A sample has a mass of 7.412 grams and a volume of 0.217 liters. What is the density of this sample?

Answer:

0.089

Explanation:

Step 1:

The balanced equation for the reaction is given below:

H2O + Cl2O <=> 2HClO

Step 2:

Data obtained from the question. This includes:

Concentration of H2O, [H2O] = 0.077 M

Concentration of Cl2O, [Cl2O] = 0.077 M

Concentration of HClO, [HClO] = 0.023 M

Equilibrium constant, K =?

Step 3:

Determination of the equilibrium constant. This is illustrated below:

The equilibrium constant for the above reaction is given below:

K = [HClO]^2 / [H2O] [Cl2O]

K = (0.023)^2 / (0.077 x 0.077)

K = 0.089

Therefore, the equilibrium constant for the above reaction is 0.089

Answer:

The answer is 0.089

I just took the test

Explanation:

Consider the reaction.

Upper H subscript 2 upper o (g) plus upper C l subscript 2 upper O (g) double-headed arrow 2 upper H upper C l upper O (g).

At equilibrium, the concentrations of the different species are as follows.

[H2O] = 0.077 M

[Cl2O] = 0.077 M

[HClO] = 0.023 M

What is the equilibrium constant for the reaction at this temperature?

0.089

0.26

3.9

11

The type of interparticle force that has the greatest influence on the physical properties of each substance are identified. Factors that result in a stronger ionic bond overall are listed.

The type of interparticle force that has the greatest influence on the physical properties of each substance are as follows:

The factors that result in a stronger ionic bond overall are:

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Cl2 and CH4 experience London dispersion forces, NH2OH has hydrogen bonding, PCl3 has dipole-dipole interactions, and CaCl2 and KI demonstrate ionic bonding. Greater absolute charges and smaller ion sizes boost the strength of ionic bonds.

The substances listed – Cl2, NH2OH, PCl3, CH4, CaCl2, and KI – have different types of interparticle forces that affect their physical properties. Cl2 and CH4 are nonpolar with London dispersion forces; NH2OH has hydrogen bonding; PCl3 has dipole-dipole interactions. CaCl2 and KI are ionic compounds, where the primary interparticle force is ionic bonding.

The factors that enhance the strength of ionic bonds are usually the greater absolute charges on the ions and the smaller size of the ions. Larger charges mean that the electrostatic attraction between the ions is stronger, leading to a stronger ionic bond. Smaller ions, on the other hand, can get closer together, which also intensifies the electrostatic attraction and hence increases the strength of the ionic bond.

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Answer:

Option B.

Explanation:

As any reaction of combustion, the O₂ is a reactant and the products are CO₂ and H₂O. Combustion reaction for ethane is:

2C₂H₆  +  7O₂   →   4CO₂  +  6H₂O

So 2 moles of ethane react with 7 moles of oxygen to make 4 moles of dioxide and 6 moles of water.

Then 2 moles of ethane will produce 4 moles of CO₂

Answer: 14.3 atm

Explanation: solution attached

The pressure of hydrogen gas in the tank is approximately 4.92 atm.

To find the pressure of the hydrogen gas, we need to use the ideal gas law equation: PV = nRT. Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

To convert the temperature from Celsius to Kelvin, we add 273 to the given temperature. Now, let's plug in the values: P(22.9 L) = (14.0 mol)(0.0821 L·atm/mol·K)(12°C + 273). Solving this equation for P, the pressure of the gas is approximately 4.92 atm.

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Answer:

The sodium chloride serves to pull out the water from the organic layer to the aqueous or water layer

Explanation:

Shaking it with sodium chloride also called brine pulls the water away from the organic composition to the water layer due to the affinity of the salt to absorb more water and become less concentrated and dense as the salt has more ability to absorb the water than to absorb organic compounds

Answer:

Option (D)

Explanation:

Antarctica is located in the south pole and covers a large area of ice-covered region. It is an extremely cold environment, having a low freezing temperature. The glaciers and ice-bergs of this region has been constantly affected due to the increasing global surface temperature. This rise in the surface temperature of the earth is due to the increasing concentration of CO₂ in the atmosphere. One of the reasons for this temperature increase is also due to the introduction of the industrial revolution, which is responsible for the emission of a large number of harmful gases, including carbon into the atmosphere. Due to this global climate change, these glaciers are melting at a faster rate, resulting in the rise of the sea level.

Thus, the Antarctic ice sheets have been calving off large pieces of icebergs over the last few decades.

Hence, the correct answer is option (D), which is the true statement.

Answer:

c. ) polyethylene with high density plasticizers added to increase density.

Explanation:

High - density polyethylene -

Polyethylene is a crystalline structure , which have a very wide area of application , it is a type of thermoplastic polymer .

It is produced in huge tons every year due to its wide range of application.

Commercially it is produced by the very famous , the Ziegler - Natta catalysts .

In the polyethylene , a high - density polyethylene is added in order increase its density , and hence this type of polyethylene is referred to as high - density polyethylene .

High-density polyethylene (HDPE) is composed of primarily linear, unbranched chains of polyethylene in a close packing arrangement. This structure gives HDPE a higher tensile strength and makes it suitable for stronger and more rigid products.

High-density polyethylene (HDPE) is composed of primarily linear, unbranched chains of polyethylene in a close packing arrangement. Therefore, the correct answer is b). HDPE has a relatively low degree of branching, which allows the polymer chains to pack closely together, resulting in stronger intermolecular forces and a higher tensile strength compared to more highly branched polymers like low-density polyethylene (LDPE). HDPE's characteristics make it suitable for use in products that require strength and rigidity, such as detergent bottles, milk jugs, and water pipes.

In contrast, LDPE has a higher degree of branching, which prevents the polymer chains from packing as tightly, leading to a more flexible material with a lower tensile strength. This type of polyethylene is used for products that need to be more flexible, such as beach balls and plastic bags. HDPE and LDPE are both types of polyethylene, but they have different properties due to their molecular structure and how the polymer chains pack together.

Answer:

phospholipids

Explanation:

As regards the lipid membrane of the first protocells, it is most likely that it initially consisted of simpler fatty acids than the phospholipids that make up the current membranes (see appendix). If the modern membranes are bilayers of glycerol phospholipids, the primitive membranes would probably be made up of simpler, single-chain molecules, also amphiphilic in nature (with a soluble part and another insoluble in water), such as monocarboxylic acids or alcohols. The origin of these compounds could be multiple. On the one hand, it has been seen that they are very abundant in meteorites of the type of carbonaceous chondrites, so they could have arrived on Earth already formed from outer space. But it is also possible that they were formed abioticly on the primitive Earth by the reaction of CO and hydrogen to give rise to various hydrocarbons, a reaction that would be viable at high temperatures in the presence of ferric catalysts, on the surface of montmorillonite clays and also in hydrothermal conditions. Regardless of the origin, the result would be the presence of fatty acids initially very diluted in an aqueous solution, but which would be concentrated by successive evaporation cycles, or by the formation of small aerosolized drops that would also transfer those vesicles to points distant from the place where The first organic membrane compounds that formed on Earth were formed and would be.

Answer:

Y=X

Explanation:

The number of moles of acid and base reacting with each other is the same in both trials.

In the neutralization reaction between HCl and NaOH, a one-to-one mole ratio leads to a predictable enthalpy change (ΔHrxn). The temperature increase signifies an exothermic reaction, and ΔHrxn can be calculated using the formula q=mcΔT, considering the reaction's exothermic nature.

The relationship between X and Y refers to the amount of heat (ΔH) released during the neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH). In this reaction, a one-to-one mole ratio is observed, meaning 1 mole of HCl reacts with 1 mole of NaOH to produce 1 mole of sodium chloride (NaCl) and water (H₂O)

When calculating the enthalpy change, the student must use the temperature change, the mass of the solution, and the specific heat capacity (which is assumed to be the same as water's for this type of experiment). The amount of heat produced (ΔHrxn) can be calculated using the formula q = mcΔT, where 'm' is the mass of the solution, 'c' is the specific heat capacity, and 'ΔT' is the change in temperature.

In a typical calorimetry experiment, the student would need to carefully measure the temperature changes and use stoichiometry to relate the amounts of the reactants to the heat released. If the temperature increases, as in the example provided, it indicates that the reaction is exothermic, and the value for ΔHrxn would be negative, signifying that heat is released into the surroundings.

The final answer is:

[tex]\[ \Delta S^\circ_{\text{rxn}} = 1716.72 \, \text{J/mol-K} \][/tex]

Let's calculate the energy change of the reaction using the given data:

Reaction equation:

[tex]\[ \text{Sn}(s) + 2\text{Cl}_2(g) \rightarrow \text{SnCl}_4(l) \][/tex]

Given:

[tex]\[ \Delta H^\circ_{\text{rxn}} = -511.3 \, \text{kJ/mol} \]\\Temperature (\( T \)) = \( 25^\circ \text{C} \) = \( 298.15 \, \text{K} \)\\Pressure (\( P \)) = \( 1.00 \, \text{bar} \)\\\\[/tex]

First, we convert[tex]\( \Delta H^\circ_{\text{rxn}} \)[/tex] from kJ/mol to J/mol:

[tex]\[ \Delta H^\circ_{\text{rxn}} = -511.3 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = -511300 \, \text{J/mol} \][/tex]

Next, we use the formula for Gibbs Free Energy [tex](\( \Delta G^\circ_{\text{rxn}} \)):[/tex]

[tex]\[ \Delta G^\circ_{\text{rxn}} = \Delta H^\circ_{\text{rxn}} - T \Delta S^\circ_{\text{rxn}} \][/tex]

At standard conditions [tex](\( 25^\circ \text{C} \) and \( 1.00 \, \text{bar} \)), \( \Delta G^\circ_{\text{rxn}} \)[/tex] is equal to zero:

[tex]\[ 0 = -511300 \, \text{J/mol} - (298.15 \, \text{K}) \times \Delta S^\circ_{\text{rxn}} \][/tex]

Solving for[tex]\( \Delta S^\circ_{\text{rxn}} \):[/tex]

[tex]\[ \Delta S^\circ_{\text{rxn}} = \frac{-(-511300 \, \text{J/mol})}{298.15 \, \text{K}} \]\[ \Delta S^\circ_{\text{rxn}} = 1716.72 \, \text{J/mol-K} \][/tex]

So, the final answer is:

[tex]\[ \Delta S^\circ_{\text{rxn}} = 1716.72 \, \text{J/mol-K} \][/tex]

1. Convert [tex]\( \Delta H^\circ_{\text{rxn}} \)[/tex]  from kJ/mol to J/mol.

2. Use the formula [tex]\( \Delta G^\circ_{\text{rxn}} = \Delta H^\circ_{\text{rxn}} - T \Delta S^\circ_{\text{rxn}} \) and solve for \( \Delta S^\circ_{\text{rxn}} \) at standard conditions where \( \Delta G^\circ_{\text{rxn}} = 0 \).[/tex]

Complete Question:

The standard internal energy change for a reaction can be symbolized as Δ U ∘ rxn or Δ E ∘ rxn . For each reaction equation, calculate the energy change of the reaction at 25 ∘ C and 1.00 bar . Sn ( s ) + 2 Cl 2 ( g ) ⟶ SnCl 4 ( l ) Δ H ∘ rxn = − 511.3 kJ/mol

The standard internal energy change for the reaction is approximately [tex]\( -511.3 \text{ kJ/mol} \).[/tex]

The standard internal energy change for the given reaction at 25°C and 1.00 bar can be calculated using the standard enthalpy change, H°rxn, provided that the reaction occurs at constant pressure and the only work done is pressure-volume work. Under these conditions, the change in internal energy (U°rxn) can be approximated by the change in enthalpy (Hrxn), because the difference between H°rxn and U°rxn is the product of the pressure, volume change, and the number of moles of gas, which is often small for reactions that do not involve a significant amount of gas.

 The reaction is as follows:

[tex]\[ \text{Sn (s)} + 2\text{Cl}_2 \text{(g)} \rightarrow \text{SnCl}_4 \text{(l)} \][/tex]

 Given:

[tex]\[ \Delta H^\circ_{\text{rxn}} = -511.3 \text{ kJ/mol} \][/tex]

At constant pressure (1.00 bar) and temperature (25°C), the relationship between Hrxn and U°rxn is given by:

[tex]\[ \Delta H^\circ_{\text{rxn}} = \Delta U^\circ_{\text{rxn}} + P\Delta V \][/tex]

 For reactions involving gases, the term [tex]\( P\Delta V \)[/tex] represents the work done by the system on the surroundings due to volume change against the constant external pressure \( P \). Since the number of moles of gas decreases during the reaction (2 moles of Cl2 gas are consumed to form 1 mole of SnCl4 liquid), \( \Delta V \) is negative, and thus \( P\Delta V \) is also negative.

However, for condensed phases (solids and liquids), the volume change is typically small, and the [tex]\( P\Delta V \)[/tex] term is often negligible compared to the enthalpy change. Therefore, for the reaction given, which involves a solid reactant and a liquid product, we can assume that [tex]\( \Delta H^\circ_{\text{rxn}} \approx \Delta U^\circ_{\text{rxn}} \).[/tex]

 Thus, the standard internal energy change for the reaction is approximately equal to the standard enthalpy change:

[tex]\[ \Delta U^\circ_{\text{rxn}} \approx \Delta H^\circ_{\text{rxn}} \][/tex]

[tex]\[ \Delta U^\circ_{\text{rxn}} \approx -511.3 \text{ kJ/mol} \][/tex]

 Therefore, the standard internal energy change for the reaction is approximately [tex]\( -511.3 \text{ kJ/mol} \).[/tex]

 In conclusion, the energy change of the reaction at 25°C and 1.00 bar is:

[tex]\[ \boxed{\Delta U^\circ_{\text{rxn}} \approx -511.3 \text{ kJ/mol}} \][/tex]

This value is a good approximation for the standard internal energy change of the reaction under the given conditions.

 The answer is: [tex]\Delta U^\circ_{\text{rxn}} \approx -511.3 \text{ kJ/mol}.[/tex]

Answer:

A) relative deprivation

Explanation:

The given example is of the relative deprivation principle of sociology.

The Theory of Relative Deprivation. Throughout sociology, the principle of relative deprivation is a concept of societal change and campaigns, whereby people act for social changes in order to gain something (for example, privileges, prestige, or wealth) that others have and that they also feel they should have the same. Here Haley is feeling deprived of brand new bicycle as has a used one.

Answer:

Q = 1267720 J

Explanation:

∴ QH2O = mCpΔT

∴ m H2O = 500 g

∴ Cp H2O = 4.186 J/g°C = 4.183 E-3 KJ/g°C

∴ ΔT = 120 - 50 = 70°C

⇒ QH2O = (500 g)(4.183 E-3 KJ/g°C)(70°C) = 146.51 KJ

∴ ΔHv H2O = 40.7 KJ/mol

moles H2O:

∴ mm H2O = 18.015 g/mol

⇒ moles H2O = (500 g)(mol/18.015 g) = 27.548 mol H2O

⇒ ΔHv H2O = (40.7 KJ/mol)(27.548 mol) = 1121.21 KJ

⇒ Qt = 146.51 KJ + 1121.21 KJ = 1267.72 KJ = 1267720 J

Answer:

Remain the same

Explanation:

The law of mass conservation states that matter cannot be created nor destroyed but can be converted from one form to another.

Essentially, what Lavoisier was trying to proof is that by heating the mixture, after all the change the mass still remains. That was why he used a sealed flask. If the flask was not sealed, it probably would have been that some of the mass will escape as vapor to the atmosphere which might be difficult to account for

Answer:

                      %age Yield  =  66.05 %

Explanation:

                     In order to check the efficiency of a reaction the %age is calculated so that it can be concluded that either the conditions provided for certain reactions are favourable or not because at large scales reactions with greater %age yields are favoured as it results in high product quantity and hence, greater economical advantages.

                   The percent yield is given as;

                  %age Yield  =  Actual Yield / Theoretical Yield × 100

Putting values,

                   %age Yield  =  253 g / 383 g × 100

                   %age Yield  =  66.05 %

Hence, for the given reaction the success of the reaction is only 65% hence, steps can be made to improve this yield by modifying the reaction conditions.

Answer:

[KOH] = 1.47 M

[KOH] = 1.22 m

KOH = 6.86 % m/m

Explanation:

Let's analyse the data

1.87 L is the volume of solution

Density is 1.29 g/mL → Solution density

155 g of KOH → Mass of solute

Moles of solute is (mass / molar mass) = 2.76 moles.

Molarity is mol/L → 2.76 mol / 1.87 L = 1.47 M

Let's determine, the mass of solvent.

Molality is mol of solute / 1kg of solvent

We can use density to find out the mass of solution

Mass of solution - Mass of solute = Mass of solvent

Density = Mass / volume

1.29 g/mL = Mass / 1870 mL

Notice, we had to convert L to mL, cause the units of density.

1.29 g/mL . 1870 mL = Mass → 2412.3 g

2412.3 g - 155 g = 2257.3 g of solvent

Let's convert the mass of solvent to kg

2257.3 g / 1000 = 2.25kg

2.76 mol / 2.25kg = 1.22 m (molality)

% percent by mass = mass of solute in 100g of solution.

(155 g / 2257.3 g) . 100g = 6.86 % m/m

The molarity, molality, and mass percent concentration of the KOH solution are approximately 1.476 M, 1.224 m, and 6.43%, respectively.

The first step in solving this problem is to determine the number of moles of KOH in the solution. The molar mass of KOH is approximately 56.11 g/mol, so you can calculate the number of moles by dividing the mass of the KOH by its molar mass: moles of KOH = 155 g / 56.11 g/mol = 2.762 mol.

Molarity (M) is defined as the number of moles of solute per liter of solution. To find the molarity, divide the number of moles by the volume of the solution in liters: M = 2.762 mol / 1.87 L = 1.476 M.

To calculate molality (m), you need the mass of the solvent in kilograms. First, find the total mass of the solution: mass = volume x density = 1.87 L x 1.29 g/mL = 2411.3 g. Then, subtract the mass of the KOH to find the mass of the solvent: mass of solvent = 2411.3 g - 155 g = 2256.3 g = 2.2563 kg. Now, you can find the molality: m = 2.762 mol / 2.2563 kg = 1.224 m.

Finally, the mass percent concentration of the solution is the mass of the solute divided by the mass of the solution, multiplied by 100: mass percent = (155 g / 2411.3 g) x 100 = 6.43%.

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Answer:

0.0078 L of the stock solution is required to make up 1.00 L of 0.13 M [tex]HNO_{3}[/tex]

Explanation:

According to laws of equivalence, [tex]C_{1}V_{1}=C_{2}V_{2}[/tex]

where, [tex]C_{1}[/tex] and [tex]C_{2}[/tex] are initial and final concentration respectively. [tex]V_{1}[/tex] and [tex]V_{2}[/tex] are initial and final volume respectively.

Here, [tex]C_{1}=16.6M[/tex], [tex]C_{2}=0.13M[/tex] and [tex]V_{2}=1.00L[/tex]

So, [tex]V_{1}=\frac{C_{2}V_{2}}{C_{1}}[/tex]

or, [tex]V_{1}=\frac{(0.13M)\times (1.00L)}{(16.6M)}[/tex]

or, [tex]V_{1}=0.0078L[/tex]

Hence 0.0078 L of the stock solution is required to make up 1.00 L of 0.13 M [tex]HNO_{3}[/tex]

Answer:

57.1 kilo Joules of heat would be released.

Explanation:

[tex]HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O(l)[/tex] ΔH°=-57.1kJ/mol

Molarity of HCl = 2.00 M

Molarity of NaOH = 1.00 M

According to reaction , 1 M of HCl reacts with 1 M of NaOH. Then 2.00 M of HCl will react with:

[tex]\frac{1}{1}\times 2.00 M= 2M [/tex] of NaOH

But according to question we only have 1.00 M NaOH .So, this means that NaOH is limiting reagent and HCl is an excessive reagent.

Heat evolved will depend upon concentration of NaOH solution :

Heat evolved when 1.00 M of NaOH reacts =

[tex]1.00\times (-57.1 kJ/mol)=-57.1 kJ[/tex]

Negative sign means that heat is released during the reaction.

57.1 kilo Joules of heat would be released.

The heat released during the repeated experiment with 2 mole of HCl is 57.1 kJ.

The given reaction,

[tex]\bold {HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O(l) \ \ \ \ \ \ \ \ \ \Delta H = - 57.1\ kJ/mol}[/tex]

Concentration of HCl is 2 mole.

So, 2 mole of HCl react with 2 mole of NaOH but only mole of NaOH is available.

So, NaOH is a Limiting factor in the reaction and HCl is excessive factor.

Only, 1mole of HCl will react with available 1 mole of NaOH,

Therefore, the heat released during the repeated experiment with 2 mole of HCl is 57.1 kJ.

To know more about Enthalpy,

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Answer:

Silicon-28

Explanation:

To estimate the isotope with the highest percentage of abundance, we look at which of the isotopes have the closest resemblance to the value of the final atomic mass.

From what is given, it can be observed that the isotope with the highest value is that of the 28 because it can be seen that it is large enough to have affected the value of the isotope near it.

Answer:

Piruvate

Explanation:

Glucose + ATP32 --------> Glucose-6-P32 ----->Fructose-6-P32

Fructose-6-P32 + ATP32 ---------> Fructose-1,6 bi-P32

Fructose-1,6 bi-P32 ----------> Glyceraldehyde-3-P32 + Dihydroxiacetone Phosphate32 (that are in equilibrium)

Glyceraldehyde-3-P32 +NADH +Pi (non radioactive) -----> Glyceraldehyde-1,3-P32 (1,3 biphosphoglycerate where only in C3, there is a P32)

After that, the phosphorus in Carbon1 is donated to ADP to for ATP (non radioactive), and we have 3-Phosphoglycerate (radioactive because its P32), then it's converted to 2-Phosphoglycerate (radioactive), then Phosphoenolpiruvate (radioactive), that donates its P32 to ADP to produce ATP, remaining Piruvate as end product at the end of glucolysis

Answer:

30.10%

Explanation:

The mass of water is:

50.00 g − 34.95 g = 15.05 g

So the percentage of water is:

15.05 g / 50.00 g × 100% = 30.10%

Final answer:

The experimental percentage of water in copper(II) sulfate pentahydrate, after heating a sample and measuring the mass loss, is calculated to be 30.10%.

Explanation:

Experimental Percentage of Water in a Hydrate

The question involves calculating the percentage of water in a hydrate, specifically copper(II) sulfate pentahydrate, which is a classic chemistry experiment. To determine the percentage of water, you have to compare the mass of water lost upon heating to the original mass of the hydrate.

First, calculate the loss of mass due to heating: 50.00 grams (initial mass) - 34.95 grams (final mass) = 15.05 grams of water lost.

Percentage of water in the hydrate is found by the formula:
Mass% of H₂O = (Mass of water lost / Original mass of hydrate) * 100%.
Substituting the known values gives us: Mass% of H₂O = (15.05 g / 50.00 g) * 100% = 30.10%.

The experimental percentage of water in the copper(II) sulfate pentahydrate is therefore 30.10%, which is determined by this experimental process of heating and weighing.

Final answer:

An exothermic dissolution process suggests that A-B attractive forces are significant enough to overcome A-A and B-B intermolecular forces, resulting in energy being released as heat.

Explanation:

When a solution is made by dissolving solute B in solvent A and the solution process is exothermic, it indicates that the energy released in forming A-B attractive forces is greater than the energy required to overcome both A-A and B-B attractive forces. Since the process is exothermic, this means that the solute-solvent attractions are strong enough to not only break the solute-solute and solvent-solvent interactions but also provide excess energy that is released as heat. If the energy required to separate the solute and solvent was greater than the energy released upon mixing, an endothermic reaction would occur, and the solutes might not dissolve.

Answer:Ba^2+ is a solid because the concerntration ofSO4 ion for the separation is 3.0×10^4M

Explanation: The equation for the reaction are:

1. BaSO4 ---->Ba^2+ SO4^2-

Ksp=1.1×10^-10

2.CaSO4 ------> Ca^2+ + SO4^2-

Ksp=2.4×10^-5

/Ba^2+/=0.150M

/Ca^2+/=0.080M

Calcuim sulfate is more soluble than barium sulfate, therefore add SO4^2- ion to leaveCa^2+ in the solution and precipitate Ba^2+ ions.

SO4^2-= 2.4×10^-5/0.080

SO4^2-=3.0×10^-4 M

ForCa^2+ ion to be left in the solution,reaction quotient must be less than the Ksp

Q=[Ca^2+][SO4^2-] <ksp

SO4^2- =Ksp/[Ca^2+]

Answer: The volume of acid and water that must be mixed will be 4.8 L and 11.2 L

Explanation:

We are given:

Volume of mixture = 16 L

Percent of acid present = 30 %

Calculating the percentage of acid present in the mixture:

[tex]\Rightarrow 16\times \frac{30}{100}=4.8L[/tex]

The mixture is made entirely of acid and water.

Volume of acid in the mixture = 4.8 L

Volume of water in the mixture = 16 - 4.8 = 11.2 L

Hence, the volume of acid and water that must be mixed will be 4.8 L and 11.2 L

Final answer:

The chemist needs 4.8 liters of pure acid and 11.2 liters of water to prepare 16 liters of a solution that is 30% acid.

Explanation:

To prepare 16 liters of a solution that is 30% acid, we start by setting up an equation to represent the mix of pure acid and water. Let the amount of pure acid needed be x liters, and therefore the amount of water would be 16 - x liters. Since the solution is 30% acid, we can write the equation as follows:

0.30 × 16 = x

When we solve for x, we get:

4.8 = x

This means we need 4.8 liters of pure acid and 16 - 4.8 = 11.2 liters of water to make the 30% acid solution.

Answer:

yes because it is a great planet

Explanation:

because of the reason for silicon

Final answer:

The dissociation reactions for acetic acid, hexaaquacobalt(III) ion, and methylammonium along with their Ka equilibrium expressions show how each acid disassociates into its constituent ions in water, and how the concentration of these ions at equilibrium can be represented.

Explanation:

Completing the dissociation reaction and the Ka equilibrium expression for each of the following acids in water:

(A) Acetic acid [tex]($\text{HC}_2\text{H}_3\text{O}_2$)[/tex]

[tex]\[\text{HC}_2\text{H}_3\text{O}_2(\text{aq}) \rightleftharpoons \text{H}^+(\text{aq}) + \text{C}_2\text{H}_3\text{O}_2^-(\text{aq})\\\\K_a = \frac{[\text{H}^+][\text{C}_2\text{H}_3\text{O}_2^-]}{\text{HC}_2\text{H}_3\text{O}_2}\][/tex]

(B) Hexaaquacobalt(III) ion [tex]($\text{Co(H}_2\text{O)}_6^{3+}$)[/tex]

[tex]\[\text{Co(H}_2\text{O)}_6^{3+}(\text{aq}) \rightleftharpoons \text{H}^+(\text{aq}) + \text{Co(H}_2\text{O)}_5\text{OH}_2^+(\text{aq})\\\\K_a = \frac{[\text{H}^+][\text{Co(H}_2\text{O)}_5\text{OH}_2^+]}{\text{Co(H}_2\text{O)}_6^{3+}}\][/tex]

(C) Methylammonium[tex]($\text{CH}_3\text{NH}_3^+$)[/tex]

[tex]\[\text{CH}_3\text{NH}_3^+(\text{aq}) \rightleftharpoons \text{H}^+(\text{aq}) + \text{CH}_3\text{NH}_2(\text{aq})\\\\K_a = \frac{[\text{H}^+][\text{CH}_3\text{NH}_2]}{\text{CH}_3\text{NH}_3^+}\][/tex]

Answer:

omework Help. Steno's laws of stratigraphy describe the patterns in which rock layers are deposited. The four laws are the law of superposition, law of original horizontality, law of cross-cutting relationships, and law of lateral continuity.

Explanation:

Answer: 451.9g

Explanation:

Weight of empty flask= 450g

Weight of sugar= 1.9g

Weight of sugar+flask=450+1.9=451.9g

The sum of the massed of both the sugar and the flask is the mass the balance will read after the addition of the sugar.

Answer:

[tex][H^{+}] = 0.761 \frac{mol}{L}[/tex]

[tex][OH^{-}]=1.33X10^{-14}\frac{mol}{L}[/tex]

[tex]pH = 0.119[/tex]

Explanation:

HCl and HNO₃ both dissociate completely in water. A simple method is to determine the number of moles of proton from both these acids and dividing it by the total volume of solution.

[tex]n_{H^{+} } from HCl = [HCl](\frac{mol}{L}). V_{HCl}(L) \\ n_{H^{+} } from HNO_{3} = [HNO_{3}](\frac{mol}{L}). V_{HNO_{3}}(L)[/tex]

Here, n is the number of moles and V is the volume. From the given data moles can be calculated as follows

[tex]n_{H^{+} } from HCl = (5.00)(0.093)[/tex]

[tex]n_{H^{+} } from HCl = 0.465 mol[/tex]

[tex]n_{H^{+} } from HNO_{3} = (8.00)(0.037)[/tex]

[tex]n_{H^{+} } from HNO_{3} = 0.296 mol[/tex]

[tex]n_{H^{+}(total) } = 0.296 + 0.465[/tex]

[tex]n_{H^{+}(total) } = 0.761 mol[/tex]

For molar concentration of hydrogen ions:

[tex][H^{+}] = \frac{n_{H^{+}}(mol)}{V(L)}[/tex]

[tex][H^{+}] = \frac{0.761}{1.00}[/tex]

[tex][H^{+}] = 0.761 \frac{mol}{L}[/tex]

From dissociation of water (Kw = 1.01 X 10⁻¹⁴ at 25°C) [OH⁻] can be determined as follows

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

[tex][OH^{-}]=\frac{Kw}{[H^{+}] }[/tex]

[tex][OH^{-}]=\frac{1.01X10-^{-14}}{0.761 }[/tex]

[tex][OH^{-}]=1.33X10^{-14}\frac{mol}{L}[/tex]

The pH of the solution can be measured by the following formula:

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

[tex]pH = -log(0.761)[/tex]

[tex]pH = 0.119[/tex]

Answer:

The rate constant for a second order reaction is

k = 0.51 dm-3 s-1.

Explanation:

In a regular second-order reaction the rate equation is given by v = k[A][B], if the reactant B concentration is constant then v = k[A][B] = k'[A], where k' the pseudo–first-order rate constant = k[B].

Also 1/|A| = Kt + 1/|Ao|

But Ao = 0.657 M and

A = 0.0981 M also

t = 17.0 s

Therefore

1/| 0.0981 M| =K × 17.0 s + 1/| 0.657 M|

→ 10.19/M = 17K + 1.52/M

10.19/M - 1.52/M = 8.67/M = 17K

K = 8.67M/17s = 0.51 dm-3 s-1.

k = 0.51 dm-3 s-1.

Answer:

The answer to your question is Volume = 214.3 ml

Explanation:

Data

mass = 120 g of FeCl₂

concentration = 0.01 M

volume = ?

Formula

Molarity = [tex]\frac{number of moles}{volume}[/tex]

Solve for volume

Volume = [tex]\frac{number of moles}{molarity}[/tex]

Process

1.- Convert grams to moles

Atomic weight = 56 g

                            56 g of Fe --------------- 1 mol

                         120 g of Fe  ----------------  x

                               x = (120 x 1) / 56

                               x = 2.14 moles

2.- Calculate the volume

Volume = [tex]\frac{2.14}{0.01}[/tex]

Volume = 214.3 ml

Answer:

We need 94.67 liters if a 0.01 M solution

Explanation:

Step 1: Data given

Molarity of the solution = 0.01 M

Mass of FeCl2 = 120.0 grams

Molar mass FeCl2 = 126.75 g/mol

Step 2: Calculate moles of FeCl2

moles FeCl2 = massFeCl2 / molar mass FeCl2

Moles FeCl2 = 120.0 grams / 126.75 g/mol

Moles FeCl2 = 0.9467 moles

Step 3: Calculate  volumes FeCl2

Molarity = moles / volume

Volume = moles / molarity

Volume = 0.9467 moles /0.01 M

Volume = 94.67 L

We need 94.67 liters if a 0.01 M solution