write complete ionic and net ionic equations for, SrCO3(s)+H2SO4(aq) -> SrSO4(s) +CO2(g)+H2O(l)

Waves produced a diffraction pattern.

Results supported the wave theory of light.

Answer:

Waves produced a diffraction pattern.

Results supported the wave theory of light.

Explanation:

i got it right

Final answer:

At 393 K, which is 119.85°C, acetic acid exists in a liquid state, as it is above its melting point and below its boiling point.

Explanation:

Acetic acid, also known by its systematic name ethanoic acid, exists in different physical states depending on the temperature. Pure acetic acid solidifies at 16.6°C and is referred to as glacial acetic acid because it often solidified in cold laboratories lending the appearance of ice. At 393 K (which is equivalent to 119.85°C), acetic acid is well above its melting point and well below its boiling point of about 118°C. Hence, at 393 K, acetic acid would exist in a liquid state. It is important to note that the compound forms dimers due to strong intermolecular hydrogen bonding, which affects its physical properties, including melting and boiling points.

Answer:

              Volume =  100 cm³

Solution:

             Density measures and compares the amount of mass contained by a substance to its volume. It is expressed as,

                                           Density  =  Mass / Volume

Data Given;

                   Density  =  1.0 g/cm³

                   Mass  =  100 g  

                   Volume  =  ?

Formula Used;

                   Density  =  Mass ÷ Volume

Solving for Volume,

                   Volume  =  Mass ÷ Density  

Putting values,

                   Volume  =  100 g ÷ 1.0 g.cm⁻³

                   Volume =  100 cm³

The balanced chemical equation for the reaction between phosphoric acid and cesium hydroxide to produce cesium dihydrogen phosphate and water is 3 CsOH(aq) + H3PO4(aq) → Cs3PO4(aq) + 3 H2O(l).

The balanced chemical equation for the reaction between aqueous solutions of phosphoric acid (H3PO4) and cesium hydroxide (CsOH) to produce aqueous cesium dihydrogen phosphate (CsH2PO4) and liquid water (H2O) is as follows:

3 CsOH(aq) + H3PO4(aq) → Cs3PO4(aq) + 3 H2O(l)

This reaction is an example of an acid-base neutralization, resulting in the formation of a salt (cesium dihydrogen phosphate) and water.

Explanation:

A pH scale is a scale which helps in identifying whether a given substance is basic, acidic or neutral in nature.

So, when a substance has pH value less then 7 then it means the substance is acidic in nature. Whereas when a substance has pH value more then 7 then it means the substance is basic in nature.

On the other hand, if pH value of a substance is equal to 7 then it means the substance is neutral in nature.

Hence, we can conclude that the pH scale runs from 0 to 14, and is an indication of how acidic or basic a solution is. A pH close to zero indicates an acidic solution, and a pH near 14 indicates a basic solution.

The hydrogen ion concentration [H⁺] is 0.1 mol/L, making the pH of the solution 1. This pH level indicates that the solution is acidic.

Let's work through the problem step by step:

A. Calculate the hydrogen ion concentration [H⁺]

We know that the product of hydrogen ion concentration [H⁺] and hydroxide ion concentration [OH⁻] in water is equal to 1 x 10⁻¹⁴ at 25°C. This is represented by the equation:

[H⁺] * [OH⁻] = 1 x 10⁻¹⁴

Given that [OH⁻] = 1 x 10⁻¹³ mol/L, we can rearrange the equation to solve for [H⁺]:

[H⁺] = 1 x 10⁻¹⁴ / [OH⁻]

Substituting the given value:

[H⁺] = 1 x 10⁻¹⁴ / 1 x 10⁻¹³ = 1 x 10⁻¹ = 0.1 mol/L

B. Calculate the pH

The pH of a solution is calculated using the formula:

pH = -log[H⁺]

Substituting the value for [H⁺]:

pH = -log(0.1) = 1

C. Identify if the solution is an acid, base, or neutral

A solution with a pH less than 7 is acidic, a pH greater than 7 is basic, and a pH of 7 is neutral.

Since the pH of this solution is 1, it is an acidic solution.

To find the molecular formula, divide the given molar mass by the empirical formula mass of the compound and multiply the empirical formula by the resulting value. In this case, the molecular formula is 2 times the empirical formula.

The molar mass of a compound can be used to determine its molecular formula. To find the molecular formula, we divide the given molar mass by the empirical formula mass of the compound. If the resulting value is a whole number, then the empirical formula is also the molecular formula. If it is not a whole number, we need to multiply the empirical formula by the integer value closest to the resulting value to obtain the molecular formula.

In this case, since the molar mass is given as 73.8 grams/mole, and the calculated empirical formula mass is 176.124 g/mol, we find that 73.8 g/mol divided by 176.124 g/mol is equal to 0.419. Since this is not a whole number, we need to multiply the empirical formula by 2 since 2 is the integer that is closest to 0.419. Therefore, the molecular formula is 2 times the empirical formula.

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Answer: The correct answer is Option d.

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as [tex]K_{eq}[/tex]

For a general chemical reaction:

[tex]aA+bB\rightarrow cC+dD[/tex]

The expression for [tex]K_{eq}[/tex] is written as:

[tex]K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}[/tex]

There are 3 conditions:

From the above expression, the equilibrium constant is directly dependent on product concentration. Thus, more is the concentration of product, more will be the equilibrium constant.

The highest values of [tex]K_{eq}[/tex] will favor the product more.

Hence, the correct answer is Option d.

The dissolved particles in a solution containing a molecular solute are the molecules of the solute themselves. When a molecular solute dissolves, its molecules distribute uniformly within the solvent, forming a homogeneous mixture at the molecular level. However, ionic solutes dissociate into individual cations and anions.

The dissolved particles in a solution containing a molecular solute like sugar (C12H22O11) would be molecules. In chemistry, a solution can be thought of as a mixture that is homogeneous at the molecular level. In these solutions, one component (the solvent) is usually present in a significantly larger concentration. Other components (solutes), are present in relatively lesser amounts. When a covalent solid like sugar dissolves in water, its molecules become uniformly distributed among the molecules of water creating an aquaeous solution.

However, not all solutions result in dissolved molecules. Solutions that include ionic compounds like sodium chloride (NaCl) will separate into individual cations and anions when dissolved in water. In contrast to molecular substances, ionic type compounds will break apart into separate ions. Hence, it's crucial to understand the type of solute in your solution.

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Answer:c mobile home

Explanation:

A mobile home is preconstructed structure, that can be run and transported to other location according to the choice of living place. This can be prepared from wood or other kind of cheap material which can keep the weight of mobile home low.

Thus mobile home is the lowest maintainace cost home for a single family.

The answer would be false because the true answer would be astronomy

A solution containing fluoride ions will be weakly basic because fluoride reacts with water to produce hydroxide ions, raising the pH slightly above 7.

Unlike chloride ions, which make a neutral solution, fluoride ions make the solution weakly basic.

When determining the pH of a solution containing fluoride ions (F¯), it is important to understand how fluoride interacts with water. Sodium fluoride (NaF) dissolves in water, producing fluoride ions, which then weakly react with water:

F¯ (aq) + H₂O (l) ⇒ HF (aq) + OH¯ (aq)

This reaction creates hydroxide ions (OH¯), making the solution weakly basic. Thus, while a chloride ion solution is neutral, a fluoride ion solution will have a pH slightly above 7 due to the weak base formed.

Answer:The green house gases are: carbon dioxide, nitrous oxide

Explanation:

Green house gases are the atmospheric gases which absorbs and emit radiant energy. These gases are responsible green house effect and rise in temperature of planet earth.For example: carbon-dioxide,nitrous oxide, methane, , water vapors etc. There also man made green house gases like: chloro-fluorocarbons , hydro-fluorocarbons etc.

The Heck reaction involves bond formation between an alkene and an aryl halide. With bromobenzene and propene, two isomers can be obtained, 1-propylbenzene and 2-propylbenzene. These differ in the attachment point of the propyl group to the benzene ring.

The Heck reaction involves a palladium-catalyzed carbon-carbon bond formation between an alkene and an aryl halide. When bromobenzene reacts with propene in a Heck reaction with pd(php3)4 and et3n, two constitutional isomers can be obtained. The difference between the two isomers would lie in the point of attachment of the propyl group to the benzene ring - it could either be at the end (1-propylbenzene) or at the middle (2-propylbenzene).

For 1-propylbenzene, the propyl group is attached to the benzene ring at one end, resulting in a straight chain configuration.

However, for 2-propylbenzene, the propyl group is attached to the benzene ring in the middle, resulting in a branched chain configuration.

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Kc is approximately 0.00362, and Kp is approximately 2.167 based on molar concentrations and gas collected data.

To accurately calculate the equilibrium constants [tex]\( K_c \)[/tex] and [tex]\( K_p \)[/tex] for the given reaction, we need to follow the steps correctly:

[tex]\[ \text{CH}_4(\text{g}) + \text{H}_2\text{O}(\text{g}) \leftrightarrow \text{CO}(\text{g}) + 3\text{H}_2(\text{g}) \][/tex]

1. Molar Masses:

- [tex]\(\text{CH}_4\):[/tex] 16.04 g/mol

- [tex]\(\text{H}_2\text{O}\)[/tex] : 18.02 g/mol

- [tex]\(\text{CO}\):[/tex] 28.01 g/mol

- [tex]\(\text{H}_2\):[/tex] 2.02 g/mol

2. Calculate Moles of Each Substance:

[tex]\(\text{CH}_4\): \( \frac{42.3 \, \text{g}}{16.04 \, \text{g/mol}} = 2.638 \, \text{moles} \)\\ \(\text{H}_2\text{O}\): \( \frac{49.2 \, \text{g}}{18.02 \, \text{g/mol}} = 2.730 \, \text{moles} \)\\\(\text{CO}\): \( \frac{8.32 \, \text{g}}{28.01 \, \text{g/mol}} = 0.297 \, \text{moles} \)\\\(\text{H}_2\): \( \frac{2.63 \, \text{g}}{2.02 \, \text{g/mol}} = 1.301 \, \text{moles} \)[/tex]

3. Calculate Concentrations:

[tex]- \([ \text{CH}_4 ] = \frac{2.638 \, \text{moles}}{5.00 \, \text{L}} = 0.528 \text{M} \)\\\\- \([ \text{H}_2\text{O} ] = \frac{2.730 \, \text{moles}}{5.00 \, \text{L}} = 0.546 \, \text{M} \) \\\\- \([ \text{CO} ] = \frac{0.297 \, \text{moles}}{5.00 \, \text{L}} = 0.0594 \, \text{M} \)\\\\- \([ \text{H}_2 ] = \frac{1.301 \, \text{moles}}{5.00 \, \text{L}} = 0.260 \, \text{M} \)[/tex]

4. Calculate [tex]K_c[/tex]

[tex]\[ K_c = \frac{[ \text{CO} ][ \text{H}_2 ]^3}{[ \text{CH}_4 ][ \text{H}_2\text{O} ]} \][/tex]

[tex]\[ K_c = \frac{(0.0594) \times (0.260)^3}{(0.528) \times (0.546)} \][/tex]

[tex]\[ K_c = \frac{0.0594 \times 0.017576}{0.288288} \][/tex]

[tex]\[ K_c = \frac{0.0010431744}{0.288288} \][/tex]

[tex]\[ K_c \approx 0.00362 \][/tex]

5. Calculate [tex]K_p[/tex]

[tex]\[ K_p = K_c \left( RT \right)^{\Delta n} \][/tex]

- [tex]\(\Delta n = \text{moles of products} - \text{moles of reactants} = (1 + 3) - (1 + 1) = 2\)[/tex]

- Assuming [tex]\( T = 298 \, \text{K} \)[/tex] (standard temperature) and [tex]( R = 0.0821 \, \text{L atm/(mol K)} \):[/tex]

[tex]\[ K_p = 0.00362 \left( 0.0821 \times 298 \right)^2 \][/tex]

[tex]\[ K_p = 0.00362 \left( 24.4658 \right)^2 \][/tex]

[tex]\[ K_p = 0.00362 \times 598.5877 \][/tex]

[tex]\[ K_p \approx 2.167 \][/tex]

Conclusion:

[tex]\( K_c \approx 0.00362 \)\\\\ \( K_p \approx 2.167 \)[/tex]

In summary, Kc is approximately 0.00362, and Kp is approximately 2.167

When 6.01 g of NH₃ reacts with 4.91 g of HCl, there are 3.71 grams of excess NH₃. HCl is the limiting reactant in this reaction based on the number of moles available.

The reaction between ammonia (NH₃) and hydrogen chloride (HCl) to form ammonium chloride (NH₄Cl) is described by the balanced equation:

NH₃ + HCl → NH₄Cl

First, we need to determine the number of moles of each reactant.

Molar mass of NH₃ = 14.01 (N) + 3 × 1.01 (H) = 17.04 g/mol

Moles of NH₃ = 6.01 g / 17.04 g/mol = 0.3526 moles

Molar mass of HCl = 1.01 (H) + 35.45 (Cl) = 36.46 g/mol

Moles of HCl = 4.91 g / 36.46 g/mol = 0.1347 moles

From the balanced equation, the mole ratio of NH₃ to HCl is 1:1, meaning they react on a 1:1 basis. Therefore, HCl is the limiting reactant because we have fewer moles of HCl than NH₃.

Since the reaction is 1:1, 0.1347 moles of HCl will fully react with 0.1347 moles of NH₃. We started with 0.3526 moles of NH₃ and used up 0.1347 moles:

Moles of NH₃ remaining = 0.3526 - 0.1347 = 0.2179 moles

To find the mass of the excess NH₃:

Mass of excess NH₃ = 0.2179 moles × 17.04 g/mol = 3.71 g

Therefore, when 6.01 g of NH₃ reacts with 4.91 g of HCl, there are 3.71 grams of excess NH₃.

The solubility of O2 in water is directly proportional to the partial pressure of the gas. To find the solubility of O2 at a pressure of 2.40 atm, we can use Henry's law and the given solubility at 0.370 atm.

Solubility of O2 in water at different pressures

The solubility of oxygen (O2) in water is directly proportional to the partial pressure of oxygen gas above the water. This relationship is described by Henry's law. According to Henry's law, the solubility of a gas is equal to the product of the Henry's law constant (kH) and the partial pressure of the gas. Therefore, to find the solubility of O2 at a pressure of 2.40 atm and 25 °C, we can use the following equation:

Solubility = kH * Pressure

Given that the solubility of O2 at 0.370 atm and 25 °C is 15.3 g/100 g H2O, we can rearrange the equation to solve for the Henry's law constant:

kH = Solubility / Pressure

Substituting the values into the equation, we get:

kH = (15.3 g/100 g H2O) / 0.370 atm

Solving this equation will give us the value of kH. Once we have the value of kH, we can use it to calculate the solubility of O2 at a pressure of 2.40 atm.