Write the formula unit equation for this reaction occurring in water: Potassium sulfate and barium chloride are mixed to form potassium chloride and barium sulfate.

1. K2SO4 (s) + BaCl2 (s) → 2 KCl (aq) + BaSO4 (s)
2. K2SO4 (aq) + BaCl2 (aq) → 2 KCl (aq) + BaSO4 (s)
3. 2 K3SO4 (aq) + 3 BaCl2 (aq) → 6 KCl (aq) + 2 Ba3(SO4)2 (s)

Answers

Answer 1
Final answer:

The correct formula unit equation for the reaction of potassium sulfate and barium chloride forming potassium chloride and barium sulfate in water is: K2SO4 (aq) + BaCl2 (aq) → 2 KCl (aq) + BaSO4 (s). This is because both potassium sulfate and barium chloride are soluble in water, but barium sulfate is not.

Explanation:

The correct formula unit equation for the reaction of potassium sulfate and barium chloride forming potassium chloride and barium sulfate in water is: K2SO4 (aq) + BaCl2 (aq) → 2 KCl (aq) + BaSO4 (s).

This can be justified as when potassium sulfate (K2SO4) and barium chloride (BaCl2) are dissolved in water they dissociate into their respective ions, making them aqueous. The formation of potassium chloride (KCl) acknowledges that potassium ions and chloride ions can be attracted to each other in the water solution, so this is also aqueous. However, barium sulfate (BaSO4) is known to be insoluble in water, and hence, precipitates as a solid (s). Hence, option 2 is correct.

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Related Questions

An aqueous potassium iodate ( KIO 3 ) solution is made by dissolving 587 grams of KIO 3 in sufficient water so that the final volume of the solution is 5.00 L. Calculate the molarity of the KIO 3 solution.

Answers

Answer:

[KIO₃] = 0.548 M

Explanation:

Molarity is a sort of concentration which involves moles of solute in 1L of solution.

Volume of solution 5L

Mass of solution: 587 g

Let's convert the mass to moles (mass / molar mass)

587 g / 214 g/mol = 2.74 moles

Molarity is mol/L → 2.74 mol / 5L = 0.548 M

Asprin is a compond with the molecular formula, C9H8O4. What is the percent composition?

Answers

Answer:

Oxygen       35.6 percent

Carbon        60 percent

Hydrogen    4.4 percent

Explanation:

Molecular mass of aspirin[tex](C_9H_8O_4)=180 gm/mol.[/tex]

Mass of carbon in 180 gm of aspirine[tex]=12\times 9=108\ gm.[/tex]

Therefore, percentage of carbon in aspirin is [tex]=\dfrac{108}{180}\times 100=60\ percent.[/tex]

Similarly, mass of hydrogen in 8 gm of aspirine[tex]=1\times 8=8\ gm[/tex].

Hydrogen's percentage in aspirin[tex]=\dfrac{8}{180}\times 100=4.4\ percent.[/tex]

Also, mass of oxygen in aspirin is [tex]=16\times 4=64\ gm.[/tex]

Oxygen's percentage in aspirin[tex]=\dfrac{64}{180}\times 100=35.6\ percent.[/tex]

Hence, this is the required solution.

Final answer:

The percent composition of aspirin (C9H8O4) is calculated based on its molecular mass of 180.15 amu. It is 60.03% carbon, 4.48% hydrogen, and 35.49% oxygen.

Explanation:

To determine the percent composition of aspirin, which has the molecular formula C9H8O4, we need to calculate the percentage of each element in the compound based on its molecular mass. The molecular mass of aspirin is 180.15 amu. This is calculated by adding the masses of nine carbon (C) atoms, eight hydrogen (H) atoms, and four oxygen (O) atoms together.

To find the percent composition, we first calculate the total mass of each type of atom in one molecule of aspirin and then divide it by the molecular mass of aspirin. To express it as a percentage, we multiply by 100%. The atomic masses are approximately 12.01 amu for carbon, 1.008 amu for hydrogen, and 16.00 amu for oxygen.

The calculations for percent composition are as follows:

Carbon: (9 atoms × 12.01 amu/atom) / 180.15 amu × 100% = 60.03%Hydrogen: (8 atoms × 1.008 amu/atom) / 180.15 amu × 100% = 4.48%Oxygen: (4 atoms × 16.00 amu/atom) / 180.15 amu × 100% = 35.49%

Therefore, the percent composition of aspirin (C9H8O4) is 60.03% carbon, 4.48% hydrogen, and 35.49% oxygen.

A certain ionic compound X has a solubility in water of 40.3 g/L at 20. degrees C. Calculate the mass X of required to prepare 500. mL of a saturated solution of X in water at this temperature. Be sure your answer has the correct unit symbol and 3 significant digits.

Answers

Answer:

20.1 g

Explanation:

The solubility indicates how much of the solute the solvent can dissolve. A solution is saturated when the solvent dissolved the maximum that it can do, so, if more solute is added, it will precipitate. The solubility varies with the temperature. Generally, it increases when the temperature increases.

So, if the solubility is 40.3 g/L, and the volume is 500 mL = 0.5 L, the mass of the solute is:

40.3 g/L = m/V

40.3 g/L = m/0.5L

m = 40.3 g/L * 0.5L

m = 20.1 g

Final answer:

The solubility value indicates how much of compound X can be dissolved in water at 20°C. Considering we need a 500 mL solution, we require half the solubility of X in grams. Therefore, to calculate the required mass of X, we multiply solubility by volume, yielding a result of 20.15 g.

Explanation:

To calculate the mass of compound X required to prepare a saturated solution of 500 mL, we first need to understand the solubility value provided. The solubility of X is given as 40.3 g/L at 20°C, meaning that 1 L of water can dissolve 40.3 g of X at this temperature.

Since we need only 500 mL (or 0.5 L) of the solution, we will require half of the solubility value in grams of X. Therefore, by simple multiplication, we get:

X mass = Solubility * Volume

= 40.3 g/L * 0.5 L = 20.15 g

So, 20.15 g of compound X is needed for a saturated solution of 500 mL at 20°C.

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A chemist mixes 71.1 g of carbon tetrachloride with 65.4 g of isopropenylbenzene and 4.02 g of 2-ethyltoluene. Calculate the percent by mass of each component of this solution. Be sure each of your answer entries has the correct number of significant digits mass percentー1 □ component 0% carbon tetrachloride isopropenylbenzene 2-ethyltoluene

Answers

Answer:

∴ wt% CCl4 = 50.597 %

∴ wt% C9H10 = 46.541 %

∴ wt% C9H12 = 2.861 %

Explanation:

wt% = ((mass solute)/(mass solution))×100

⇒ mass sln = 71.1 g CCl4 + 65.4 C9H10 + 4.02 g C9H12

⇒ mass sln = 140.52 g

∴ CCl4:

⇒ wt% CCl4 = ((71.1 g CCl4)/(140.82 g sln))×100

⇒ wt% CCl4 = 50.597 %

∴ C9H10:

⇒ wt% C9H10 = ((65.4 g C9H10)/(140.52 g sln))×100

⇒ wt% C9H10 = 46.541 %

∴ C9H12:

⇒ wt% C9H12 = ((4.02 g C9H12)/(140.52 g sln))×100

⇒ wt% C9H12 = 2.861 %

Final answer:

The percent by mass of each component in the solution is calculated by dividing each component's mass by the total mass and multiplying by 100. The results are 50.62% for carbon tetrachloride, 46.53% for isopropenylbenzene, and 2.86% for 2-ethyltoluene.

Explanation:

In order to calculate the percent by mass of each component of the solution, the total mass of the solution needs to be determined first. The total mass would be obtained by summing the individual masses of the carbon tetrachloride, isopropenylbenzene, and 2-ethyltoluene. That is, 71.1 g + 65.4 g + 4.02 g = 140.52 g.

Then, the mass of each component is divided by the total mass and multiplied by 100 to obtain the percentage.

For carbon tetrachloride, the calculation would be (71.1 / 140.52) x 100 = 50.62%. For isopropenylbenzene, the calculation would be (65.4 / 140.52) x 100 = 46.53%. For 2-ethyltoluene, the calculation would be (4.02 / 140.52) x 100 = 2.86%.

Our three significant figures in each of the percentages are due to the fact that our least precise measurement, 4.02 g of 2-ethyltoluene, has four significant figures.

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A 0.1014 g sample of a purified CHO compound was burned in a combustion apparatus and produced 0.1486 g CO2 and 0.0609 g of H2O. Mass spectrometry analysis revealed that the CHO compound had a molar mass of 180 g/mol. What is the molecular formula for this compound?

Answers

Answer:

[tex]C_6H_{12}O_6[/tex]

Explanation:

Mass of water obtained = 0.0609 g

Molar mass of water = 18 g/mol

Moles of [tex]H_2O[/tex] = 0.0609 g /18 g/mol = 0.00338 moles

2 moles of hydrogen atoms are present in 1 mole of water. So,

Moles of H = 2 x 0.00338 = 0.00676 moles

Molar mass of H atom = 1.008 g/mol

Mass of H in molecule = 0.00676 x 1.008 = 0.00681 g

Mass of carbon dioxide obtained = 0.1486 g

Molar mass of carbon dioxide = 44.01 g/mol

Moles of [tex]CO_2[/tex] = 0.1486 g  /44.01 g/mol = 0.00337 moles

1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,

Moles of C = 0.00337 moles

Molar mass of C atom = 12.0107 g/mol

Mass of C in molecule = 0.00337 x 12.0107 = 0.04047 g

Given that the compound only contains hydrogen, oxygen and carbon. So,

Mass of O in the sample = Total mass - Mass of C  - Mass of H

Mass of the sample = 0.1014 g

Mass of O in sample = 0.1014 - 0.04047 - 0.00681 = 0.05412 g  

Molar mass of O = 15.999 g/mol

Moles of O  = 0.05412  / 15.999  = 0.00338 moles

Taking the simplest ratio for H, O and C as:

0.00676 : 0.00338 : 0.00337

= 2 : 1 : 1

The empirical formula is = [tex]CH_2O[/tex]

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 1×12 + 2×1 + 16= 30 g/mol

Molar mass = 180 g/mol

So,  

Molecular mass = n × Empirical mass

180 = n × 30

⇒ n = 6

The formula of compound = [tex]C_6H_{12}O_6[/tex]

Final answer:

The molecular formula of the compound is (CHO)6.

Explanation:

To determine the molecular formula of the compound, we need to find the empirical formula first. We can do this by finding the moles of carbon and hydrogen in the sample using the masses of CO2 and H2O produced.

Mass of CO2 = 0.1486 g
Mass of H2O = 0.0609 g

Now, we need to convert the masses of CO2 and H2O to moles and divide by the smallest value to find the ratio of moles of carbon to hydrogen.

After finding the ratio, we multiply it by a common factor to get whole numbers. In this case, the ratio is 1:1, so the empirical formula is CHO.

The empirical formula has a molar mass of 30 g/mol.

To find the molecular formula, we divide the given molar mass of 180 g/mol by the empirical formula molar mass. The result is 6, which means the molecular formula is (CHO)6.

When a 0.089M aqueous solution of a certain acid is prepared, the acid is 12.% dissociated. Calculate the pH of the solution. Round your answer to 2 decimal places.

Answers

Answer:

pH = 1.98

Explanation:

Given a general acid dissociation:

HA + H₂O  ⇆  H₃O⁺   + A⁻    

The pH is -log[ H₃O⁺ ]

Therefore we need to determine [ H₃O⁺ ] to answer this question, and we should use the data of % dissociation of the acid.

Percent dissociation is

% dissociation = [ H₃O⁺ ] / [ HA ]₀ x 100

where [ HA ]₀  is the original acid concentration, so we can calculate [ H₃O⁺ ] , and then the pH.

12 =  [ H₃O⁺ ] /0.089 M ⇒  [ H₃O⁺ ]  = (12 x 0.089 /100) M

                                                          = 1.07 x 10⁻² M

and pH = - log ( 1.07 x 10⁻² ) = 1.98

Final answer:

To determine the pH of a 0.089M solution of an acid that is 12% dissociated, we first calculate the hydronium ion concentration and then use the formula for pH. After calculation, the pH is found to be 2.97.

Explanation:

To calculate the pH of the solution where the acid is 12% dissociated, we first need to determine the concentration of hydronium ions ([H+]) in the solution.

Given a 0.089M solution of an acid that is 12% dissociated, the concentration of dissociated hydronium ions is:


 

This concentration has three significant figures. Using the formula for pH which is:


 

The pH is then calculated as:

pH = -log(0.01068) = 2.97

The result is in the acidic pH range, and it has been rounded to two decimal places to match the number of significant figures from the initial concentration given.

For each of the following reactions, give the balanced equation for the reaction and state the meaning of the equation in terms of the numbers of individual molecules and in terms of moles of molecules.

a. MnO2(s) + Al(s)-Mn(s) + Al2O3(s)
b. B203(s) + CaF2(s)-BF3(g) + CaO(s)
c. NO2(g) + H2O)> HNO;(aq) + NO(g)
d. C6H6(g) + H2(g)â C6H12(g) 12)

Answers

Answer:

a. 3MnO2 + 4Al —> 3Mn+ 2Al2O3

3 moles of MnO2 required 4 moles of Al to produce 3 moles of Mn and 2moles of 2Al2O3

b. B203 + 3CaF2 —> 2BF3 + 3CaO

1mole of B203 requires 3 moles of CaF2 to produce 2moles of BF3 and 3 moles of CaO

c. 3NO2 + H2O —> 2HNO3 + NO

3 moles of NO2 requires 1mole of H2O to produce 2moles of HNO3 and 1mole of NO

d. C6H6 + 3H2 —> C6H12

1mole of C6H6 requires 3 moles of H2 to produce 1mole of C6H12

Final answer:

To balance the equations, ensure there are equal numbers of each atom. The balanced equations and their meanings in terms of molecules and moles are...

Explanation:

In order to balance the equations, you need to ensure that there are equal numbers of each type of atom on both sides of the equation. Here are the balanced equations for each reaction:

a. MnO2(s) + 2Al(s) -> Mn(s) + Al2O3(s)

b. B2O3(s) + 3CaF2(s) -> 2BF3(g) + 3CaO(s)

c. 2NO2(g) + H2O(l) -> HNO2(aq) + NO(g)

d. C6H6(g) + 3H2(g) -> C6H12(g)

In terms of the number of individual molecules, the balanced equation shows the ratio in which the reactants combine to form the products. In terms of moles of molecules, the balanced equation allows you to calculate the amount of each substance involved in the reaction using the mole ratio.

A 8-kg plastic tank that has a volume of 0.18 m3 is filled with liquid water. Assuming the density of water is 1000 kg/m3, determine the weight of the combined system.

Answers

Answer:

1.84 × 10³ N

Explanation:

To determine the weight of the system, we will determine its mass which is equal to the sum of the masses of the plastic tank (8 kg) and the water.

We have 0.18 m³ of water with a density of 1000 kg/m³. Its mass is:

0.18 m³ × 1000 kg/m³ = 180 kg

The mass of the system is 180 kg + 8 kg = 188 kg.

We can find the weight (w) of the system using Newton's second law of motion.

w = m × g = 188 kg × 9.81 m/s² = 1.84 × 10³ N

where

g is the gravity

Final answer:

The weight of the combined system, including the plastic tank and the water, is 180 kg.

Explanation:

The weight of an object depends on its mass and the gravitational force acting on it. To find the weight of the combined system, we need to determine the mass of the plastic tank and the water it contains. The 8-kg plastic tank does not contribute to the weight of the system, but the weight of the water can be calculated using its density.

The volume of the water is given as 0.18 m³, and the density of water is 1000 kg/m³. Multiplying these values together will give us the mass of the water:

Weight of the water = density × volume = 1000 kg/m³ × 0.18 m³ = 180 kg

Therefore, the weight of the combined system, including the plastic tank and the water, is 180 kg.

Formation of which type of product--byproduct, side product, both, or neither--always coincides with a loss in percent yield of the desired product?

Answers

Answer:

Side products coincides with a loss in percentage yield of the product

Explanation:

- The product yield is defined as

product yield= actual yield/theoretical yield

- Byproducts are undesired products that come from the same reaction pathway that our products → even if the reaction procedes ideally , we will always have byproducts → byproducts do not alter the Yield by themselves

- Side products , however , come from undesired products from side reactions  and consume reactants due to these alternative pathways without producing our product → side products do alter the yield of the product

If the pH of a solution is 2.78 with an absolute uncertainty of ±0.05, what is the LaTeX: \left[H^+\right][ H + ] and its uncertainty?

Answers

Answer:

[H⁺] = 16.6 (±1.9) x10⁻⁴ M

Explanation:

The pH is defined as:

pH = -log[H⁺]

So we can calculate [H⁺]:

2.78 = -log[H⁺]-2.78 = log[H⁺][H⁺] = [tex]10^{-2.78}[/tex]

[H⁺] = 1.66x10⁻³ M

The relative uncertainty in [H⁺] is

uR/R = 2.303 x AbsoluteUncertaintyuR/R = 2.303 * 0.05 = 0.115

Thus the uncertainty in the concentration is:

1.66x10⁻³ M * 0.115 = 1.91x10⁻⁴ M

You need to prepare an acetate buffer of pH 5.31 5.31 from a 0.784 M 0.784 M acetic acid solution and a 2.85 M KOH 2.85 M KOH solution. If you have 930 mL 930 mL of the acetic acid solution, how many milliliters of the KOH KOH solution do you need to add to make a buffer of pH 5.31 5.31 ? The p K a pKa of acetic acid is 4.76. 4.76.

Answers

Answer:

214 milliliters of KOH needs to be added in 1 litre of 0.784 M acetic acid to make a acetate buffer of 5.31

Explanation:

To solve the problem, let us first use the Henderson-Hasselbalch equation to determine the amount of acetate needed to make a buffer of pH 5.31.

Henderson-Hasselbalch equation:

[tex]pH=pKa + log(\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]})[/tex]

Here, let us consider the moles of both species instead of the molar concentration, as the volume for both is the same. Also, acetate will be formed by the neutralization of acetic acid, hence the final moles of acetic acid will be the difference of initial moles of acetic acid and the moles of acetate formed. Now the equation becomes as follows:

[tex]pH=pKa + log(\frac{n_{ CH_{3}COO^{-}}}{n_{iCH_{3}COOH}-n_{ CH_{3}COO^{-}}}})[/tex]

From given data

pH = 5.31

pKa = 4.76

n(CH₃COO⁻) = ?

ni(CH₃COOH) = 0.784 mol (initial moles of acetic acid)

Placing the data in the equation, we get:

[tex]5.31=4.76 + log(\frac{n_{ CH_{3}COO^{-}}}{0.784-n_{ CH_{3}COO^{-}}}})\\ \\ n_{ CH_{3}COO^{-}}=10^{5.31-4.76}(0.784-(n_{ CH_{3}COO^{-}}))\\ \\ n_{ CH_{3}COO^{-}}= 2.78 mol-3.55(n_{ CH_{3}COO^{-}})\\ \\ n_{ CH_{3}COO^{-}}= 0.61mol[/tex]

The molar ratio of KOH and CH₃COOH is 1:1, i.e 1 mol of KOH will react with CH₃COOH and give 1 mol of acetate (CH₃COO⁻). Hence, 0.61 mol of KOH will give 0.61 mol of KOH. Now to determine the volume of 2.85 M KOH that contains 0.61 moles:

[tex]M_{KOH} =\frac{n_{KOH} }{V_{KOH} (L)}[/tex]

[tex]2.85=\frac{0.61}{V_{KOH} (L)}\\ \\ V_{KOH} (L)=\frac{0.61}{2.85}\\ \\ V_{KOH}=0.214 litre[/tex]

Finally convert liter into milliliter dividing by 1000 (mL/L)

Volume of KOH required = 214 milliliters

Halogenated compounds are particularly easy to identify by their mass spectra because chlorine and bromine occur naturally as mixtures of two abundant isotopes. Chlorine occurs as 35Cl (75.8%) and 37Cl (24.2%); bromine occurs as 79Br (50.7%) and 81Br (49.3%). For the compound Chlorocyclohexane, C6H11Cl: At what masses do the molecular ions occur? (List in order of increasing mass separated by commas, e.g. 120,122.) What are the percentages of each molecular ion?

Answers

Answer:

The increasing order of masses of molecule ions:

118 g/mol(75.8%) ,  120 g/mol(24.2%)

Explanation:

Chlorine occurs as 35-Cl (75.8%) and 37-Cl (24.2%).

Atomic mass of 35-Cl = 35 g/mol

Atomic mass of 37-Cl = 37 g/mol

Mass of Chlorocyclohexane in which 35-cl is present as a chlorine atom: [tex]C_6H_{11}Cl[/tex]

[tex]=6\times 12 g/mol+11\times 1 g/mol+1\times 35 g/mol=118 g/mol[/tex]

Mass of Chlorocyclohexane in which 37-Cl is present as a chlorine atom: [tex]C_6H_{11}Cl[/tex]

[tex]=6\times 12 g/mol+11\times 1 g/mol+1\times 37 g/mol=120 g/mol[/tex]

The increasing order of masses of molecule ions:

118 g/mol(75.8%) < 120 g/mol(24.2%)

Given one mole of each substance, which of the following will produce the FEWEST particles in aqueous solution? 1. sodium nitrate 2. CH2Cl2 3. K2SO4 4. sodium phosphate

Answers

Answer: The substance that produces fewest particles is [tex]CH_2Cl_2[/tex]

Explanation:

Ionization reaction is defined as the reaction in which an ionic compound dissociates into its ions when dissolved in aqueous solution.

Covalent compounds do not dissociate into ions when dissolved in aqueous solution.

For the given options:

Option 1:  Sodium nitrate

The chemical formula of sodium nitrate is [tex]NaNO_3[/tex]

The ionization reaction for the given compound follows:

[tex]NaNO_3(aq.)\rightarrow Na^+(aq.)+NO_3^-(aq.)[/tex]

This produces in total of 2 ions.

Option 2:  [tex]CH_2Cl_2[/tex]

The given compound is a covalent compound and do not dissociate into its ions. It remains as such as a single unit.

Option 3:  [tex]K_2SO_4[/tex]

The chemical name for the given compound is potassium sulfate.

The ionization reaction for the given compound follows:

[tex]K_2SO_4(aq.)\rightarrow 2K^+(aq.)+SO_4^{2-}(aq.)[/tex]

This produces in total of 3 ions.

Option 4:  Sodium phosphate

The chemical formula of sodium phosphate is [tex]Na_3PO_4[/tex]

The ionization reaction for the given compound follows:

[tex]Na_3PO_4(aq.)\rightarrow 3Na^+(aq.)+PO_4^{3-}(aq.)[/tex]

This produces in total of 4 ions.

Hence, the substance that produces fewest particles is [tex]CH_2Cl_2[/tex]

Ne ( g ) effuses at a rate that is ______ times that of Cl 2 ( g ) under the same conditions.

Answers

Answer: 1.88times as that of Cl2

Explanation:

According to Graham law of effusion , the rate of effusion is inversely proportional to the square root of the molar mass

Rate= 1/√M

R1/R2 =√M2/M1

Let the rate of diffusion of Ne= R1

And rate of diffusion of Cl2 = R2

M1 ,molar mass of Ne= 20g/mol

M2,molar mass of Cl2 =71g/mol

R1/R2 = √ (71/20)

R1/R2 = 1.88

R1= 1.88R2

Therefore the Ne effuses at rate that is 1.88times than that of Cl2 at the same condition.

Using Graham's law of effusion, Ne gas is found to effuse approximately 1.88 times faster than Cl₂ gas under the same conditions.

This is because the rate of effusion is inversely proportional to the square root of molar mass.

To determine the rate at which Neon (Ne) effuses compared to Chlorine (Cl₂) under the same conditions, we can use Graham's law of effusion. According to Graham's law, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

The molar mass of Ne is approximately 20 g/mol, and the molar mass of Cl₂ is approximately 71 g/mol. Applying Graham's law:

(Rate of Ne) / (Rate of Cl₂) = [tex]\sqrt{sqrt(Molar mass of Cl_{2} / Molar mass of Ne)}[/tex]

This means Ne gas effuses approximately 1.88 times faster than Cl₂ gas under the same conditions.

Suppose that the root-mean-square velocity vrms of water molecules (molecular mass is equal to 18.0 g/mol) in a flame is found to be 1150 m/s. What temperature does this represent

Answers

The temperature represented by a root-mean-square velocity of 1150 m/s for water molecules in a flame is approximately [tex]5.74 \cdot 10^2^4 Kelvin.[/tex]

The temperature represented by a root-mean-square velocity of water molecules can be calculated using the formula:

[tex]\[ v_{rms} = \sqrt{\frac{3kT}{m}} \][/tex]

Where:

- [tex]\( v_{rms} \)[/tex] = root-mean-square velocity of the molecules (1150 m/s in this case)

[tex]\( k \) = Boltzmann constant (1.38 x 10^-23 J/K)\\\( T \) = temperature in Kelvin (what we're trying to find)\\\( m \) = molecular mass of the water (18.0 g/mol or 0.018 kg/mol)[/tex]

Now, we can rearrange the formula to solve for \( T \):

[tex]\[ T = \frac{m v_{rms}^2}{3k} \][/tex]

Substitute the given values:

[tex]\[ T = \frac{0.018 kg/mol \times (1150 m/s)^2}{3 \times 1.38 \times 10^{-23} J/K} \][/tex]

Calculating:

[tex]\[ T = \frac{0.018 \times 1322500}{4.14 \times 10^{-23}} \]\[ T \approx \frac{23745}{4.14 \times 10^{-23}} \]\[ T \approx 5.74 \times 10^{24} K \][/tex]

The temperature represented by a root-mean-square velocity of 1150 m/s for water molecules in a flame is approximately [tex]5.74 \cdot 10^2^4 Kelvin.[/tex]

To calculate the temperature, we used the formula for root-mean-square velocity and rearranged it to solve for temperature. Plugging in the given values, we obtained the temperature in Kelvin. This extremely high temperature result suggests that either the calculation or the assumptions about the system may not be accurate or applicable in a real-world scenario.

Complete Question:

Suppose that the root-mean-square velocity vrms of water molecules (molecular mass is equal to 18.0 g/mol) in a flame is found to be 1150 m/s. What temperature does this represent?

Give the balanced equation for the following.
The combustion of liquid butanol (C4H9OH) forms carbon dioxide and water vapor. A combustion reaction refers to a reaction of a substance with oxygen gas. (Use the lowest possible coefficients. Include states-of-matter under the given conditions in your answer.)

Answers

Answer:

The balanced chemical reaction is given as:

[tex]C_4H_9OH(l)+6O_2(g)\rightarrow 4CO_2(g)+5H_2O(g)[/tex]

Explanation:

Combustion is defined as chemical reaction which an organic compounds reacts with oxygen gas to to give water and carbon dioxide as a products along with releases of heat energy.

The combustion of liquid butanol gives water vapor and carbon dioxide as a product and this reaction is given as:

[tex]C_4H_9OH(l)+6O_2(g)\rightarrow 4CO_2(g)+5H_2O(g)[/tex]

According to reaction, 1 mole of butanol reacts with 6 moles of oxygen gas to gives 4 moles of carbon dioxide gas and 5 moles of water vapors.

Final answer:

The combustion of liquid butanol, in the presence of oxygen, forms carbon dioxide and water vapor. The balanced chemical equation is: C4H9OH(l) + 6O2(g) → 4CO2(g) + 5H2O(g).

Explanation:

The combustion of liquid butanol (C4H9OH) in the presence of oxygen (O2) produces carbon dioxide (CO2) and water (H2O). The balanced chemical equation for the combustion of liquid butanol can be written as:  

C4H9OH(l) + 6O2(g) → 4CO2(g) + 5H2O(g)

Here, C4H9OH(l) represents liquid butanol, O2(g) represents oxygen gas, CO2(g) represents carbon dioxide gas and H2O(g) represents water vapor. This equation states that one mole of butanol reacts with six moles of oxygen to form four moles of carbon dioxide and five moles of water vapor.

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The results from an iodine-deficient diet, which causes a hyperplasia or overgrowth of the thyroid gland ______

Answers

Answer:

Goiter.

Explanation:

The thyroid gland is responsible for the secretion of the triidotyronine and thyroxine hormone in the body. The thyroid hormone regulates the metabolism and amount of particular ions in the body.

The iodine acts as the precursor for the maintenance of the thyroxine. The deficiency of the thyroid hormone can cause goiter.  The thyroid gland gets swollen and cause problem in the breathing. The production of mucus and cough is common in goiter.

Thus, the answer is goiter.

Concentrated hydrochloric acid is an aqueous solution that is 34.70 % HCl. The density of the solution is 1.19 g/mL. What mass of HCl is contained in 0.356 L of solution?

Answers

The mass of HCl that is contained in the solution is 147 g HCl

Why?

To find the mass of HCl we have to apply what is called a conversion factor. In a conversion factor we put the units we don't want at the bottom, and the ones we want at the top.

For this question, we want to go from liters of solution to mass of HCl, and the conversion factor is laid out as follows:

[tex]0.356Lsolution*\frac{1000mL}{1L}*\frac{1.19 g solution}{1 mL solution}*\frac{34.70 g HCl}{100 g solution}=147 g HCl[/tex]

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A solid substance has a vapor pressure of 100 mm Hg at the melting point (100 °C).
Assuming an atmospheric pressure of 760 mm Hg, describe the behavior of this solid as the temperature is raised from room temperature to its melting point.

Answers

Answer:

The melting point would be 250. That is the answer I think

Explanation:

The solid substance will turn into vapor when the temperature is increased from room temperature into meeting point temperature.

The given parameters;

Vapor pressure of the solid, [tex]P_{vap}[/tex] = 100 mmHgMelting point of the solid, T = 100 °C.Atmospheric pressure, [tex]P_{atm}[/tex] = 760 mmHgRoom temperature, [tex]T_0[/tex] = 20 °C

A substance boils when the atmospheric pressure is equal to the vapor pressure.

Since the vapor pressure of the solid substance occurs at the meeting temperature of the substance, once the temperature of the solid substance is increased to the meeting point temperature, the solid substance will turn into vapor (gas).

Thus, we can conclude that the solid substance will turn into vapor when the temperature is increased from room temperature into meeting point temperature.

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What challenges would you encounter with the titration if you had used apple cider vinegar or balsamic vinegar as the analyte instead of white vinegar?

Answers

Explanation:

Apple cider vinegar or balsamic vinegar might contain various synthetics which might  interact with or nullified the findings, whereas white vinegar comprises of acidic acid only. In fact, apple cider vinegar and balsamic vinegar are deep in appearance this would make it very difficult to determine the color. This is why it is preferable to use white vinegar in the Titration process.

The challenge that would be encountered with the titration if you had used balsamic vinegar as an analyte is the inability to determine the endpoint of the titration with your naked eye.

An analyte, also known as a titrand is a solution whose concentration is needed to be determined. During titration, it will be difficult and impossible to determine the endpoint of titration because it will be difficult to catch the color changes with visible eyes.

To eliminate this difficulty, balsamic vinegar needs to be diluted with enough water to detect the color change, or the use of white vinegar is used.

Similarly, apple cider, as well as balsamic vinegar, comprises additional acid than acetic acid, this will increase the level of acidity during the titration, unlike white vinegar that contains only acetic acid.

Thus, the acid in apple cider vinegar or balsamic vinegar will be more and higher during titration compared to that of white vinegar.

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A chemist prepares a solution of mercury(II) iodide by weighing out of mercury(II) iodide into a volumetric flask and filling the flask to the mark with water. Calculate the concentration in of the chemist's mercury(II) iodide solution. Be sure your answer has the correct number of significant digits.

Answers

The question is incomplete, here is the complete question:

A chemist prepares a solution of mercury(II) iodide [tex](HgI_2)[/tex] by measuring out 0.0122 µmol of mercury(II) iodide into a 400 mL volumetric flask and filling the flask to the mark with water.

Calculate the concentration in mol/L of the chemist's mercury(II) iodide solution. Be sure your answer has the correct number of significant digits.

Answer: The molarity of chemist's mercury (II) iodide solution is [tex]3.05\times 10^{-8}mol/L[/tex]

Explanation:

To calculate the molarity of solution, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]

We are given:

Moles of mercury (II) iodide = [tex]0.0122\mu mol=0.0122\times 10^{-6}mol[/tex]    (Conversion factor:  [tex]1mol=10^6\mu mol[/tex] )

Volume of solution = 400. mL

Putting values in above equation, we get:

[tex]\text{Molarity of }HgI_2=\frac{0.0122\times 10^{-6}\times 1000}{400.}\\\\\text{Molarity of }HgI_2=3.05\times 10^{-8}mol/L[/tex]

Hence, the molarity of chemist's mercury (II) iodide solution is [tex]3.05\times 10^{-8}mol/L[/tex]

Imagine you created a toxin such that when a neuron fired an action potential, the toxin would bind immediately to the sodium-potassium pump at the top of the action potential, but the sodium and potassium channels would still function. What process(es) would be affected

Answers

Answer:Maintaining resting potential and returning to resting potential after the hyperpolarization phase of an action potential

Explanation:TOXINS are chemical substances which are known to be POISONOUS produced with living organisms that causes harm to other organisms, examples include Venom from snakes which when a person is bitten by a Snake it will possibly lead to death if not adequate treated.

HYPERPOLARIZATION is a term that explains the change in membrane potential due to toxin,it make the membrane more electronegative. When the toxin has hyped the level of Sodium-Potassium level returning to a rest state will be most affected.

Nitrogen monoxide and hydrogen react to form nitrogen and water, like this: 2NO+2 H2(g)-N29)+2H2O(9) Write the pressure equilibrium constant expression for this reaction

Answers

Answer:

Partial pressure N₂ . (Partial pressure H₂O)² / (Partial pressure H₂)² . (Partial pressure NO)² = Kp

Explanation:

The reaction is:

2NO + 2H₂ → N₂ + 2H₂O

The expression for Kp (pressure equilibrium constant) would be:

Partial pressure N₂ . (Partial pressure H₂O)² / (Partial pressure H₂)² . (Partial pressure NO)²

There is another expression for Kp, where you work with Kc (equilibrium constant)

Kp = Kc (R.T)^Δn

where R is the Ideal Gases constant

T° is absolute temperature

Δn = moles of gases formed - moles of gases, I had initially

Final answer:

The pressure equilibrium constant expression for the reaction 2NO + 2H2 -> N2 + 2H2O is Kp = (p(N2))^1 * (p(H2O))^2 / (p(NO))^2 * (p(H2))^2.

Explanation:

The pressure equilibrium constant expression is defined as the ratio of the product of the partial pressures of the products raised to their stoichiometric coefficients divided by the product of the partial pressures of the reactants raised to their stoichiometric coefficients.

For the reaction 2NO + 2H2 -> N2 + 2H2O, the pressure equilibrium constant expression can be written as:

Kp = (p(N2))^1 * (p(H2O))^2 / (p(NO))^2 * (p(H2))^2

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Question 25 An aqueous solution at has a concentration of . Calculate the concentration. Be sure your answer has the correct number of significant digits.

Answers

The given question is lacking some details, the complete question is following

Question:

An aqueous solution at 25 °C has a OH⁻ concentration of 2.5 x 10⁻⁴ M . Calculate the H₃O⁺ concentration. Be sure your answer has the correct number of significant digits

Answer:

Concentration of H₃O⁺ is:

[tex][H_{3}O^{+}]=4.0X10^{-9} M[/tex]

Explanation:

In aqueous solutions the product of the concentration of hydronium ions H₃O⁺ and hydroxide ions OH⁻ is 1.0 x 10⁻¹⁴. This value is the dissociation or ionization constant of water at 25 °C. Its formula is given as:

[tex]Kw = [H_{3} O^{+}][OH^{-} ][/tex]

[tex]1.0 X 10^{-14}= [H_{3}O^{+}](2.5 X 10^{-4})[/tex]

[tex][H_{3}O^{+}]= \frac{1.0X10^{-14}}{2.5 X 10^{-4}}[/tex]

[tex][H_{3}O^{+}]=4.0X10^{-9} M[/tex]

P.S: As the smallest number of significant figure in the ratio was two, so the answer contains two significant figures.

Final answer:

To calculate the hydronium ion concentration from the hydroxide ion concentration of 0.001 M at 25 °C, use the water ion-product constant, Kw (1.0 × 10^-14), and the inverse relationship between [H3O+] and [OH-] to find [H3O+] = 1.0 × 10^-11 M.

Explanation:

To calculate the hydronium ion concentration in an aqueous solution with a hydroxide ion concentration of 0.001 M at 25 °C, we use the ion-product constant for water (Kw), which is 1.0 × 10-14 at 25 °C. The concentration of hydronium ions [H3O+] and hydroxide ions [OH-] are inversely proportional, which means as the concentration of one goes up, the other goes down. Therefore, the calculation for the hydronium ion concentration can be done using the equation:

Kw = [H3O+] × [OH-]

Substituting in the values we have:

1.0 × 10-14 = [H3O+] × 0.001

Therefore, the concentration of hydronium ions [H3O+] is:

[H3O+] = ⅖{1.0 × 10-14}{0.001}

[H3O+] = 1.0 × 10-11 M

Make sure that your final answer has the correct number of significant figures, which is guided by the number of significant figures in the given hydroxide ion concentration (0.001 M has one significant figure).

luminum and oxygen react according to the following equation: 4Al(s) +3O2(g) --> 2Al2O3(s) What mass of Al2O3, in grams, can be made by reacting 4.6 g Al with excess oxygen?

Answers

Answer: 8.7 grams

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.  

To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{4.6g}{27g/mol}=0.17moles[/tex]

[tex]4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)[/tex]

As oxygen is in excess, Aluminium is the limiting reagent and limits the formation of products.

According to stoichiometry:

4 moles of aluminium give = 2 moles of [tex]Al_2O_3(s)[/tex]

Thus 0.17 moles of aluminium give=[tex]\frac{2}{4}\times 0.17=0.085mol[/tex]

Mass of [tex]Al_2O_3=moles\times {\text {molar mass}}=0.085\times 102g/mol=8.7g[/tex]

Thus the mass of [tex]Al_2O_3(s)[/tex]  is 8.7 grams

Draw a Lewis structure for C2H3Cl. Show all unshared electron pairs. None of the atoms bears a formal charge, and all atoms have octets (except for hydrogen atoms, which have duets).

Answers

Answer:

See picture below

Explanation:

To do this, we first need to know how are the bonds in this molecule are. To do so, let's calculate the number of insaturations in the molecule:

n°I = C+1 - (H-N+X)/2

These numbers indicate if the molecule has double bond, triple bond, ring, cyclo, among other options:

n°I = 3 - (3+1)/2

n°I = 1

As it's a very small molecule, we can assume this molecule only have a double bond, and it's an alkene.

So the lewis structure, shows the electrons and the bonding, and also shows the unshared electron pairs, depending on how much electron have each molecule.

In the case of carbon:

[C] = [He] 2s2 2p2 ----> 4 electrons

[H] = 1s1 ----> 1 electron

[Cl] = [Ne] 3s2 3p5 ----> 7 electrons.

Therefore, we also know that Carbon has a double bond, so, the main molecule would have something like this:

C = C

so next to the carbons, we can put two hydrogens and in the other carbon, the chlorine and the remaining hydrogen.

See picture below for structure:

The Lewis structure of C₂H₃Cl is attached in the image below.

Lewis structures are also known as Lewis dot structures or electron dot structures. They are diagrams that represent the arrangement of atoms and valence electrons in a molecule or ion.

In a Lewis structure, the symbol of each atom is used to represent the nucleus and inner-shell electrons, while dots or lines are used to represent the valence electrons. Valence electrons are the outermost electrons involved in bonding and determining the chemical properties of an atom. The image is attached below.

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Consider the steps in coal gasification: C(coal) + H2O(g) → CO(g) + H2(g) ΔH°rxn = 129.7 kJ CO(g) + H2O(g) → CO2(g) + H2(g) ΔH°rxn = -41 kJ CO(g) + 3H2(g) → CH4(g) + H2O(g) ΔH°rxn = -206 kJ Calculate the heat of reaction, ΔH°rxn, for overall reaction for the production of methane, CH4.

Answers

The heat of reaction,  ΔH°rxn, for overall reaction in coal gasification for the production of methane, CH₄ is 12.4 kJ

From the question,

We are to determine the heat of reaction for overall reaction for the production of methane in coal gasification

The equation for the reaction of coal gasification is

2C(coal) + 2H₂O → CO₂ + CH₄  

From the question,

We have the following equations of reactions

(1)        C(coal) + H₂O(g) → CO(g) + H₂(g)               ΔH°rxn = 129.7 kJ

(2)       CO(g) + H₂O(g) → CO₂(g) + H₂(g)               ΔH°rxn = -41 kJ

(3)       CO(g) + 3H₂(g) → CH₄(g) + H₂O(g)              ΔH°rxn = -206 kJ

Multiply (1) by 2 to get

(4)       2C(coal) + 2H₂O(g) → 2CO(g) + 2H₂(g)       ΔH°rxn = 259.4 kJ

Now, adding equations (2), (3), and (4), we get

(2)       CO(g) + H₂O(g) → CO₂(g) + H₂(g)               ΔH°rxn = -41 kJ

(3)       CO(g) + 3H₂(g) → CH₄(g) + H₂O(g)              ΔH°rxn = -206 kJ

(4)       2C(coal) + 2H₂O(g) → 2CO(g) + 2H₂(g)       ΔH°rxn = 259.4 kJ

--------------------------------------------------------------------------------------------------

          2C(coal) + 2H₂O(g) → CH₄(g) + CO₂(g)        ΔH°rxn = 12.4 kJ

Hence, the heat of reaction,  ΔH°rxn, for overall reaction in coal gasification for the production of methane, CH₄ is 12.4 kJ

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Final answer:

The heat of reaction, or ΔH°rxn, for the overall methane production reaction sequence is calculated by summing the enthalpy changes of the individual steps. This value comes out as -117.3 kJ per the application of Hess's Law.

Explanation:

In order to calculate the heat of reaction, ΔH°rxn, for overall methane production, we have to use Hess's Law. According to Hess's Law, if a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps. The first reaction has an enthalpy change of 129.7 kJ, the second -41 kJ, and the third -206 kJ.

Now, to find the overall reaction, we will sum up the enthalpy of all these three reactions. So, ΔH°rxn for the overall reaction would be calculated as 129.7 kJ - 41 kJ - 206 kJ = -117.3 kJ. Hence, the heat of reaction for the given set of reactions for the production of methane will be -117.3 kJ.

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The percent yield of a reaction between elemental zinc and an aqueous solution of 0.50 M hydro-chloric acid is known to be 78.0%. We need to produce 35.5 g of zinc chloride, what is the minimum amount in mL of hydrochloric acid that are required, given that zinc is in excess

Answers

Answer:

1.3 × 10³ mL

Explanation:

Let's consider the following reaction.

Zn + 2 HCl → ZnCl₂ + H₂

The percent yield is 78.0%. The real yield (R) of zinc chloride is 35.5 g. The theoretical yield (T) of zinc chloride is:

35.5 g (R) × (100 g T/ 78.0 g R) = 45.5 g T

The molar mass of zinc chloride is 136.29 g/mol. The moles corresponding to 45.5 g of zinc chloride is:

45.5 g × (1 mol/ 136.29 g) = 0.334 mol

The molar ratio of HCl to ZnCl₂ is 2:1. The moles of HCl that react with 0.334 moles of ZnCl₂ are 2 × 0.334 mol = 0.668 mol.

We need 0.668 moles of a 0.50 M HCl solution. The volume required is:

0.668 mol × (1000 mL/0.50 mol) = 1.3 × 10³ mL

A noble gas atom in the liquid phase is on average surrounded by 10 other atoms. Each pair of atoms contribute g

Answers

Answer: noble gases are in reactive.

Explanation: noble gases are present in the right most corner of the periodic table in the 8th group. So their outermost shells are complete. Their boiling point, mass increases down the group. the have strong forces of interaction. their ionization energy decreases down the group

If heat is added to ice and liquid water in a closed container and after the addition of the heat, ice and liquid water remain, _____

Answers

Complete Question:

If heat is added to ice and liquid water in a closed container and after the addition of the heat, ice and liquid water remain, (A) the vapor pressure of the water will decrease. (B) the temperature will increase somewhat. (C) the temperature will decrease somewhat. (D) the vapor pressure of the water will remain constant.

Answer:

A

Explanation:

When heat is added to a system, the internal energy of the molecules increases, and they become more agitated, because of that the temperature intends to increase. But when exists a liquid-vapor equilibrium this increase of temperature may be balanced by the vapor pressure.

The vapor pressure is the pressure that the vapor does when it is in equilibrium with the liquid. So, as higher is it, as easy it will be to the liquid to evaporate. When the temperature increases more liquid will evaporate, because the molecules are more agitated, and so the vapor pressure must increase.

But, if the ice and liquid remain, it indicates that no liquid was evaporated, so, the pressure decreased, to avoid the effect of the temperature, which will remain constant.

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