10}y=\frac{5}{16}\times 8[/tex]       ..........(1)

[tex]\frac{3}{5}x+\frac{7}{10}y=\frac{11}{16}\times 8[/tex]       ..........(2)

Solving both the equations, we get:

[tex]4x+3y=25[/tex]       ..........(3)

[tex]6x+7y=55[/tex]       ..........(4)

Now we are multiplying equation 3 by 6 and equation 4 by 4, we get:

[tex]24x+18y=150[/tex]       ..........(5)

[tex]24x+28y=220[/tex]       ..........(6)

Now we are subtracting equation 5 from 6, we get the value of 'y'.

y = 7

Now put the value of 'y' in equation 5, we get the value of 'x'.

x = 1

Thus, the weight of first bar and second bar is, 1 kg and 7 kg respectively.

Providing lights and air for the schools in your district.

Explanation:

The major cause of the pollutant entering the atmosphere is through the use of fossil fuels for various activities. These fuels when burnt releases Carbon dioxide, Carbon monoxide, nitrogen dioxide etc gases.

Among all the given options, providing lights and air for the school in your district would consume the least power. Consuming least power would translate into the need for producing less energy hence less fossil fuel consumption. As the fossil fuels consumption would be less, less amount of harmful gases would be emitted in atmosphere thus causing less air pollution.

The scenario LEAST likely to contribute to air pollution from the use of fossil fuels is raising livestock in West Texas as it is not directly associated with the combustion of fossil fuels.

The student's question asks which scenario is LEAST likely to contribute to air pollution from the use of fossil fuels in the options provided. While burning fossil fuels releases pollutants like carbon monoxide, carbon dioxide, nitrogen dioxide, and sulfur dioxide into the atmosphere, raising livestock in West Texas (option A) is not primarily associated with burning fossil fuels. Instead, it's more related to methane emissions from the digestive processes of the cattle, which is a greenhouse gas but not a direct result of fossil fuel combustion like the other options are. Taking a cross-country trip in a charter bus (option B), providing power for the city (option C), and providing lights and air for schools in your district (option D) all involve the direct combustion of fossil fuels, and thus, contribute to air pollution related to those activities.

Answer:

Partial pressure for each of the three gases, in the mixture is 15 atm

Explanation:

Remember that the total pressure of a mixture, is the sum of partial pressures from the gases contained in the mixture.

Our total pressure = 45 atm

The 3 gases have the same pressure, so we can propose this equation:

3x = 45 atm

where x is the partial pressure for each of the three gases.

x = 45/3 → 15 atm

43.56 grams of  are produced if 16g of  CH4 reacts with 64g of O2.

Explanation:

Balance equation for the reaction:

CH4 + 2O2⇒ CO2 +2H2O

Data given : mass of CH4 =16 grams  atomic mass = 16.04 grams/mole

                    mass of water 36 gram  atomic mass = 18 grams/moles

                   mass of CO2=?                atomic mass = 44.01 grams/mole

number of moles = [tex]\frac{mass}{atomic mass of one mole}[/tex]    equation 1

number of moles in CH4

   n = [tex]\frac{16}{16.04}[/tex]

       = 0.99 moles

Since combustion is done in presence of oxygen hence it is an excess reagent and methane is limiting reagent so production of CO2 depends on it.

From the equation

1 mole of CH4 gave 1 mole of CO2

O.99 moles of CH4 will give x moles of CO2

[tex]\frac{1}{1}[/tex] = [tex]\frac{x}{0.99}[/tex]

x = 0.99 moles of carbon dioxide

grams of CO2 = number of moles x atomic mass

                         = 0.99 x 44.01

                          = 43.56 grams of CO2 is produced.

Answer:

0.066mol/dm^3

Explanation:

The molarity is simply calculating the number of moles in 1L or 1000ml

To get this, we need to know the amount in grammes that would be present in 1L.

Since 2g is sufficient for 200ml, then for 1000ml, the amount sufficient would be 5 * 2 = 10g

Now to get the number of moles needed, we need to know the molar mass of the compound. That is molar mass of FeSO4 = 56 + 32 + 4(16) = 152g/mol

The number of moles is thus 10/152 = 0.066 mol/dm^3

Answer:. 4. dialog metaphor

Explanation:

This is a metaphor of HCI in which interacting with the computer is much like carrying on a conversation or dialog. The user asks the computer for something, and the computer responds. The computer might then ask the user for something, and the user responds. The text provides an example that describes a manager and an assistant carrying on a conversation about messages .

Answer:

The sequence is: 3,1,2,4

Explanation:

It is understood as the potential of action to the electric wave that originates from the changes that the neuronal membrane undergoes due to the electrical variations and its relation between the external and internal means of the neuron. It begins in the axon cone, where a large number of sodium channels are observed. Its phases are as follows:

-resting potential

-depolarization

-repolarization

-hyperpolarization

-resting potential

-the potential for action and release of neurotransmitters

The correct steps in the generation of an action potential is 3, 1, 2, 4.

• A brief reversal of membrane potential where the membrane potential varies from -70 millivolts to +30 millivolts is termed as the action potential.  

• The action potential possess three main phases, that is, depolarization, repolarization, and hyperpolarization.  

• When the positively charged sodium ions moves into a neuron with the opening of voltage-gated sodium channels it is known as depolarization.  

• The closing of the sodium ion channels and the opening of the potassium ion channels results in repolarization.  

• Due to an excess of open potassium channels and the efflux of potassium from the cell hyperpolarization takes place.  

• The depolarization is also known as the rising phase, which results when the positively charged sodium ions suddenly rush in via the open voltage-gated sodium channels.  

• The repolarization also known as the falling phase due to slow closing of the sodium channels and the opening of the potassium channels.  

• Hyperpolarization is a phase of increased permeability of potassium resulting in excessive potassium efflux before the closing of the potassium channels.  

Thus, the correct sequence of the generation of an action potential is 3, 1, 2, 4.

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Answer:

1.38 g H₂

Explanation:

Zn  +  2 HCl  ⇒  ZnCl₂  +  H₂

To solve, you need to first find the limiting reagent.  Convert grams to moles and use the mole ratios in the chemical equation to convert from the reagent to the product.  The reagent that produces the least is the limiting reagent.

Zn

(50 g)/(65.38 g/mol) = 0.7648 mol Zn

(0.7648 mol Zn) × (1 mol H₂/1 mol Zn) = 0.7648 mol H₂

HCl

(50 g)/(36.46 g/mol) = 1.371 mol HCl

(1.371 mol HCl) × (1 mol H₂/2 mol HCl) = 0.6857 mol H₂

HCl produces less product so this is the limiting reagent.  Since you have already converted to moles, you simply need to convert from moles to grams to find the amount of hydrogen made.

(0.6857 mol) × (2.016 g/mol) = 1.38 g H₂

You will have 1.38 g of H₂.

Answer:

Hi

Explanation:

1.38 g H₂

Explanation:

Zn  +  2 HCl  ⇒  ZnCl₂  +  H₂

To solve, you need to first find the limiting reagent.  Convert grams to moles and use the mole ratios in the chemical equation to convert from the reagent to the product.  The reagent that produces the least is the limiting reagent.

Zn

(50 g)/(65.38 g/mol) = 0.7648 mol Zn

(0.7648 mol Zn) × (1 mol H₂/1 mol Zn) = 0.7648 mol H₂

HCl

(50 g)/(36.46 g/mol) = 1.371 mol HCl

(1.371 mol HCl) × (1 mol H₂/2 mol HCl) = 0.6857 mol H₂

HCl produces less product so this is the limiting reagent.  Since you have already converted to moles, you simply need to convert from moles to grams to find the amount of hydrogen made.

(0.6857 mol) × (2.016 g/mol) = 1.38 g H₂

So you will end up with 1.38 g of H₂.

Answer:

The two elements are POLONIUM and RADIUM.

Explanation:

Maria Curie is a French physicist and chemist, though she was of a Polish naturals. She was the first woman to receive a Noble Price which she earned for conducting leading and head way research on radioactivity. She discovered the theory of radioactivity; also the techniques isolating radioactive isotopes. These helped her and her husband discover Polium and Radium.

Answer: She discovered polonium and radium.

Explanation: In 1896, intrigued by the physicist Henri Becquerel’s accidental discovery of radioactivity, Curie began studying uranium rays; Pierre soon joined her in her research. Two years later, the Curies discovered polonium—named after Marie’s homeland—and radium. In 1903 they shared the Nobel Prize in physics with Becquerel for their groundbreaking work on radioactivity.

Answer:

may be exchanged for equity securities.

Explanation:

Convertible bonds -

It refers to the type of bond which can easily be converted to stocks .

It refers to as the fixed - income debt security which can give the interest payments and can be converted to some predetermined number of equity shares and common stock , is referred to as convertible bonds .

The process of conversion can be done at any time period of the bond .

Hence , from the given information of the question ,

The correct answer is may be exchanged for equity securities.

The enthalpy of formation for C6H12O6 in the fermentation of glucose can be calculated using Hess's Law and the enthalpies of formation of C2H5OH (l) and CO2 (g). The enthalpy of formation for C6H12O6 is -1273.5 kJ/mol.

The enthalpy of formation (ΔHf) for a compound is the heat released or absorbed when one mole of the compound is formed from its constituent elements in their standard states. In this case, we are given the enthalpies of formation for C2H5OH (l) and CO2 (g), and we need to calculate the enthalpy of formation for C6H12O6.

Since the balanced equation for the fermentation of glucose to produce ethyl alcohol and carbon dioxide is given, we can use Hess's Law to solve this problem. By manipulating the given equation and using the enthalpies of formation, we can determine the enthalpy of formation for C6H12O6.

Using the enthalpies of formation for C2H5OH (l) and CO2 (g), which are -277.7 kJ/mol and -393.5 kJ/mol respectively, we can calculate the enthalpy of formation for C6H12O6 to be -1273.5 kJ/mol.

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Answer :

Exothermic reactions : A, B, D

Endothermic reaction : C

Explanation :

Endothermic reaction : It is defined as the chemical reaction in which the energy is absorbed from the surrounding.

In the endothermic reaction, the energy of reactant are less than the energy of product.

Exothermic reaction : It is defined as the chemical reaction in which the energy is released into the surrounding.

In the exothermic reaction, the energy of reactant are more than the energy of product.

As we know that, heat is released in the bond formation and heat is required in the bond breaking.

Reaction (A):

[tex]N_2(g)+3H_2(g)\rightarrow 2NH_3(g)[/tex]

In this reaction, the more bonds are formed than the bond broken. That means, heat will be released. So, It is an exothermic reaction.

Reaction (B):

[tex]S(g)+O_2(g)\rightarrow SO_2(g)[/tex]

In this reaction, burning of sulfur with oxygen takes place. That means, heat will be released. So, It is an exothermic reaction.

Reaction (C):

[tex]2H_2O(g)\rightarrow 2H_2(g)+O_2(g)[/tex]

This reaction is a decomposition reaction. That means, heat will be required. So, It is an endothermic reaction.

Reaction (D):

[tex]2F(g)\rightarrow F_2(g)[/tex]

In this reaction, formation of fluorine molecule from its atoms releases heat. So, It is an exothermic reaction.

Hence, the exothermic reactions are, A, B, D and endothermic reaction is, C

Endothermic reactions absorb heat energy from their surroundings, while exothermic reactions release heat energy. The type of reaction can sometimes be predicted, but not always. Examples are provided for each reaction type.

The subject matter in focus is predicting whether reactions, specifically:

Reaction A. N2 ( g ) + 3 H2 ( g ) ⟶ 2 NH3 ( g )

Reaction B. S ( g ) + O2 ( g ) ⟶ SO2 ( g )

Reaction C. 2 H2O ( g ) ⟶ 2 H2 ( g ) + O2 ( g )

<

Reaction D. 2 F ( g ) ⟶ F2 ( g ),

will be endothermic or exothermic.

An exothermic reaction is a chemical reaction that releases energy by light or heat. It is the opposite of an endothermic reaction. Expressed in a chemical equation: reactants → products + energy.

An example of an exothermic reaction is the mixture of sodium and chlorine to yield table salt. This reaction produces a bright yellow light.

An endothermic reaction is a process or reaction that absorbs energy in the form of heat. Its enthalpy change is positive, and it takes in heat energy directly from its surroundings.

An example of an endothermic reaction is photosynthesis. Plants absorb sunlight (energy) to convert carbon dioxide and water into glucose (food) and oxygen.

Often in Chemistry, is not possible to predict directly if a reaction is exothermic or endothermic just by looking at it but there are a few cases where a good guess can be made.

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If a negatively charged ion is more concentrated inside the cell, the forces required to balance the chemical gradient would be directed Inward. Thus, the equilibrium potential for this ion would be Positively charged.

Explanation:

The measurement of potential of resting membrane is distributed unequally in the form of ions or the charged particles, which are consist between both the cell's internal structure and external structure, by the membrane's changing permeability to various ion forms.

Like in most of the neurons the potassium and organic ions which are common in amino acids are present more in internal portion of cell than its outer portion. By comparison sodium and chloride ions are normally present in the cell externally at higher concentrations. This implies there are balanced gradients of concentration around the membrane for all the most concentrated forms of ions.

Answer : The enthalpy of this reaction is, 0.975 kJ/mol

Explanation :

First we have to calculate the heat produced.

[tex]q=m\times c\times (T_2-T_1)[/tex]

where,

q = heat produced = ?

m = mass of solution = 126 g

c = specific heat capacity of water = [tex]4.18J/g^oC[/tex]

[tex]T_1[/tex] = initial temperature = [tex]21.00^oC[/tex]

[tex]T_2[/tex] = final temperature = [tex]24.70^oC[/tex]

Now put all the given values in the above formula, we get:

[tex]q=126g\times 4.18J/g^oC\times (24.70-21.00)^oC[/tex]

[tex]q=1948.716J=1.95kJ[/tex]

Now we have to calculate the enthalpy of this reaction.

[tex]\Delta H=\frac{q}{n}[/tex]

where,

[tex]\Delta H[/tex] = enthalpy change = ?

q = heat released = 1.95 kJ

n = moles of compound = 2.00 mol

Now put all the given values in the above formula, we get:

[tex]\Delta H=\frac{1.95kJ}{2.00mole}[/tex]

[tex]\Delta H=0.975kJ/mol[/tex]

Thus, the enthalpy of this reaction is, 0.975 kJ/mol

Answer:

The molecular formula is C3H6

Explanation:

Step 1: Data given

Suppose the compound has a mass of 100 grams

The compound contains:

85.63 % C = 85.63 grams C

14.37 % H = 14.37 grams H

Molar mass C = 12.01 g/mol

Molar mass H = 1.01 g/mol

Step 2: Calculate moles

Moles = grams / molar mass

Moles C = 85.63 grams / 12.01 g/mol

Moles C = 7.130 moles

Moles H = 14.37 grams / 1.01 g/mol

Moles H = 14.2 moles

Step 3: Calculate the mol ratio

We divide by the smallest amount of moles

C: 7.130 moles / 7.130 moles = 1

H = 14.2 moles / 7.130 moles = 2

The empirical formula is CH2

The molar mass of CH2 = 14 g/mol

Step 4: Calculate molecular formula

We have to multiply the empirical formula by n

n = 42 / 14 = 3

n*(CH2) = C3H6

The molecular formula is C3H6

The empirical formula of the compound is [tex]CH_2[/tex], and the molecular formula is [tex]C_2H_4[/tex].

To determine the empirical formula, we start by assuming a 100 g sample of the compound. Given the percent composition, we have 85.63 g of carbon and 14.37 g of hydrogen. We then convert these masses to moles by dividing by the molar mass of each element:

For carbon (C), the molar mass is approximately 12.01 g/mol:

Moles of C = 85.63 g / 12.01 g/mol  7.13 moles of C

For hydrogen (H), the molar mass is approximately 1.008 g/mol:

Moles of H = 14.37 g / 1.008 g/mol 14.26 moles of H

Next, we find the simplest whole-number ratio of moles of C to moles of H by dividing both by the smallest number of moles:

Dividing by the smaller number of moles (7.13 moles of C):

Ratio of C to H  7.13/7.13 : 14.26/7.13  1 : 2

Thus, the empirical formula is [tex]CH_2[/tex].

To find the molecular formula, we need the molar mass of the empirical formula and compare it to the given molar mass of the compound. The molar mass of the empirical formula [tex]CH_2[/tex] is:

Molar mass of [tex]CH_2[/tex] = (12.01 g/mol for C) + (2 × 1.008 g/mol for H) 14.026 g/mol

Now, we calculate the molecular formula by finding the ratio of the molar mass of the compound to the molar mass of the empirical formula:

Ratio = Molar mass of compound / Molar mass of empirical formula

Ratio = 42.0 g/mol / 14.026 g/mol  3

This ratio tells us that the molecular formula is three times the empirical formula, so:

Molecular formula = [tex](CH_2)_3[/tex] = [tex]C_3H_6[/tex]

However, we must check if there is a smaller whole number ratio that would give us the correct molar mass. In this case, the empirical formula [tex]CH_2[/tex] already gives us the simplest ratio, and the molecular formula must be an integer multiple of the empirical formula. The correct molecular formula that is an integer multiple and has the correct molar mass is [tex]C_2H_4[/tex] (which is also an alkene, consistent with the given percent composition and molar mass).

Answer:

Explanation:

Cells differ in the concentration of Na+  and many other chemicals inside and out side of the cell, so diffusion of Na+ ions into the cell is facilitated by the Na+ concentration gradient across the membrane.

The diffusion of K+ ions out of the cell is also prevented by the electrical gradient across the plasma membrane.

In the cell, the electro chemical gradient is larger for Na+ than for K+ and many other substances.

Ions move across the cell membrane through channels, driven by the combined forces of the concentration gradient (chemical) and the electrical gradient. This process, known as diffusion, is influenced by the electrochemical gradient. The correct option is C.

The driving forces for the diffusion of Na+ and K+ ions through their respective channels are dependent on both the concentration gradient and the electrical gradient, which together make up the electrochemical gradient. For Na+ ions, the concentration gradient facilitates their diffusion into the cell, as there are more Na+ ions outside the cell than inside (option a). However, the inside of the cell is typically negatively charged compared to the outside and there is an opposite electrical gradient, that impedes the transport of Na+ ions into the cell (option b). For the K+ ions, there are more inside the cell than outside, creating a concentration gradient that facilitates their diffusion out of the cell. However, the electrical gradient impedes this diffusion (option c).

Thus, ions move across the membrane via channels in a way that balances both the concentration (chemical) gradient and the electrical gradient to establish the electrochemical equilibrium.

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The radius of a barium atom is  r = 2.19  [tex]\times[/tex] 10^-8

Explanation:

The atomic weight of barium is 137.34.

The body-centered cubic structure has two atoms per unit cell.

Therefore, the mass of Ba in a unit cell is calculated as,

                        [tex]m =[/tex]  [tex]\frac{2 \times 137.34}{6.023 \times 10^2^3}[/tex]

                       [tex]m = 4.56 \times 10^{-22} g[/tex]

              volume = mass / density

                            [tex]= \frac{4.56 \times 10^{-22} }{3.50}[/tex] 

               volume  = [tex]1.30 \times 10^{-22} cm^3[/tex]

The edge length of a cube then is the cube root of[tex]1.30 \times 10^{-22} cm^3[/tex]  or

                     [tex]a = 5.06 \times 10^{-8}[/tex]

The body diagonal is 4 x the radius and equals a  1.732,

Therefore       r = (a [tex]\times[/tex] 1.732) / 4

                           = (5.06 [tex]\times[/tex] 10^-8

                         [tex]r = 2.19 \times10^{-8}[/tex]

The radius of a barium atom is  [tex]r = 2.19 \times10^{-8}[/tex]cm.

Answer: The average atomic mass of copper is 63.546 amu

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex]  .....(1)

Mass of isotope 1 = 62.93 amu

Percentage abundance of isotope 1 = 69.2 %

Fractional abundance of isotope 1 = 0.692

Mass of isotope 2 = 64.93 amu

Percentage abundance of isotope 2 = 30.8 %

Fractional abundance of isotope 2 = 0.308

Putting values in equation 1, we get:

[tex]\text{Average atomic mass of Copper}=[(62.93\times 0.692)+(64.93\times 0.308)]\\\\\text{Average atomic mass of Copper}=63.546amu[/tex]

Hence, the average atomic mass of copper is 63.546 amu

Answer:

Volume required from standard solution = 4675 mL

Explanation:

Given data:

Final volume = 75.0 mL

Final molarity = 130 M

Molarity of standard solution = 2.000 M

Volume required from standard solution = ?

Solution:

We use the formula,

C₁V₁ = C₂V₂

here,

C₁ = Molarity of standard solution

V₁ = Volume required from standard solution

C₂ = Final molarity

V₂ = Final volume

Now we will put the values in formula,

C₁V₁ = C₂V₂

2.000 M × V₁ = 130 M × 75.0 mL

V₁  = 9750 M. mL / 2.000 M

V₁  = 4675 mL

Answer:

We have to add 9.82 grams of calcium acetate

Explanation:

Step 1: Data given

Molarity of the calcium acetate solution = 0.207 M

Volume = 300 mL = 0.300 L

Molar mass calcium acetate = 158.17 g/mol

Step 2: Calculate moles calcium acetate

Moles calcium acetate = molarity * volume

Moles calcium acetate = 0.207 M * 0.300 L

Moles calcium acetate = 0.0621 moles

Step 3: Calculate mass calcium acetate

Mass calcium acetate = moles * molar mass

Mass calcium acetate = 0.0621 moles * 158.17 g/mol

Mass calcium acetate = 9.82 grams

We have to add 9.82 grams of calcium acetate

Answer:

The partial pressure of He = 1.8 atm

Explanation:

Step 1: Data given

The total mass = 6.0 atm

Partial pressure of Xe = 2.7 atm

Mol fraction of Ne = 0.2500

Step 2: Calculate partial pressure of Ne

Partial pressure Ne = total pressure * mol fraction

Partial pressure Ne = 6.0 atm * 0.2500

Partial pressure Ne = 1.5 atm

Step 3: Calculate the partial pressure of He

Total pressure = partial pressure of Xe + partial pressure of Ne + partial pressure of He

6.0 atm = 2.7 atm + 1.5 atm + pHe

pHe = 6.0 - 2.7 - 1.5

pHe = 1.8 atm

The partial pressure of He = 1.8 atm