Name two "Storage Polysaccharides" and two "Structural Polysaccharides"

Answer:

a) HCN  - hybridization sp

b) C(CH₃)₄ - hybridization sp³

c) H₃O⁺  - hybridization sp³

d) - CH₃ - hybridization sp³

Explanation:

Hybridization occurs to allow an atom to make more covalent bonds than the original electronic distribution would allow or to allocate ligands in an energetically stable geometry.

Carbon can have thre hybridization states: sp³ , sp² and sp.

Oxygen usualluy has an sp³ hybridization.

In order to determine the hybridization, we need to consider the number of atoms attached to the central atom and the number of lone pairs.

The figure attached shows the species and the hybridization of their central atoms.

Answer:

0.1933 M

Explanation:

Data provided in the question:

Number of Moles of Na₂SO₄ initially taken = 0.145 mol

Volume of solution, V = 750 mL = 0.750 L

Now, the molarity is given as:

Molarity = [tex]\frac{\textup{moles of solute}}{\textup{volume of solution in liter}}[/tex]

or

Molarity = [tex]\frac{0.145}{0.750}[/tex]

or

Molarity = 0.1933 M

The molarity of the solution is 0.1933 M.

To calculate the molarity of a solution, you need to know the moles of solute and the volume of solution. In this case, we have 0.145 mol of Na2SO4 and a volume of 750 mL. First, convert the volume to liters by dividing by 1000: 750 mL = 0.75 L. Now, calculate the molarity using the formula: Molarity (M) = moles of solute / volume of solution in liters.

So, Molarity = 0.145 mol / 0.75 L = 0.1933 M

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Answer: The concentration of water at equilibrium is 0.3677 mol/L

Explanation:

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric coefficients. It is represented by [tex]K_{c}[/tex]

For a general chemical reaction:

[tex]aA+bB\rightarrow cC+dD[/tex]

The [tex]K_{c}[/tex] is written as:

[tex]K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}[/tex]

The chemical equation for the conversion of methane to carbon monoxide and hydrogen gas follows:

[tex]CH_4+H_2O\rightleftharpoons 3H_2+CO[/tex]

The [tex]K_{c}[/tex] is represented as:

[tex]K_{c}=\frac{[H_2]^3[CO]}{[CH_4][H_2O]}[/tex]      ....(1)

To calculate the concentration, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]

We are given:

[tex][CO]=\frac{0.145mol}{0.379L}=0.383mol/L[/tex]

[tex][H_2]=\frac{0.218mol}{0.379L}=0.575mol/L[/tex]

[tex][CH_4]=\frac{0.25mol}{0.379L}=0.660mol/L[/tex]

[tex]K_c=0.30[/tex]

Putting values in equation 1, we get:

[tex]0.30=\frac{(0.575)^3\times 0.383}{0.660\times [H_2O]}[/tex]

[tex][H_2O]=\frac{(0.575)^3\times 0.383}{0.660\times 0.30}=0.3677[/tex]

Hence, the concentration of water at equilibrium is 0.3677 mol/L

Answer:

The answer is the option "A"

Explanation:

The reaction rate is the amount of substance formed or transformed per unit of time. The catalytic action allows to increase the reaction rate or the selectivity. The units are moles / time (mol / s)

mol = moles

s = seconds

So the correct answer is "A"

Answer: Option (a) is the correct answer.

Explanation:

As the given situation shows that water is being added to water. This will liberate heat into the surround as the reaction will be exothermic in nature.

It is known that chemical reactions in which heat is released are known as exothermic reactions.

Hence, when more water is added to this tank the temperature of the system will increase due to release of heat.

Thus, we can conclude that if more water is added to this tank the temperature of the system will increase.

Answer:

HCI (aq) + NaOH (aq) --> H2O (l) + NaCl (aq)

Explanation:

A double displacement reaction is a type of reaction where two reactants exchange ions to form two new compounds. Here we can see that both the acid and the base exchanged an ion each to for salt and water, 2 new products.

When 5.0 moles of C₄H₁₀ react with an excess of O2, 20.0 moles of CO₂ are formed.

According to the balanced chemical equation:

2C₄H₁₀(g) + 13O₂(g) ==> 8CO₂(g) + 10H₂O(L)

The stoichiometric coefficient of CO₂ is 8. This means that for every 2 moles of C₄H₁₀ reacted, 8 moles of CO₂ are formed. Therefore, when 5.0 moles of C₄H₁₀  react, the number of moles of CO₂ formed can be calculated using the stoichiometric ratio:

Number of moles of CO₂ = (5.0 moles C₄H₁₀) × (8 moles CO₂ / 2 moles C₄H₁₀ )

                       = 20.0 moles CO₂

So, when 5.0 moles of C₄H₁₀  react with an excess of O₂, 20.0 moles of CO₂ are formed.

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In the combustion of butane, 2 moles of C4H10 react to produce 8 moles of CO2, making a 1:4 ratio. Therefore, when 5.0 moles of butane reacts it produces 20 moles of CO2 due to this stoichiometric ratio.

In the balanced chemical equation for the combustion of butane, we see that '2 moles of C4H10' react to produce '8 moles of CO2'. This is a ratio of 1:4. So, for each mole of butane that reacts, 4 moles of carbon dioxide are produced.

If in a particular reaction, we have '5.0 moles of C4H10', then according to the stoichiometric ratio (1:4) we can calculate the number of moles of CO2 formed. By multiplying the moles of C4H10 by 4, i.e., 5.0 x 4, we get '20 moles of CO2'. Therefore, when 5.0 moles of butane reacts with excess oxygen, 20 moles of CO2 are produced.

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Answer: The wavelength of radiation is [tex]4.34\times 10^{21}m[/tex]

Explanation:

The relationship between wavelength of light and frequency of light is given by the equation:

[tex]\nu=\frac{c}{\lambda}[/tex]

where,

[tex]\nu[/tex] = frequency of radiation = [tex]6.912\times 10^{-14}s^{-1}[/tex]

c = speed of light = [tex]3\times 10^8m/s[/tex]

[tex]\lambda[/tex] = wavelength of radiation = ?

Putting values in above equation, we get:

[tex]6.912\times 10^{-14}s^{-1}=\frac{3\times 10^8m/s}{\lambda}\\\\\lambda=4.34\times 10^{21}m[/tex]

Hence, the wavelength of radiation is [tex]4.34\times 10^{21}m[/tex]

Answer:

The final pH is 7

Explanation:

KOH + HCl ------> KCl + H2O

KOH ----> K+  + OH-

0,3 M        0,3 M  +  0,3 M

HCl -----> H+  +  Cl-

0,2 M         0,2M   +  0,2M

In both cases they are strong acid and base because they dissociate completely; The initial molarity is the same, at the end of the dissociation.

As we have an acid next to a strong base, the final solution will have a neutralization reaction where the OH- of the base react with the H + of the acid to give water. The excess of OH- or H + defines whether the solution will have an acid or alkaline pH.

Molarity = moles / volume (L)

Moles of OH- = 0,3 M x 0,020 L = 0,006 moles

(remember volume in L, so 20mL = 0,020 L)

Moles of H+ = 0,2 M x 0,030 L = 0,006 moles

As we have the same quantity for OH- and H+, the solution will have neutral pH.

Answer : All of the above are valid expressions of the reaction rate.

Explanation :

The given rate of reaction is,

[tex]4NH_3+7O_2\rightarrow 4NO_2+6H_2O[/tex]

The expression for rate of reaction for the reactant :

[tex]\text{Rate of disappearance of }NH_3=-\frac{1}{4}\times \frac{d[NH_3]}{dt}[/tex]

[tex]\text{Rate of disappearance of }O_2=-\frac{1}{7}\times \frac{d[O_2]}{dt}[/tex]

The expression for rate of reaction for the product :

[tex]\text{Rate of formation of }NO_2=+\frac{1}{4}\times \frac{d[NO_2]}{dt}[/tex]

[tex]\text{Rate of formation of }H_2O=+\frac{1}{6}\times \frac{d[H_2O]}{dt}[/tex]

From this we conclude that, all the options are correct.

Based on the equation of the reaction, all of the given expressions are valid expressions of the reaction rate.

The rate of a chemical reaction is the rate at which reactants are used up or the rate at which products are formed.

The rate of a reaction is obtained from the balanced equation of the reaction.

The equation of the reaction is given below:

Rate of disappearance of NH3 = - 1/4 × Δ[NH3]/Δt

Rate of disappearance of O2 = - 1/7 × Δ[O2]/Δt

Rate of formation of NO2 = 1/4 × Δ[NO2]/Δt

Rate of formation of H2O = 1/6 × Δ[H2O]/Δt

Therefore, all of the above are valid expressions of the reaction rate.

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Answer:

The sample tube holds 3.0 cc which is equal to 0.0030 Liters.

Explanation:

Volume of the blood sample in the sample tube = 3.0 cc

cc = cubic centimeter ; A unit to measure a volume

1 cubic centimeter is equal to 0.001 Liters of volume.

[tex]1 cm^3=0.001 L[/tex]

So in 3.0 cubic centimeter there will be:

[tex]3.0 cm^3=3.0\times 0.001 L=0.0030 L[/tex]

The sample tube holds 3.0 cc which is equal to 0.0030 Liters.

Answer:

(a) 17,178 mg/m3

(b) 11,625 mg/m3

Explanation:

The concentration of CO in mg/m3 can be calculated as

[tex]C (mg/m3) =(P/RT)*MW*C_{ppm}[/tex]

For standard conditions (1 atm and 25°C), P/RT is 0.0409.

Concentration of 1.5% percent by volume of CO is equivalent to 1.5*10,000 ppm= 15,000 ppm CO.

The molecular weigth of CO is 28 g/mol.

(1) For 25°C and 1 atm conditions

[tex]C=(P/RT)*MW*C_{ppm}\\\\C=0.0409*28*15,000=17,178[/tex]

(b) For 200°C and 1.1 atm,

[tex]P/RT=0.0409*(P/P_{std})*(T_{std}/T)\\P/RT=0.0409*(1.1atm/1atm)*(273+15K/273+200K)=0.0277[/tex]

Then the concentration in mg/m3 is

[tex]C=(P/RT)*MW*C_{ppm}\\\\C=0.0277*28*15,000=11,625[/tex]

Answer:

x = 0.01503 = 1.503 × 10⁻² , four significant figures

Explanation:

Significant figures refer to the digits of a number that have meaning and contribute to the precision of the given number.

Since the given equation is:

[tex]\frac{1}{x}=\:66.54[/tex]

Cross-multiplying to solve for the value of x:

[tex]x \times 66.54 = 1[/tex]

⇒ [tex]x = \frac{1}{66.54}[/tex]

⇒ x = 0.01503 = 1.503 × 10⁻² , has four significant figures.

Explanation:

Actual value = 102.0 mg/dL

Calculation of absolute error and relative error for the measured value, 98.4 mg/dL

Absolute error = |Actual value - measured value|

                         = | 102.0 - 98.4 | = 3.6

Relative error = [tex]\frac{Absolute\ error}{Actual\ value}[/tex]

                       = [tex]\frac{3.6}{102.0} = 0.03529[/tex]

Calculation of absolute error and relative error for the measured value, 104.3 mg/dL

Absolute error = |Actual value - measured value|

                         = | 102.0 - 104.3 | = 2.3

Relative error = [tex]\frac{Absolute\ error}{Actual\ value}[/tex]

                       = [tex]\frac{2.3}{102.0} = 0.02255[/tex]

Calculation of absolute error and relative error for the measured value, 97.4 mg/dL

Absolute error = |Actual value - measured value|

                         = | 102.0 - 97.4 | = 4.6

Relative error = [tex]\frac{Absolute\ error}{Actual\ value}[/tex]

                       = [tex]\frac{4.6}{102.0} = 0.04509[/tex]

Calculation of absolute error and relative error for the measured value, 106.7 mg/dL

Absolute error = |Actual value - measured value|

                         = | 102.0 - 106.7 | = 4.7

Relative error = [tex]\frac{Absolute\ error}{Actual\ value}[/tex]

                       = [tex]\frac{4.7}{102.0} = 0.04607[/tex]

Calculation of absolute error and relative error for the measured value, 93.0 mg/dL

Absolute error = |Actual value - measured value|

                         = | 102.0 - 93 | = 9

Relative error = [tex]\frac{Absolute\ error}{Actual\ value}[/tex]

                       = [tex]\frac{9}{102.0} = 0.08823[/tex]

Answer:

[tex]Mg(s) + HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)[/tex]

[tex]MgO(s) + HCl(aq) \rightarrow MgCl_2(aq) + H_2O(l)[/tex]

Explanation:

When Mg reacts with HCl, magnesium chloride and hydrogen is formed. Mg is an active element and displaces hydrogen from HCl. So, this is a type of single displacement reaction.

[tex]Mg(s) + HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)[/tex]

When magnesium oxide (MgO) reacts with HCl, magnesium chloride and water is formed. This reaction is a type of neutralization reaction. MgO is a water insoluble base and HCl is acid. So. in this reaction, acid reacts with base to form salt [tex]MgCl_2[/tex]

[tex]MgO(s) + HCl(aq) \rightarrow MgCl_2(aq) + H_2O(l)[/tex]

The equation for the reaction between HCl and MgO is 2HCl(aq) + MgO(s) → MgCl2(aq) + H2O(l). The equation for the reaction between HCl and Mg is 2HCl(aq) + Mg(s) → MgCl2(aq) + H2(g). The reactions differ in terms of the reactants involved.

The equation for the reaction between HCl and MgO is:

2HCl(aq) + MgO(s) → MgCl2(aq) + H2O(l)

The equation for the reaction between HCl and Mg is:

2HCl(aq) + Mg(s) → MgCl2(aq) + H2(g)

The reactions differ in terms of the reactants involved. In the reaction with MgO, the reactant is an oxide compound (MgO) while in the reaction with Mg, the reactant is a pure metal (Mg).

Answer:

Binding affinity measures the strength of the interaction between a molecule to its ligand; it is expressed in terms of the equilibrium dissociation constant; and the higher value of this constant, the more weaker the binding between the molecule and the ligand is. On the other hand, small constans means that the interaction is tight. So "C" binds most tightly to the enzyme and "D" binds least tightly.

Answer:

No, it doesn't.

Explanation:

To convert ppb to mg/m³ we first need to convert to ppm, by just divide the amount by 1,000, so the concentration in the sample is 4.8 ppm.

mg/m³ = (ppm x molar mass)/molar volume

Using the molar mass in gram and the molar volume in liters, multiplying by the parts per million, we will get the concentration in mg/m³.

Molar mass of C = 12 g/mol; molar mass of Cl = 35.5 g/mol

Molar mass of CCl4 = 12 + 4x35.5 = 154 g/mol

Assuming, 25ºC and 1 atm, the molar volume of an ideal gas is 24.45 L, so:

mg/m³ = (4.8 x 154)/24.45

mg/m³ = 30.2

Which is higher than the limit of 12.6 mg/m³

Answer:

Sample is greater than the standard (32.96 mg/m³ > 12.6 mg/m³)

Explanation:

Please look at the solution in the attached Word file

Answer : The time taken will be 25 min.

Explanation :

First we have to determine the amount of fluid.

As, 1 mL contains 30 gtt

So, 500 mL contains [tex]\frac{500mL}{1mL}\times 30gtt=1500gtt[/tex]

Now we have to determine the time taken.

As, 60 gtt takes time = 1 min

So, 1500 gtt takes time = [tex]\frac{1500gtt}{60gtt}\times 1min=25min[/tex]

Therefore, the time taken will be 25 min.

Final answer:

To administer 500 mL of fluid with a drop factor of 30 gtt/mL at a drip rate of 60 gtt/min, the calculation reveals it will take 250 minutes.

Explanation:

To calculate the time required to administer exactly 500 mL of fluid through an IV with a drop factor of 30 gtt/mL at a drip rate of 60 gtt/min, we must first understand the given terms. The drop factor (gtt/mL) is a measure indicating the number of drops (gtt) that make up 1 mL of fluid. The drip rate (gtt/min) specifies how many drops of fluid are administered per minute.

First, we calculate the total number of drops in 500 mL of fluid, using the drop factor:

Next, to find out how long it will take to administer these 15000 drops at a rate of 60 drops per minute, we divide the total number of drops by the drip rate:

Therefore, it will take 250 minutes to administer exactly 500 mL of fluid through the IV at the specified conditions.

Answer: The cell potential of the cell is 0.31 V

Explanation:

We know that:

[tex]E^o_{(Br_2/2Br^-)}=1.07V\\E^o_{(MnO_4^-/Mn^{2+})}=1.51V[/tex]

The substance having highest positive [tex]E^o[/tex] potential will always get reduced and will undergo reduction reaction. Here, [tex]MnO_4^-[/tex] will undergo reduction reaction will get reduced. And, bromine will get oxidized.

Oxidation half reaction: [tex]2Br^-(aq.)+2e^-\rightarrow Br_2(l);E^o_{(Br_2^/2Br^-)}=1.07V[/tex]      ( ×  5)

Reduction half reaction: [tex]MnO_4^-(aq.)+8H^+(aq.)+5e^-\rightarrow Mn^{2+}(aq.)+4H_2O(l);E^o_{(MnO_4^-/Mn^{2+})}=1.51V[/tex]  ( ×  2)

The net reaction follows:

[tex]2MnO_4^-(aq.)+10Br^-(aq.)+16H^+(aq.)\rightarrow 2Mn^{2+}(aq.)+5Br_2(l)+8H_2O(l)[/tex]

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:

[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]

Putting values in above equation, we get:

[tex]E^o_{cell}=1.51-(1.07)=0.44V[/tex]

To calculate the EMF of the cell, we use the Nernst equation, which is:

[tex]E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]^2}{[MnO_4^{-}]^2\times [Br^-]^{10}\times [H^+]^{16}}[/tex]

where,

[tex]E_{cell}[/tex] = electrode potential of the cell = ?V

[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = 0.44 V

n = number of electrons exchanged = 10

[tex][H^{+}]=1M[/tex]

[tex][Mn^{2+}]=0.15M[/tex]

[tex][MnO_4^{-}]=0.01M[/tex]

[tex][Br^{-}]=0.01M[/tex]

Putting values in above equation, we get:

[tex]E_{cell}=0.44-\frac{0.059}{10}\times \log(\frac{(0.15)^2}{(0.01)^2\times (0.01)^{10}\times (1)^{16}})\\\\E_{cell}=0.31V[/tex]

Hence, the cell potential of the cell is 0.31 V

Answer:

10,93% of a carrot is carbohydrate.

Explanation:

In order to solve this you just have to create a rule of three where 64g is the 100% and you want to calculate how much is 7g in percentage:

[tex]\frac{64}{100}= \frac{7}{x}\\ x=\frac{7*100}{64}\\ x=10,9375[/tex]

So there are 10.9375 % carbohydrates in a piece of carrot.

The percent, by mass, of carbohydrate in a medium-sized carrot is approximately 10.9%.

To find the percent, by mass, of carbohydrate in a medium-sized carrot, we can use the following formula:

Percent mass of carbohydrate = (mass of carbohydrate / total mass) × 100

In this case, the mass of carbohydrate is given as 7.0 g and the total mass of the carrot is 64 g. Plugging these values into the formula:

Percent mass of carbohydrate = (7.0 g / 64 g) × 100 = 10.9%

Therefore, approximately 10.9% of the carrot's mass is carbohydrate.

Answer:

(a) Utilization

Explanation:

Thermodynamic energy are:

Kinetic energy is that type of energy of a body which occurs due to the motion of body. Kinetic energy is always positive. In a bound system, the system remains bound as long as the kinetic energy is less than the potential energy due to the interaction of the body.

                            [tex]K.E = \frac{1}{2}mv^{2}[/tex]

Potential energy is defined as the energy in which the energy is possessed by a body or a system for doing some work, by virtue of its position above the ground level.

Therefore,

Potential energy = PE = mgh

This potential energy is a result of gravity pulling downwards. The gravitational constant, g, is the acceleration of an object due to gravity. This acceleration is about 9.8 m s⁻².

Stored energy is a Potential energy.

Hence, the only energy which is not a thermodynamic energy is option (a) Utilization energy.

Answer:

Ion-ion force between Na+ and Cl− ions

London dispersion force between two hexane molecules

Explanation:

"Ion-dipole force between Na+ ions and a hexane molecule " does not exist since hexane has only non-polar bonds and therefore no dipole.

"Ion-ion force between Na+ and Cl− ions " exists since both are ions.

"Dipole-dipole force between two hexane molecules " does not exist since hexane molecules do not have a dipole.

"Hydrogen bonding between Na+ ions and a hexane molecule " does not exist since the hydrogen in the hydrogen bond must be bonded directly to an electronegative atom, which hexane does not have since it is a hydrocarbon.

"London dispersion force between two hexane molecules" exist since hexane is a molecular compound.

Answer:

Ion-ion force between Na⁺ and Cl⁻ ions.

London dispersion force between two hexane molecules.

Explanation:

Suppose that NaCl is added to hexane (C₆H₁₄) instead of water. Which of the following intermolecular forces will exist in the system?

Check all that apply.

Explanation:

Silly Putty -

It is an inorganic polymer , in which the molecules are linked via covalent bonding . And the bonds can easily be broken , by the application of even a small amount of pressure .

The composition of silly putty is , the Elmer's glue solution and the sodium borate in the ration , 4 : 1 .

silly putty is polar , hydrophilic and binding in nature .

Candle Wax -

The Wax of the candle is an ester of the ethylene glycol and fatty acids , and the chemical properties of was is , it is hydrophobic in nature and is insoluble in water as well .

The principles affecting the properties of silly putty and candle wax involve their hydrophobic, nonpolar nature, making them insoluble in water and more soluble in nonpolar solvents like heptane. Solubility depends on the balance of intermolecular forces and molecule polarity, with polar and ionic substances being hydrophilic. Heat transfer during solvation is dictated by the relative strength of solute-solvent interactions compared to original solute-solute and solvent-solvent interactions.

The chemistry principle for substances such as silly putty and candle wax as it relates to hydrophilic/hydrophobic, polar/nonpolar, and soluble/non-soluble properties is rooted in their intermolecular forces and molecular structure. Silly putty, which is a silicone polymer, is primarily hydrophobic and non-polar, making it insoluble in water and more soluble in nonpolar solvents. On the other hand, candle wax, typically composed of long-chain hydrocarbons, is also nonpolar and hydrophobic, displaying similar solubility characteristics. Things that are polar and can dissolve in water are referred to as being hydrophilic, which allows them to form hydrogen bonds with water. This property is not present in nonpolar substances like silly putty and candle wax, which resist mixing with water.

In terms of solubility predictions for various substances:
Vegetable oil is nonpolar and more soluble in a nonpolar solvent like heptane.

Isopropyl alcohol is polar due to the presence of an -OH group and more soluble in water.

Potassium bromide is ionic and highly soluble in water which is a polar solvent.

When solutions form, heat may be released or absorbed depending on the intermolecular forces involved. In solvation where the solute-solvent interactions are stronger than the solute-solute and solvent-solvent interactions, heat is released (exothermic). Conversely, if breaking the original interactions requires more energy than is released during new interaction formation, the process is endothermic and absorbs heat.

Answer:

Explanation:

The equilibrium that takes place is:

H₂PO₄⁻ ↔ HPO₄⁻² + H⁺    pka= 7.21 (we know this from literature)

To solve this problem we use the Henderson–Hasselbalch (H-H) equation:

pH = pka + [tex]log\frac{[A^{-} ]}{[HA]}[/tex]

In this case [A⁻] is [HPO₄⁻²], [HA] is [H₂PO₄⁻], pH=7.0, and pka = 7.21

If we use put data in the H-H equation, and solve for [HPO₄⁻²], we're left with:

[tex]7.0=7.21+log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\ -0.21=log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\\\10^{-0.21} =\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\0.616 * [H2PO4^{-}] = [HPO4^{-2}][/tex]

From the problem, we know that [HPO₄⁻²] + [H₂PO₄⁻] = 0.1 M

We replace the value of [HPO₄⁻²] in this equation:

0.616 * [H₂PO₄⁻] + [H₂PO₄⁻] = 0.1 M

1.616 * [H₂PO₄⁻] = 0.1 M

[H₂PO₄⁻] = 0.0619 M

With the value of [H₂PO₄⁻]  we can calculate [HPO₄⁻²]:

[HPO₄⁻²] + 0.0619 M = 0.1 M

[HPO₄⁻²] = 0.0381 M

With the concentrations, the volume and the molecular weights, we can calculate the masses:

Answer:

a) k = [tex]\frac{[C][D]}{[A][B]}[/tex]

b) A value of 45000 means that tendency of your reaction is to have more products in a ratio of 45000:1

c) No, the products are more stable than reactants.

Explanation:

a) For a reaction:

A + B → C + D

The equlibrium constant (k) is:

k = [tex]\frac{[C][D]}{[A][B]}[/tex]

Where [x] is the molar concentration of x. Always the expression of equilibrium constant is molar concentration of products over molar concentration of reagents.

b) Having in mind the expression of equilibrium constant when k>1 the concentration of products is higher than concentration of reagents. Thus, when k<1 concentration of reagents is higher than concentration of products.

A value of 45000 means that tendency of your reaction is to have more products in a ratio of 45000:1

c) Again, a value of 45000 means that tendency of your reactants is react to produce products. Thus, the products are more stable than reactants.

I hope it helps!

Answer:

1-condensation of ethanol

2-evaporation of ethanol

Explanation:

According to the question ,

Answer : The value of the rate constant 'k' for this reaction is [tex]1.9375\times 10^{-3}M^{-2}s^{-1}[/tex]

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

[tex]A+B+C\rightarrow D+E[/tex]

Rate law expression for the reaction:

[tex]\text{Rate}=k[A]^a[B]^b[C]^c[/tex]

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

[tex]1.24\times 10^{-4}=k(0.40)^a(0.40)^b(0.40)^c[/tex] ....(1)

Expression for rate law for second observation:

[tex]3.6\times 10^{-4}=k(0.40)^a(0.40)^b(1.20)^c[/tex] ....(2)

Expression for rate law for third observation:

[tex]4.8\times 10^{-4}=k(0.80)^a(0.40)^b(0.40)^c[/tex] ....(3)

Expression for rate law for fourth observation:

[tex]4.8\times 10^{-4}=k(0.80)^a(0.80)^b(0.40)^c[/tex] ....(4)

Dividing 1 from 2, we get:

[tex]\frac{3.6\times 10^{-4}}{1.24\times 10^{-4}}=\frac{k(0.40)^a(0.40)^b(1.20)^c}{k(0.40)^a(0.40)^b(0.40)^c}\\\\3=3^c\\c=1[/tex]

Dividing 1 from 3, we get:

[tex]\frac{4.8\times 10^{-4}}{1.24\times 10^{-4}}=\frac{k(0.80)^a(0.40)^b(0.40)^c}{k(0.40)^a(0.40)^b(0.40)^c}\\\\4=2^a\\a=2[/tex]

Dividing 3 from 4, we get:

[tex]\frac{4.8\times 10^{-4}}{4.8\times 10^{-4}}=\frac{k(0.80)^a(0.40)^b(0.40)^c}{k(0.80)^a(0.80)^b(0.40)^c}\\\\1=2^b\\b=0[/tex]

Thus, the rate law becomes:

[tex]\text{Rate}=k[A]^2[B]^0[C]^1[/tex]

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

[tex]1.24\times 10^{-4}=k(0.40)^2(0.40)^0(0.40)^1[/tex]

[tex]k=1.9375\times 10^{-3}M^{-2}s^{-1}[/tex]

Hence, the value of the rate constant 'k' for this reaction is [tex]1.9375\times 10^{-3}M^{-2}s^{-1}[/tex]

Answer:

Yes, there is a loss of 8350 J of energy in the form of heat

Explanation:

The principle of energy conservation is described mathematically

as the energy conservation equation as follows:

                            ΔK+ΔU=Q+W        

where:

Analyzing each term of the equation:

   [tex]ΔK=\frac{1}{2}mv_{f} ^{2}  - \x]frac{1}{2}mv_{i} ^{2}=\frac{1}{2}*100*15^{2} - \frac{1}{2}*100*0^{2}=11250 J[/tex]

ΔU=[tex]mgh_{f}-mgh_{i}=100*9,8*0-100*9,8*20=-19600[J][/tex]

Q=?

W=0 [J]

Replacing in the main equation:

11250-19600=Q+0

Q= -8350 [J]

So, the answer is YES, there is a loss of 8350 J of  energy in the form of heat.

Answer:

True

Explanation:

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Hence , the statement is correct .