When a double replacement reaction produced one compound that is soluble and one compounds that is insoluble,a. only the solution of dissolved ion resultsb. only a solid is formedc. the soluble compound exists as dissolved ions in a solution while insoluble compound forms as a precipitated. none

Answer:

ΔG°rxn = -72.9 kJ

Explanation:

Let's consider the following reaction.

HCN(g) + 2 H₂(g) → CH₃NH₂(g)

We can calculate the standard Gibbs free energy of the reaction (ΔG°rxn) using the following expression:

ΔG°rxn = ΔH° - T.ΔS°

where,

ΔH° is the standard enthalpy of the reaction

T is the absolute temperature

ΔS° is the standard entropy of the reaction

ΔG°rxn = -158.0 KJ - 387 K × (-219.9 × 10⁻³ J/K)

ΔG°rxn = -72.9 kJ

The estimated ΔG°rxn for the given reaction HCN(g) + 2 H2(g) → CH3NH2(g) at 387K, calculated using the equation ΔG = ΔH - TΔS and the given values ΔH° = −158.0 kJ and ΔS° = −219.9 J/K, is -72.86 kJ.

To estimate ΔG ° rxn for the given reaction HCN(g) + 2 H2(g) → CH3NH2(g) at 387K, we need to use the equation ΔG = ΔH - TΔS. Given that ΔH° = −158.0 kJ and ΔS° = −219.9 J/K (remember to convert kJ to J, so ΔH is -158000 J), we substitute the values into the equation: ΔG = (-158000 J) - (387K * -219.9 J/K) =  -158 kJ + 85.14 kJ = -72.86 kJ. So, the estimated ΔG°rxn for the reaction under these conditions is -72.86 kJ.

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Answer:

0,07448M of phosphate buffer

Explanation:

sodium monohydrogenphosphate (Na₂HP) and sodium dihydrogenphosphate (NaH₂P) react with HCl thus:

Na₂HP + HCl ⇄ NaH₂P + NaCl (1)

NaH₂P + HCl ⇄ H₃P + NaCl (2)

The first endpoint is due the reaction (1), When all phosphate buffer is as NaH₂P form, begins the second reaction. That means that the second endpoint is due the total concentration of phosphate that is obtained thus:

0,01862L of HCl×[tex]\frac{0,1000mol}{L}[/tex]= 1,862x10⁻³moles of HCl ≡ moles of phosphate buffer.

The concentration is:

[tex]\frac{1,862x10^{-3}moles}{0,02500L}[/tex] = 0,07448M of phosphate buffer

I hope it helps!

Explanation:

[tex]4 Fe(s) + 3O_2(g)\rightarrow 2Fe_2O_3(s) ,\Delta H = -1652 kJ[/tex]

a) Heat released when 4.48 moles of iron is reacted with excessive oxygen:

According to reaction, when 4 moles of an iron reacts with 3 moles of oxygen it gives 1625 kilo Joules of heat.

Then 4.48 moles of iron will give:

[tex]\frac{1652 kJ}{4}\times 4.48=1850.24 kJ[/tex]

1850.24 kJ of heat is released when 4.48 moles of an iron is reacted with excess oxygen.

b) Heat released when 1.76 mol [tex]Fe_2O_3[/tex] is produced.

According to reaction, when 2 moles of an [tex]Fe_2O_3[/tex] are produced 1625 kilo Joules of heat is released

Then heat released on production of 1.76 mol [tex]Fe_2O_3[/tex] :

[tex]\frac{1652 kJ}{2}\times 1.76=1453.76 kJ[/tex]

1453.76 kJ heat is released when 1.76 mol [tex]Fe_2O_3[/tex] is produced.

c) Heat released when 2.66 grams of iron is reacted with excessive oxygen:

Moles of iron = [tex]\frac{2.66 g}{56 g/mol}=0.0475 mol[/tex]

According to reaction, when 4 moles of an iron reacts with 3 moles of oxygen it gives 1625 kilo Joules of heat.

Then 0.0475 moles of iron will give:

[tex]\frac{1652 kJ}{4}\times 0.0475=19.62 kJ[/tex]

19.62 kJ of heat is released when 2.66 grams of iron is reacted with excessive oxygen.

d) Heat released when 12.8 g Fe and 1.49 g oxygen gas are reacted:

Moles of iron = [tex]\frac{12.8 g}{56 g/mol}=0.2286 mol[/tex]

According to reaction, 4 moles of iron reacts with 3 moles of oxygen.Then 0.2286 mol will react with:

[tex]\frac{3}{4}\times 0.2286 mol=0.17145 mol[/tex] of oxygen

Moles of oxygen = [tex]\frac{1.49 g}{56 g/mol}=0.04656 mol[/tex]

According to reaction, 3  moles of oxygen gas reacts with 4 moles of iron .Then 0.04656 mol will react with:

[tex]\frac{4}{3}\times 0.04656 mol=0.06208 mol[/tex] of iron.

From this we can conclude that oxygen is in limiting amount. So, the amount of energy release will depend upon moles of oxygen gas.

According to reaction, when 4 moles of an iron reacts with 3 moles of oxygen it gives 1625 kilo Joules of heat.

Then 0.04656 moles of oxygen gas will give:

[tex]\frac{1652 kJ}{3}\times 0.04656 =25.7011 kJ[/tex]

25.7011 kJ of Heat is released when 12.8 g Fe and 1.49 g oxygen gas are reacted

The heat released in an exothermic reaction can be calculated using stoichiometry based on the ΔH value. The released heat varies with the amount of reactant or product involved. By calculating the molar ratios, the heat released is computed for each sub-question accordingly.

The reaction you provided loosens heat, hence it is an exothermic reaction. The heat released can be calculated using stoichiometry if we know the amount of reactants or products. ΔH (-1652 kJ) is the amount of heat released for the reaction of 4 moles of Fe with 3 moles of O2 to form 2 moles of Fe2O3.

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Answer:

Correct option -D

Explanation:

Mass defect:

The difference between the sum of the masses of individual nucleons that form an atomic nucleus and the mass of the nucleus.

Hence, i- is correct

Nuclear fission:

It is the process of splitting a nucleus into two nuclei with smaler masses.

Hence, ii- is correct

Nuclear fission involved bombarding with nuclei [tex]^{235}_{92}U[/tex] with [tex]^{4}_{2}He[/tex]

Nuclear fission of [tex]^{235}_{92}U[/tex] with neutrons forms two smaller mass nuclei.

[tex]^{1}_{0}n+^{235}_{92}U \rightarrow ^{91}_{36}Kr+^{142}_{56}B+3^{1}_{0}n+Energy[/tex]

Hence, iii- is incorrect

Therefore, option -D is correct

Statements i) and ii) are correct. Mass defect refers to the difference in mass between that of a nucleus and the sum of the masses of its component nucleons. Nuclear fission refers to the splitting of a heavier nucleus into two nuclei with smaller mass numbers.

The correct statement(s) from the ones provided are: i) Mass defect is the difference in mass between that of a nucleus and the sum of the masses of its component nucleons. This is because the mass of a nucleus is not just the sum of the individual masses of the protons and neutrons. Some mass is converted into energy to provide the binding force that holds the nucleus together, so the mass of the nucleus has a slight deficit (or defect) compared to the combined masses of its components.

ii) Nuclear fission involves the splitting of a heavier nucleus into two nuclei with smaller mass numbers. This process usually does not occur naturally, but is induced by bombardment with neutrons. The first reported nuclear fission occurred in 1939 when uranium-235 atoms were bombarded with slow-moving neutrons that split the uranium nuclei into smaller fragments.

Thus, the correct option is D. i) and ii) only.

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Answer:

7.43 × 10²⁴ m⁻³

Explanation:

Data provided in the question:

Conductivity of a semiconductor specimen, σ = 2.8 × 10⁴ (Ω-m)⁻¹

Electron concentration, n = 2.9 × 10²² m⁻³

Electron mobility, [tex]\mu_n[/tex] = 0.14 m²/V-s

Hole mobility, [tex]\mu_p[/tex]= 0.023 m²/V-s

Now,

σ = [tex] nq\mu_n+pq\mu_p[/tex]

or

σ = [tex] q(n\mu_n+p\mu_p)[/tex]

here,

q is the charge on electron = 1.6 × 10⁻¹⁹ C

p is the hole density

thus,

2.8 × 10⁴ = 1.6 × 10⁻¹⁹( 2.9 × 10²² × 0.14 +  p × 0.023 )

or

1.75 × 10²³ = 0.406 × 10²² + 0.023p

or

17.094 × 10²² = 0.023p

or

p = 743.217 ×  10²²

or

p = 7.43 × 10²⁴ m⁻³

Answer:

The standard potential for this galvanic cell is 1.25 V

Explanation:

Step 1

Oxidation: Fe(s) → Fe^2+ (aq) + 2e-     E° = 0.447 V

Reduction: Ag+(aq) + e- → Ag(s)          E° = 0.7996 V

The electron number for the oxidation is 2; the electron number for the reduction is 1.

By multiplying the reduction reaction by 2, we make the electron number the same between the two half-equations

2 Ag+(aq) + 2e- → 2Ag(s)          E° = 0.7996 V

Step 2: The overall balanced equation

Fe(s) + 2Ag+(aq) → Fe^2+(aq) + 2ag(s)   E° = 0.447 V + 0.7996 V = 1.25V

The standard potential for this galvanic cell is 1.25 V

Answer:

Complete ionic equation:

2H²⁺(aq) + SO₄²⁻(aq) + CaCO₃(s)   →  H₂O(l) + CO₂(g) + Ca²⁺(aq) + SO₄²⁻(aq)

Net ionic equation:

2H²⁺(aq) + CaCO₃(s)   →  H₂O(l) + CO₂(g) + Ca²⁺(aq)

Explanation:

Chemical equation:

H₂SO₄(aq) + CaCO₃(s)   →  H₂O(l) + CO₂(g) + CaSO₄(aq)

Balanced chemical equation:

H₂SO₄(aq) + CaCO₃(s)   →  H₂O(l) + CO₂(g) + CaSO₄(aq)

Complete ionic equation:

2H²⁺(aq) + SO₄²⁻(aq) + CaCO₃(s)   →  H₂O(l) + CO₂(g) + Ca²⁺(aq) + SO₄²⁻(aq)

Net ionic equation:

2H²⁺(aq) + CaCO₃(s)   →  H₂O(l) + CO₂(g) + Ca²⁺(aq)

Answer:

Complete ionic equation: 2 H⁺(aq) + SO₄²⁻(aq) + CaCO₃(s) → H₂O(I) + CO₂(g) + Ca²⁺(aq) + SO₄²⁻(aq)

Net ionic equation: 2 H⁺(aq) + CaCO₃(s) → H₂O(I) + CO₂(g) + Ca²⁺(aq)

Explanation:

The molecular equation includes all the species in the molecular form.

H₂SO₄(aq) + CaCO₃(s) → H₂O(I) + CO₂(g) + CaSO₄(aq)

The complete ionic equation includes all the ions and the molecular species.

2 H⁺(aq) + SO₄²⁻(aq) + CaCO₃(s) → H₂O(I) + CO₂(g) + Ca²⁺(aq) + SO₄²⁻(aq)

The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and the molecular species.

2 H⁺(aq) + CaCO₃(s) → H₂O(I) + CO₂(g) + Ca²⁺(aq)

Answer:

119 kCal per serving.

Explanation:

The heat energy necessary to elevates water's temperature from 23.4°C to 37.9°C can be calculated by the equation below:

Q = mcΔT

Q: heat energy

m: mass in g

c: specific heat capacity in cal/g°C

ΔT = temperature variation in °C

m is the mass of water, considering the density of water to be 1g/mL, 100 mL of water weights  100g. Therefore:

Q = 100 g x 1.00 cal/g°C x (37.9 - 23.4)°C

Q = 1450 cal

1450 cal ____ 0.341 g peanuts

x             ____  28 g peanuts

x = 119061.58 cal

This means that the cal from fat per serving of peanuts is at least 119 kCal.

The temperature change of the water indicated 1.45 Calories for 0.341 grams of peanut. Therefore, a serving size of 28.0 grams contains option 2) 119 Calories from fat.

To determine the Calories from fat per serving, we must first calculate the energy released by the peanut during combustion and then scale it to a full serving size.

Thus, the Calories from fat per serving is option 2) 119 Cal/serving.

Answer:

a) 40 %

b) [tex]4.04~g~CO_2[/tex]

c) [tex]5.53x10^2^3~molecules~of~O_2[/tex]

Explanation:

For a) we will have to calculate the molar mass of [tex]C_6H_1_2O_6[/tex], so the first step is to find the atomic mass of each atom and multiply by the amount of atoms in the molecule.

C => 12*(6) = 72

H => 1*(12) = 12

O => 6*(16) = 96

Molar mass = 180 g/mol

Then we can calculate the percentage by mass:

[tex]Percentage~=~\frac{72}{180}*100=40[/tex]

For b) we have to start with the reaction of glucose:

[tex]C_6H_1_2O_6~+~6O_2~->~6CO_2~+~6H_2O[/tex]

Then we have to convert the grams of glucose to moles, the moles of glucose to moles of carbon dioxide and finally the moles of carbon dioxide to grams. To do this we have to take into account the following conversion ratios:

-) 180 g of glucose = 1 mol glucose

-) 1 mol glucose = 6 mol carbon dioxide

-) 1 mol carbon dioxide = 44 g carbon dioxide

[tex]16.55~g~C_6H_1_2O_6\frac{1~mol~C_6H_1_2O_6}{180~g~C_6H_1_2O_6}\frac{6~mol~CO_2}{1~mol~C_6H_1_2O_6}\frac{44~g~CO_2}{1~mol~CO_2}=4.04~g~CO_2[/tex]

For C, we have to start with the conversion from grams of glucose to moles, the moles of glucose to moles of oxygen and finally the moles of oxygen to molecules. To do this we have to take into account the following conversion ratios:

-) 180 g of glucose = 1 mol glucose

-) 1 mol glucose = 6 mol oxygen

-) 1 mol oxygen = 6.023x10^23 molecules of O2

[tex]16.55~g~C_6H_1_2O_6\frac{1~mol~C_6H_1_2O_6}{180~g~C_6H_1_2O_6}\frac{6~mol~O_2}{1~mol~C_6H_1_2O_6}\frac{6.023x10^2^3~molecules~O_2}{1~mol~O_2}=~5.53x10^2^3~molecules~of~O_2[/tex]

This detailed answer explains the mass percent of carbon in glucose, the mass of CO2 produced by its combustion, and the number of oxygen molecules needed for complete combustion.

a) Mass Percent of Carbon in Glucose: Glucose has a molar mass of 180.16 g/mol. The molar mass of the carbon in one mole of glucose is 72.06 g (6 carbons in a molecule of glucose). Therefore, the mass percent of carbon in glucose is 40%.

b) Mass of CO2 Produced: The combustion of 16.55 g of glucose will produce 44.01 g of CO2.

c) Oxygen Molecules Needed: For the complete combustion of 16.55 g of glucose, 23 molecules of O2 are required.

Answer: The chemical equation is written below.

Explanation:

Transmutation is defined as the process in which one chemical isotope gets converted to another chemical isotope. The number of protons or neutrons in the isotope gets changed.

The chemical equation for the reaction of curium-242 nucleus with alpha particle (helium nucleus) follows:

[tex]_{96}^{242}\textrm{Cm}+_4^2\textrm{He}\rightarrow _{98}^{245}\textrm{Cf}+_0^1\textrm{n}[/tex]

The product formed in the nuclear reaction are californium-245 nucleus and a neutron particle.

Answer:

1. Strong electrolytes: HCl and NaOH.

Weak electrolytes: CH₃COOH and NH₃.

Nonelectrolytes: (NH₂)₂CO (urea) and CH₃OH (methanol).

2. C) [tex]C) Sr(C₂H₃O₂)₂ (aq) + Na₂SO₄ (aq) → SrSO₄ (s) + 2NaC₂H₃O₂ (aq)[/tex]

3. The spectator ions are Na⁺ and C₂H₃O₂⁻.

4. Written in explanation section.

Explanation:

A characteristic of strong electrolytes is that the solute is assumed to be 100 percent dissociated  into ions in solution, therefore they are great conductors of electricity. Eg: HCl and NaOH.

On the other hand, weak electrolytes are not completely dissociated in solution, therefore they are poor conductors of electricity. Eg: CH₃COOH and NH₃.

(By dissociation we mean the breaking up of the compound into cations and anions).

     A nonelectrolyte does not conduct electricity when dissolved in water. Eg: (NH₂)₂CO (urea) and CH₃OH (methanol).

     2. The products formed on each reaction are (always remember to balance the equations):

A) [tex]2 LiOH (aq) + Na₂S (aq) → Li₂S (aq) + 2NaOH (aq)[/tex]

B) [tex](NH₄)₂SO4 (aq) +2 LiCl (aq) → 2NH₄Cl (aq) + Li₂SO₄ (aq)[/tex]

C) [tex]C) Sr(C₂H₃O₂)₂ (aq) + Na₂SO₄ (aq) → SrSO₄ (s) + 2NaC₂H₃O₂ (aq)[/tex]

D) [tex]D) KNO₃ (aq) + NaOH (aq) → NaNO₃ (aq) + KOH (aq) [/tex]

The reaction C will produce SrSO₄, a white color precipitate.

    3. When ionic compounds dissolve in water, they break apart into their component cations and anions. To be more realistic, the equations should show the dissociation of dissolved ionic compounds into ions. This is called a ionic equation which shows dissolved species as free ions. To see whether a precipitate might form from this solution, we first combine the cation and anion from different compounds, and refer to the solubility rules. The spectator ions are ions that are not involved in the overall reaction.

Therefore, for the equation chosen above:

[tex]Sr²⁺ + 2C₂H₃O₂⁻ + 2Na⁺ + SO₄²⁻ → SrSO₄ + 2Na⁺ + 2C₂H₃O₂⁻[/tex]

Because spectator ions appear on both sides of an equation, they can be eliminated from the ionic equation.

[tex]Sr²⁺ + SO₄²⁻ → SrSO₄[/tex]

Finally, we end up with the net ionic equation, which shows only the species that actually take part in the reaction.

In this reaction, the spectator ions are Na⁺ and C₂H₃O₂⁻.

     4.  Molecular equations:

A) [tex]2 LiOH (aq) + Na₂S (aq) → Li₂S (aq) + 2NaOH (aq)[/tex]

B) [tex](NH₄)₂SO4 (aq) +2 LiCl (aq) → 2NH₄Cl (aq) + Li₂SO₄ (aq)[/tex]

C) [tex]C) Sr(C₂H₃O₂)₂ (aq) + Na₂SO₄ (aq) → SrSO₄ (s) + 2NaC₂H₃O₂ (aq)[/tex]

D) [tex]D) KNO₃ (aq) + NaOH (aq) → NaNO₃ (aq) + KOH (aq) [/tex]

         Complete ionic equations:

A) [tex]2Li⁺ + OH⁻ + 2Na⁺ + S²⁻ → 2Li⁺ + S²⁻ + 2Na⁺ +OH⁻[/tex]

B) [tex]2NH₄⁺ + SO₄²⁻ + 2Li⁺ + 2Cl⁻ → 2Li⁺ +S²⁻ +2NH₄⁺ +2Cl⁻ + 2Li⁺ + SO₄²⁻[/tex]

C) [tex]Sr²⁺ + 2C₂H₃O₂⁻ + 2Na⁺ + SO₄²⁻ → SrSO₄ + 2Na⁺ + 2C₂H₃O₂⁻[/tex]

D) [tex]K⁺ + NO₃⁻ + Na⁺ +OH⁻ → Na⁺ + NO₃⁻ + K⁺ + OH⁻[/tex]

        If all products are aqueous, a net ionic equation cannot be written because all ions are canceled out as spectator ions. Therefore, no precipitation reaction occurs. The only net equation can be written for reaction C):

C) [tex]Sr²⁺ + SO₄²⁻ → SrSO₄[/tex]

Answer:

a) 2-chloro-2,3-dimethylbutane

b) 1-chloro-2,3-dimethylbutane

Explanation:

The monochlorination is a reaction in which we have to add only 1 Cl to the molecule. In this case, we will have to add a Cl to a primary carbon (a) and to a tertiary carbon (b).

In the monochlorination of the primary carbon, we can choose any methyl carbon. For the monochlorination of the terciary carbon we have to choose an CH carbon.

(See the figure)

Final answer:

When 2,3-dimethylbutane undergoes monochlorination under radical substitution conditions, multiple constitutional isomers can be formed due to the presence of primary and tertiary carbon atoms.

Explanation:

When 2,3-dimethylbutane undergoes monochlorination under radical substitution conditions, multiple constitutional isomers can be formed. The monochlorination reaction involves the substitution of a hydrogen atom on the carbon chain with a chlorine atom. Since 2,3-dimethylbutane has two different types of carbon atoms (primary and tertiary), the resulting constitutional isomers will also differ.

(a) Primary alkyl halide: The primary carbon atom is bonded to one other carbon atom and one hydrogen atom. So, a constitutional isomer could be formed by substituting the hydrogen atom with a chlorine atom.

(b) Tertiary alkyl halide: The tertiary carbon atom is bonded to three other carbon atoms. So, a constitutional isomer could be formed by substituting one of the carbon atoms with a chlorine atom.

Answer:

The molecular formula for the organic compound is C₈H₈O₂

Explanation:

70.58 % C , 5.9 % H , and 23.50 % O

This is a percent by mass, so:

in 100 g of compound I have 70.58 g of C, 5.9 g of H and 23.5 g of O.

We need to apply the formula for freezing point depression, the colligative property.

ΔT = Kf . m

Where ΔT means T°F (pure solvent) − T°F (solution)

Kf, the cryoscopic constant and m is molality (mol of solute in 1kg of solvent); Kf for cyclohexane is 20.8 °C/m

6.59°C - 3.29°C = 20.8 °C/m . m

3.3°C = 20.8 °C/m . m

3.3°C /20.8 m/°C = molal

0.158 m/kg

In 1kg of solvent I have 0.158 moles, but I have 718.1 g of cyclohexane

1000 g _____ 0.158 moles of solute

718.1 g _____ 0.114 moles of solute

To find out the molar mass of my compound I have to divide

mass/mol → 15.5 g /0.114 m = 136.04 g/m

Now I have to work with the percent and think this rules of three:

100 g of compound has ____ 70.58 g of C __5.9 g of H __23.5 g of O

136.04 g of compound has ___  96.02 g of C __8.02 g of H __ 32 g of O

Molar mass of C = 12 g/m

Molar mass of H = 1 g/m

Molar mass of O = 16 g/m

g / molar mass = moles

C: 96 g / 12 g/m = 8

H: 8 g / 1 g/m =8

O: 32 g/ 16 g/m = 2

The molecular formula for the organic compound is C₈H₈O₂

To find the molecular formula of the organic compound, calculate the empirical formula first, which is CH₃O. Then calculate the molar mass of the compound, which is 29g/mol. Divide the molar mass of the compound by the molar mass of the empirical formula to find the molecular formula ratio. The molecular formula is CH₃O.

To determine the molecular formula of the organic compound, we need to calculate the empirical formula first.

The percentages of carbon, hydrogen, and oxygen in the compound represent the mass ratios of these elements. Assume we have 100g of the compound, which means we have 70.58g of carbon, 5.9g of hydrogen, and 23.50g of oxygen. Now we convert these masses to moles using the molar masses: 12g/mol for carbon, 1g/mol for hydrogen, and 16g/mol for oxygen.

Next, we divide the moles of each element by the smallest number of moles to get the simplest mole ratio. The empirical formula of the compound is CH3O.

To find the molecular formula, we need to know the molar mass of the compound. This can be calculated by adding the atomic masses of the elements in the empirical formula: 12g/mol for carbon, 1g/mol for hydrogen, 16g/mol for oxygen. The sum is 29g/mol.

Since the molar mass of the compound is 29g/mol and 15.5g of the compound is dissolved in 718.1g of cyclohexane, we can calculate the moles of the compound: 15.5g / 29g/mol = 0.534moles.

The molecular formula is found by multiplying the empirical formula by an integer value to match the molar mass. To do this, divide the molar mass of the compound by the molar mass of the empirical formula: 29g/mol / 29g/mol = 1.

Therefore, the molecular formula of the organic compound is CH3O.

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Answer:

A carboxylate salt and water

Explanation:

A carboxylic acid is an organic compound that has general formula RCOOH, where R is a carbon chain. Because it's an acid, the neutralization will happen when it reacts with a base, such as NaOH.

When this reaction occurs, the base will dissociate in Na⁺ and OH⁻, and the acid will ionize in RCOO⁻ and H⁺, so the products will be RCOO⁻Na⁺ (a carboxylate salt) and H₂O (water).

Answer:

A. tetrahedral

Explanation:

With the above information in mind, the correct answer would then have to be either a) or b), however the complex  [Zn(NH₃)₂Cl₂]²⁺ has 4 ligands. This means that the correct answer is a), because the trigonal bipyramidal geometry has 5 ligands.

Answer:

[tex]X_{H_2}=\0.3584[/tex]

[tex]X_{CO_2}=\0.6416[/tex]

Explanation:

According to the Dalton's law of partial pressure, the total pressure of the gaseous mixture is equal to the sum of the pressure of the individual gases.

Partial pressure of carbon dioxide = 401 mmHg

Partial pressure of hydrogen gas = 224 mmHg

Total pressure,P = sum of the partial pressure  of the gases = 401 + 224 mmHg = 625 mmHg

Also, the partial pressure of the gas is equal to the product of the mole fraction and total pressure.

So,

[tex]P_{CO_2}=X_{CO_2}\times P[/tex]

[tex]X_{CO_2}=\frac{P_{CO_2}}{P}[/tex]

[tex]X_{CO_2}=\frac{401\ mmHg}{625\ mmHg}[/tex]

[tex]X_{CO_2}=\0.6416[/tex]

Similarly,

[tex]X_{H_2}=\frac{P_{H_2}}{P}[/tex]

[tex]X_{H_2}=\frac{224\ mmHg}{625\ mmHg}[/tex]

[tex]X_{H_2}=\0.3584[/tex]

Answer:

c) spontaneous

Explanation:

According the equation of Gibb's free energy -

∆G = ∆H -T∆S

∆G = is the change in gibb's free energy

∆H = is the change in enthalpy

T = temperature

∆S = is the change in entropy .

And , the sign of the  ΔG , determines whether the reaction is Spontaneous or non Spontaneous or at equilibrium ,

i.e. ,

if

From the question ,

The value for ΔG is negative ,

Hence ,

ΔG < 0 , the reaction is Spontaneous

Answer:

The molarity of the solution is 0.386 M

Explanation:

Step 1: Data given

Mass of NaCl = 7.88 grams

Volume of the solution = 350 mL

Molar mass of NaCl = 58.5 g/mol

Step 2: Calculate number of moles

Number of moles NaCl = mass of NaCl / molar mass of NaCl

Number of moles NaCl = 7.88 grams / 58.5 g/mol

Number of moles = 0.135 moles

Step 3: Calculate molarity of the solution

Molarity = Number of moles NaCl / volume

Molarity = 0.135 moles / 0.350 L

Molarity = 0.386 M

The molarity of the solution is 0.386M

Answer:

F⁻(aq) + H⁺(aq) ⇄ HF(aq)

Explanation:

When aqueous solutions of potassium fluoride and hydrochloric acid are mixed, an aqueous solution of potassium chloride and hydrofluoric acid results. The corresponding molecular equation is:

KF(aq) + HCl(aq) ⇄ KCl(aq) + HF(aq)

The full ionic equation includes all the ions and the molecular species. HF is a weak acid so it exists mainly in the molecular form.

K⁺(aq) + F⁻(aq) + H⁺(aq) + Cl⁻(aq) ⇄ K⁺(aq) + Cl⁻(aq) + HF(aq)

The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and the molecular species.

F⁻(aq) + H⁺(aq) ⇄ HF(aq)

Answer:

B

Explanation:

Diamagnetism , paramagnetism and ferromagnetism are terms which are used to describe the magnetic activity of transition metals. These terms helps to know the response the metal will have when placed in a magnetic field. These activities can be discerned from the d-block electronic configuration. If the number of electrons are even, this means they all form a pair and the metal is diamagnetic. If otherwise, there is an unpaired electron which causes the paramagnetic activity of the metal in magnetic field.

To the question, options A-D are diamagnetic but configurations with d8 are the ones that form square planar complexes

Answer:

Explanation:

Use gas equation to calculate mol of H2O gas:  

P = pressure = 389/760 = 0.512atm  

V = volume = 4L  

n=???

R = 0.082057

T = 121+273 = 394K  

0.512 * 4 = n*0.082057*394

n = (0.512*4) / (0.082057*394)  

n = 2.048 / 32.33

n = 0.063 mol H2O  

Molar5 mass H2O = 18g/mol  

Mass of water = 0.063*18

Mass of water valour = 1.14g

Answer:

Mass = 12.82 g

Explanation:

Given data:

Mass of oxygen = 8.5 g

Mass of H₂S = 10.0 g

Mass of SO₂ = ?

Solution:

Chemical equation;

2H₂S + 3O₂ →  2SO₂ + 2H₂O

Number of moles of H₂S:

Number of moles = mass/ molar mass

Number of moles = 10.0 g / 34 g/mol

Number of moles =0.3 mol

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 8.5 g / 32 g/mol

Number of moles = 0.3 mol

Now we will compare the moles of SO2 with oxygen and hydrogen sulfide.

                         O₂             :         SO₂

                           3             :            2

                           0.3          :            2/3×0.3=0.2 mol

                         H₂S           :            SO₂

                           2              :            2

                         0.3             :           0.3

The number of moles of SO₂  produced by oxygen are less so it will limiting reactant.

Mass of SO₂:

Mass = number of moles × molar mass

Mass = 0.2 mol × 64.1 g/mol

Mass = 12.82 g

Answer:

a. 750Hz, b. 4.0ppm, c. 600Hz

Explanation:

The Downfield Shift (Hz) is given by the formula

Downfield Shift (Hz) = Chemical Shift (ppm) x Spectrometer Frequency (Hz)

Using the above formula we can solve all three parts easily

a. fspec = 300 MHz, Chem. Shift = 2.5ppm, 1MHz = 10⁶ Hz, 1ppm (parts per million) = 10⁻⁶

Downfield Shift (Hz) = 2.5ppm x 300MHz x (1Hz/10⁶MHz) x (10⁻⁶/1ppm)

Downfield Shift = 750 Hz

The signal is at 750Hz Downfield from TMS

b. Downfield Shift = 1200 Hz, Chemical Shift = ?

Chemical Shift = Downfield shift/Spectrometer Frequency

Chemical Shift = (1200Hz/300MHz) x (1ppm/10⁻⁶) = 4.0 ppm

The signal comes at 4.0 ppm

c. Separation of 2ppm, Downfield Shift = ?

Downfield Shift (Hz) = 2(ppm) x 300 (MHz) x (1Hz/10⁶MHz) x (10⁻⁶/1ppm) = 600 Hz

The two peaks are separated by 600Hz

Answer:

The minimum mass of acetaminophen that the new system must be able to detect is 0.02 milligrams.

Explanation:

Range of the equipment to detect acetaminophen in blood = 10.0  to 30.0 μg/mL

Minimum amount of acetaminophen that can be detected= 10.0 μg/mL

Volume of blood sample = 2 mL

Amount of acetaminophen in 2 ml sample of blood =

[tex]2 mL\times 10.0 \mu g/mL=20.0 \mu g[/tex]

20.0 μg = 0.02 mg (1 μg = 0.001 mg)

The minimum mass of acetaminophen that the new system must be able to detect is 0.02 milligrams.

Answer:

(a) rate =  4.82 x 10⁻³s⁻¹  [N2O5]

(b) rate =   1.16 x 10⁻⁴  M/s

(c) rate =   2.32 x 10⁻⁴ M/s

(d) rate =   5.80 x 10⁻⁵ M/s

Explanation:  

We are told the rate law is first order in N₂O₅, and its rate constant is 4.82 x 10⁻³s⁻¹ . This means the rate is proportional to the molar concentration   of   N₂O₅, so

(a) rate = k [N2O5] = 4.82 x 10⁻³s⁻¹ x [N2O5]

(b) rate = 4.82×10⁻³s⁻¹  x 0.0240 M =  1.16 x 10⁻⁴ M/s

(c) Since the reaction is first order if the concentration of  N₂O₅ is double the rate will double too:  2 x   1.16 x 10⁻⁴ M/s = 2.32 x 10⁻⁴ M/s

(d) Again since the reaction is halved to 0.0120 M, the rate will be halved to

1.16 x 10⁻⁴ M/s / 2 =  5.80 x 10⁻⁵ M/s

Answer:

a) r = 4.82x10⁻³*[N2O5]

b) 1.16x10⁻⁴ M/s

c) The rate is doubled too (2.32x10⁻⁴ M/s)

d) The rate is halved too (5.78x10⁻⁴ M/s)

Explanation:

a) The rate law of a generic reaction (A + B → C + D) can be expressed by:

r = k*[A]ᵃ*[B]ᵇ

Where k is the rate constant, [X] is the concentration of the compound X, and a and b are the coefficients of the reaction (which can be different from the ones of the chemical equation).

In this case, there is only one reactant, and the reaction is first order, which means that a = 1. So, the rate law is:

r = k*[N2O5]

r = 4.82x10⁻³*[N2O5]

b) Substituing the value of the concentration in the rate law:

r = 4.82x10⁻³*0.0240

r = 1.16x10⁻⁴ M/s

c) When [N2O5] = 0.0480 M,

r = 4.82x10⁻³*0.0480

r = 2.32x10⁻⁴ M/s

So, the rate is doubled too.

d) When [N2O5] = 0.0120 M,

r = 4.82x10⁻³*0.0120

r = 5.78x10⁻⁴ M/s

So, the rate is halved too.

Answer: 0.94 V

Explanation:

For the given chemical reaction :

[tex]Zn(s)+Cu^{2+}(aq)\rightarrow Cu(s)+Zn^{2+}[/tex]

Using Nernst equation :

[tex]E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Cu^{2+}]}[/tex]

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = [tex]298K[/tex]

n = number of electrons in oxidation-reduction reaction = 2

[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = +1.10 V

[tex]E_{cell}[/tex] = emf of the cell = ?

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

[tex]E_{cell}=+1.10-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{2.5}{1.0\times 10^{-5}}[/tex]

[tex]E_{cell}=+1.10-0.16V=0.94V[/tex]

The cell potential for this reaction is 0.94 V

The cell potential for the given reaction depends on the concentration of the reactants. We can use the Nernst equation to calculate the cell potential when the concentration of [Cu2+]=1.0⋅10−5M and [Zn2+]=2.5M.

The standard cell potential (E°cell) for the given reaction is +1.10V. To find the cell potential for this reaction when the concentrations of [Cu2+]=1.0⋅10−5M and [Zn2+] = 2.5M, we need to use the Nernst equation:

Ecell = E°cell - (0.0592/n) x log(Q)

Here, n is the number of electrons involved in the reaction, and Q is the reaction quotient, which is equal to [Zn2+]/[Cu2+]. In this case, n = 2. Plugging in the values, we get:

Ecell = 1.10V - (0.0592/2) x log((2.5M)/ (1.0⋅10−5M))

Solving this equation will give us the cell potential for the given concentrations.

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Answer:

A) CaCO₃

C) No precipitate

Explanation:

To answer these questions we need to consider the solubility rules.

Identify the precipitate or lack thereof for the following:

CaCl₂(aq) + K₂CO₃(aq) ⇒ CaCO₃(s) + 2 KCl(aq)

FeCl₂(aq) + (NH₄)₂SO₄(aq) ⇒ FeSO₄(aq) + 2 NH₄Cl(aq)

Final answer:

The concentration of the chemist's mercury(II) iodide solution is 0.019 μmol/L after rounding to two significant digits.

Explanation:

The concentration of the chemist's mercury(II) iodide solution can be calculated using the formula for molarity, which is the number of moles of solute divided by the volume of the solution in liters.

In this case, we are given 0.0067μmol of HgI2 and a volume of 0.350 L:

Molarity (μM) = 0.0067μmol / 0.350 L = 0.01914μmol/L

After rounding to two significant digits, the concentration of the solution is 0.019 μM.

Answer:

27.7 g

Explanation:

To solve this problem we use the definition of molarity:

From the problem we're given M = 2.5 M and volume = 0.100 L. We use the above formula and calculate the required moles of calcium chloride (CaCl₂):

Finally we use the molecular weight of calcium chloride to calculate the required mass:

Rounding to the nearest tenth of a gram the answer is 27.7 g.

To make a 2.5 M stock solution of calcium chloride, you would need to weigh out approximately 276.2 grams of calcium chloride.

To make a 2.5 M stock solution of calcium chloride, you need to calculate the number of grams of calcium chloride required. The molecular weight of calcium chloride is 110.98 g/mol. The formula to calculate the number of grams is:

Grams = Moles x Molecular Weight

Since you want to make 100 mL of a 2.5 M solution, you first need to convert mL to moles using the formula:

Moles = Volume (L) x Molarity (M)

Finally, you can substitute the calculated moles into the first formula to find the grams, and round to the nearest tenth of a gram.

Using these calculations, you would need to weigh out approximately 276.2 grams of calcium chloride to make the 2.5 M stock solution.

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Answer:

34.7mL

Explanation:

First we have to convert our grams of Zinc to moles of zinc so we can relate that number to our chemical equation.

So: 6.25g Zn x (1 mol / 65.39 g) = 0.0956 mol Zn

All that was done above was multiplying the grams of zinc by the reciprocal of zincs molar mass so our units would cancel and leave us with moles of zinc.

So now we need to go to HCl!

To do that we multiply by the molar coefficients in the chemical equation:

[tex]\frac{0.0956g Zn}{1 } (\frac{2 mol HCl}{1molZn})[/tex]

This leaves us with 2(0.0956) = 0.1912 mol HCl

Now we use the relationship M= moles / volume , to calculate our volume

Rearranging we get that V = moles / M

Now we plug in: V = 0.1912 mol HCl / 5.50 M HCl

V= 0.0347 L

To change this to milliliters we multiply by 1000 so:

34.7 mL