Consider the reaction C2H4(g) + H2O(g)CH3CH2OH(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.90 moles of C2H4(g) react at standard conditions. S°surroundings = J/K

Answer:

[tex]\large \boxed{84.7 \, \%}[/tex]

Explanation:

Mᵣ:                          58.44      278.11

           Pb(NO₃)₂ + 2NaCl ⟶ PbCl₂ + 2NaNO₃

m/g:                         26.3

1. Moles of NaCl

[tex]\text{Moles of NaCl} = \text{26.3 g NaCl} \times \dfrac{\text{1 mol NaCl}}{\text{58.44 g NaCl}} = \text{0.4505 mol NaCl}[/tex]

(b) Moles of PbCl₂

[tex]\text{Moles of PbCl${_2}$} = \text{0.4505 mol NaCl} \times \dfrac{\text{1 mol PbCl${_2}$}}{\text{2 mol NaCl}} = \text{0.2253 mol PbCl${_2}$}[/tex]

(c) Theoretical yield of PbCl₂

[tex]\text{Mass of PbCl${_2}$} = \text{0.2253 mol PbCl${_2}$} \times \dfrac{\text{278.11 g PbCl${_2}$}}{\text{1 mol PbCl${_2}$}} = \text{61.52 g PbCl${_2}$}[/tex]

(d) Percent yield

[tex]\text{Percent yield} = \dfrac{\text{ actual yield}}{\text{ theoretical yield}} \times 100 \,\% = \dfrac{\text{52.1 g}}{\text{61.52 g}} \times 100 \, \% = \mathbf{84.7 \,\%}\\\\\text{The percent yield is $\large \boxed{\mathbf{84.7 \, \% }}$}[/tex]

Final answer:

In a reaction where lead (II) nitrate reacts with sodium chloride to form lead (II) chloride, given 26.3 g of NaCl and 52.1 g of PbCl2 produced, the percent yield is calculated to be 83.27%.

Explanation:

The student is performing a reaction where lead (II) nitrate reacts with sodium chloride to produce lead (II) chloride and sodium nitrate. To calculate the percent yield, we use the actual mass of PbCl2 obtained from the experiment (52.1 g), and compare it with the theoretical mass that should have been produced if the reaction were 100% efficient.

First, you need to calculate the moles of NaCl:

Based on the stoichiometry of the balanced equation, 2 moles of NaCl will produce 1 mole of PbCl2 (2:1 ratio). Therefore, 0.45 moles of NaCl should theoretically produce 0.225 moles of PbCl2.

Now, calculate the theoretical yield:

To find the percent yield:

The percent yield of the reaction is therefore 83.27%.

Answer: The net cell equation for the given electrochemical cell is given below.

Explanation:

The given chemical cell follows:

[tex]Co(s)|Co^{2+}(aq.,0.0155M)||Ag^{+}(aq,3.50M)|Ag(s)[/tex]

Oxidation half reaction: [tex]Co(s)\rightarrow Co^{2+}(aq,0.0155M)+2e^-[/tex]

Reduction half reaction: [tex]Ag^{+}(aq,3.50M)+e^-\rightarrow Ag(s)[/tex]       ( × 2)

Net cell reaction: [tex]Co+2Ag^{+}(aq,3.50M)\rightarrow Co^{2+}(aq,0.0155M)+2Ag(s)[/tex]

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

Hence, the net cell equation for the given electrochemical cell is given above.

Explanation:

Relation between heat energy and specific heat is as follows.

            [tex]q_{bomb} = c \times \Delta t[/tex]

or,            c = [tex]\frac{q_{bomb}}{\Delta t}[/tex]

where,     c = specific heat

           [tex]q_{bomb}[/tex] = heat energy

           [tex]\Delta t[/tex] = change in temperature

Putting the given values into the above formula we will calculate the specific heat as follows.

                c = [tex]\frac{q_{bomb}}{\Delta t}[/tex]

                   = [tex]\frac{5.23 kJ}{7.33^{o}C}[/tex]

                   = 0.713 [tex]kJ^{o}C[/tex]

Thus, we can conclude that heat capacity C of the calorimeter is 0.713 [tex]kJ^{o}C[/tex].

Final answer:

The heat capacity C of the calorimeter is calculated using the formula C = q / ΔT, with the given values resulting in a C of 0.7138 kJ/°C.

Explanation:

The heat capacity C of the calorimeter can be calculated using the amount of heat energy released and the resultant temperature change. The formula to calculate heat capacity is C = q / ΔT, where q is the heat energy released and ΔT is the change in temperature.

In this case, 5.23 kJ of heat energy was released, causing a temperature rise of 7.33 °C. Therefore, the heat capacity of the calorimeter is calculated as follows:

C = 5.23 kJ / 7.33 °C = 0.7138 kJ/°C

It's crucial to note that the heat capacity is commonly expressed in kJ/°C for bomb calorimeters, reflecting the amount of energy required to raise the temperature of the entire calorimeter setup by one degree Celsius.

Explanation:

Let us assume that the concentration of [[tex]OH^{-}[/tex] and [tex]H^{+}[/tex] is equal to x. Then expression for [tex]K_{w}[/tex] for the given reaction is as follows.

          [tex]K_{w} = [OH^{-}][H^{+}][/tex]

          [tex]K_{w} = x^{2}[/tex]

      [tex]6.8 \times 10^{-15} = x^{2}[/tex]

Now, we will take square root on both the sides as follows.

          [tex]\sqrt{6.8 \times 10^{-15}} = \sqrt{x^{2}}[/tex]

          [tex][H^{+}] = 8.2 \times 10^{-8}[/tex] M

Thus, we can conclude that the [tex]H_{3}O^{+}[/tex] concentration in neutral water at this temperature is [tex]8.2 \times 10^{-8}[/tex] M.

Answer: The concentration of [tex]H_3O^+[/tex] in neutral water is [tex]8.2\times 10^{-8}M[/tex]

Explanation:

The chemical equation for the ionization of water follows:

[tex]2H_2O\rightleftharpoons H_3O^++OH^-[/tex]

The expression of [tex]K_w[/tex] for above equation, we get:

[tex]K_w=[H_3O^+]\times [OH^-][/tex]

We are given:

[tex]K_w=6.8\times 10^{-15}[/tex]

[tex][H^+]=[OH^-]=x[/tex]

Putting values in above equation, we get:  

[tex]6.8\times 10^{-15}=x\times x\\\\x=8.2\times 10^{-8}M[/tex]

Hence, the concentration of [tex]H_3O^+[/tex] in neutral water is [tex]8.2\times 10^{-8}M[/tex]

Using Henry's Law, the solubility of CO2 at 1.8 bar and 298 K is found to be 2970 g/m3. Since there is 1 L of water, this equates to 2.97 g of CO2 dissolved in the 1.00 L bottle of carbonated water.

The question asks how many grams of CO2 are dissolved in a 1.00 L bottle of carbonated water at a pressure of 1.8 bar and 298 K using the Henry's Law constant for CO2. To find the concentration of CO2 in the water, we use Henry's Law which states that the solubility of a gas in a liquid is proportional to the pressure of that gas above the liquid. The Henry's Law equation is Sg = kH * Pg, where Sg is the solubility of the gas, kH is the Henry's Law constant and Pg is the partial pressure of the gas.

The density of water is given as 998 kg/m3, which we use to convert the volume of water to mass. Since 1 L of water has a mass of approximately 998 g and the Henry's Law constant for CO2 at 298 K is 1.65×103 bar, we can plug in our values: Sg = (1.65×103 bar)(1.8 bar) = 2970 g/m3 of CO2. Since we have 1 L of water, or 0.998 kg, we can convert to grams: 0.998 kg × 1000 g/kg = 998 g.

To find the grams of CO2 dissolved, we use the solubility in g/m3: 2970 g/m3 × 0.001 m3 = 2.97 g of CO2.

Answer:

She used 50 mL of NaOH to reach the endpoint.

Explanation:

Assuming the burette is filled to the point marked 3.30 ml, You would record the initial point as 3.30 ml:

If at the end of the titration the level of the NaOH is at 20.30 mL; Subtract the initial reading from the final burette reading to get how many mL was used to reach an end point.

That is  20.3 - 3.3 = 17.00 mL

Therefore, the titration would have required 17.00 mL.

Remember that you should read the number that is at the bottom of the meniscus and at an eye level in order to avoid error.

Initial mark = 2.5 mL

Final mark = 52.5 mL.

volume used = 52.5 - 2.5

                            = 50 mL

Explanation:

Answer:

The answer is 3-Phenylpropanoic acid (see attached structure)

Explanation:

From spectral data:

3005 cm-1 ⇒ carboxylic acid (broad band)

1670 cm-1 ⇒ C=C

1603 cm-1 ⇒ Aromatic C-C bond

H NMR frequency at 2.6 ppm, singlet, ⇒ OH with no surrounding protons, possible deshielding  (clearer investigation of  spectrum would be expedient).

Hence, our C9H10O2 compound has an aromatic ring and carboxylic acid group attached to it.

The compound is probably a molecule with an aromatic ring (benzene) with an attached ester group (COOCH3) that satisfies all the given spectral data.

The spectral data given corresponds to a compound with molecular formula C9H10O2. From the IR data, the bands at 3005 cm-1 suggests C-H sp2 bond (alkene or aromatic), at 1676 cm-1 indicates carbonyl group (C=O), and 1603 cm-1 suggests a carbon-carbon double bond (C=C) which might be in an aromatic ring. The 1H NMR data: 2.6 ppm singlet signifies the protons of a methyl group (CH3) attached to an electronegative atom like a carbonyl carbon.

Based on these data, a probable structure for the compound could be a molecule, which is an aromatic ring (benzene) with an attached ester group (COOCH3). That gives the right molecular formula, the required carbonyl, alkene and sp2 hybridized C-H bonds for the IR, and the singlet in the NMR for the methyl group (CH3) of the ester.

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This is an incomplete question, here is a complete question.

Hydrogen peroxide decomposes to water and oxygen according to the following reaction:

[tex]H_2O_2(aq)\rightarrow H_2O+\frac{1}{2}O_2(g)[/tex]

It’s rate of decomposition is measured by titrating samples of the solution with potassium permanganate at certain intervals.

Initial rate determinations at 40°C for the decomposition give the following data:

[H₂O₂] (M)      Initial Rate (mol/L min)

 0.10                  1.93 × 10⁻⁴

 0.20                 3.86 × 10⁻⁴

 0.30                 5.79 × 10⁻⁴

Calculate the half-life for the reaction at 40°C?

Answer : The half life for the reaction is, 3590.7 minutes

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]H_2O_2(aq)\rightarrow H_2O+\frac{1}{2}O_2(g)[/tex]

Rate law expression for the reaction:

[tex]\text{Rate}=k[H_2O_2]^a[/tex]

where,

a = order with respect to [tex]H_2O_2[/tex]

Expression for rate law for first observation:

[tex]1.93\times 10^{-4}=k(0.10)^a[/tex] ....(1)

Expression for rate law for second observation:

[tex]3.86\times 10^{-4}=k(0.20)^a[/tex] ....(2)

Expression for rate law for third observation:

[tex]5.79\times 10^{-4}=k(0.30)^a[/tex] ....(3)

Dividing 2 by 1, we get:

[tex]\frac{3.86\times 10^{-4}}{1.93\times 10^{-4}}=\frac{k(0.20)^a}{k(0.10)^a}\\\\2=2^a\\a=1[/tex]

Thus, the rate law becomes:

[tex]\text{Rate}=k[H_2O_2]^1[/tex]

[tex]\text{Rate}=k[H_2O_2][/tex]

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

Putting values in equation 1, we get:

[tex]1.93\times 10^{-4}=k(0.10)^1[/tex]

[tex]k=1.93\times 10^{-3}min^{-1}[/tex]

Now we have to calculate the half-life for the reaction.

The expression used  is:

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

[tex]1.93\times 10^{-3}min^{-1}=\frac{0.693}{t_{1/2}}[/tex]

[tex]t_{1/2}=3590.7min[/tex]

Thus, the half life for the reaction is, 3590.7 minutes

Answer:

A solution that is 0.10 M HCN and 0.10 M LiCN

. A solution that is 0.10 M NH3 and 0.10 M NH4Cl

Explanation:

A buffer consists of a weak acid and its conjugate base counterpart. HCN is a weak acid and the salt LiCN contains its counterpart conjugate base which is the cyanide ion. A buffer maintains the pH by guarding against changes in acidity or alkalinity of the solution.

A solution of ammonium chloride and ammonia will also act as a basic buffer. A buffer may also contain a weak base and its conjugate acid.

Answer:

Good buffer systems are:  

A) NH3 + NH4Cl

C) HCN + LiCN

D) HF + NaF

Explanation:

Buffers consist in a mixture of a weak acid with its salt or a weak alkaly with its salt. All buffer systems are conformed by:

1) Weak acid + salt

or

2) Weak alkaly + salt

It is very important these salts come from the weak acid or weak alkaly. It means, the anion of the acid must be the anion in the salt which is going to be part of the buffer system. On the other hand, the cation of the weak alkaly must be the cation of the salt which is going to form the salt in the buffer system.

Then, when we evaluate all options in this exercise, answers are the following:

A) 0.10 M NH3 and 0.10 M NH4Cl. It is a buffer because NH3 (ammonia) is a weak alkaly and NH4Cl is a salt coming from NH3.

Buffer component reactions:

Reaction weak alkaly:   NH3 + H2O <-----> NH4+ + OH-

Reaction salt in water:  NH4Cl ---> NH4+ + Cl-

NH4+ is the cation of the weak alkaly so it must be part of the salt in the buffer system. Then NH4Cl is a salt from NH3.

C) 0.10 M HCN and 0.10 M LiCN. It is a buffer because HCN is a weak acid and LiCN is a salt which is coming from HCN.

Buffer component reactions:

Reaction weak acid:     HCN + H2O <-----> H3O+ + CN-  

Reaction salt in water:  LiCN --> Li+ + CN-

CN- is the anion of the acid, so it must be part of the salt in the buffer system. Then LiCN is a salt from HCN.

D) 0.10 M HF and 0.10 M NaF. It is a buffer because HF is a weak acid and NaF is a salt which is coming from HF.

Buffer component reactions:

Reaction weak acid:      HF + H2O <------> H3O+ + F-

Reaction salt in water:   NaF ---> Na+ + F-

F- is the anion of the weak acid (HF), so it must be part of the salt in th buffer systema. Then NaF is a salt coming from HF.

However option B, it is not a buffer, because it is a mixture of 0.10 M HCN and 0.10 M NaF.    Salt is not coming from the weak acid.

Reaction weak acid:    HCN + H2O <-----> H3O+ + CN-  (anion of the acid is CN-)

Rection salt in water:   NaF --> Na+ + F-  (anion in the salt is F-, not CN-)

Anion of the acid is CN- and the anion in the salt is F- so it is not a salt coming from the weak acid. Then option B it is not a buffer system.

Explanation:

CO(g) + Cl2(g) ⇄ COCl2(g)

This question is based on Le Chatelier's principle.

Le Chatelier's principle is an observation about chemical equilibria of reactions. It states that changes in the temperature, pressure, volume, or concentration of a system will result in predictable and opposing changes in the system in order to achieve a new equilibrium state.

a) COCl2 is added to the reaction mixture

COCL2 is a product in the reaction. If we add additional product to a system, the equilibrium will shift to the left, in order to produce more reactants. The reaction would shift to the left.

b) Cl2 is added to the reaction mixture

if we add reactants to the system, equilibrium will be shifted to the right to in order to maintain equilibrium by producing more products.

c) COCl2 is removed from the reaction mixture

if we remove products from the system, equilibrium will be shifted to the right to in order to maintain equilibrium by producing more products.

Chemical reactions at equilibrium, such as CO(g) + Cl2(g) ⇌ COCl2(g), shift in response to changes to re-establish equilibrium. If a product or reactant is added, the reaction will respectively shift towards reactants or products. Similarly, if a product or reactant is removed, the reaction will respectively shift towards products or reactants.

In chemistry, chemical reactions at equilibrium respond to disturbances according to Le Châtelier's principle: the system shifts in a way that counters the disturbance and re-establishes equilibrium. Now, let's consider the reaction: CO(g) + Cl2(g) ⇌ COCl2(g).

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The question is incomplete, here is the complete question:

The decay constant for 14-C is [tex]0.00012yr^{-1}[/tex] In 1947, the famous cave paintings in Lascaux, France were discovered and testing revealed that charcoal in the cave contained 20% of the 14-C found in living trees. Write a formula for the age of the charcoal (hence of the associated paintings). Show your work to find this formula.

Answer: The formula for the age of the charcoal is [tex]t=\frac{2.303}{1.2\times 10^{-4}yr^{-1}}\log \frac{100}{20}[/tex]

Explanation:

Carbon-14 isotope is a radioisotope and its decay process follows first order kinetics.

Rate law expression for first order kinetics is given by the equation:

[tex]k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}[/tex]

where,

k = rate constant  = [tex]0.00012yr^{-1}=1.2\times 10^{-4}yr^{-1}[/tex]

t = time taken for decay process = ? yr

[tex][A_o][/tex] = initial amount of the sample = 100 grams

[A] = amount left after decay process = (100 - 20) = 80 grams

Putting values in above equation, we get:

[tex]1.2\times 10^{-4}=\frac{2.303}{t}\log\frac{100}{20}\\\\t=\frac{2.303}{1.2\times 10^{-4}yr^{-1}}\log \frac{100}{20}[/tex]

Hence, the formula for the age of the charcoal is [tex]t=\frac{2.303}{1.2\times 10^{-4}yr^{-1}}\log \frac{100}{20}[/tex]

Answer:

0.169

Explanation:

Let's consider the following reaction.

A(g) + 2B(g) ⇄ C(g) + D(g)

We can find the pressures at equilibrium using an ICE chart.

       A(g) + 2 B(g) ⇄ C(g) + D(g)

I       1.00     1.00        0        0

C       -x        -2x        +x       +x

E    1.00-x  1.00-2x     x         x

The pressure at equilibrium of C is 0.211 atm, so x = 0.211.

The pressures at equilibrium are:

pA = 1.00-x = 1.00-0.211 = 0.789 atm

pB = 1.00-2x = 1.00-2(0.211) = 0.578 atm

pC = x = 0.211 atm

pD = x = 0.211 atm

The pressure equilibrium constant (Kp) is:

Kp = pC × pD / pA × pB²

Kp = 0.211 × 0.211 / 0.789 × 0.578²

Kp = 0.169

Final answer:

ΔH∘rxn is equal to ΔH∘f of the product(s) for the reactions CO(g) + 1/2O2(g) → CO2(g) and 2Li(s) + F2(g) → 2LiF(s).

Explanation:

In chemistry, the enthalpy change of a reaction (ΔH°rxn) is equivalent to the enthalpy of formation (ΔH°f) of the product(s) when the reaction involves the formation of a compound from its elements in their standard states.

The reactions for which ΔH∘rxn is equal to ΔH∘f of the product(s) are:

For these reactions, the enthalpy change of the reaction is equal to the standard enthalpy of formation of the product(s). The enthalpy change of the reaction can be calculated using Hess's law and the enthalpy changes of the individual reactions.

Answer:

The pH of the solution is 9.46

Explanation:

         C₅H₅N + H₂O ⇌ C₅H₅NH+OH⁻

I          0.46

C         - x                           +x +x

E         0.46 - x                        

-LogKb = Pkb

[tex]Kb =10^{[-PKb]} = 10^{[-8.75]} = 1.778 X 10^{-9}[/tex]

[tex]Kb = \frac{[C5H5NH][OH^-]}{[C5H5N]}[/tex]

[tex]1.778 X10^{-9}= \frac{X^2}{0.46-X} \\\\X^2 = 8.1788 X 10^{-10} - 1.778 X10^{-9}X\\\\X^2 + 1.778 X10^{-9}X -8.1788 X 10^{-10}\\\\X = 2.85977 X 10^{-5} = [OH^-][/tex]

pOH = -Log[OH⁻]

pOH = -Log [2.85977 x 10⁻⁵]

pOH = 4.54

pOH + pH = 14

pH = 14 - pOH

pH = 14 - 4.54

pH = 9.46

Therefore, the pH = 9.46

Answer:

Equilibrium Concentration of H₂O(g)  = 0.803

Equilibrium Concentration of Cl₂O(g)  = 0.540

Equilibrium Concentration of HOCl (g) = 0.214

Explanation:

Given;

                H₂O(g)  +   Cl₂O(g) <-------> 2HOCl (g)

I                 0.577        0.314                   0.666                

C               - x              -x                        +2x

E             0.577 - x     0.314 - x               0.666 +2x

[tex]K_c = \frac{[HOCL]^2}{[H_2O][CL_2O]} \\\\0.09 = \frac{[0.666+2x]^2}{[0.577-x][0.314-x]}\\\\0.09(0.1812 -0.891x+x^2) = (0.666+2x)(0.666+2x)\\\\0.0163-0.0802x+0.09x^2 = 0.4436+2.664x+4x^2\\\\3.91x^2+2.7742x+0.4273 =0\\\\x = -0.226, or -0.483[/tex]

Equilibrium Concentration of H₂O(g) = 0.577 - (- 0.226) = 0.803

Equilibrium Concentration of Cl₂O(g) = 0.314 - (- 0.226) = 0.540

Equilibrium Concentration of HOCl (g)  = 0.666 +2(- 0.226) = 0.214

Thus, from the result it can be seen that at equilibrium, the reactants are favored.

Answer:

a) Na2SiO3(s) + 8HF(aq) → H2SiF6(aq) + 2NaF(aq) + 3H2O (l)

b) 2.40 moles HF

c) 5.25 grams NaF

d)0.609 grams Na2SiO3

e) 0.89 grams Na2SiO3

Explanation:

Step 1: Data given

Step 2: The balanced equation

Na2SiO3(s) + 8HF(aq) → H2SiF6(aq) + 2NaF(aq) + 3H2O (l)

b) How many moles of HF are needed to react with 0.300 mol of Na2SiO3?

For 1 mol Na2SiO3 we need 8 moles HF to produce 1 mol H2SiF6, 2 moles NaF and 3 moles H2O

For 0.300 moles Na2SiO3 we'll need 8*0.300 = 2.40 moles HF

c. How many grams of NaF form when 0.500 mol of HF reacts with excess Na2SiO3?

For 1 mol Na2SiO3 we need 8 moles HF to produce 1 mol H2SiF6, 2 moles NaF and 3 moles H2O

For 0.500 moles we'll have 0.500 / 4 = 0.125 moles NaF

Mass NaF = 0.125 moles * 41.99 g/mol

Mass NaF = 5.25 grams NaF

d. How many grams of Na2SiO3 can react with 0.800 g of HF?

Moles HF = 0.800 grams / 20.01 g/mol

Moles HF = 0.0399 moles

For 1 mol Na2SiO3 we need 8 moles HF to produce 1 mol H2SiF6, 2 moles NaF and 3 moles H2O

For 8 moles we need 1 moles Na2SiO3 to react

For 0.0399 moles we need 0.0399/8 = 0.00499 moles Na2SiO3

Mass Na2SiO3 = 0.00499 moles * 122.06 g/mol  

Mass Na2SiO3 = 0.609 grams Na2SiO3

Using your answer to part d-If you started with 1.5 g of Na2SiO3how many grams would be left after the reaction is complete?

Number of moles HF = 0.0399 moles

Number of moles Na2SiO3 = 1.5 grams / 122.06 g/mol = 0.0123 moles

For 1 mol Na2SiO3 we need 8 moles HF to produce 1 mol H2SiF6, 2 moles NaF and 3 moles H2O

For 8 moles we need 1 moles Na2SiO3 to react

For 0.0399 moles we need 0.0399/8 = 0.00499 moles Na2SiO3

There will remain 0.0123 - 0.00499 = 0.00731 moles  Na2SiO3

Mass Na2SiO3 remaining = 0.00731 * 122.06 g/mol = 0.89 grams Na2SiO3

Final answer:

The balanced chemical equation requires 6 moles of HF for every mole of Na2SiO3. For different amounts of reactants, stoichiometry is used to calculate the moles and grams of other substances involved in the reaction.

Explanation:

The balanced chemical equation for the reaction between sodium metasilicate (Na2SiO3) and hydrofluoric acid (HF) is as follows:

Na2SiO3(s) + 6HF(aq) → H2SiF6(aq) + 2NaF(aq) + 3H2O(l)

b. According to the balanced equation, 6 moles of HF are needed to react with 1 mole of Na2SiO3. Therefore, to react with 0.300 mol of Na2SiO3, you would need 1.8 moles of HF (0.300 mol x 6).

c. Again according to the balanced equation, every 6 moles of HF produce 2 moles of NaF. So when 0.500 mol of HF reacts, it produces (0.500 mol HF x 2 mol NaF / 6 mol HF) = 0.167 moles of NaF. The molar mass of NaF is approximately 41.99 g/mol, so the mass of NaF formed is (0.167 mol x 41.99 g/mol) = 7.01 grams.

d. Given the molar mass of HF is approximately 20.01 g/mol, 0.800 g of HF corresponds to (0.800 g / 20.01 g/mol) = 0.0400 moles of HF. Using the balanced equation, we can determine the amount of Na2SiO3 that can react: (0.0400 mol HF x 1 mol Na2SiO3 / 6 mol HF) = 0.00667 moles of Na2SiO3. The molar mass of Na2SiO3 is approximately 122.06 g/mol, so the mass of Na2SiO3 that can react is (0.00667 mol x 122.06 g/mol) = 0.814 grams.

e. If you start with 1.5 g of Na2SiO3 and only 0.814 grams can react with the given 0.800 g of HF, then the amount of Na2SiO3 left after the reaction is (1.5 g - 0.814 g) = 0.686 grams.

Answer:

The procedure you will use in this exercise exploits the difference in acidity and solubility just described.

(a) you will dissolve your unknown in ethyl acetate (an organic solvent). All of the possible compounds are soluble in ethyl acetate.

(b) you will extract with sodium bicarbonate to remove any carboxylic acid that is present.

(c) you will extract with sodium hydroxide to remove any phenol that is present.

(d) you will acidify both of the resulting aqueous solutions to cause any compounds that were extracted to precipitate.

Answer:

ion travelled : 2.00cm -0.40cm

                     = 1.6cm

∴ Rf = 1.6/2.0

        = 0.80

Explanation:

Retention factor is the ratio of distance travelled by solute divided by distance travelled by solvent.

since you given distance travelled by solvent then the distance required by solute is need in this case the ion.

To design a high energy meal bar, ingredients should be chosen based on their potential chemical energy, with high-energy phosphates, lipids, proteins, and carbohydrates ranked from highest to lowest energy contents.

Ranking Atom Interactions by Chemical Energy

When designing a meal bar with high amounts of potential energy for busy students, your friend should consider the chemical structure of the ingredients. Chemical energy is stored within the bonds of atoms in food molecules. This chemical potential energy is what our bodies metabolize to generate energy, measured in Calories (calorimeter units). Atom interactions in food molecules can be broadly ranked from highest to lowest potential chemical energy as follows:

High-energy phosphates (such as in ATP)

Lipid molecules/fats (like triglycerides)

Protein (amino acid chains)

Carbohydrates (like glucose)

It is important to note that while lipids generally contain more energy per gram than proteins or carbohydrates, the balance and quality of nutrients also play a crucial role in the healthfulness of a meal bar. Therefore, your friend should focus on including a mix of macronutrients that release energy over time and sustain a student's activity levels efficiently.

To create a meal bar with a high amount of potential chemical energy, the macronutrients should be ranked as follows: lipids (fats) with the most chemical energy, followed by proteins, and then carbohydrates.

When designing a superfood meal bar with a high amount of potential chemical energy, we need to consider the atom interactions that store the most energy. The rank of atom interactions from highest to lowest chemical energy typically aligns with the macronutrient composition, as these are the main carriers of chemical energy in food.

Macronutrients Ranked by Chemical Energy:

These rankings are based on the caloric content obtained through oxidation processes like metabolism. In designing a high-energy food bar, your friend should focus on incorporating healthy fats, complete proteins, and complex carbohydrates while considering the overall nutritional balance.

The estimated volume occupied by 18 kg of ethylene at 55℃ and 35 bar is approximately 476.6 liters. The mass of ethylene contained in a [tex]0.25m^3[/tex] cylinder at 50℃ and 115 bar is approximately 3.03 kg.

To estimate the volume occupied by 18 kg of ethylene at 55℃ and 35 bar, we can use the ideal gas law expressed as PV = nRT. First, we need to calculate the number of moles (n) using the molar mass of ethylene ([tex]C_2H_4[/tex]) which is approximately 28 g/mol. Therefore, n = 18000 g / 28 g/mol = 642.86 moles. The gas constant (R) is 0.08314 L*bar/mol*K. Converting the temperature to Kelvin, T = 55 + 273 = 328 K. Now, we can solve for V:

V = (nRT) / P = (642.86 mol * 0.08314 L*bar/mol*K * 328 K) / 35 bar = 476.6 L

To determine the mass of ethylene in a 0.25-[tex]m^3[/tex] cylinder at 50℃ and 115 bar, we apply the ideal gas law again. We first convert the volume to liters (V = 0.25 [tex]m^3[/tex] * 1000 L/[tex]m^3[/tex] = 250 L), and temperature to Kelvin (T = 50 + 273 = 323 K). The pressure is already in bar. Now, we calculate the number of moles, solving for n:

n = PV / RT = (115 bar * 250 L) / (0.08314 L*bar/mol*K * 323 K) = 108.14 moles

The mass of ethylene is then m = n * molar mass = 108.14 moles * 28 g/mol = 3027.92 g or approximately 3.03 kg of ethylene.

Answer:

1. H3PO4 + 3NaOH —> Na3PO4 + 3H2O

2. 38.5mL

Explanation:

1. We'll begin by writing a balanced equation for the reaction. This is illustrated below:

H3PO4 + 3NaOH —> Na3PO4 + 3H2O

2. H3PO4 + 3NaOH —> Na3PO4 + 3H2O

From the equation above, the following data were obtained:

nA (mole of the acid) = 1

nB (mole of the base) = 3

Data obtained from the question include:

Vb (volume of base) =?

Mb (Molarity of base) = 0.120 M

Va (volume of acid) = 35.0 mL

Ma (Molarity of acid) = 0.0440 M

Using the formula MaVa/MbVb = nA/nB, the volume of the base (i.e NaOH) can be obtained as follow:

MaVa/MbVb = nA/nB

0.0440 x 35/ 0.120 x Vb = 1/3

Cross multiply to express in linear form as shown below:

0.120 x Vb = 0.0440 x 35 x 3

Divide both side by 0.120

Vb = (0.0440 x 35 x 3) /0.120

Vb = 38.5mL

Therefore, 38.5mL of 0.120 M NaOH is needed for the complete neutralization.

Answer:

We need 38.5 mL of NaOH to neutralize the H3PO4 solution

Explanation:

Step 1: Data given

Molarity of NaOH = 0.120 M

Volume of H3PO4 = 35.0 mL = 0.035 L

Molarity of H3PO4 = 0.0440 M

Step 2: The balanced equation

H3PO4 + 3NaOH —> Na3PO4 + 3H2O

Step 3: Calculate the volume of NaOH

b*Ca*Va = a *Cb*Vb

⇒with b = the coefficient of NaOH = 3

⇒with Ca = the concentration of H3PO4 = 0.0440 M

⇒with Va = the volume of H3PO4 = 35.0 mL = 0.0350 L

⇒with a = the coefficient of H3PO4 = 1

⇒with Cb = the concentration of NaOH = 0.120 M

⇒with Vb = the volume of NaOH = TO BE DETERMINED

3*0.0440 * 0.0350 = 0.120 * Vb

Vb = 0.0385 L = 38.5 mL

We need 38.5 mL of NaOH to neutralize the H3PO4 solution

Answer:

Ar < Cl - < S2-

Explanation:

All the species written above are isoelectronic. This means that they all possess the same number of electrons. All the species above possess 18 electrons, the noble gas electron configuration.

However, for isoelectronic species, the greater the atomic number of the specie, the smaller it is. This is because, greater atomic number implies that their are more protons in the nucleus exerting a greater attractive force on the electrons thereby making the specie smaller in size due to high electrostatic attraction.

The correct order of increasing expected size among Cl-, S2-, and Ar based on the simple shell model and their respective nuclear charges—considering they are all isoelectronic—is Ar < Cl- < S2-.

The question asks to arrange the atoms and/or ions Cl-, S2-, and Ar in order of increasing expected size based on the simple shell model of atomic radius. In the simple shell model, atomic radius increases with the addition of electron shells and decreases with an increase in nuclear charge for isoelectronic species, meaning species with the same number of electrons. Cl-, S2-, and Ar are all isoelectronic with the closed-shell electron configuration of [Ar]. The Cl- ion has 17 protons, S2- has 16 protons, and Ar has 18 protons.

Following this reasoning:

Therefore, the correct order from smallest to largest is: Ar < Cl- < S2-.

Answer:

0.4g/mL

Explanation:

The following data were obtained from the question:

Mass = 2g

Volume = 75 — 70 = 5mL

Density =?

Density = Mass /volume

Density = 2/5

Density = 0.4g/mL

The density of the substance is 0.4g/mL

The density of a substance that has a mass of 2.0 g, and when placed in a graduated cylinder the volume rose from 70 mL to 75 mL is 0.4 g/mL.

To find the density of the substance, we need to calculate the volume it displaces and then divide its mass by that volume.

The volume displaced by the substance can be found by subtracting the initial volume in the graduated cylinder from the final volume after the substance is added. The initial volume is 70 mL and the final volume is 75 mL. Therefore, the volume displaced by the substance is:

Volume displaced = Final volume - Initial volume

Volume displaced = 75 mL - 70 mL

Volume displaced = 5 mL

Now, we have the mass of the substance, which is 2.0 g, and the volume it displaces, which is 5 mL. The density [tex]\rho[/tex] is calculated by dividing the mass (m) by the volume (V):

[tex]\rho = \frac{m}{V}[/tex]

[tex]\rho = \frac{2.0 \text{ g}}{5 \text{ mL}}[/tex]

[tex]\rho = 0.4 g/mL[/tex]

Answer:

The Brönsted-Lowry model focuses on the transfer of H⁺ in an acid-base reaction.

Explanation:

According to the theory of Brönsted-Lowry , an acid is a chemical substance capable of releasing hydrogen ions, while a base is that chemical substance capable of accepting hydrogen ions. That is, acids are substances capable of yielding protons (hydrogen H + ions) and substance bases capable of accepting them.

On the other hand, the conjugate base of a Brønsted-Lowry acid is the species that forms after an acid donated a proton. The conjugate acid of a Brønsted-Lowry base is the species that forms when a base accepts a proton.

Thus, the acid-base reaction is one in which the acid transfers a proton to a base (proton transfer H⁺).

This theory, unlike another theory like Arrhenius, does not require the presence of water as a means of reaction for the transfer of H⁺.

The Brönsted-Lowry model focuses on the transfer of H⁺ in an acid-base reaction.

Answer:

the correct answer:

C, B, D, A.

Explanation:

The chemical compounds named are compounds that have strong chemical bonds, forming cubic structures and crystals with very high boiling and melting points to less solid structures.

some are oxides, others salts, others are even used to emit radiation.

Answer:

true

Explanation:

This an incomplete question, here is a complete question.

Calculate the empirical formula for each stimulant based on its elemental mass percent composition.

a. nicotine (found in tobacco leaves): C 74.03%, H 8.70%, N 17.27%

b. caffeine (found in coffee beans): C 49.48%, H 5.19 %, N 28.85% and O 16.48%

Answer:

(a) The empirical formula for the given compound is [tex]C_5H_7N[/tex]

(b) The empirical formula for the given compound is [tex]C_4H_5N_2O[/tex]

Explanation:

Part A: nicotine

We are given:

Percentage of C = 74.03 %

Percentage of H = 8.70 %

Percentage of N = 17.27 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 74.03 g

Mass of H = 8.70 g

Mass of N = 17.27 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =[tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{74.03g}{12g/mole}=6.17moles[/tex]

Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{8.70g}{1g/mole}=8.70moles[/tex]

Moles of Nitrogen = [tex]\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{17.27g}{14g/mole}=1.23moles[/tex]

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.23 moles.

For Carbon = [tex]\frac{6.17}{1.23}=5.01\approx 5[/tex]

For Hydrogen  = [tex]\frac{8.70}{1.23}=7.07\approx 7[/tex]

For Nitrogen = [tex]\frac{1.23}{1.23}=1[/tex]

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

The empirical formula for the given compound is [tex]C_5H_7N_1=C_5H_7N[/tex]

Part B: caffeine

We are given:

Percentage of C = 49.48 %

Percentage of H = 5.19 %

Percentage of N = 28.85 %

Percentage of O = 16.48 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 49.48 g

Mass of H = 5.19 g

Mass of N = 28.85 g

Mass of O = 16.48 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = [tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{49.48g}{12g/mole}=4.12moles[/tex]

Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{5.19g}{1g/mole}=5.19moles[/tex]

Moles of Nitrogen = [tex]\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{28.85g}{14g/mole}=2.06moles[/tex]

Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{16.48g}{16g/mole}=1.03moles[/tex]

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.03 moles.

For Carbon = [tex]\frac{4.12}{1.03}=4[/tex]

For Hydrogen  = [tex]\frac{5.19}{1.03}=5.03\approx 5[/tex]

For Nitrogen = [tex]\frac{2.06}{1.03}=2[/tex]

For Nitrogen = [tex]\frac{1.03}{1.03}=1[/tex]

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N : O = 4 : 5 : 2 : 1

The empirical formula for the given compound is [tex]C_4H_5N_2O_1=C_4H_5N_2O[/tex]