1.50 × 104 J of energy is transferred thermally into a huge tank filled with liquid water. The water temperature remains constant at 10.0 ∘C during the process.Part ABy how much does the entropy of the water change?

Answer: The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

Explanation:

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

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]       ....(1)

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:  

[tex]\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol[/tex]

For the given chemical reaction:

[tex]2Fe(s)+O_2(g)\rightarrow 2FeO(s)[/tex]

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = [tex]\frac{2}{2}\times 0.0771=0.0771mol[/tex] of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:  

[tex]0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g[/tex]

To calculate the percentage yield of iron (ii) oxide, we use the equation:

[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

[tex]\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%[/tex]

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

The question relates to calculating the theoretical and percent yield of iron(II) oxide from a chemical reaction between iron and oxygen. The theoretical yield has been stated as 21.6 grams, and the percent yield as 85%. However, without a balanced chemical equation and proper stoichiometric calculations provided, the theoretical yield cannot be confirmed, and the actual percent yield calculation with the given numbers does not match the stated 85%.

The question is asking for the theoretical and percent yield of iron(II) oxide when iron reacts with excess oxygen. In stoichiometry, the theoretical yield is the maximum amount of product that can be produced from a given amount of reactants, while the percent yield is a comparison of the actual yield to the theoretical yield, calculated as (actual yield/theoretical yield) × 100%.

To calculate the theoretical yield of iron(II) oxide, we would need to use the balanced equation for the reaction between iron and oxygen. However, the information provided is incomplete for this calculation. The student has provided the value of 21.6 grams as the theoretical yield without the necessary calculations or a balanced equation presented, and a percent yield of 85%. To determine the percent yield, you would divide the actual yield of iron(II) oxide (5.17 grams) by the theoretical yield (21.6 grams) and multiply by 100%.

However, without a balanced chemical equation and complete stoichiometric calculations, we cannot confirm the provided theoretical yield. If the provided theoretical yield of 21.6 grams is correct, then the percent yield calculation would be (5.17 grams / 21.6 grams) × 100%, giving an actual percent yield of 23.94%, not 85%.

Answer:

8 to 1.

Explanation:

O₂ + 2H₂ → 2H₂O.

It is clear that one mole of oxygen combines with two moles of hydrogen atoms to form 2 moles of water.

So, the molar ratio of oxygen to hydrogen is (1 to 2).

So, the mass ratio of oxygen to hydrogen (32.0 g/4.0 g) = (8: 1).

Final answer:

To determine which reactant is in excess, compare the number of moles of each reactant with the stoichiometric ratio in the balanced equation. In this case, NO2 is in excess.

Explanation:

To determine which reactant is in excess, we need to compare the number of moles of each reactant with the stoichiometric ratio in the balanced equation.

The balanced equation is:

3NO2(g) + H2O(l) -> 2HNO3(l) + NO(g)

According to the equation, the ratio of NO2 to H2O is 3:1. So, for every 3 moles of NO2, we need 1 mole of H2O.

Given that there are 4.2 mol of NO2 and 0.50 mol of H2O, we can calculate the mole ratio of NO2 to H2O:

4.2 mol NO2 / 3 mol NO2 = 1.4 mol NO2

0.50 mol H2O / 1 mol H2O = 0.50 mol H2O

From the calculations, it is clear that the amount of NO2 is more than the required amount based on the stoichiometric ratio. Therefore, NO2 is in excess.

Answer:

Buffer 1.

Explanation:

Ammonia is a weak base. It acts like a Bronsted-Lowry Base when it reacts with hydrogen ions.

[tex]\rm NH_3\; (aq) + H^{+}\; (aq) \to {NH_4}^{+}\; (aq)[/tex].

[tex]\rm NH_3[/tex] gains one hydrogen ion to produce the ammonium ion [tex]\rm {NH_4}^{+}[/tex]. In other words, [tex]\rm {NH_4}^{+}[/tex] is the conjugate acid of the weak base [tex]\rm NH_3[/tex].

Both buffer 1 and 2 include

The ammonia [tex]\rm NH_3[/tex] in the solution will react with hydrogen ions as they are added to the solution:

[tex]\rm NH_3\; (aq) + H^{+}\; (aq) \to {NH_4}^{+}\; (aq)[/tex].

There are more [tex]\rm NH_3[/tex] in the buffer 1 than in buffer 2. It will take more strong acid to react with the majority of [tex]\rm NH_3[/tex] in the solution. Conversely, the pH of buffer 1 will be more steady than that in buffer 2 when the same amount of acid has been added.

Answer: The concentration of [tex]HNO_3[/tex] solution will be 0.094 M.

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HNO_3[/tex]

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is [tex]Ba(OH)_2[/tex]

We are given:

Conversion factor: 1L = 1000 mL

[tex]n_1=1\\M_1=?M\\V_1=0.110L=110mL\\n_2=2\\M_2=0.100M\\V_2=51.9mL[/tex]

Putting values in above equation, we get:

[tex]1\times M_1\times 110=2\times 0.100\times 51.9\\\\M_1=0.094M[/tex]

Hence, the concentration of [tex]HNO_3[/tex] solution will be 0.094 M.

The concentration of the HNO3 solution can be found by calculating the moles of Ba(OH)2 used in neutralization, which gives us the moles of HNO3 due to reaction stoichiometry. The HNO3 concentration (molarity) is then found by dividing its moles by its volume.

In the given neutralization reaction, two moles of HNO3 react with one mole of Ba(OH)2. Given that 51.9 mL of 0.100 M Ba(OH)2 is used to completely neutralize the solution, the moles of Ba(OH)2 used can be obtained by using the formula Molarity (M) = Moles/Liter and converting the mL to L. The reaction stoichiometry implies that the moles of HNO3 must be twice that of Ba(OH)2.

Therefore, the total moles of HNO3 can be calculated by doubling the moles of Ba(OH)2. Divide these moles by the volume of the HNO3 solution (0.110 L) to find its molarity.

To summarize, the calculation process involves finding out the moles of Ba(OH)2 used, using this to find the moles of HNO3, and then dividing this by the volume of HNO3 solution to yield the molarity. Please carry out these calculations to get your final answer.

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Answer:

1,4-dihydro = 113 kJ·mol⁻¹

1,2-dihydro = 101 kJ·mol⁻¹

Explanation:

In 1,4-dihydronaphthalene, the 2,3-double bond is isolated from the benzene ring.

In 1,2-dihydronaphthalene, the 3,4-double bond is conjugated with the benzene ring.

Thus, 1,2-dihydronaphthalene is partially stabilized by resonance interactions between the ring and the double bond (think, styrene).

1,2-Dihydronaphthalene is at a lower energy level because of this stabilization.

The heat of hydrogenation of 1,2-dihydronaphthalene is therefore less than that of the 1,4-isomer when each is hydrogenated to the common product, 1,2,3,4-tetrahydronaphthalene.

Answer:

Here's what I get.

Explanation:

(a) The buffer equilibrium

The equation for the buffer equilibrium is

[tex]\rm CH_{3}COOH(aq) + H$_{2}$O(l) $\, \rightleftharpoons \,$ CH$_{3}$COO$^{-}$(aq) + H$_{3}$O$^{+}$(aq)[/tex]

(b) Addition of acid

If you add a strong acid like HNO₃, you are increasing the concentration of hydronium ion.

Per Le Châtelier's Principle, the system will respond in such a way as to decrease the concentration of hydronium ion.

The position of equilibrium will shift to the left.

(c) Addition of base.

If you add a strong base like KOH, The hydroxide ions will react with the hydronium ions to form water.

The concentration of hydronium ions will decrease.

Per Le Châtelier's Principle, the system will respond in such a way as to increase the concentration of hydronium ions.

The position of equilibrium will shift to the right.

When a strong acid is added to a buffer solution containing CH3COOH and CH3COO-, the equilibrium shifts to the left. When a strong base is added, the equilibrium shifts to the right.

A buffer solution containing CH3COOH and CH3COO- can resist changes in pH when small amounts of a strong acid or base are added. When a strong acid such as HNO3 is added to the buffer system, the equilibrium will shift to the left, favoring the formation of more CH3COOH to consume the added H3O+ ions. This maintains the pH of the buffer solution. On the other hand, when a strong base such as KOH is added to the buffer system, the equilibrium will shift to the right, favoring the formation of more CH3COO- ions to consume the added OH- ions. This also helps to maintain the pH of the buffer solution.

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Answer:

0.05 moles of sodium sulfate are required.

Explanation:

[tex]Na_2SO_4(s) \rightarrow 2Na^+ (aq) SO_4^{2-} (aq)[/tex]

Concentration of sodium ions in 1 L = 0.10 M

[tex]0.10 M=\frac{\text{Moles of sodium ions}}{1.0 L}[/tex]

Moles of sodium ions = 0.10 mol

1 mole of sodium  sulfate gives 2 moles of sodium ions.

Then 0.10 mol of sodium ions will be given by:

[tex]\frac{1}{2}\times 0.10 mol=0.05 mol[/tex]

0.05 moles of sodium sulfate are required.

Explanation:

A precipitate is defined as an insoluble substance that emerges upon mixing of two aqueous solutions.

For example, [tex]2NaOH(aq) + Sr(NO_{3})_{2}(aq) \rightarrow Sr(OH)_{2}(s) + 2NaNO_{3}(aq)[/tex]

As precipitate is a solid and it is represented by (s). And, an aqueous solution is represented by (aq).

So, the products of the given reactants will be as follows.

[tex]KOH(aq) + HNO_{3} \rightarrow KNO_{3}(aq) + H_{2}O(aq)[/tex]

[tex]AgC_{2}H_{3}O_{2}(aq) + HC_{2}H_{3}O_{2}(aq) \rightarrow \text{No reaction}[/tex]

[tex]Pb(NO_{3})_{2}(aq) + HCl(aq) \rightarrow PbCl_{2}(s) + 2HNO_{3}(aq)[/tex]

[tex]Cu(NO)_{3}_{2}(aq) + CH_{3}COOK(aq) \rightarrow Cu(CH_{3}COO)_{2})(aq) + 2KNO_{3}(aq)[/tex]

[tex]2NaOH(aq) + Sr(NO_{3})_{2}(aq) \rightarrow Sr(OH)_{2}(s) + 2NaNO_{3}(aq)[/tex]

Hence, we can conclude that out of the given options, these two equations will produce a precipitate.

[tex]Pb(NO_{3})_{2}(aq) + HCl(aq) \rightarrow PbCl_{2}(s) + 2HNO_{3}(aq)[/tex]

[tex]2NaOH(aq) + Sr(NO_{3})_{2}(aq) \rightarrow Sr(OH)_{2}(s) + 2NaNO_{3}(aq)[/tex]

The combinations of Pb(NO3)2 (aq) and HCl (aq), and NaOH (aq) and Sr(NO3)2 (aq) will produce precipitates, as the resultant compounds (PbCl2 and Sr(OH)2 respectively) are insoluble in solutions.

In the context of chemistry, a precipitate is a solid that forms in a solution during a chemical reaction. The combination that will produce a precipitate can be predicted using solubility rules, which detail the solubility of different compounds.

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Answer:

[tex]\boxed{\text{-194 kJ/mol}}[/tex]

Explanation:

Mn + 2HCl ⟶ MnCl₂ + H₂

There are two energy flows in this reaction.

[tex]\begin{array}{cccl}\text{Heat from reaction} & + &\text{Heat absorbed by water} & = 0\\q_{1} & + & q_{2} & =0\\n\Delta_{r}H & + & mC\Delta T & =0\\\end{array}[/tex]

Data:

Mass of Mn = 0.625 g  

                V = 100.0 mL

               T₁ = 23.5 °C

               T₂ = 28.8 °C

Calculations:

(a) Moles of Mn

[tex]n = \text{0.625 g Mn} \times \dfrac{\text{1 mol Mn }}{\text{54.94 g Mn}} = \text{0.011 38 mol Mn}[/tex]

(b) Mass of solution

[tex]m = \text{100.0 mL} \times \dfrac{\text{1.00 g}}{\text{1 mL}} = \text{100.0 g}[/tex]

(c) ΔT

ΔT = T₂ - T₁ = 28.8 °C – 23.5 °C = 5.3 °C

(d) q₁

[tex]q_{1} = \text{0.011 38 mol Mn} \times \Delta_{r}H = 0.01138 \Delta_{r}H \text{ mol}[/tex]

(e) q₂

q₂ = 100.0 × 4.184 × 5.3  = 2220 J

(f) ΔH

[tex]\begin{array}{rcl}0.01138 \Delta_{r}H + 2220 & = & 0\\0.01138 \Delta_{r}H & = & -2220\\\\\Delta_{r}H & = & \dfrac{-2220}{0.01138}\\\\ & = & \text{-194000 J/mol}\\ & = & \boxed{\textbf{-194 kJ/mol}}\\\end{array}\\\\[/tex]

The ΔHrxn for the reaction as written is -201.4 KJ/mol.

From the information provided in the question;

Mass of Mn = 0.625 g

Volume of solution = 100.0 mL

Initial temperature = 23.5 °C

Final temperature =  28.8 °C

Now;

The equation of the reaction is;

Mn(s) + 2HCl(aq) ------> MnCl2(aq) + H2(g)

Number of moles of Mn =  0.625 g /55 g/mol = 0.011 moles

Temperature rise = Final temperature - Initial temperature =  28.8 °C - 23.5 °C = 5.3 °C

Mass of the solution = Density of solution × volume of solution = 1.00 g/mL ×  100.0 mL = 100 g

From the formula;

ΔHrxn =- mcθ

ΔHrxn  is negative because heat is evolved.

m = mass of solution

c = specific heat capacity of the solution

θ= temperature rise

ΔHrxn = mcθ/number of moles

ΔHrxn =-(100 g × 4.18 J/g∘C  × 5.3 °C)/0.011 moles

ΔHrxn = -201.4 KJ/mol

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When 0.620 g Mn is combined with enough hydrochloric acid to make 100.0 mL of solution in a coffee-cup calorimeter, all of the Mn reacts, raising the temperature of the solution from 23.1 ∘C to 28.9 ∘C. Find ΔHrxn for the reaction as written. (Assume that the specific heat capacity of the solution is 4.18 J/g∘C and the density is 1.00 g/mL.)

1. Double replacement but no reaction

2. Double replacement but no reaction

3. Combustion

4. Combustion

5. Single replacement

6. Double replacement

I didn't solve the ones that are repeated! Hope it helped;)

Answer:

1. Double displacement.

2. Combustion.

3. Simple displacement.

4.Double displacement.

Explanation:

Hello,

1. [tex]2HBr(aq)+Ba(OH)_2(aq) -->2H_2O(l)+BaBr_2(aq)[/tex]

In this case, it is about a double displacement reaction since all the cations (H and Ba) and anions (Br and OH) are exchanged.

2. [tex]C_2H_4(g)+3O_2(g)-->2CO_2(g)+2H_2O(l)[/tex]

In this case, it is about the combustion of ethene.

3. [tex]Cu(s)+FeCl_2(aq)-->Fe(s)+CuCl_2(aq)[/tex]

In this case, it is about a simple displacement reaction since the iron (II) cations become solid iron and on the contrary for copper.

4. [tex]Na_2S(aq)+2AgNO_3(aq)-->2NaNO_3(aq)+Ag_2S(s)[/tex]

Finally, it is about a double displacement chemical reaction since the sodium and silver cations are exchanged with the sulfide and nitrate anions.

Best regards.

Answer: The half-life of the reaction in 2.24 seconds.

Explanation:

We are given a reaction which follows first order kinetics.

The formula used to calculate the half -life of the reaction for first order kinetics follows:

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

where,

[tex]t_{1/2}[/tex] = half-life of the reaction

k = rate constant of the reaction = [tex]0.310s^{-1}[/tex]

Putting values in above equation, we get:

[tex]t_{1/2}=\frac{0.693}{0.310s^{-1}}\\\\t_{1/2}=2.235sec\approx 2.24sec[/tex]

The rule which is applied for multiplication and division problems is that the least number of significant figures in any number of a problem will determine the number of significant figures in the solution.

In the problem, the least precise significant figures are 3. Thus, the answer will also have 3 significant figures.

Hence, the half-life of the reaction in 2.24 seconds.

Answer:

Solubility of O₂(g) in 4L water = 3.42 x 10⁻² grams O₂(g)

Explanation:

Graham's Law => Solubility(S) ∝ Applied Pressure(P) => S =k·P

Given P = 0.209Atm & k = 1.28 x 10⁻³mol/L·Atm

=> S = k·P = (1.28 x 10⁻³ mole/L·Atm)0.209Atm = 2.68 x 10⁻³ mol O₂/L water.

∴Solubility of O₂(g) in 4L water at 0.209Atm = (2.68 x 10⁻³mole O₂(g)/L)(4L)(32 g O₂(g)/mol O₂(g)) = 3.45 x 10⁻² grams O₂(g) in 4L water.

0.164 grams of oxygen gas was dissolved in 4.00 Liters of water.

0.0343 grams of oxygen gas was dissolved in 4.00 Liters of water.

Explanation:

         [tex]S=k_H\times p_o[/tex]

      Where:

            S = Solubility of gas in liquid

           [tex]k_H[/tex]= Henry's law constant

            [tex]p_o[/tex]= Partial pressure of a gas

Given:

The Henry's law constant for oxygen gas in water at 20°C is [tex]1.28 \times 10^{-3} mol/(Latm)[/tex]

To find:

a) Mass of oxygen gas in 4.00 Liter of water with pure oxygen at 1.00 atm.

b) Mass of oxygen gas in 4.00 Liter of water with oxygen gas at a partial pressure of 0.209 atm.

Solution:

a)

The Henry's law constant for oxygen gas in water at 20°C = [tex]k_H[/tex]= [tex]1.28 \times 10^{-3} mol/(Latm)[/tex]

The pressure of pure oxygen gas above water = [tex]p_o=1.00 atm[/tex]

The solubility of the oxygen gas in water:

[tex]S=1.28\times 10^{-3} mol/(Latm)\times 1.00 atm\\S=1.28\times 10^{-3} mol/L[/tex]

There are [tex]1.28\times 10^{-3}[/tex] moles of oxygen gas in 1 liter of water

The volume of the water = 4.00 L

Moles of oxygen gas in 4.00 L of water:

[tex]=1.28\times 10^{-3}\times 4.00 mol=5.12\times 10^{-3}mol[/tex]

Mass of [tex]5.12\times 10^{-3}[/tex] moles of oxygen gas:

[tex]=5.12\times 10^{-3}mol\times 31.998 g/mol=0.164 g[/tex]

0.164 grams of oxygen gas was dissolved in 4.00 Liters of water.

b)

The Henry's law constant for oxygen gas in water at 20°C = [tex]k_H[/tex]= [tex]1.28 \times 10^{-3} mol/(Latm)[/tex]

The partial pressure of oxygen gas above water = [tex]p_o=0.209atm[/tex]

The solubility of the oxygen gas in water:

[tex]S=1.28\times 10^{-3} mol/(Latm)\times 0.209atm\\S=2.68\times 10^{-4} mol/L[/tex]

There are [tex]2.68\times 10^{-4}[/tex] moles of oxygen gas in 1 liter of water

The volume of the water = 4.00 L

Moles of oxygen gas in 4.00 L of water:

[tex]=2.68\times 10^{-4}\times 4.00 mol=1.072\times 10^{-3}mol[/tex]

Mass of [tex]1.072\times 10^{-3}[/tex] moles of oxygen gas:

[tex]=1.072\times 10^{-3}mol\times 31.998 g/mol=0.0343g[/tex]

0.0343 grams of oxygen gas was dissolved in 4.00 Liters of water.

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Answer:

  4:5

Explanation:

Let x represent the fraction of the mix that is hot water. Then the temperature of the mix is ...

  60x +15(1-x) = 40·1

  45x = 25 . . . . . . . . . subtract 15

  x = 25/45 = 5/9 . . . divide by the coefficient of x

This is the fraction that is hot water, so the fraction that is cold water is ...

  1-5/9 = 4/9

The ratio of cold to hot is ...

  cold : hot = (4/9) : (5/9) = 4 : 5

_____

Additional comments

The problem assumes that the energy contained in a given mass of water is proportional to its temperature. That is almost true, sufficiently so that we can reasonably use that approximation.

If heat loss is figured into the problem, then additional information is needed regarding the energy content of water at temperatures in the range of interest. That is not provided by this problem statement, so we have ignored the heat loss.

Answer:

A. One or two letters are used to represent the name of each element

APEX

One or two letters are used to represent the name of each

element

The name of each element in the periodic table is symbolized by one or two letters.

There are different ways of achieving this. Sometimes;

These are some of the ways of symbolizing the name of each element.

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Kp calculated using equilibrium partial pressures of reactants and products is 0.206198. An ICE table helps determine the changes in pressures of H₂, Cl₂, and HCl during the reaction. After finding the equilibrium pressures, the Kp value is determined using the equilibrium constant expression.

To calculate the equilibrium constant Kp for the reaction H₂(g) + Cl₂(g) ⇌ 2 HCl(g), we can use the equilibrium partial pressures of the reactants and products. Based on the reaction stoichiometry, the change in pressure for H₂ and Cl₂ should be equal and opposite to the change for HCl, since two moles of HCl are produced for each mole of H₂ and Cl₂ that react.

Using the equilibrium pressure of HCl (1.418 atm) and the initial pressures of H₂ (4.06 atm) and Cl₂ (3.5 atm), we can set up an ICE table to find the changes in pressure (Δ) during the reaction and then the equilibrium pressures of H₂ and Cl₂.

Let x represent the change in pressure for H₂ and Cl₂. Since the ratio of HCl to H₂ and Cl₂ in the balanced equation is 2:1, the change in pressure for HCl will be 2x. If 1.418 atm is the equilibrium pressure of HCl, the change is the same amount (since HCl starts at 0 atm), and thus x is half of this value, which is 0.709 atm.

We can now calculate the equilibrium pressures of H₂ and Cl₂ by subtracting x from their initial pressures:

After determining the equilibrium pressures for all species, we apply the equilibrium expression for the reaction to find Kp:

Kp = [P(HCl)] ²/ [P(H₂) * P(Cl₂)]

Kp = [1.418]² / [3.351 * 2.791]

Kp = [2.010724] / [3.351 * 2.791]

Kp = [2.010724] / [9.75141]

Kp = 0.206198

By calculating this expression we get the value of Kp for the reaction at the given temperature is 0.206198.

Answer:

Explanation:

For reaction stoichiometry problems like this, convert given data to moles and set up a ratio expression that relates to the balanced equation. That is ...

Given Rxn =>      Mg(OH)₂ + 2HCl => 2H₂O + MgCl₂

Given mass =>      3.26g = (3.26g)/(58g/mol) = 0.056 mole Mg(OH)₂

If from equation  1 mole Mg(OH)₂ reacts with 2 moles HCl

then,            0.056 mole Mg(OH)₂ reacts with x moles HCl

Setting up ratio and proportion expression ...

=> (1 mole Mg(OH)₂) /(0.056 mole Mg(OH)₂) = (2 moles HCl)/x

=> x = [2(0.056)/(1)] mole HCl neutralized = 0.112mole HCl (36g/mol)

= 4.05 grams of HCl neutralized.

4.08 grams of HCl are neutralized by a dose of milk of magnesia containing 3.26 g of Mg(OH)₂.

Let's consider the neutralization reaction between HCl and Mg(OH)₂.

Mg(OH)₂(aq) + 2 HCl(aq) → 2 H₂O(l) + MgCl₂(aq)

First, we will convert 3.26 g of Mg(OH)₂ to moles using its molar mass (58.32 g/mol).

[tex]3.26 g \times \frac{1mol}{58.32g} = 0.0559 mol[/tex]

The molar ratio of Mg(OH)₂ to HCl is 1:2. The moles of HCl that react with 0.0559 moles of Mg(OH)₂ are:

[tex]0.0559 mol Mg(OH)_2 \times \frac{2molHCl}{1mol Mg(OH)_2} = 0.112molHCl[/tex]

Finally, we will convert 0.112 moles of HCl to grams using its molar mass (36.46 g/mol).

[tex]0.112 mol \times \frac{36.46g}{mol} = 4.08 g[/tex]

4.08 grams of HCl are neutralized by a dose of milk of magnesia containing 3.26 g of Mg(OH)₂.

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Answer:

91.16% has decayed & 8.84% remains

Explanation:

A = A₀e⁻ᵏᵗ => ln(A/A₀) = ln(e⁻ᵏᵗ) => lnA - lnA₀ = -kt => lnA = lnA₀ - kt

Rate Constant (k) = 0.693/half-life = 0.693/10³yrs = 6.93 x 10ˉ⁴yrsˉ¹

Time (t) = 1000yrs  

A = fraction of nuclide remaining after 1000yrs

A₀ = original amount of nuclide = 1.00 (= 100%)  

lnA = lnA₀ - kt

lnA = ln(1) – (6.93 x 10ˉ⁴yrsˉ¹)(3500yrs) = -2.426

A = eˉ²∙⁴²⁶ = 0.0884 = fraction of nuclide remaining after 3500 years

Amount of nuclide decayed = 1 – 0.0884 = 0.9116 or 91.16% has decayed.

Answer: The excess reagent for the given chemical reaction is calcium hydroxide and the amount left after the completion of reaction is 0.115375 moles. The amount of calcium chloride formed in the reaction is 1.068 grams.  

Explanation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]   ....(1)

For calcium hydroxide:

Given mass of calcium hydroxide = 9.27 g

Molar mass of calcium hydroxide = 74.093 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of calcium hydroxide}=\frac{9.27g}{74.093g/mol}=0.125mol[/tex]

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

We are given:

Volume of hydrochloric acid = 38.5mL = 0.0385 L   (Conversion factor: 1 L = 1000 mL)

Molarity of the solution = 0.500 moles/ L

Putting values in above equation, we get:

[tex]0.500mol/L=\frac{\text{Moles of hydrochloric acid}}{0.0385L}\\\\\text{Moles of hydrochloric acid}=0.01925mol[/tex]

[tex]2HCl(aq.)+Ca(OH)_2(s)\rightarrow CaCl_2(s)+2H_2O(l)[/tex]

Here, the solid salt is calcium chloride.

By Stoichiometry of the reaction:

2 moles of hydrochloric acid reacts with 1 mole of calcium hydroxide.

So, 0.01925 moles of hydrochloric acid will react with = [tex]\frac{1}{2}\times 0.01925=0.009625moles[/tex] of calcium hydroxide.

As, given amount of calcium hydroxide is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrochloric acid is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of hydrochloric acid produces 1 mole of calcium chloride.

So, 0.01925 moles of hydrochloric acid will produce = [tex]\frac{1}{2}\times 0.01925=0.009625moles[/tex] of calcium chloride.

Now, calculating the mass of calcium chloride from equation 1, we get:

Molar mass of calcium chloride = 110.98 g/mol

Moles of calcium chloride = 0.009625 moles

Putting values in equation 1, we get:

[tex]0.009625mol=\frac{\text{Mass of calcium chloride}}{110.98g/mol}\\\\\text{Mass of calcium chloride}=1.068g[/tex]

Hence, the excess reagent for the given chemical reaction is calcium hydroxide and the amount left after the completion of reaction is 0.115375 moles. The amount of calcium chloride formed in the reaction is 1.068 grams.

HCl is a limiting reactant

mass of salt: 1.068375 g

8.559 g  of the excess reactant (Ca(OH)₂)remain after the reaction is complete

The reaction equation is the chemical formula of reagents and product substances

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products

Terms used:

Mole

The mole itself is the number of particles contained in a substance amounting to 6.02.10²³

[tex] \large {\boxed {\boxed {\bold {mol = \frac {mass} {molar \: mass}}}} [/tex]

We determine the mole of each reactant to determine the limiting reactant

then:

mol Ca (OH₂ = mass: molar mass

mole of Ca (OH)₂ = 9.27 g: 74

mole of Ca (OH)₂ = 0.1253

mole HCl: 38.5 ml x 0.5 M = 19.25 mlmol = 0.01925 mol

From the number of moles, it can be seen that HCl is a limiting reactant

Reaction:

                     Ca(OH)₂  +    2HCl         ⇒    CaCl₂      +         2H₂O

initial mole    0.1253         0.01925

reaction        0.009625   0.01925        0.009625           0.01925

remaining     0.115675       -                  0.009625           0.01925

Remaining unreacted Ca (OH)₂ mole: 0.115675

Amount of mass of CaCl₂ salt formed:

CaCl₂ mass = mole x molar mass

CaCl₂ mass = 0.009625 x 111

CaCl₂ mass = 1.068375

Remaining Ca(OH)₂ = 0.115675 x 74 = 8.559 g

The mass of one mole of raindrops

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moles of NaOH

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moles of water you can produce

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Answer:

Wt%H₂SO₄ = 10.2% (w/w)

Explanation:

1.61 molal H₂SO₄ = 1.61 mole H₂SO₄/Kg Solvent = 1.61 mole(98 g/mole)/Kg Solvent = 113.68 g H₂SO₄/Kg Solvent

Solution weight = 1000 g solvent + 113.68 g solute = 1113.68 g Solution

Wt%H₂SO₄ = (113.68 g/1113.68g)100% = 10.2% (w/w)

The percentage of sulfuric acid = 10.2% (w/w)

Given:

Molality of solution = 1.61m

It is a measure of the number of moles of solute in a solution corresponding to 1 kg or 1000 g of solvent.

1.61 molal H₂SO₄ = 1.61 mole H₂SO₄/Kg Solvent

= 1.61 mole(98 g/mole)/Kg Solvent = 113.68 g H₂SO₄/Kg Solvent

Solution is made up of solute and solvent. Thus,

Solution weight = 1000 g solvent + 113.68 g solute = 1113.68 g Solution

[tex]Wt\% \text{ of } H_2SO_4 = \frac{113.68 g}{1113.68g}*100\% = 10.2\% (w/w)[/tex]

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Answer : The value of q, w and U for the reversible, isothermal compression are, -52271.69 J, 52271.69 J and 0 J respectively.

Explanation : Given,

Moles of gas = 10 mole

Initial pressure of gas = 1 atm

Final pressure of the gas = 10 atm

Temperature of the gas = [tex]0^oC=273+0=273K[/tex]

According to the question, this is the case of isothermal reversible compression of gas.

As per first law of thermodynamic,

[tex]\Delta U=q+w[/tex]

where,

[tex]\Delta U[/tex] = internal energy

q = heat

w = work done

As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.

So, at constant temperature the internal energy is equal to zero.

The expression used for work done will be,

[tex]w=-2.303nRT\log (\frac{P_1}{P_2})[/tex]

where,

w = work done on the gas

n = number of moles of gas

R = gas constant = 8.314 J/mole K

T = temperature of gas

n = moles of the gas

[tex]P_1[/tex] = initial pressure of gas

[tex]P_2[/tex] = final pressure of gas

Now put all the given values in the above formula, we get the work done.

[tex]w=-2.303\times 10mole\times 8.314J/moleK\times 273K\times \log (\frac{1atm}{10atm})[/tex]

[tex]w=52271.69J[/tex]

And we know that, the heat is equal to the work done with opposite sign convention.

So, [tex]q=-52271.69J[/tex]

Therefore, the value of q, w and U for the reversible, isothermal compression are, -52271.69 J, 52271.69 J and 0 J respectively.

Fe2O3+3CO→2Fe+3CO2, 3CO is oxidized in this reaction, as oxygen is added to it. 2HCl+2KMnO4+3H2C2O4→6CO2+2MnO2+2KCl+4H2O, 3H2C2O4 is reduced as hydrogen is added and 2KMnO4 is reduced as oxygen is removed.

Oxidation is defined as the process in which loss of electrons take place during the reaction by a molecule atom or ion.

Oxidizing agent is defined as a substance that causes oxidation by accepting electron therefore it gat reduced.

Reduction is defined as the transfer of electrons between spices in a chemical reaction. It shows loss of oxygen.

Reducing agent is defined as one of the reactant of redox reaction which reduces the another reactant by giving out electrons to the reactant.

Thus, Fe2O3+3CO→2Fe+3CO2, 3CO is oxidized in this reaction, as oxygen is added to it. 2HCl+2KMnO4+3H2C2O4→6CO2+2MnO2+2KCl+4H2O, 3H2C2O4 is reduced as hydrogen is added and 2KMnO4 is reduced as oxygen is removed.

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Answer:

False

Explanation:

For different diameter pipes flow rate remains constant but velocity is different. (from continuity A1V1= A2V2). As area changes the velocity of fluid changes

Answer:

Helium

Explanation:

Answer:

-238.54 kJ/mol.

Explanation:

C(graphite) + 2H₂(g) + ½ O₂(g) → CH3OH(l) ΔHf° = ? ?? kJ/mol.

(1) C(graphite) + O₂(g) → CO₂(g),                         ΔHf₁° = -393.5 kJ/mol .

(2) H2(g) + ½ O₂(g) → H₂O(l),                               ΔHf₂° = -285.8 kJ/mol .

(3) CH₃OH(l) + 3/2 O₂(g) → CO₂(g) + 2H₂O(l),    ΔH₃° = -726.56 kJ/mol .

(1) C(graphite) + O₂(g) → CO₂(g),                         ΔHf₁° = -393.5 kJ/mol .

(2) 2H2(g) + O₂(g) → 2H₂O(l),                               ΔHf₂° = 2x(-285.8 kJ/mol ).

(3) CO₂(g) + 2H₂O(l) → CH₃OH(l) + 3/2 O₂(g),     ΔH₃° = 726.56 kJ/mol .

The standard enthalpy of formation of liquid methanol, CH₃OH(l) =  ΔHf₁° + 2(ΔHf₂°) - ΔH₃° = (-393.5 kJ/mol ) + 2(-285.8 kJ/mol ) - (- 726.56 kJ/mol) = -238.54 kJ/mol.

The standard enthalpy change of formation of liquid methanol, CH3OH(l), is calculated by manipulating and summing the reactions provided, considering the rules of Hess's law. The resulting ΔH° of the methanol formation from its elements is +976.0 kJ/mol.

To calculate the standard enthalpy of formation ΔH° of liquid methanol, CH3OH(l), we will use the concept of Hess's law and the provided chemical equations. Hess's law states that the enthalpy change of a reaction depends only on the initial and final states, not on the pathway or steps taken to achieve the conversion.

The standard enthalpy formation reaction is defined as the formation of 1 mol of a compound from its elemental forms under standard state conditions. For methanol, the reaction would be: C(graphite) + 2H2(g) + 1/2O2 ⟶ CH3OH(l)

None of the provided reactions exactly match this equation. However, their combination, while keeping in mind the stoichiometry, gives the desired reaction. Note that as we know enthalpy is a state function, we could 'add' or 'subtract' reactions to get the desired one.

The reverse of the CO2 formation reaction is: CO2(g) ⟶ C(graphite) + O2, ΔH° = +393.5 kJ/mol

The reaction for the formation of 2 mol of H2O needs to be halved: 1/2H2(g) + 1/2O2 ⟶ 1H2O(l), ΔH° = -285.8 kJ/mol/2 = -142.9 kJ/mol

The combustion of methanol reaction needs to be reversed: CO2(g) + 2H2O(l) ⟶ CH3OH(l) + O2(g), ΔH° = +726.4 kJ/mol

Adding these revised equations together gives the desired equation for the formation of methanol from its elements and the ΔH formation is the sum of the ΔH of these revised reactions δH°(formation of CH3OH) = +393.5 kJ/mol - 142.9 kJ/mol + 726.4 kJ/mol = +976. Omega kJ/mol.

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Answer:

Need to add 178ml of 2.41M KOH into the 580ml of 0.872M HOAc => Buffer Solution with pH = 5.50.

Explanation:

1. Determine the Base to Acid Ratio using the Henderson-Hasselbalch Equation

Given pH(Bfr) = 5.50; pKa(HOAc) = -logKa = -log(1.8x10ˉ⁵) = 4.76

pH(Bfr) = pKa(HOAc) + log([OAcˉ]/[HOAc]) => 5.5 = 4.76 + log([OAcˉ]/[HOAc])

=> log([OAcˉ]/[HOAc]) = 5.50 – 4.76 = 0.74 => [OAcˉ]/[HOAc] = 10^0.74 = 5.495*

*For an HOAc/OAcˉ Buffer to have a pH = 5.50 the [OAcˉ] concentration must be 5.495 times greater than the [HOAc] concentration.  

2. Determine moles of KOH need be added to 580ml of 0.872M HOAc such that the moles of OAc⁻ is 5.495 times greater than moles of HOAc. The Acid/Base Rxn is HOAc + KOH => KOAc + H₂O.ˉ

That is, given 580ml(0.872M HOAc) + V(L)∙2.41M KOH => 5.495 x moles HOAc

=> 0.58(0.872)mole HOAc + X moles KOH => 5.495 x 0.58(0.872)mole KOAc (=OAcˉ)

=> 0.506 mole HOAc + X moles KOH* => 2.7805 mole KOAc ( = 2.7805 mole OAcˉ)

*note => KOH is the limiting reactant and will be consumed when added into HOAc solution leaving HOAc with OAcˉ that constitutes the buffer solution. The moles of KOH added is equal to the moles of OAcˉ produced. [HOAc] will decrease and [OAcˉ] will increase until the OAcˉ concentration is 5.495 times grater than the HOAc concentration.  

3. After adding X moles of KOH, the following solution results …  

=> (0.506 mole – X mole)HOAc + X mole OAcˉ

=> Since the moles OAcˉ is 5.495 x moles HOAc, the following linear expression in one unknown is generated …

=> moles OAcˉ produced/moles HOAc remaining = X / 0.506 – X = 5.495 where X = OAcˉ produced from rxn and 0.506 – X is the HOAc remaining. The moles of OAcˉ must be 5.495 times greater than moles of HOAc.  

Solving for X => X = 5.495(0.506 – X) = 2.7805 – 5.495X => X = 2.7805/6.495 = 0.428 mole OAcˉ produced. The 0.428 mole OAcˉ is 5.495 time greater than (0.506 - 0.428) mole HOAc remaining.  

4. This means that 0.428 mole KOH is needed for reaction with 580ml of 0.872M HOAc to give 0.428 mole OAcˉ with (0.506 – 0.482)mole HOAc remaining after mixing solution.

The volume of 2.41M KOH needed to deliver 0.428 mole KOH is V(KOH)∙2.41M = 0.428 mole => V(KOH) = (0.428/2.41)Liters = 0.178 Liter = 178ml of 2.41M KOH.

5. Verification of Results …

580ml(0.872M HOAc) + 178ml(2.41M KOH)

=> 0.58(0.872)mole HOAc + 0.178(2.41)mole KOH  

=> 0.506 mole HOAc + 0.428 mole KOH  

=> (0.506 – 0.428)mole HOAc + 0.428mole OAc⁻

=> 0.078 mole HOAc + 0.428mole OAcˉ

=> (0.078mol HOAc)/(0.58 + 0.178)Liter Soln + (0.428 mole OAcˉ)/(0.58 + 0.178)Liter Soln

=> (0.078mole/0.758L)HOAc + (0.428mole/0.872L)OAcˉ

=> 0.1029M HOAc + 0.5646M OAcˉ  (Buffer Solution with pH = 5.50)

Checking pH of this buffer solution…

                HOAc     ⇄    H⁺       +      OAcˉ

C(eq)     0.1029M          [H⁺]           0.5646M

Ka = [H⁺][OAcˉ]/[HOAc] => [H⁺] = Ka[HOAc]/[OAcˉ]  

= [(1.8 x 10ˉ⁵)(0.1029)/(0.5646)]M = 3.28 x 10ˉ⁶M

pH = -log[H⁺] -log(3.28 x 10ˉ⁶) = 5.50

Answer:

The correct answer is :

A proline residue in an alpha-helix is destabilizing because it cannot contribute to hydrogen bonding needed for helix stabilization and it introduces conformational rigidity by locking the peptide backbone into a specific conformation.So,option A,E are correct.

The presence of a proline residue in the middle of an ">α-helix is predicted to destabilize the structure for a couple of reasons. First, proline lacks the ">α-NH group that would typically act as an H-bond donor, which is crucial for stabilizing the helical structure (option A). Second, due to the ring structure of proline, it restricts the backbone conformation and disrupts the ideal ">φ and ">ψ angles necessary for maintaining an α-helix, causing conformational rigidity (option E). As proline's nitrogen is part of a rigid ring and not available for hydrogen bonding, it prevents the formation of the periodic hydrogen bonds that give the α-helix its characteristic stability.

Answer: The volume of CO formed is 254.43 L.

Explanation:

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

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

Given mass of [tex]Ni(CO)_4[/tex] = 444 g

Molar mass of [tex]Ni(CO)_4[/tex] = 170.73 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of }Ni(CO)_4=\frac{444g}{170.73g/mol}=2.60mol[/tex]

For the given chemical reaction:

[tex]Ni(CO)_4(l)\rightarrow Ni(s)+4CO(g)[/tex]

By stoichiometry of the reaction:

1 mole of nickel tetracarbonyl produces 4 moles of carbon monoxide.

So, 2.60 moles of nickel tetracarbonyl will produce = [tex]\frac{4}{1}\times 2.60=10.4mol[/tex] of carbon monoxide.

Now, to calculate the volume of the gas, we use ideal gas equation, which is:

PV = nRT

where,

P = Pressure of the gas = 752 torr

V = Volume of the gas = ? L

n = Number of moles of gas = 10.4 mol

R = Gas constant = [tex]62.364\text{ L Torr }mol^{-1}K^{-1}[/tex]

T = Temperature of the gas = [tex]22^oC=(273+22)K=295K[/tex]

Putting values in above equation, we get:

[tex]752torr\times V=10.4mol\times 62.364\text{ L Torr }mol^{-1}K^{-1}\times 295K\\\\V=254.43L[/tex]

Hence, the volume of CO formed is 254.43 L.

The volume of CO gas formed from the complete decomposition of 444 g of Ni(CO)4 can be found using the ideal gas law.

The given equation shows the thermal decomposition of nickel tetracarbonyl, Ni(CO)4: Ni(CO)4 (l) → Ni (s) + 4CO (g)

To find the volume of CO gas formed, we need to use the ideal gas law:

PV = nRT

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

We can use the given information to find the number of moles of CO, and then use the ideal gas law to find the volume. First, we need to convert the mass of Ni(CO)4 to moles using the molar mass:

Moles of Ni(CO)4 = mass of Ni(CO)4 / molar mass of Ni(CO)4

Next, we can use the stoichiometry of the balanced equation to relate the moles of Ni(CO)4 to moles of CO:

Moles of CO = moles of Ni(CO)4 × (4 moles of CO / 1 mole of Ni(CO)4)

Finally, we can use the ideal gas law to find the volume of CO:

V = (n × R × T) / P

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Answer:

Temperature at which molybdenum becomes superconducting is-272.25°C

Explanation:

Conductor are those hard substances which allows path of electric current through them. And super conductors are those hard substances which have resistance against the flow of electric current through them.

As given, molybdenum becomes superconducting at temperatures below 0.90 K.

Temperature in Kelvins can be converted in °C by relation:

T(°C)=273.15-T(K)

Molybdenum becomes superconducting in degrees Celsius.

T(°C)=273.15-0.90= -272.25 °C

Temperature at which molybdenum becomes superconducting is -272.25 °C

Mixed molarity = 0.25 M

Osmotic pressure solution: 6.15 atm

Osmotic pressure is the minimum pressure given to the solution so that there is no osmotic displacement from a more dilute solution to a more concentrated solution.

General formula:

[tex]\large{\boxed {\bold {\pi \: = \: M \: x \: R \: x \: T}}}[/tex]

π = osmotic pressure (atm)

M = solution concentration (mol / l)

R = constant = 0.08205 L atm mol-1 K-1

T = Temperature (Kelvin)

Mixing solution

To find the molarity of the mixed solution we can use the following formula:

Vc. Mc = V1.M1 + V2.M2

where

Mc =Mixed molarity

Vc = mixed volume

Mr. NaCl = 58.5

Mr. Dextrose = C₆H₁₂O₆ = 180

mole NaCl = gram / Mr

mole NaCl = 1.75 / 58.5 = 0.03

Dextrose mole = 40/180 = 0.22

Assuming 1 liter of solution

M mixture = mole NaCl + mole dextrose / 1 liter

M mixture = 0.03 + 0.22 / 1 L

M mix = 0.25 M

T = 25 + 273 = 298 K

The osmosis pressure of the solution becomes:

π = 0.25. 0.0825. 298 = 6.15 atm

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Grade: Senior High School

Subject: Chemistry

Chapter: Colligative property

Keywords: osmotic pressure, molarity, mole,dextrose, NaCl