A chemist determines by measurements that 0.0500 moles of oxygen gas participate in a chemical reaction.
Calculate the mass of oxygen gas that participates. Round your answer to significant digits.

The question is incomplete, here is the complete question:

A solution is prepared at 25°C that is initially 0.075 M in chlorous acid [tex](HClO_2)[/tex] , a weak acid with [tex]K_a=1.1\times 10^{-2}[/tex], and 0.34 M in potassium chloride [tex](KClO_2)[/tex] . Calculate the pH of the solution. Round your answer to 2 decimal places.

Answer: The pH of the solution is 2.62

Explanation:

The chemical equation for the reaction of chlorous acid and potassium hydroxide follows:

[tex]HClO_2+KOH\rightarrow KClO_2+H_2O[/tex]

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]

[tex]pH=pK_a+\log(\frac{[NaHCO_3]}{[H_2CO_3]})[/tex]

We are given:

[tex]pK_a[/tex] = negative logarithm of acid dissociation constant of chlorous acid = 1.96

[tex][KClO_2]=0.34M[/tex]

[tex][HClO_2]=0.075M[/tex]

pH = ?

Putting values in above equation, we get:

[tex]pH=1.96+\log(\frac{0.34}{0.075})\\\\pH=2.62[/tex]

Hence, the pH of the solution is 2.62

Answer:

0.63 is the fraction of bonding

Explanation:

This is the formula for fraction of bonding

covalent =  1 - e ^( - 0.25 . ΔEn²)

First of all, let's determine ΔEn, where you subtract the two values of electronegativity for each element

En O = 3.5

En Ti = 1.5

ΔEn = 3.4 - 1.5 = 2

covalent = 1 - e ^( - 0.25 2²)

covalent = 1 - e ^(- 0.25 . 4)

covalent = 1 - e ^-1 → 0.63

bonding is mostly covalent

Calcium chloride (CaCl2) is more effective at lowering the freezing point of ice than sodium chloride (NaCl) because it produces more ions when it dissociates, leading to greater freezing point depression.

To determine which salt would help you eat your homemade ice cream faster by lowering the freezing point of ice more effectively, we need to consider the colligative properties of each salt. The property of interest here is freezing point depression, which is a colligative property that states the freezing point of a solution is lower than that of the pure solvent due to the presence of solute particles.

Calcium chloride (CaCl2) is more effective at lowering the freezing point than sodium chloride (NaCl) because it dissociates into three ions (one Ca2+ and two Cl-), while NaCl dissociates into only two ions (one Na+ and one Cl-). Since the extent of freezing point depression is dependent on the number of solute particles in solution, CaCl2 will lower the freezing point more than NaCl for the same mass.

Given that the freezing points of saturated solutions are −22°C for NaCl and −30°C for CaCl2, using 200 g of CaCl2 would result in a lower freezing temperature compared to the same amount of NaCl, thus allowing your ice cream to freeze quicker.

Answer: The formula does not correlate with the general molecular formula of any known homologous series of hydrocarbon compounds

Explanation:

Hydrocarbons are arranged in families of compounds known as homologous series, each having its unique molecular formula.

Alkanes CnH2n+2

Alkenes CnH2n

Alkynes CnHn

C2H8 does not fit into any of these homologous series. The compound cannot be cyclic because it has only two carbon atoms. Considering all these, the existence of this hypothetical compound is simply an impossibility.

Chemists would be skeptical of a molecule with the formula C2H8 because it contradicts the octet rule. This rule outlines that carbon typically forms four bonds and hydrogen forms a single bond. A C2H8 molecule suggests more than four bonds per carbon, which does not usually occur in stable compounds.

The skepticism among chemists regarding a molecule with the formula C2H8 stems from the very basics of chemistry and molecular structures. According to the octet rule, carbon (C) forms four bonds, and hydrogen (H) forms a single bond. So a molecule with two carbon atoms would usually only have six hydrogen atoms attached (as in the case of ethane, C2H6), allowing each carbon atom to form the maximum four bonds (three with hydrogen, one with the other carbon). A molecule with two carbons and eight hydrogens would suggest more than four bonds per carbon atom, which is generally not observed in stable compounds.

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When presented with a concentration of 1 mg/mL, one can express it through molarity (if the molar mass of the solute is known) or mass-volume percent concentration. In this case, the solution's mass-volume percent concentration would be 0.1%.

The two concentration expressions that can be used to describe a solution with a concentration of 1 mg/mL are molarity and mass-volume percent concentration.

Molarity is defined as the amount of solute in moles divided by the volume of the solution in liters. Assuming the molar mass of the solute is known, the molarity can be calculated using the provided concentration.

Mass-volume percent concentration (% m/v) is another applicable concept. It is defined as the mass of solute in grams divided by the volume of the solution in milliliters, multiplied by 100%. In your case, as the solution is 1 mg/mL, it translates to a mass-volume percent concentration of 0.1%.

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Final answer:

To identify a concentration expression compatible with 1 mg/mL for a solution with density 1.00 g/mL, molarity can be used if the molar mass of the solute is known. Molality is not directly applicable without knowing the weight of the solvent separately.

Explanation:

When considering a solution with a concentration of 1 mg/mL of solute and a density of 1.00 g/mL, it's important to note that molality cannot be used directly, because it requires the weight of the solvent (not the solution), and it's expressed in kilograms. The question pertains to a solution that has its density equivalent to that of water, thus allowing for certain concentration expressions like molarity to provide a good approximation to molality for dilute aqueous solutions.

To convert from mg/mL to molarity (M), we need the molar mass of the solute. For example, if we have potassium bromide (KBr), which has a molar mass of 119.0 g/mole, a 1 mg/mL solution would be 1 g/L or 1000 mg/L. To find the molarity, divide the concentration in grams per liter (1 g/L) by the molar mass of KBr, yielding approximately 0.0084 M. This would be similar to the molality for a dilute KBr solution because the molality and molarity converge when the density is close to that of water (1.00 g/mL).

Answer:

The extinction coefficient is 1.15 x 10⁴ M⁻¹.cm⁻¹ (value is not rounded off)

Explanation:

According to Lambert-Beer law

[tex]A = ε.b.C[/tex]

Here, A is absorbance, ε is extinction coefficient, b is the length of the cuvette and C is the molar concentration of substance X.

This equation is used for the relation between concentration and absorbance of electromagnetic radiation absorbing species. It is a linear equation and can be used for making a calibration curve, which is used for the analysis of an unknown concentration solution. The slope of this curve according to the equation is the product of extinction coefficient (M⁻¹.cm⁻¹) and the length of the cuvette in cm.

In this problem, the slope is provided, which can be mathematically represented as:

[tex]slope =  ε.b[/tex]

[tex]2.3 X 10^{3}M^{-1} = ε.(0.2 cm)[/tex]

[tex]ε = 1.15 X 10^{4} M^{-1}.cm^{-1}[/tex] (not rounded off)

[tex]ε = 1 X 10^{4} M^{-1}.cm^{-1}[/tex] (rounded off)

To move a solid compound to the bottom of a melting point capillary tube, one should tap the open end on a hard surface. A needle can be used to adjust the position of the compound if necessary, without compacting it too much.

To force a solid compound to the bottom of a melting point capillary tube, you should tap the open end of the capillary tube on a hard surface, such as a bench top. This action utilizes gravity to help settle the compound into the closed end of the tube. If the compound does not move down to the bottom, you can use a long, thin object like a needle to carefully push the compound down without compacting it too tightly, which could affect its melting behavior. Alternatively, tapping the closed end might cause some compacting and is not usually recommended for settling the compound.

While using the capillary tube, it's important to understand capillary action and the characteristics of liquid-glass interactions. For example, if a capillary tube is placed into a beaker of ethylene glycol, the ethylene glycol will rise into the tube by capillary action due to the strong adhesive forces between the polar Si-OH groups on the surface of glass and the molecules of the ethylene glycol, creating a concave meniscus. This is opposite to what happens with a nonpolar liquid like SAE 20 motor oil, which cannot form strong interactions with the polar Si-OH groups, resulting in the oil having a convex meniscus and a reduction in the capillary action.

Answer:

Please refer to the attachment below for answer and explanation.

Explanation:

Please refer to the attachment below for answer and explanation.

Answer:

1. 24.7g of Br2

2. 19.9g of HCl

Explanation:Please see attachment for explanation

Answer:

Kc = 4.774 * 10¹³

Explanation:

the desired reaction is

2 NO₂(g) ⇋ N₂(g) + 2 O₂(g)

Kc =[N₂]*[O₂]² /[NO₂]²

Since

1/2 N₂(g) + 1/2 O₂(g) ⇋ NO(g)

Kc₁= [NO]/(√[N₂]√[O₂]) →  Kc₁²= [NO]²/([N₂][O₂])

and

2 NO₂(g) ⇋ 2 NO(g) + O₂(g)

Kc₂= [NO]²*[O₂]/[NO₂]² →  1/Kc₂= [NO₂]²/([NO]²[O₂])

then

Kc₁²* (1/Kc₂) = [NO]²/([N₂]*[O₂]) *[NO₂]²/([NO]²[O₂])  = [NO₂]²/([N₂]*[O₂]²) = 1/Kc

Kc₁² /Kc₂ = 1/Kc

Kc= Kc₂/Kc₁² =1.1*10⁻⁵/(4.8*10⁻¹⁰)² = 4.774 * 10¹³

Final answer:

To find the Kc for the reaction 2NO²(g) ⇌ N²(g) + 2O²(g), we reverse and manipulate the given equilibria, taking reciprocals and squaring as necessary, and then multiplying the constants to get the overall Kc.

Explanation:

To determine the equilibrium constant Kc for the given reaction 2NO²(g) ⇌ N²(g) + 2O²(g), we can manipulate the given equilibria (1) and (2) to yield the desired reaction.

First, we reverse the equilibrium (1), 1/2 N²(g) + 1/2 O²(g) ⇌ NO(g), whose Kc is 4.8 x 10-10. When an equation is reversed, the equilibrium constant becomes the reciprocal of the original. Thus, the new Kc for NO(g) ⇌ 1/2 N²(g) + 1/2 O²(g) is 1 / (4.8 x 10-10). To find the Kc for N²(g) + O²(g) ⇌ 2NO(g), we need to multiply the new reaction by two, which means squaring the Kc.

Second, we consider reaction (2), which is 2NO²(g) ⇌ 2NO(g) + O²(g), with a Kc of 1.1 x 10-5.

By adding the modified equilibrium (1) with the reaction (2), we get the desired equilibrium: 2NO(g) + O2(g) ⇌ 2NO2(g). We then multiply the Kc values of these individual steps to obtain the overall Kc for the desired reaction. This is because the equilibrium constant of the overall reaction is the product of the equilibrium constants of the individual steps.

In summary:

The dissolution of ammonium nitrate is an endothermic process. In terms of calculating the final temperature of an activated cold pack containing water and ammonium nitrate, you would use a specific formula with the given data. The details of this calculation were outlined in the detailed answer.

a. The dissolution of ammonium nitrate is an endothermic process. This is because it absorbs heat from its surroundings in the process of dissolving, leading to a drop in temperature. Hence, endothermic reactions cool their surroundings. The given heat of solution for ammonium nitrate is positive (+25.4 kJ/mol), which is characteristic of endothermic reactions.

b. To calculate the final temperature of the activated cold pack, you would need to use the formula q = mcΔT, where q is the heat absorbed or released, m is the mass, c is the specific heat, and ΔT is the change in temperature (final-initial). Given data can be used to calculate the heat absorbed by the reaction, then using that value along with the given initial temperature and specific heat, we can find the final temperature. Please note that the calculations involved may require quite a few steps and it may be best to consult your chemistry teacher or textbook for the specific mathematical operation.

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Final answer:

After 10.0 g of ammonia reacts with 10.0 g of hydrogen chloride, 31.5 g of ammonium chloride is produced with leftover hydrogen chloride remaining.

Explanation:

When ammonia is mixed with hydrogen chloride (HCl), there is a chemical reaction in which ammonium chloride (NH4Cl) is formed. This reaction can be represented by the balanced chemical equation: NH3 (g) + HCl (g) → NH4Cl (s). The molar mass of ammonia (NH3) is approximately 17 g/mol and that of hydrogen chloride (HCl) is approximately 36.5 g/mol. When 10.0 g of each reactant is mixed, stoichiometry is used to determine the products and their masses after the reaction goes to completion.

Form the given masses, we can calculate the moles of each reactant: 10.0 g NH3 ÷ 17 g/mol ≈ 0.588 mol NH3 and 10.0 g HCl ÷ 36.5 g/mol ≈ 0.274 mol HCl. Since the reaction occurs in a 1:1 molar ratio, ammonia is the limiting reactant because we have less moles of it compared to moles of HCl. Therefore, all the ammonia will react with HCl to produce 0.588 mol of NH4Cl. The mass of the produced NH4Cl can be found by multiplying the moles of NH4Cl by its molar mass (approximately 53.5 g/mol), so 0.588 mol × 53.5 g/mol ≈ 31.5 g of NH4Cl. There will be no remaining ammonia. Since HCl is in excess, after the reaction the system will contain 31.5 g of NH4Cl and some unreacted HCl.

Final answer:

To find the pH of the final solution after reacting Ca(OH)2 and HNO3, calculate moles of reactants, determine moles needed for the reaction, and use the formula for a strong base to find the final pH.

Explanation:

Step 1: Calculate the moles of HNO3 used: 0.045 L × 1.00 mol/L = 0.045 mol.

Step 2: Determine the moles of Ca(OH)2 needed to react with the HNO3: 2 moles of HNO3 react with 1 mole of Ca(OH)2, so 0.045 mol HNO3 requires 0.0225 mol Ca(OH)2.

Step 3: Use the volume and molarity of the Ca(OH)2 to find the pH using the formula for a strong base: pH = 14 - pOH. From the given data, we can calculate pOH = -log(0.672) = 0.173. Therefore, pH = 14 - 0.173 = 13.827.

The pH of the final solution is 0.324.

To determine the pH of the final solution, follow these steps:

Calculate moles of Ca(OH)₂:

Molar mass of Ca(OH)₂ is 74.10 g/mol. Moles of Ca(OH)₂:

(0.350 g) / (74.10 g/mol) = 0.00472 mol

Calculate moles of HNO₃:

Molarity (M) = moles/volume in L. For HNO₃, moles:

(1.00 M) x (0.0450 L) = 0.0450 mol

Determine reaction and limiting reagent:

The balanced equation for the reaction is:

Ca(OH)₂ + 2 HNO₃ ➞ Ca(NO3)₂ + 2 H₂O

Moles of HNO₃ required to react with Ca(OH)₂:

0.00472 mol Ca(OH)₂ x 2 = 0.00944 mol HNO₃

Excess HNO₃:

0.0450 mol - 0.00944 mol = 0.03556 mol

Final volume of solution = 75.0 mL = 0.0750 L. H⁺ concentration:

if there is no carbon dioxide your test tube will be blue

if there is a medium amount of carbon dioxide your test tube is green

if there are high amounts of CO2 it will be

yellow

In the context of bromothymol blue as an indicator, if no CO2 is present, the solution will be blue. If a medium amount of CO2 is present, the solution turns green. A large amount of CO2 would turn the solution yellow due to the acidic conditions created by carbonic acid.

The colors in the test tube depend on the identification or indicator test you are performing. However, if we were referring to the bicarbonate buffer system and the formation of carbonic acid, we could consider the pH indicators' reaction to acidity caused by carbon dioxide concentration. One common pH indicator is bromothymol blue. ~ This works as follows:

It's important to remember, this answer assumes the use of bromothymol blue, a common pH indicator. Different indicators may produce different color changes.

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Answer:

For a material to be molecular, it must possess the following properties

1. Molecular substances is made up of non metals atoms

2. Molecular substances are always unable to conducy electricity

3. Molecular substances have strong intramolecular forces and weak intermolecular forces

4. Molecular substance has low boiling and melting points

The properties of a substance that makes a material molecular are listed below:

Molecular substances are substances usually containing two or more atoms combined together by a covalent bond.

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Answer:

42,650 kg of calcium carbonate will be produced everyday.

13,600.5 kg of magnesium hydroxide will be produced everyday.

Explanation:

Concentration of calcium ions = 46.9 mg/L

Concentration of magnesium ions = 14.8  mg/L

Volume of solution treated everyday , V= 80 million gal

= [tex]80\times 10^6 gal=4.546\times 80\time 10^6 L=3.637\times 10^8 L[/tex]

1 gallon = 4.546 Liter

Mass of  calcium ion in V = [tex]46.9 mg/L\times 3.637\times 10^8 L[/tex]

=  [tex]1.706\times 10^{10} mg[/tex]

1 mg = 0.001 g

[tex]1.706\times 10^{7} g[/tex]

Moles of calcium ions = [tex]\frac{1.706\times 10^{7} g}{40 g/mol}=426,500 mol[/tex]

From 1 mole of calcium ion 1mol of carbonate is formed . then from 426,500 moles of calcium ion will form :

[tex]\frac{1}{1}\times 426,500 mol=426,500 mol[/tex] of calcium carbonate

Mass of 426,500 moles of calcium carbonate:

426,500 mol × 100 g/mol  = 42,650,000 g = 42,650 kg

Mass of  magnesium ion in V = [tex]14.8 mg/L\times 3.637\times 10^8 L[/tex]

= [tex]5.382\times 10^{9} mg[/tex]

=  [tex]5.382\times 10^{6} g[/tex]

Moles of magnesium ions = [tex]\frac{5.382\times 10^{6} g}{24 g/mol}=224,250 mol[/tex]

From 1 mole of magnesium ion 1 mol of magnesium hydroxide is formed . then from 224,250 moles of magnesium ion will form :

[tex]\frac{1}{1}\times 224,250 mol=224,250 mol[/tex] of magnesium hydroxide

Mass of 224,250 moles of magnesium hydroxide:

224,250 mol × 58 g/mol  = 13,006,500 g = 13,006.5 kg

42,650 kg of calcium carbonate will be produced everyday.

13,600.5 kg of magnesium hydroxide will be produced everyday.

Answer:

C₃H₄N₂ is the molecular formula

Explanation:

The molecular ion peak is obtained at 68 units, which is the molecular mass of the compound. The even number of nitrogen atoms gives even mass of the molecule. In this problem, the mass is even so there is an even number of nitrogen atoms. It can be either two or four, from which it must be two nitrogen, as four nitrogen atoms will give higher mass. Now, we know that 28 out of 68 units belongs to nitrogen

The number of carbon atoms can also be determined in the same manner. Only one or two carbon atoms will leave much of the remaining units (28 or 16 units respectively) to hydrogen atoms. This leaves the option of three carbon atoms as any number higher than it will give molecular mass more than 68 units.

For hydrogen atoms, only 4 units are remaining this means four hydrogen atoms are present in the molecule. This is possible, if we consider that the compound is cyclic and contains two double bonds.

PS: If we consider two carbon atoms then there will be 16 units left which means sixteen hydrogen atoms. The valency of two carbon and two nitrogen atoms will only allow a maximum of eight hydrogen atoms in the molecule (giving only 60 units).

Imidazole, with a molecular ion of 68, has the molecular formula C3H4N2, determined through understanding the application of the nitrogen rule in mass spectrometry.

The molecular formula for the compound imidazole, which contains carbon (C), hydrogen (H), and nitrogen (N), and has a molecular ion of 68, is C3H4N2. According to the nitrogen rule in mass spectrometry, nitrogen contributes either +1 or -1 to the mass, depending on whether the number of nitrogens in the compound is odd or even. Imidazole has an even number of nitrogens, hence the even-mass molecular ion of 68. The makeup of the remaining 68 mass units results from the carbon and hydrogen in the molecule: 3 carbons (each with a mass of 12, for a total of 36) and 4 hydrogens (each with a mass of 1, for a total of 4), and 2 nitrogens (each with a mass of 14, for a total of 28). So, the total mass (36 + 4 + 28) equals 68, aligning with the M+ value.

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Answer:

Xe will have the highest partial pressure

Explanation:

Using Dalton's law of partial pressures for ideal gases

p=P*x

where

p= partial pressure , P= total pressure and x = mole fraction = n / ∑n

since the number of moles is related with mass through

n= m/M

where

m= mass and M= molecular weight

then if m is the same for all the gases

x = m*M/ ∑ (m*M) = m*M/ m∑ M  = M/∑ M

thus

p=P*x = P*M/ ∑ M

for the 3 gases

p₁=P*x₁ = P*M₁/ (M₁+M₂+M₃)

p₂=P*x₃ = P*M₂/ (M₁+M₂+M₃)

p₂=P*x₃ = P*M₃/ (M₁+M₂+M₃)

then for gasses under the same pressure (P=constant) and same mass (m=constant) , p is higher when the molecular weight is higher . Therefore Xe will have the highest partial pressure

Answer:

He

Explanation:

The partial pressure of a gas is given by mole fraction of the gas multiplied by the total pressure of the gas. The mole fraction of each gas is the number of moles of that gas divided by the total number of moles present. The number of moles of each gas depends on its relative atomic mass since they are all of the same mass. The smaller the relative atomic mass, the greater number of moles of a gas and the greater mole fraction of that gas and the greater its partial pressure. Hence, helium is the lightest gas in the list hence it will have the highest partial pressure.

The frequency of radio waves with a wavelength of 40nm is approximately [tex]7.5 x 10^15 Hz.[/tex]

The frequency of radio waves can be calculated using the formula:

Frequency = Speed of light / Wavelength

Given that the wavelength is 40nm, we need to convert it into meters by dividing it by 10^9. Therefore, the wavelength is [tex]40 x 10^-9 m.[/tex]

The speed of light is approximately [tex]3 x 10^8 m/s.[/tex]Plugging in these values into the formula:

Frequency =[tex](3 x 10^8 m/s) / (40 x 10^-9 m) = 7.5 x 10^15 Hz[/tex]

Therefore, the frequency of these radio waves is approximately [tex]7.5 x 10^15 Hz.[/tex]

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To calculate the frequency, use the equation speed of light = wavelength x frequency, rearrange the equation to find frequency = speed of light / wavelength, and plug in the values to find the frequency.

To calculate the frequency of radio waves with a wavelength near 40nm, we can use the equation: speed of light = wavelength x frequency. The speed of light is approximately 3 x 10^8 m/s. Rearranging the equation gives us frequency = speed of light / wavelength. Converting the wavelength from nm to m, we get 40nm = 40 x 10^-9 m. Plugging the values into the equation, we find that the frequency is approximately 7.5 x 10^15 Hz.

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Complete Question:

A small sphere of initial volume V is filled with n moles of helium at initial pressure and temperature and T. Which of the following statements is true?

a) The volume decreases to V/2, and the pressure increases to 4P when the temperature is T/2

b) n/2 moles of gas are removed, the volume is decreased to V/2, and the pressure decreases to P/4 with a drop in temperature of T/2

c) n moles of gas are added, the total sample is heated to 2T, and the pressure drops to P/2 when the volume increases to 8V

d) The amount of gas is doubled to 2n, the pressure is doubled to 2P, and the volume is doubled to 2V, with a corresponding temperature drop to T/2

Answer:

c

Explanation:

Let's consider the helium as an ideal gas, so it can be studied by the ideal gas law, which states:

PV = nRT

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

PV/nT = R. Thus the initial state must be equal to the final state.

So, let's check the statements:

a) Let's indicate the final state as P₂, V₂, n₂ and T₂. So, if T₂ = T/2:

PV/nT = P₂V₂/n₂T₂

PV/nT = P₂V₂/n₂(T/2)

PV/nT = 2P₂V₂/n₂T

So, if V₂ = V/2 and P₂ = 4P:

PV/nT = 2*(V/2 * 4P)/n2T

PV/nT = 4VP/n2T

Which is not correct!

b) Now, if T₂ = T/2:

PV/nT = 2P₂V₂/n₂T

If n/2 is removes, n₂ = n/2. And, V₂ = V/2 and P₂ = P/4:

PV/nT = 2*(V/2 * P/4)/(n/2)*T

PV/nT = 4*(V/2 *P/4)/nT

PV/nT = PV/2nT

Which is not corret!

c) Now, if V₂ = 8V:

PV/nT = P₂*8V/n₂T₂

And n₂ = n +n = 2n, T₂ = 2T and P₂ = P/2:

PV/nT = (P/2)*8V/2n*2T

PV/nT = 8*(PV)/2*2n*2T

PV/nT = 8*(PV)/8*(nT)

PV/nT = PV/nT

So, it's correct!

d) Now, T₂ = T/2, n₂ = 2n, P₂ = 2P, and V₂ = 2V:

PV/nT = 2P*2V/2n*(T/2)

PV/nT = 4PV/nT

Which is not correct!

The given question is incomplete. The complete question is as follows.

The solubility product of calcium fluoride () is  at 25 degrees C. Will a fluoride concentration of 1.0 mg/L be soluble in a water containing 200 mg/L of calcium?

Explanation:

Reaction equation for the given chemical reaction is as follows.

       [tex]CaF_{2} \rightleftharpoons Ca^{2+} + 2F^{-}[/tex]

Therefore, expression for [tex]K_{sp}[/tex] will be as follows.

        [tex]K_{sp} = [Ca^{2+}][F^{-}]^{2}[/tex]

                     =

Also, moles of  per liter = \frac{\text{mass of F^{-} per L}}{\text{molar mass of F}}[/tex]

                = [tex]\frac{1.0 \times 10^{-3}}{19.0}[/tex]

               = [tex]5.263 \times 10^{-5} mol[/tex]

Hence,    [tex][F^{-}] = \frac{\text{moles of F^{-}}{volume}[/tex]

                       = [tex]\frac{5.263 \times 10^{-5}}{1}[/tex]

                      =  M

Now, moles of  per L = \frac{\text{mass of Ca^{2+} per L}}{\text{molar mass of Ca}}[/tex]

            = [tex]\frac{200 \times 10^{-3}}{40.1}[/tex]

           =  M

Also,   [tex][Ca^{2+}] = \frac{moles of Ca^{2+}}{volume}[/tex]

                      = [tex]\frac{4.988 \times 10^{-3}}{1}[/tex]

                     =  M

Hence, ionic product =

                 = [tex](4.988 \times 10^{-3}) \times (5.263 \times 10^{-5})^{2}[/tex]

                = [tex]1.38 \times 10^{-11}[/tex]

As, the ionic product is less than the [tex]K_{sp}[/tex], this means that the fluoride will be soluble in water containing the calcium.

Answer:

164 g/mol

Explanation:

According to Graham's law, the rate of effusion of a gas (r) is inversely proportional to the square root of its molar mass (M).

rH₂/rX = √[M(X)/ M(H₂)]

(rH₂/rX)² = M(X)/ M(H₂)

M(X) = (rH₂/rX)² × M(H₂)

M(X) = (9)² × 2.02 g/mol

M(X) = 164 g/mol

The molar mass of the unknown gas is 164 g/mol.

The molar mass of the unknown gas is 64 g/mol.

The rate of effusion of a gas (r) is inversely proportional to the square root of its molar mass (M).

It is given by:

[tex]rH_2/rX = \sqrt{[M(X)/ M(H_2)]} \\\\(rH_2/rX)^2 = M(X)/ M(H_2)\\\\M(X) = (rH_2/rX)^2 * M(H₂)\\\\M(X) = (9)^2 * 2.02 g/mol\\\\M(X) = 164 g/mol[/tex]

Thus, the molar mass of the unknown gas is 164 g/mol.

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Answer: [tex]CrF_3[/tex] has greater impact on the freezing point depression of ice.

Explanation:

Depression in freezing point is given by:

[tex]\Delta T_f=i\times K_f\times m[/tex]

[tex]\Delta T_f=T_f-T_f^0[/tex] = Depression in freezing point

i= vant hoff factor

[tex]K_f[/tex] = freezing point constant

m= molality

[tex]\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]

a) i = 4 for [tex]CrF_3[/tex] as it dissociates to give 4 ions in water.

[tex]CrF_3\rightarrow Cr^{3+}+3F^-[/tex]

b) i = 3 for [tex]CaF_2[/tex] as it dissociates to give 3 ions in water.

[tex]CaF_2\rightarrow Ca^{2+}+2F^-[/tex]

As the vant hoff factor is higher for [tex]CrF_3[/tex] , it has greater impact on the freezing point depression of ice.

Final answer:

CrF3 would have a greater impact on the freezing point depression of ice than CaF2 because it dissociates into more particles, four compared to CaF2 which dissociates into three.

Explanation:

In determining which substance would have a greater impact on the freezing point depression of ice, we need to consider the number of particles each substance releases into solution when dissolved. CrF3 will ionize to form one Cr3+ ion and three F- ions, for a total of four particles. Conversely, CaF2 will dissociate to form one Ca2+ and two F- ions, resulting in three particles. According to colligative properties, CrF3 will therefore have a greater effect on freezing point depression than CaF2 because it produces a larger number of dissociated particles.

Answer:

Partial pressure O₂  = 2.284 atm

Explanation:

This is problem with unit conversion. Let's convert everything to atm

760 mmHg ___ 1 atm

540 mmHg ___ (540 / 760) = 0.710 atm

101325 Pa ____ 1 atm

203303 Pa ____ (203303 / 101325) = 2.006 atm

Total pressure of the mixture = Sum of partial pressure of each gas

0.710 atm (He) + 2.006 (N₂) + Partial Pressure O₂ = 5 atm

Partial pressure O₂ = 5 atm - 2.006 atm - 0.710 atm = 2.284 atm

Answer:

(a)  0.72 lbm· ft/s²

(b)  20.3 kPa,  2.94 lbf / in²

(c)  98.6 ºF,  310 K

(d)  1.5 x 10⁻² J,  6.1 x 10⁻² cal

Explanation:

Our strategy here will be to find the conversion factors for the quantities we are asked in each part, and perform the calculations.

(a) 10,000 dynes to lbm ·ft/s²

here we are asked to convert  the  force of 10,000 dynes to lbm ·ft/s². Recall that F= ma ( m= mass, a = acceleration), thus

10,000 dynes = 10 g cm/s²

converting the force

10,000 g cm/s² x (1 lbm/454 g) x (1 ft / 30.48 cm ) /s² = 0.72 lbm· ft/s²

(b)

1 atm = 101.33 pa

0.2 atm x ( 101.33 kPa ) = 20.3 kPa

1 atm = 14.7 lbf / in²

0.2 atm x ( 14.7 lbf / in² /atm ) = 2.94 lbf / in²

(c) The formula for the conversion from ºC to ºF is:

ºF = 9/5 ºC +32

ºF = 9/5 ( 37ºC) + 32 = 98.6 ºF

K = ºC + 273

K = (37 + 273) K = 310 K

(d) 50 in²·lbm/s² to joules and calories

Since the unit in² ·lbm/s² is not that common, lets convert it using their definition.

These are energy units, and we know the energy is the force times distance. In turn force is mass times acceleration so that the units of energy are mass time distance per time squared.

Joules is the unit of energy  in the metric system.

50 in² lbm/s² = 50 in²x ( 2.54 cm/in x 1m /100cm)² x (1lbm x 0.454 Kg/lbm)/s²

= 1.5 x 10⁻² Kg m²/² =  1.5 x 10⁻² J

To convert to cal it wilñl be easier to use the value in joules just calculated:

1.5 x 10⁻² J x  (4.184 cal/J) = 6.1 x 10⁻² cal

Final answer:

This response provides detailed conversions for different units in physics including dynes, lbm, atm, and temperature measurements. It explains how to convert between various units and provides step-by-step examples for each conversion.

Explanation:

Unit conversions:

(a) To convert 10,000 dynes to lbm ∙ ft/s² and lbf:
1 dyne = 1 g cm/s² = 10⁻⁵ N, therefore, 10,000 dynes = 0.1 N. Using 1 lb = 453.59 g, 1 ft = 0.3048 m, 1 lb = 4.448 N, the conversion is: 10,000 dynes = 0.1 N = 0.0225 lb or 0.1 N = 0.0225 lbf.

(b) To convert 0.2 atm to Kpa and lbf:
1 atm = 101.3 Kpa, therefore, 0.2 atm = 20.26 Kpa. Also, 1 atm = 14.7 lb/in², so 0.2 atm = 2.94 lb/in².

(c) Converting 37 °C to °F and K:
To convert °C to °F: °F = (°C × 9/5) + 32. So, 37 °C = 98.6 °F. To obtain K from °C, use K = °C + 273. Thus, 37 °C = 310 K.

Answer:

When the temperature lowers to 264K the volume lowers to 50.95 L

Explanation:

Step 1: Data given

The initial volume of a gas = 57.9 L

The initial temperature = 300 K

The temperature lowers to 264 K

Step 2: Charles's law

V1 / T1 = V2 / T2

⇒ with V1 = the initial volume of the gas = 57.9 L

⇒ with T1 = the initial temperature = 300K

⇒ with V2 the new volume = TO BE DETERMINED

⇒ with T2 = the final temperature = 264K

57.9L /300K = V2 / 264K

V2 = (57.9*264)/300

V2 = 50.95 L

When the temperature lowers to 264K the volume lowers to 50.95 L

Some Potassium bromide precipitates out of solution.

Option B.

Solubility is defined as the tendency of a substance to get mixed into a solvent at a particular temperature and pressure. The amount of solubility is defined as the amount of substance in grams which will make a saturated solution of 100ml at a temperature and pressure.

Potassium bromide is a salt and nitrogen is a gas. The solubility of the salts generally increase with temperature in water, and decreases with decrease in temperature. So in case of potassium bromide, the solubility of the salt will decrease, leaving some precipitate in the room temperature.

While in case of gases, the solubility of them do decrease with increase in temperature. So at room temperature, solubility of nitrogen will be more than that in 750°C. So no gas will bubble off.

Some of the potassium bromide precipitates out of the solution.  

• The solubility of the gas is inversely proportional to the temperature. With the decrease in temperature, the solubility of the gas increases.  

• Therefore, in the given case, nitrogen will not move out, in spite of that opposite will take place, that is, more atmospheric nitrogen will get dissolve in the solution.  

• The solubility of ionic salts like potassium bromide is directly corelated to the temperature, that is, the solubility of the salt decreases when the reduction in temperature takes place.  

• The precipitation of some of the salt takes place because solubility is low at lower temperature and the excess salt stays in solid form.  

Thus, some of the potassium bromide precipitates out of the solution .

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Given the balanced equation 2A + 3B = 3C, with 6 moles of A and 2 moles of B, A is the limiting reactant. The maximum moles of C formed is 9 (Option D).

First, let's identify the limiting reactant by comparing the mole ratios of A and B in the balanced equation:

[tex]\[2A + 3B \rightarrow 3C\][/tex]

The mole ratio between A and B is 2:3. If you have 6 moles of A and 2 moles of B, you can compare the actual ratio (6 moles A to 2 moles B) with the ratio in the balanced equation (2 moles A to 3 moles B).

The actual ratio is 6:2, which simplifies to 3:1. The balanced ratio is 2:3. The limiting reactant will be the one with the smaller ratio, so in this case, A is the limiting reactant.

Now, let's determine the maximum number of moles of C that can be formed using the stoichiometry of the balanced equation:

For every 2 moles of A, 3 moles of C are formed.

Since you have 6 moles of A (the limiting reactant), you can use the ratio to find the moles of C:

[tex]\[\text{Moles of C} = \left( \frac{3 \text{ moles C}}{2 \text{ moles A}} \right) \times 6 \text{ moles A} = 9 \text{ moles C}\][/tex]

Therefore, the correct answer is:

D. 9 moles

The probable question may be:

Consider the chemical equation: 2A+3B=3C. If 6 moles of A and 2 moles of  B are reacted, what is the maximum number of moles of C that can be formed?

A. 4 moles

B. 2 moles

C. 3 moles

D. 9 moles

Answer: - 894.6 kJ/mol.

Explanation:

Hess law is states that the changes in enthalpies in a chemical reactions is independent of the pathway between the initial and final states.

∆H is the change in the sum of the internal energy of a system.

We are to find the Value of ΔH°(rxn) for the equation:

NH3 (g) + 2 O2 (g) ⟶ HNO3 (g) + H2O (g). ----------------------------------(**).

From the series of equations given;

==> 4NH3 (g) + 5O2 (g) -------->4 NO(g) + 6H2O (l). ∆H = -1166.0 kJ/mol.--------------------------------------(1).

===> 2NO(g) + O2 (g) ------> 2NO2 (g). ∆H = -116.2 kJ/mol.---------------(2).

===> 3NO2 (g) + H2O (l) ---------> 2HNO3 (aq) + NO (g). ∆H = -137.3 kJ/mol.-------------------------------------(3).

The first thing to do is to multiply equation (2) by 3. Also, multiply equation (3) by 2. This will give us equation (4) and (5) respectively.

6NO + 3 O2 ----------------> 6NO2. ∆H= 3 × (-116.2 kJ/mol) = -348.6 kJ/mol. ------------------------------------(4).

6NO2 + 2 H2O ----------------> 4HNO3 + 2 NO. ∆H= 2 × (-137.3kj/mol) = -274.6 kJ/mol ---------------------------(5).

Next, add equations (4) and (5) to give;

4NO +3O2 +2H2O -------------> 4HNO3. ∆H = -623.2 kJ/mol. -----(6).

Add this equation to the equation (1) from above, we have;

4NH3 + 8O2 --------------> 4HNO3 + 4H2O. ∆H= -1789.2 kJ/mol. --------(7).

Then, divide the equation (7) above by 2 to give us back the equation (**).

NH3 (g) + 2 O2 (g) ⟶ HNO3 (g) + H2O (g). ∆H= -894.6 kJ/mol.

Δ H^∘ (rxn)= - 894.6 kJ/mol.

Answer : The mass percentage of barium in the compound is, 53.8 %

Explanation : Given,

Mass of barium compound = 441 mg

Mass of barium sulfate = 403 mg = 0.403 g       (1 mg = 0.001 g)

The balanced chemical reaction will be:

[tex]Ba^{2+}(aq)+Na_2SO_4(aq)\rightarrow BaSO_4(s)+2Na^+(aq)[/tex]

First we have to calculate the moles of [tex]BaSO_4[/tex]

[tex]\text{Moles of }BaSO_4=\frac{\text{Mass of }BaSO_4}{\text{Molar mass of }BaSO_4}[/tex]

Molar mass of [tex]BaSO_4[/tex] = 233.38 g/mole

[tex]\text{Moles of }BaSO_4=\frac{0.403g}{233.38g/mole}=0.001727mole[/tex]

Now we have to calculate the moles of barium ion.

From the balanced chemical reaction, we conclude that

As, 1 mole of barium sulfate produced from 1 mole of barium ion

So, 0.001727 mole of barium sulfate produced from 0.001727 mole of barium ion

Now we have to calculate the mass of barium ion.

[tex]\text{ Mass of }Ba^{2+}=\text{ Moles of }Ba^{2+}\times \text{ Molar mass of }Ba^{2+}[/tex]

Molar mass of barium = 137.3 g/mol

[tex]\text{ Mass of }Ba^{2+}=(0.001727moles)\times (137.3g/mole)=0.2371g[/tex]

Now we convert the mass of barium ion from gram to mg.

Conversion used : (1 g = 1000 mg)

Mass of barium ion = 0.2371 g = 237.1 mg

Now we have to calculate the mass percentage of barium in the compound.

Mass percent of barium = [tex]\frac{237.1mg}{441mg}\times 100=53.8\%[/tex]

Thus, the mass percentage of barium in the compound is, 53.8 %