a. Draw 2,3-dichloro octane.
b. Write the lewis structure for H20 molecule.

Answer:

The aspirin could be determined if pure or not by testing it with FeCI3 (aq)

The sample of aspirin can be tested for it's purity by treating it with compound of ferric chloride.

Compound is defined as a chemical substance made up of identical molecules containing atoms from more than one type of chemical element.

Molecule consisting atoms of only one element is not called compound.It is transformed into new substances during chemical reactions. There are four major types of compounds depending on chemical bonding present in them.They are:

1)Molecular compounds where in atoms are joined by covalent bonds.

2) ionic compounds where atoms are joined by ionic bond.

3)Inter-metallic compounds where atoms are held by metallic bonds

4) co-ordination complexes where atoms are held by co-ordinate bonds.

They have a unique chemical structure held together by chemical bonds Compounds have different properties as those of elements because when a compound is formed the properties of the substance are totally altered.

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Answer:

ΔH = -2446J

Explanation:

Based on the reaction:

2 Na(s) + Cl2(g) → 2NaCl

We can find the enthalpy of this reaction using Hess's law:

The enthalpy of a reaction is equal to the sum of the enthalpy of products times their reaction quotient subtracting the enthalpy of reactants times their reaction quotient. For the reaction of the problem:

ΔH = 2ΔH(NaCl) - [2ΔH(Na) + ΔHCl2)]

ΔH(NaCl) = 670J

ΔH(Na) = 235cal * (4.184J/1cal) = 983J

ΔHCl2 = 435cal * (4.184J/1cal) = 1820J

ΔH = 2*670J - [2*983J + 1820J]

ΔH = 1340J - [3786J]

Answer:

the heat content of a system at constant pressure

Explanation:

The reduction of the species defines the gain of electrons. The iron is most likely to be reduced when reacts with zinc. Thus option A is correct.

Oxidizing agents are the species that gain electrons and get reduced, their oxidation number gets reduced when the metal reacts.

In the reactivity, series zinc is placed before iron and hence is a reducing agent that gets oxidized. Down the series, the reducing ability decreases while the oxidizing increases.

Therefore, option A. iron will be reduced when reacts with zinc.

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Answer:

a hydrogen bond between the hydrogen of water and the nitrogen of the amine

a hydrogen bond between the oxygen of water and a hydrogen from the -NH2 group

Explanation:

A hydrogen bond is formed between molecules in which hydrogen is bonded to a highly electronegative element.

In amines, hydrogen is bonded to nitrogen while in water, hydrogen is bonded to oxygen. Both are highly electronegative elements hence hydrogen bonding is possible between amines and water.

This hydrogen bond may involve;

The hydrogen of water and the nitrogen of the amine

Or

The oxygen of water and a hydrogen from the -NH2 group

Answer:

verdadero/a

falso/b

verdadero/c

Explanation:

Answer:

11

1. 6.02×10 23

this is the answer Hope it helps you

The question is incomplete, the complete question is;

. A chemist determined by measurements that 0.015 moles of hydrogen chloride participate in a chemical reaction. Calculate the mass of hydrogen chloride that participates Round your answer to 2 significant digits. x S. ?

Answer:

0.54 g

Explanation:

Recall that;

Number of moles = mass/molar mass

Molar mass of HCl =36.5 g/mol

Mass= number of moles × molar mass

Mass= 0.015 moles × 36 g/mol

Mass= 0.54 g

Explanation:

This is a decomposition reaction. Firstly, you will want to write the chemical equation out and balance it.

[tex]2Hg_2O->4Hg+O_2[/tex] (The -> is supposed to be an arrow, sorry!)

We see that there's only 1mol of Oxygen made in the products, we can do some simple math to solve for the amount of grams of Oxygen produced according to the amount of the reactant (Hg2O).

[tex]32.2gHg_2O*\frac{1molHg_2O}{417.18gHg_2O}*\frac{1molO_2}{2molHg_2O}*\frac{32gO_2}{1molO_2}[/tex]

I want to break this down, just in case:

The 417.18gHg2O is the molecular mass of the molecule (so I doubled Hg and added 16 to it to get this number).

As we can see in the chemical equation, 1mol Hg2O produces 2mol O because Oxygen is a diatomic molecule (so there will always be two of it when it's by itself).

And finally, in 1mol O2 there are 32g of O2.

** When you do math like this, always make sure that all of your units cancel out except for the units you're looking for. For example, here we're looking for the grams of Oxygen, so after everything else cancels out, we should only have grams O2.

So, 1.23gO2 should be your answer.

Answer:

The volume will be "2.95 L".

Explanation:

Given:

n = 0.104

p = 0.91 atm

T = 314 K

Now,

The Volume (V) will be:

= [tex]\frac{nRT}{P}[/tex]

By putting the values, we get

= [tex]\frac{0.104\times 0.0821\times 314}{0.91}[/tex]

= [tex]\frac{2.6810}{0.91}[/tex]

= [tex]2.95 \ L[/tex]

Answer:

e. the principal quantum number (n).

Explanation:

The size of the orbital is governed and decided by the principal quantum number n, which is dependent on the overall average distance between the number of electrons as well as the nucleus. The orbital's shape is explained by the angular quantum number. The magnetic quantum number is concerned with the orbital's orientation in space. The quantum number's spin explains the spin of the electrons.

Answer:

Explanation:

Target organ toxins are chemicals that can cause adverse effects or disease states manifested in specific organs of the body. Toxins do not affect all organs in the body to the same extent due to their different cell structures.

Answer:

dissolved in the solute

Explanation:

A solvent is the component that dissolves the solute and is present in larger amount. The type of solution is determined by the state of the solute and solvent. If you have NaCl, a solid, dissolved in water, a liquid, the type of solution is a solid/liquid solution.

Answer: It is true about the type of reaction that it is a single replacement reaction, and the cations in the two ionic compounds are different.

Explanation:

When one element in a compound is replaced by another element in a chemical reaction then it is called a single replacement reaction.

For example, [tex]K + NaCl \rightarrow KCl + Na[/tex]

Here, potassium metal is replacing the sodium metal in the sodium chloride compound.

As metals become cation by losing an electron in a chemical reaction.

Thus, we can conclude that it is true about the type of reaction that it is a single replacement reaction, and the cations in the two ionic compounds are different.

Answer: Its A

Explanation:

a single replacement reactions, and the ANIONS in the two ionic compounds are different

Answer:

Option B. The reaction will shift to the left in the direction of the reactants.

Explanation:

The equation for the reaction is given below:

CO₂ + 2H₂O <=> CH₄ + O₂

Enthalpy change (ΔH) = +890 KJ

The reaction illustrated by the equation is endothermic reaction since the enthalpy change (ΔH) is positive.

Increasing the temperature of an endothermic reaction will shift the equilibrium position to the right and decrease the temperature will shift the equilibrium position to the left.

Therefore, decreasing the temperature of the system illustrated by the equation above, will shift the reaction to the left in the direction of the reactants.

Thus, option B gives the right answer to the question.

Based on the calculations, the Gibbs's free energy for this chemical reaction is equal to -1,680.906 kJ/mol.

Given the following data:

Gibbs's free energy simply refers to the quantity of energy that is associated with a particular chemical reaction.

Mathematically, the Gibbs's free energy for this chemical reaction can be calculated by using this formula:

ΔG° = ΔH° - ΔS°

Substituting the given parameters into the formula, we have;

ΔG° = -1652 × 10³ - (298 × 0.097)

ΔG° = -1652 × 10³ - 28.906

ΔG° = -1,680.906 kJ/mol.

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Answer:

C

Explanation:

Think of 'exo' as exit and 'thermic' as relating to thermal energy/ heat. Thus, an exothermic release thermal energy as the reaction proceeds.

In an exothermic reaction, the total energy of the products is lesser than that of the reactants and ΔH (change in energy) is less than zero.

When heat is absorbed as the reaction proceeds, the chemical reaction is an endothermic reaction.

Answer:

4.00 is the pH of the mixture

Explanation:

The ethyl amine reacts with HNO3 as follows:

C2H5NH2 + HNO3 → C2H5NH3⁺ + NO3⁻

To solve this question we need to find the moles of ethyl amine and the moles of HNO3:

Moles C2H5NH2:

0.0500L * (0.100mol/L) = 0.00500 moles ethyl amine

Moles HNO3:

0.201L * (0.025mol/L) = 0.005025 moles HNO3

That means HNO3 is in excess. The moles in excess are:

0.005025 moles HNO3 - 0.00500 moles ethyl amine =

2.5x10⁻⁵ moles HNO₃

In 50 + 201mL = 251mL = 0.251L:

2.5x10⁻⁵ moles HNO₃ / 0.251L = 9.96x10⁻⁵M = [H+]

As pH = -log [H+]

pH = -log 9.96x10⁻⁵M

pH = 4.00 is the pH of the mixture

Answer:

True I think

hope this helps

Answer:

true

Explanation:

because it reacts with the water molecules by giving a proton to them

Answer:

747.5 mL

Explanation:

Assuming ideal behaviour, we can solve this problem by using Boyle's law, which states that at constant temperature:

Where in this case:

We input the data given by the problem:

And solve for V₁:

Answer:

vague symptoms are characteristic of an acute toxin, because of the of the lack of well defined consistency that these symptoms have in relation to the course of the disease progress.

a. Br, Cl, F

b. Na, K, Ba

c. He, Ar, Ne

d. Ca, Ba, Mg

Answer:

a. halogen : F ,Cl ,Br l ,At

b an alkali metal: Na,Li, Rb, Cs

c. a noble gas: He, Ne, Kr, Ar

d. an alkaline earth metal: Be,Mg,Ca, Sr

hope it helps

Answer:

A positively charged object will exert a repulsive force upon a second positively charged. This repulsive force will push the two objects apart while a negatively charged object will exert a repulsive force upon a second negatively charged object. Objects with like charge repel each other

The interaction between objects with positive and negative electric charges can be analogously modeled using magnets. The Types of Forces Involved are; Attractive Magnetic Force, Repulsive Magnetic Force and the Energy Transformations are; Potential Energy Transformation, and Kinetic Energy Transformation.

In this analogy, magnets can represent the charges, and magnetic forces can represent the electric forces.

Interaction Between Magnets

Imagine we have two magnets: one with a north pole (N) and the other with a south pole (S). When you bring the north pole of one magnet close to the south pole of the other magnet, they are attracted to each other. Conversely, if you bring the north pole of one magnet near the north pole of the other magnet, they repel each other.

Types of Forces Involved:

Attractive Magnetic Force (Analogous to Electric Attraction):

When the north pole of one magnet is brought close to the south pole of another magnet, they experience an attractive magnetic force. Similarly, when objects with opposite electric charges were brought close together, then they will experience an attractive electric force.

Repulsive Magnetic Force (Analogous to Electric Repulsion):

When two magnets with the same pole (both north or both south) are brought close to each other, they experience a repulsive magnetic force. This is analogous to the repulsion between objects with like electric charges (both positive or both negative).

Energy Transformations;

When you bring the magnets closer together or move them apart, energy transformations occur:

Potential Energy Transformation;

As the magnets are moved closer together, the potential energy of the magnetic interaction decreases. This is because the magnets' magnetic fields interact more strongly, and they tend to move toward each other due to the attractive or repulsive forces.

Kinetic Energy Transformation;

If you let the magnets go after bringing them close together, they will move towards each other (in the case of attraction) or move apart (in the case of repulsion). This movement involves a transformation of potential energy into kinetic energy. The kinetic energy increases as the magnets move, and it's at its maximum when the magnets are farthest apart (in the case of repulsion) or when they collide (in the case of attraction).

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Answer:

d.

Explanation:

H2 + Cl2 = 2HCL

From the equation 2g hydrogen combine with 71g of chlorine.

So 35.5 g Cl2 combines with 1g of H2

There are 100g of Cl2 so this will, by proportion, react with 100/35.5 g hydrogen.

This is 2.8 g hydrogen so the mass of HCl formed = 102.8 g.

The true statement is that d. between 100 and 190 g of HCl is produced.

To find mass of HCL:

H2 + Cl2 = 2HCL

From the equation, 2g of hydrogen combines with 71g of chlorine.

So 35.5 g Cl2 combines with 1g of H2

There are 100g of Cl2 so this will, by proportion, react with 100/35.5 g of hydrogen.

This is 2.8 g hydrogen so the mass of HCl formed = 102.8 g.

Hydrogen chloride may be formed by the direct combination of chlorine (Cl2) gas and hydrogen (H2) gas.

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The question is incomplete, the complete question is shown in the image attached to this answer

Answer:

A

Explanation:

We can see from the conditions of the reaction that the incoming nucleophile is -SCH3 and there are two possible leaving groups in the substrate.

First of all, we have to look at the conditions of the reaction. We can see that the reaction is taking place in DMF, a polar aprotoc solvent. This condition favours the SN2 synchronous mechanism over the SN1 ionic mechanism.

Hence, the nucleophile at the 1-position is preferentially substituted owing to the conditions of the reaction.

Thus, option A is the major product of the reaction.

Answer:

a)10.87

b)9.66

c)9.15

d)7.71

e) 5.56

f) 3.43

Explanation:

tep 1: Data given

Volume of 0.030 M NH3 solution = 30 mL = 0.030 L

Molarity of the HCl solution = 0.025 M

Step 2: Adding 0 mL of HCl

The reaction:    NH3 + H2O ⇔ NH4+ + OH-

The initial concentration:  

[NH3] = 0.030M    [NH4+] = 0M    [OH-] = OM

The concentration at the equilibrium:

[NH3] = 0.030 - XM

[NH4+] = [OH-] = XM

Kb = ([NH4+][OH-])/[NH3]

1.8*10^-5 = x² / 0.030-x

1.8*10^-5 = x² / 0.030

x = 7.35 * 10^-4 = [OH-]

pOH = -log [7.35 * 10^-4]

pOH = 3.13

pH = 14-3.13 = 10.87

Step 3: After adding 10 mL of HCl

The reaction:

NH3 + HCl ⇔ NH4+ + Cl-

NH3 + H3O+ ⇔ NH4+ + H2O

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.010 L = 0.00025 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.00025 =0.00065 moles

Moles HCl = 0

Moles NH4+ = 0.00025 moles

Concentration at the equilibrium:

[NH3]= 0.00065 moles / 0.040 L = 0.01625M

[NH4+] = 0.00625 M

pOH = pKb + log [NH4+]/[NH3]

pOH =  4.75 + log (0.00625/0.01625)

pOH = 4.34

pH = 9.66

Step 3: Adding 20 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.020 L = 0.00050 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.00050 =0.00040 moles

Moles HCl = 0

Moles NH4+ = 0.00050 moles

Concentration at the equilibrium:

[NH3]= 0.00040 moles / 0.050 L = 0.008M

[NH4+] = 0.01 M

pOH = pKb + log [NH4+]/[NH3]

pOH =  4.75 + log (0.01/0.008)

pOH = 4.85

pH = 14 - 4.85 = 9.15

Step 4: Adding 35 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.035 L = 0.000875 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.000875 =0.000025 moles

Moles HCl = 0

Moles NH4+ = 0.000875 moles

Concentration at the equilibrium:

[NH3]= 0.000025 moles / 0.065 L = 3.85*10^-4M

[NH4+] = 0.000875 M / 0.065 L = 0.0135 M

pOH = pKb + log [NH4+]/[NH3]

pOH =  4.75 + log (0.0135/3.85*10^-4)

pOH = 6.29

pH = 14 - 6.29 = 7.71

Step 5: adding 36 mL HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.036 L = 0.0009 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.0009 =0 moles

Moles HCl = 0

Moles NH4+ = 0.0009 moles

[NH4+] = 0.0009 moles / 0.066 L = 0.0136 M

Kw = Ka * Kb

Ka = 10^-14 / 1.8*10^-5

Ka = 5.6 * 10^-10

Ka = [NH3][H3O+] / [NH4+]

Ka =5.6 * 10^-10 =  x² / 0.0136

x = 2.76 * 10^-6 = [H3O+]

pH = -log(2.76 * 10^-6)

pH = 5.56

Step 6: Adding 37 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.037 L = 0.000925 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.000925 =0 moles

Moles HCl = 0.000025 moles

Concentration of HCl = 0.000025 moles / 0.067 L = 3.73 * 10^-4 M

pH = -log 3.73*10^-4= 3.43

The pH of the solution in the titration of 30 mL of 0.030 M NH₃ with 0.025 M HCl, is:

a) pH = 10.86

b) pH = 9.66

c) pH = 9.15

d) pH = 7.70

e) pH = 5.56

f) pH = 3.43          

Initially, the pH of the solution is given by the dissociation of NH₃ in water.  

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻     (1)

The constant of the above reaction is:

[tex] Kb = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]} = 1.76\cdot 10^{-5} [/tex]   (2)

At the equilibrium, we have:  

   NH₃    +    H₂O   ⇄   NH₄⁺    +    OH⁻     (3)  

0.030 M - x                      x               x

[tex] 1.76\cdot 10^{-5}*(0.030 - x) - x^{2} = 0 [/tex]

After solving for x and taking the positive value:

x = 7.18x10⁻⁴ = [OH⁻]  

Now, we can calculate the pH of the solution as follows:

[tex] pH = 14 - pOH = 14 + log(7.18\cdot 10^{-4}) = 10.86 [/tex]

Hence, the initial pH is 10.86.

After the addition of HCl, the following reaction takes place:

NH₃ + HCl ⇄ NH₄⁺ + Cl⁻  (4)  

We can calculate the pH of the solution from the equilibrium reaction (3).            

[tex] 1.76\cdot 10^{-5}(Cb - x) - (Ca + x)*x = 0 [/tex] (5)  

The number of moles of NH₃ (nb) and NH₄⁺ (na) are given by:

[tex] n_{b} = n_{i} - n_{HCl} [/tex]     (6)

[tex] n_{b} = 0.030 mol/L*0.030 L - 0.025 mol/L*0.010 L = 6.5\cdot 10^{-4} moles [/tex]          

[tex] n_{a} = n_{HCl} [/tex]   (7)

[tex] n_{a} = 0.025 mol/L*0.010 L = 2.5 \cdot 10^{-4} moles [/tex]

The concentrations are given by:

[tex] Cb = \frac{6.5\cdot 10^{-4} moles}{(0.030 L + 0.010 L)} = 0.0163 M [/tex]   (8)

[tex] Ca = \frac{2.5 \cdot 10^{-4} mole}{(0.030 L + 0.010 L)} = 6.25 \cdot 10^{-3} M [/tex]      (9)

After entering the values of Ca and Cb into equation (5) and solving for x, we have:  

[tex] 1.76\cdot 10^{-5}(0.0163 - x) - (6.25 \cdot 10^{-3} + x)*x = 0 [/tex]

x = 4.54x10⁻⁵ = [OH⁻]

Then, the pH is:

[tex] pH = 14 + log(4.54\cdot 10^{-5}) = 9.66 [/tex]

Hence, the pH is 9.66.

We can find the pH of the solution from the reaction of equilibrium (3).

The concentrations are (eq 8 and 9):

[tex] Cb = \frac{0.030 mol/L*0.030 L - 0.025 mol/L*0.020 L}{(0.030 L + 0.020 L)} = 8.0\cdot 10^{-3} M [/tex]    

[tex] Ca = \frac{0.025 mol/L*0.020 L}{(0.030 L + 0.020 L)} = 0.01 M [/tex]    

After solving the equation (5) for x, we have:

[tex] 1.76\cdot 10^{-5}(8.0\cdot 10^{-3} - x) - (0.01 + x)*x = 0 [/tex]

x = 1.40x10⁻⁵ = [OH⁻]

Then, the pH is:  

[tex] pH = 14 + log(1.40\cdot 10^{-5}) = 9.15 [/tex]

So, the pH is 9.15.

We can find the pH of the solution from reaction (3).

[tex] Cb = \frac{0.030 mol/L*0.030 L - 0.025 mol/L*0.035 L}{(0.030 L + 0.035 L)} = 3.85\cdot 10^{-4} M [/tex]      

[tex] Ca = \frac{0.025 mol/L*0.035 L}{(0.030 L + 0.035 L)} = 0.0135 M [/tex]      

After solving the equation (5) for x, we have:

[tex] 1.76\cdot 10^{-5}(3.85\cdot 10^{-4} - x) - (0.0135 + x)*x = 0 [/tex]

x = 5.013x10⁻⁷ = [OH⁻]      

Then, the pH is:  

[tex] pH = 14 + log(5.013\cdot 10^{-7}) = 7.70 [/tex]  

So, the pH is 7.70.

[tex] n_{b} = 0.030 mol/L*0.030 L - 0.025 mol/L*0.036 L = 0 [/tex]    

[tex] n_{a} = 0.025 mol/L*0.036 L = 9.0 \cdot 10^{-4} moles [/tex]

Since all the NH₃ reacts with the HCl added, the pH of the solution is given by the dissociation reaction of the NH₄⁺ produced in water.

At the equilibrium, we have:                

NH₄⁺    +    H₂O   ⇄   NH₃    +    H₃O⁺

Ca - x                             x               x

[tex] Ka = \frac{x^{2}}{Ca - x} [/tex]  

[tex] Ka(Ca - x) - x^{2} = 0 [/tex]   (10)          

We can find the acid constant as follows:

[tex] Kw = Ka*Kb [/tex]

Where Kw is the constant of water = 10⁻¹⁴

[tex] Ka = \frac{1\cdot 10^{-14}}{1.76 \cdot 10^{-5}} = 5.68 \cdot 10^{-10} [/tex]  

The concentration of NH₄⁺ is:

[tex] Ca = \frac{9.0 \cdot 10^{-4} moles}{(0.030 L + 0.036 L)} = 0.0136 M [/tex]      

After solving the equation (10) for x, we have:

x = 2.78x10⁻⁶ = [H₃O⁺]

Then, the pH is:  

[tex] pH = -log(H_{3}O^{+}) = -log(2.78\cdot 10^{-6}) = 5.56 [/tex]

Hence, the pH is 5.56.

Now, the pH is given by the concentration of HCl that remain in solution after reacting with NH₃ (HCl is in excess).

[tex] C_{HCl} = \frac{0.025 mol/L*0.037 L - 0.030 mol/L*0.030 L}{(0.030 L + 0.037 L)} = 3.73 \cdot 10^{-4} M = [H_{3}O^{+}] [/tex]      

[tex] pH = -log(H_{3}O^{+}) = -log(3.73 \cdot 10^{-4}) = 3.43 [/tex]

Therefore, the pH is 3.43.

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Answer: The total partial pressure of the solution is 131.37 torr.

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

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

Given mass of glucose = 46.8 g

Molar mass of glucose = 180 g/mol

Plugging values in equation 1:

[tex]\text{Moles of glucose}=\frac{46.8g}{180g/mol}=0.26 mol[/tex]

Given mass of methanol = 117 g

Molar mass of methanol = 32 g/mol

Plugging values in equation 1:

[tex]\text{Moles of methanol}=\frac{117g}{32g/mol}=3.66 mol[/tex]

Mole fraction is defined as the moles of a component present in the total moles of a solution. It is given by the equation:

[tex]\chi_A=\frac{n_A}{n_A+n_B}[/tex] .....(2)

where n is the number of moles

Putting values in equation 2:

[tex]\chi_{methanol}=\frac{3.66}{0.26+3.66}=0.934[/tex]

Raoult's law is the law used to calculate the partial pressure of the individual gases present in the mixture. The equation for Raoult's law follows:

[tex]p_A=\chi_A\times p_T[/tex] .....(3)

where [tex]p_A[/tex] is the partial pressure of component A in the mixture and [tex]p_T[/tex] is the total partial pressure of the mixture

We are given:

[tex]p_{methanol}=122.7torr\\\chi_{methanol}=0.934[/tex]

Putting values in equation 3, we get:

[tex]122.7torr=0.066\times p_T\\\\p_T=\frac{122.7torr}{0.934}=131.37torr[/tex]

Hence, the total partial pressure of the solution is 131.37 torr.

Answer:

Arrange the following compounds in order of acidity (highest to lowest): H2O, CH3COOH , HCl

A. CH3COOH > HCl > H2O

B. H2O > CH3COOH > HCl

C. HCl > H2O > CH3COOH

D. HCl > CH3COOH > H2O

Explanation:

The given substances are acetic acid, hydrochloric acid, and water.

Since HCl is a strong acid and it undergoes complete ionization.

CH3COOH acetic acid is a weak acid and it undergoes partial dissociation in water.

Pure water is a neutral substance.

Hence, the order of acidity is shown below:

HCl > CH3COOH > H2O.

Among the given options, option D is the correct answer.

Answer:

9.5x10¹⁷ s

Explanation:

First we convert 3.41 cubic micrometers (um³) to liters (L), as such:

With the converted rate of 3.41x10⁻¹⁵ L/min, we can calculate how many minutes it would take to fill a 54 L vessel:

Finally we convert 1.58x10¹⁶ minutes to seconds:

Answer:

CHCI3

Explanation:

there are no free CI ions hence it doesnt precipitate with an aqeous solution of AQUO33

Answer:

C₂H₆(g) + 2.5 O₂(g) ⇒ 2 CO(g) + 3 H₂O(g)

Explanation:

Let's consider the unbalanced equation in which gaseous ethane reacts with gaseous oxygen to form carbon monoxide gas and water vapor. This is an incomplete combustion reaction.

C₂H₆(g) + O₂(g) ⇒ CO(g) + H₂O(g)

We will balance it using the trial and error method.

First, we will balance C atoms by multiplying CO by 2 and H atoms by multiplying H₂O by 3.

C₂H₆(g) + O₂(g) ⇒ 2 CO(g) + 3 H₂O(g)

Then, we get the balanced equation by multiplying O₂ by 2.5.

C₂H₆(g) + 2.5 O₂(g) ⇒ 2 CO(g) + 3 H₂O(g)