Concluding Remarks. Reflect on the experimental procedures you have undertaken and the possible sources of error. Then write a summary paragraph comparing your results, commenting upon the common ion effect and its influence on solubility, and commenting upon activity effects and their influence on the Ksp value.

Answer:

What does a particular point on a line of a phase diagram represent ?

Explanation:

The lines on the phase diagram represent combinations of pressures and temperatures at which two phases coexist in equilibrium.

The phase diagram's lines depict mixtures of pressures as well as temperatures where two phases can peacefully coexist.

A phase diagram shows a visual representation of particular chemical-physical states across various pressures and temperatures.

A phase would be an area of space where a molecule's physical characteristics have been all essentially uniform.

The phase diagram's lines depict mixtures of pressures as well as temperatures where two phases can peacefully coexist.

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Explanation:

its asking how will you explain to the people on what made the water rust? what have you learned about how water rust in your class? just explain it what you already know!!

Answer:

11 electrons

Explanation:

The number of electrons is the same as the number of protona in a ATOM.

They will most likely be different in a ion, simple bond or a convalent bond.

Answer:

To produce energy from food

Explanation:

It's pretty much an energy cycle

The number of moles of  Al₂ O₃ : 0.2

The reaction equation is the chemical formula of reagents and product substances  

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products  

Reaction

2Al(s) + 3FeO(s) ⇒ 3Fe(s) + Al₂ O₃(s)

0.6 mol of Fe = 0.6 mol of FeO

mol of Al₂ O₃ :  mol of FeO = 1 : 3

\(\tt \dfrac{1}{3}\times 0.6=0.2\)

Answer:

0.00011 JK.

The process does NOT violate the second law of thermodynamics

Explanation:

The following parameters are given which are going to help in solving for the change in entropy of the system. The term "entropy'' simply means the degree of disorderliness of a system.

=> The temperature of container A = 305 K, the temperature of container B = 295 K and the amount of heat generated when the containers are placed in contact with each other = 1. 1 J.

The change in entropy of the hot container = -(1/305) = - 0.00328 J/K.

The change in entropy of the cold container = 1/295 = 0.00339 J/K.

Therefore, the change in the entropy of the system = - 0.00328 J/K + 0.00339 J/K = 0.00011 JK.

Note that the change in entropy of the system gives a positive value. Hence, this process does not violate the second law of thermodynamics.

The process does NOT violate the second law of thermodynamics.

The pressure of the chlorine gas at the given condition is 1.7 atm.

The pressure of the chlorine gas at the given condition is calculated by applying ideal gas law.

PV = nRT

where;

The number of moles of 25 g of chlorine is calculated as follows;

n = m/M

n = 25/71

n = 0.352

The pressure of the chlorine gas at the given condition is calculated as;

P = nRT/V

P = (0.352 x 0.0821 x 263) / (4.5)

P = 1.7 atm

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Answer:

Ionic bonds

Explanation:

The force of attraction between the opposite charges of the ions in an ionic compound is called an ionic bond.

An ionic bond results from the electrostatic force of attraction between oppositely charged ions.

This attraction creates an ionic or electrovalent bond.

In the following experiment, a coffee-cup calorimeter containing 100 mL of \(H_{ 2} O\) is used. The initial temperature of the calorimeter is 23.0 ∘C. If 6.60 g of \(CaCl_{2}\) is added to the calorimeter, Final temperature of the solution in the calorimeter = 11.

The first step in solving this problem is to calculate the number of moles of \(CaCl_{2}\\\) added to the calorimeter.

Moles of \(CaCl_{2}\) = mass of \(CaCl_{2}\) / molar mass of \(CaCl_{2}\)

Moles of\(CaCl_{2}\) = 6.60 g / 110.98 g/mol (molar mass of \(CaCl_{2}\)

Moles of\(CaCl_{2}\) = 0.0594 mol

We can use the equation for heat transfer to find the change in temperature of the solution. q = mCsΔT, where q is the heat transferred, m is the mass of the solution, Cs is the specific heat of the solution, and ΔT is the change in temperature.

We know that the initial temperature of the calorimeter is 23.0 ∘C and the mass of the solution is 100 g (since the density of water is 1 g/mL). We can solve for ΔT: ΔT = q / mCs

To find q, we can use the enthalpy change of solution (ΔHsoln) and the number of moles of\(CaCl_{2}\)added: q = ΔHsoln x moles of\(CaCl_{2}\)

q = -82.8 kJ/mol x 0.0594 mol

q = -4.92 kJ

Now we can solve for ΔT: ΔT = (-4.92 kJ) / (100 g x 4.184 J/g⋅∘C)

ΔT = -11.8 ∘C

We can find the final temperature of the solution by adding the change in temperature to the initial temperature: Final temperature = 23.0 ∘C - 11.8 ∘C =11 ∘C.

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The percent by mass of the lithium in the compound is 18.78%. Option C

We know that the percentage has to do with the amount of the element that can be found in the compound. We can be able to obtain this when we  find the molar mass of the compound and then obtain the mass of the element in the compound.

Hence;

Molar mass of the compound = [2(7) + 12 + 3(16)]

= 14 + 12 + 48

= 74 g/mol

The we have the mass of the lithium in the compound as 14

Thus we then have;

Percent of lithium = 14/74 * 100/1

= 18.78%

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Answer:

1930g

Explanation:

We have been given that the salinity of sea water is 3.5%;

  Now the mass of sea water sample is 2000g

So, problem here is to find the quantity of pure water in the sample = ?

Solution:

Let us find the quantity of the salt in the sample;

       Salt content  = 3.5% x 2000  = 70g

Amount of pure water  = 2000 - 70 = 1930g

Answer:

69

Explanation:

Answer: KE =20000000 J

Explanation:

KE=1/2 MV^2

1/2 ( 1000) (200)^2

KE =20000000 J

Answer:

#1 : C

Explanation:

The velocity vs. time graph that corresponds to the info shown by the position vs. time graph is that the graph that has continuous acceleration (straight line).

The above statement of continuous acceleration means a relentless acceleration. A straight lines is just known to be a continuing acceleration that's said to be unchanging. Note that when there's a curved lines in a very graph, it implies that there's a non-constant acceleration. Therefore an object that's undergoing constant acceleration is one that may be seen in an exceedingly line. So therefore, the speed vs. time graph is one that matches with the information represented by the continual acceleration (straight line) on the graph.

Constant acceleration means that velocity changes at a constant rate with respect to time. If acceleration is made zero, velocity does not change. If acceleration is positive, the magnitude of velocity increases correspondingly.

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Answer:

your answer is c your welcome

Explanation:

Answer:

(25°C × 9/5) + 32 = 77°F

Explanation:

hope this helps

Answer:

Im 98% sure it's 77 F

the maximum amount of product that can be formed is 124.39 g SO₃, and there will be 36.8 g of excess O₂ left over.

To find the amount of excess reagent, you need to first determine which reactant is limiting and which is in excess.

Determine the limiting reagent:

Use stoichiometry to determine how much product can be formed from each reactant:

mol SO2:

2 SO₂ + O₂ -> 2 SO₃

2 mol SO₃/2 mol SO₂ = 1 mol SO₃/mol SO₂

1 mol SO₃ = 80.06 g SO₍₃₎

From 2.7 mol O₂

2 SO₂ + O₂ -> 2 SO₃

1 mol SO₃/1 mol O₂ = 1 mol SO₃/mol O₂

1 mol SO₃ = 80.06 g SO₃

2.7 mol O₂ x (1 mol SO₂/1 mol O₂) x (80.06 g SO₂/mol SO₂) = 216.45 g SO₂

Since the amount of SO₂ produced from 3.1 mol of SO₂ is less than the amount produced from 2.7 mol of O₂, SO₂ is the limiting reagent.

Calculate the amount of excess reagent:

To find the amount of excess O₂, use the balanced equation to determine how much O₂ is required to react with all of the SO₂:

2 SO₂ + O₂ -> 2 SO

3.1 mol SO2 x (1 mol O₂/2 mol SO2) = 1.55 mol O₂

Subtract the amount of O₂ used from the initial amount of O₂:

2.7 mol O₂ - 1.55 mol O2 = 1.15 mol O₂

Finally, convert the excess O₂ to mass:

1.15 mol O₂ x 32.00 g/mol = 36.8 g O₂

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my answer

Explanation:

Answer. Filtration is the method of separating a solid from a liquid. A sieve sets a threshold calibration through which all undersized materials pass through. Filtration differs from sieving, where separation occurs at a single perforated layer (a sieve).

Answer:

Explanation:

Caffeine is a weak base with pKb = 10.4

Kb = 10⁻¹⁰°⁴ = 3.98 x 10⁻¹¹

molecular weight of caffeine = 194.2

455 x 10⁻³ g / L = 455 x 10⁻³ / 194.2 moles / L

concentration of given solution a = 2.343 x 10⁻³ M

Let the caffeine be represented by B .

B    +   H₂O =  BH + OH⁻

a - x                   x        x  

x² / ( a - x ) = Kb

x² / ( a - x ) = 3.98 x 10⁻¹¹

x is far less than a so a -x is almost equal to a

x² = 3.98 x 10⁻¹¹ x 2.343 x 10⁻³ = 9.32  x 10⁻¹⁴

x = 3.05 x 10⁻⁷

[ OH⁻ ] = 3.05 x 10⁻⁷

pOH = - log ( 3.05 x 10⁻⁷ )

= 7 - log 3.05

= 7 - 0.484 = 6.5

pH = 14 - 6.5 = 7.5  

The pH of 455 mg/L of caffeine is 7.5

Using the formula;

Mass concentration = molar concentration × molar mass

Molar mass of caffeine = 194 g/mol

Mass concentration of caffeine = 455 mg/L

Molar concentration = Mass concentration/molar mass

Molar concentration = 455 × 10^-3g/L/194 g/mol

= 0.00235 M

Let Caffeine by depicted by the general formula BH

We can now set up the ICE table as follows;

           :B + H2O   ⇄ BH       +       OH^-

I      0.00235              0                      0

C           - x                  +x                    +x

E    0.00235  - x           x                      x

Note that water is present in large excess

Again; pKb of caffeine =10.4

Kb = Antilog[-pKb]

Kb = Antilog [-10.4]

Kb = 3.98 × 10^-11

Kb = [BH] [OH^-]/[:B]

3.98 × 10^-11 = [x] [x]/[ 0.00235  - x  ]

3.98 × 10^-11 [ 0.00235  - x  ] = [x] [x]

9.4 × 10^-14 - 3.98 × 10^-11x = x^2

x^2 + 3.98 × 10^-11x  - 9.4 × 10^-14 = 0

x = 3.1 × 10^-7 M

Recall  [BH] = [OH^-] = 3.1 × 10^-7 M

Now;

pOH = - log [OH^-]

pOH = log [3.1 × 10^-7 M]

pOH = 6.5

But;

pH + pOH = 14

pH = 14 - pOH

pH = 14 - 6.5

pH = 7.5

The pH of 455 mg/L of caffeine is 7.5

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Missing parts

Caffeine (C8H10N4O2) is a weak base with a pKb of 10.4.  Calculate the pH of a solution containing a caffeine concentration  of 455 mg/L.

Answer:

Explanation:

All three of these elements fall in group 7 of the periodic table and are called Halogens. Chlorine is the most reactive of these elements because it comes before the other two in the table then followed by Bromine and then Iodine.

Reactivity decreases in the group 7 elements as one moves to the latter elements due to the increase in shells surrounding the nucleus.

This is because reactivity happens as the protons in the nucleus attract electrons but when more shells are piled on, the strength of this attraction decreases as the distance increases such that they will not be as reactive anymore.

Answer:

79

Explanation:

When hydrogen bromide is added to 2-methylbut-2-ene, two products are expected to be produced: 2-bromo-2-methylbutane (major product) and 3-bromo-2-methylbutane (minor product).

The addition of HBr to 2-methylbut-2-ene follows the Markovnikov addition rule. This means that the hydrogen atom and the bromine atom will add to the carbon atoms in the double bond, such that the hydrogen atom adds to the carbon with the greater number of hydrogen atoms, and the bromine atom adds to the carbon with the lesser number of hydrogen atoms.

In this case, the hydrogen atom will attach to the second carbon atom, which has three hydrogen atoms, while the bromine atom will attach to the third carbon atom, which has only one hydrogen atom. This produces the major product, 2-bromo-2-methylbutane.

The formation of the minor product, 3-bromo-2-methylbutane, occurs due to the rearrangement of the carbocation intermediate formed during the addition reaction. The carbocation can rearrange either by shifting a methyl group from the second to the third carbon, or by shifting a hydrogen atom from the third to the second carbon. This rearrangement produces the minor product, 3-bromo-2-methylbutane.

In conclusion, the addition of HBr to 2-methylbut-2-ene produces two products: 2-bromo-2-methylbutane (major product) and 3-bromo-2-methylbutane (minor product). The major product forms due to Markovnikov addition rule, while the minor product forms due to carbocation rearrangement.

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Answer:

It is D low precision, high accuracy

Explanation:

I got it correct

The Xs show the location of beanbags a student tossed at a target. The student was aiming for the center circle. Low precision, high accuracy​  best describe the student's results. Therefore, option D is correct.

There are two ways to assess observational error: accuracy and precision. Precision measures how closely two measurements are to one another, whereas accuracy measures how close a group of measurements is to its actual value. In other words, precision is a measure of statistical variability and a description of random errors.

The proximity of two or more measurements to one another is referred to as precision. Using the aforementioned example, your measurement is extremely accurate if you weigh a certain substance five times and obtain 3.2 kg each time. Accuracy is not necessary for precision.

Regardless of whether or whether two or more measurements are correct, precision is described as "the property of being exact" and relates to how close two or more measurements are to one another.

Thus, option D is correct.

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Answer:

Beta emission measn -1e0

Here reactant is phosphorus-32.

phosphorus-32 --- >    product + -1e^0

product = 16 S^32

P-32 beta decays to S-32. On the beta emission the product is formed which has same number of mass but the atomic number more than 1 of the reactant.

The density of ammonia gas in the 6.0 L container at a pressure of 820 mm Hg is approximately 0.805 g/L.

To determine the density of ammonia gas (NH3) in a 6.0 L container at a pressure of 820 mm Hg, we need to use the ideal gas law equation, which relates pressure, volume, number of moles, and temperature for a given gas.

The ideal gas law equation is:

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 in Kelvin.

Since we are given the pressure (820 mm Hg), volume (6.0 L), and assuming standard temperature and pressure (STP), we can use the values for R (0.0821 L·atm/(mol·K)) and convert the pressure to atm by dividing by 760 (1 atm = 760 mm Hg).

820 mm Hg / 760 mm Hg/atm = 1.08 atm

Now we can rearrange the ideal gas law equation to solve for density (d):

d = (P * M) / (RT)

Where M is the molar mass of ammonia (NH3), which is approximately 17.03 g/mol.

Substituting the values, we have:

d = (1.08 atm * 17.03 g/mol) / (0.0821 L·atm/(mol·K) * 298 K)

Simplifying the equation, we find:

d ≈ 0.805 g/L

Therefore, the density of ammonia gas in the 6.0 L container at a pressure of 820 mm Hg is approximately 0.805 g/L.

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Answer:

See explanation

Explanation:

The molecular geometry of an atom is connected to the number of electron pairs that surround it(whether lone pairs or bonding pairs) as well as its hybridization state. We shall now examine the N, P, or S atoms in each of the following compounds.

a)

In H3PO4, P has a tetrahedral molecular geometry and is sp3 hybridized.

b) In NH4NO3

N is sp3 hybridized in NH4^+ and sp2 hybridized in NO3^-. Also, N is tetrahedral in NH4^+ but trigonal planar in NO3^-.

c) In S2Cl2, we expect a tetrahedral geometry but as a result of the presence of two lone pairs on each sulphur atom, the molecular geometry is bent. The sulphur is sp3 hybridized.

d) In K4[O3POPO3], each phosphorus atom is in a tetrahedral molecular geometry and is sp3 hybridized.