Solution Concentrations Making salt solutions at different concentration For the following problems, calculate the mass of solid needed to make the solution, and then prepare the solutions in the Virtual Lab. Use the solution viewer to verify that the solutions you created have the target concentrations. Explain the steps you took in making each solution and in verifying that it has the correct concentration.

1). 100 ml of solution that is 1.0 M in sodium cation, using sodium chloride.
2). 100 ml of a solution that is 0.7 M in chloride anion, using magnesium chloride.
3). 250 ml of a solution that is 0.34 M in sodium cation, using sodium sulfate decahydrate.
4). 500 ml of a solution that is 0.25 m in sulfate anion, using sodium sulfate decahydrate.

Answer:

Use a stethoscope....?

what are the answer choices?

1) The theoretical yield is 18.02 g

2) The percent yield is 67%

3) The percent yield is  55.8 %

The balanced reaction equation is;

N2 + 3H2 ----> 2NH3

Number of moles of N2 = 15g/28 g/mol = 0.53 moles

If 1 moles of N2 produced 2 moles of NH3

0.53 moles of N2 will produce 0.53 * 2/1

= 1.06 moles

Theoretical yield = 1.06 moles * 17 g/mol

= 18.02 g

The percent yield = 12.1/18.02 * 100/1

= 67%

The percent yield = 19.8 g/35.5 g * 100/1

= 55.8 % as shown

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After 8 hours, approximately 15,729 foodborne bacteria will be present in the culture.

To find the growth rate, k, we can use the given information. Using the uninhibited growth formula:

A = A₀ · \(e^(kt)\),

where A is the final amount, A₀ is the initial amount, t is the time, and e is the base of the natural logarithm.

Given that the initial amount is 100 bacteria (A₀ = 100) and the amount after 1 hour is 350 bacteria (A = 350), we can plug these values into the equation:

350 = 100 · \(e^(kt)\).

Simplifying the equation, we have:

\(e^(kt)\) = 3.5.   [Equation 1]

Now, we need to solve for k. Taking the natural logarithm (ln) of both sides of Equation 1:

ln(\(e^k\)) = ln(3.5),

k = ln(3.5).

Using a calculator, we find that k ≈ 1.2528.

Now, we can use the growth rate k to answer the question. Plugging the values into the uninhibited growth formula with t = 8 hours:

A = 100 · \(e^(1.2528 · 8)\).

Evaluating the equation, we find:

A ≈ 15,729.

Therefore, after 8 hours, approximately 15,729 foodborne bacteria will be present in the culture.

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The temperature at which the gas will reach a pressure of 7.30 atm is 1627 K.

We can use the combined gas law equation to solve this problem. The combined gas law states that for a fixed amount of gas, the pressure times the volume divided by the temperature is a constant:

P₁V₁/T₁ = P₂V₂/T₂

where P₁, V₁, and T₁ are the initial pressure, volume, and temperature, respectively, and P₂, V₂, and T₂ are the final pressure, volume, and temperature, respectively.

Because the volume of the gas remains constant, we can simplify the equation to:

P₁/T₁ = P₂/T₂

Rearranging for T₂, we get:

T₂ = (P₂/P₁) * T₁

Substituting the given values, we get:

T₂ = (7.30 atm / 3.30 atm) * 293 K

T₂ = 1627 K

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The temperature of the gas must be 649.47 K in order to reach a pressure of 7.30 atm at a constant volume of 3.0 L.

What is Temperature?

Temperature is a measure of the average kinetic energy of the particles in a substance or system. It determines the direction of heat flow between two objects in contact, with heat flowing from the object with a higher temperature to the one with a lower temperature until they reach thermal equilibrium. The SI unit of temperature is the kelvin (K), which is defined based on the triple point of water, where the temperature is 273.16 K. Other common temperature scales include Celsius and Fahrenheit.

We can use the combined gas law to solve this problem:

(P1/T1) = (P2/T2)

where P1, T1 are the initial pressure and temperature and P2, T2 are the final pressure and temperature.

We are given that P1 = 3.30 atm, T1 = 20°C + 273.15 = 293.15 K, P2 = 7.30 atm, and V is constant at 3.0 L.

Plugging in these values and solving for T2, we get:

(3.30/293.15) = (7.30/T2)

T2 = (7.30 * 293.15) / 3.30

T2 = 649.47 K

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The reaction between pottasium metal and chlorine gas is an example of combination reaction and the balanced equation is as follows: 2K + Cl₂ → 2KCl

A chemical equation is a symbolic representation of a chemical reaction where reactants are represented on the left, and products on the right.

A chemical equation is said to be balanced when the number of atoms of each element on both sides of the equation are the same.

According to this question, potassium metal reacts with chlorine gas to form solid potassium chloride. The balanced equation is as follows:

2K + Cl₂ → 2KCl

Based on the above equation, pottasium combines with chlorine chemically to form pottasium chloride compound, hence, it is an example of combination reaction.

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Answer:

By writing an experiment report

Explanation:

Answer: The initial temperature was 588 Kelvin

Explanation:

The combined gas equation is,

\(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\)

where,

\(P_1\) = initial pressure of gas = 1.49 atm

\(P_2\) = final pressure of gas = 657 torr = 0.864 atm   (760 torr = 1atm)

\(V_1\) = initial volume of gas = 6400 ml

\(V_2\) = final volume of gas = 9660 ml

\(T_1\) = initial temperature of gas = ?

\(T_2\) = final temperature of gas = \(245^0C=(245+273)K=515K\)

Now put all the given values in the above equation, we get:

\(\frac{1.49\times 6400}{T_1}=\frac{0.864\times 9660}{515}\)

\(T_1=588K\)

The initial temperature was 588 Kelvin

Answer:

D

Explanation:

The offspring of the RRYY x rryy cross, which is called the F1 generation, were all heterozygous plants with round, yellow seeds and the genotype RrYy. As you see, the letters (RrYy) gave uppercase, then lowercase

The molar solubility of zinc hydroxide is 1.73 × 10⁻⁸ M. This means that at equilibrium, the concentration of Zn²⁺ and OH⁻ ions in a saturated solution of Zn(OH)₂ is 1.73 × 10⁻⁸ M.

The solubility product constant (K_sp) for zinc hydroxide, Zn(OH)₂, can be expressed as:

K_sp = [Zn²⁺][OH⁻]²

At equilibrium, the concentration of Zn²⁺ and OH⁻ ions can be expressed as "s", so the K_sp expression becomes:

K_sp = s²(4s) = 4s³

Substituting the given K_sp value of 3.00 × 10⁻¹⁷ M³

into this equation gives:

3.00 × 10⁻¹⁷ = 4s³

Solving for "s" gives:

s = 1.73 × 10⁻⁸ M.

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Answer:

joules per gram per degree (J/goC) or calories per gram per degree (cal/goC).

Explanation:

Answer:

Joules, Kelvin, or calories

Explanation:

Answer:

option D is correct answer of this question

option D is correct answer of this questionbecause the rainforest environment is different than the desert environment

option D is correct answer of this questionbecause the rainforest environment is different than the desert environment hope it will help

The reason of why rainforest plants adapted to be a different shape  as compared to the desert plants is because of option d. as the rainforest environment is different than the desert environment.

It can be like:

hence,  The reason of why rainforest plants adapted to be a different shape as compared to the desert plants is because of option d. as the rainforest environment is different than the desert environment.

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Answer:

An insect would have an easier time walking on the surface of water than on the surface of ethanol.

Water's stronger intermolecular forces lead to higher surface tension.

Higher surface tension allows water to support the insect.

Answer:

An insect would have an easier time walking on the surface of water than on the surface of ethanol.

Water's stronger intermolecular forces lead to higher surface tension.

Higher surface tension allows water to support the insect.

Explanation:

Answer:

Benifits of scientific modeling

Explanation:

When students are engaged in scientific modeling, they are able to notice patterns and develop and revise representations that become useful models to predict and explain--making their own scientific knowledge stronger, helping them to think critically, and helping them know more about the nature of science.

Because calcium chloride pellets absorb moisture from the air rather than the reaction mixture, they cannot be employed to remove water from the esterification reaction.

Desiccant calcium chloride pellets are routinely used to absorb moisture from the air. They cannot, however, be utilized to remove water from an esterification reaction because their water absorption is not selective.

It means that, calcium pellets will absorb the moisture from the surroundings including the air. In contrast, azeotropic distillation employs a solvent to generate an azeotrope with water that can be removed from the reaction mixture, essentially removing water from the reaction.

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The Wilke-Chang method is used to predict the diffusion coefficient of Phosphoric acid in water. The Wilke-Chang method is a widely used empirical equation to estimate the diffusion coefficient of a solute in a solvent.

Fick's Law of diffusion states that the mass transfer rate of a solute across a porous membrane is proportional to the concentration gradient and the diffusion coefficient.

a. It takes into account the molecular weight, viscosity, and density of the solute and solvent. By plugging in the relevant values for Phosphoric acid and water, we can calculate the diffusion coefficient.

To calculate the diffusion coefficient of Phosphoric acid in water using the Wilke-Chang method, we need to know the molecular weight of Phosphoric acid (H3PO4) and water. The molecular weight of H3PO4 is 98 g/mol, and the molecular weight of water is 18 g/mol.

The Wilke-Chang equation is given by:

D = (1 / Φ) * [(1/M1 + 1/M2) / (√(1/μ1) + √(1/μ2))] * (T / P)

where D is the diffusion coefficient, Φ is the void fraction of the ceramic filter, M1 and M2 are the molecular weights of the solute and solvent, μ1 and μ2 are the viscosities of the solute and solvent, and T is the temperature in Kelvin.

b. We have the mass transfer rate and the thickness of the ceramic filter, so by rearranging Fick's Law equation, we can calculate the concentration at the other side of the filter.

The equation is given by:

J = -D * ∆C/∆x

where J is the mass transfer rate, D is the diffusion coefficient, ∆C is the change in concentration, and ∆x is the thickness of the ceramic filter.

We have the mass transfer rate (1.4 x 10^(-9) kg mol H3PO4/s.m²) and the thickness of the ceramic filter (4.2 mm = 0.0042 m), so by rearranging the equation, we can calculate the change in concentration (∆C) and then use it to find the concentration at the other side of the ceramic filter.

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According to the research, the correct option is C. A scatterplot identifies the problem and its variables showing quantitative data that relate two variables.

It is a control and support tool that shows a geometric design that presents the existing links and the level of correlation between variables and how one variable influences another.

In this sense, given that it is a tool or graphic representation it is widely used in statistics, that helps to identify the possible association between two related sets of data, it seeks to correlate said variables in order to better control the process and improve it.

Therefore, we can conclude that according to the research, the correct option is C. A scatterplot identifies the problem and its variables showing quantitative data that relate two variables.

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The final temperature of the combined metals is approximately 31.7°C.

To solve this problem, we can use the principle of heat transfer between two objects in thermal contact, known as the heat equation:

q = m*c*ΔT

where q is the amount of heat transferred, m is the mass of the object, c is its specific heat capacity, and ΔT is the change in temperature.

Assuming that no heat is lost to the surroundings, we can set the heat gained by the iron equal to the heat lost by the gold:

mc*ΔT = m*c*ΔT

where the subscripts 'i' and 'g' refer to iron and gold, respectively.

\(final temperature = \frac{(mi ciTi+mgcgtg)}{(mici+mgcg)}\)

We get

\(final temperature = \frac{(1908*0.45*51.1+10.4*0.13*16.3)}{(19.8*0.45+10.4*0.13)}\)

                                = 31.7°C

As a result, the final temperature of the metals is approximately 31.7°C.

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The mean of the data is 4.

The formula of the mean is

Sum of all the observation÷ Total number of observations

So, According to the formula :

Sum of all the observations= 4+1+3+9+6+3+2+4 = 32

The number of observations is 8

So,

32÷8

= 4

Hence, the mean of the data is 4.

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Answer:

The answer is false.

Explanation:

Precipitation is the process where water released from clouds in the form of rain, freezing rain, sleet, snow, or hail. It is the primary connection in the water cycle that provides for the delivery of atmospheric water to the Earth.

Grams per liter of sulfur

To determine the concentration in terms of grams per liter of sulfur, we need to consider the atomic mass of sulfur in the sulfate ion (SO4^2-). The atomic mass of sulfur (S) is approximately 32.06 g/mol.

The molar ratio between sulfate and sulfur is 1:1, which means that one sulfate ion contains one sulfur atom. Therefore, the concentration of sulfur in grams per liter is also 1.5 g/L.

Molar concentration of sulfate:

To calculate the molar concentration of sulfate, we need to convert the given mass concentration (1.5 g/L) to moles per liter. The molar mass of sulfate (SO4^2-) is approximately 96.06 g/mol.

Molar concentration (in mol/L) = Mass concentration (in g/L) / Molar mass (in g/mol)

Molar concentration of sulfate = 1.5 g/L / 96.06 g/mol ≈ 0.0156 mol/L

Therefore, the molar concentration of sulfate in the estuarine water sample is approximately 0.0156 mol/L.

Normality:

Normality is a measure of the concentration of reactive species in a solution, taking into account the number of equivalents of the species. In the case of sulfate (SO4^2-), it has a charge of 2-.

Normality (in N) = Molar concentration (in mol/L) × Equivalent factor

Since sulfate has a charge of 2-, the equivalent factor is 2. Therefore, the normality of the sulfate ion is:

Normality = 0.0156 mol/L × 2 = 0.0312 N

Hence, the normality of the sulfate in the estuarine water sample is approximately 0.0312 N.

Parts per million (ppm) sulfate:

Parts per million is a ratio of the mass of a solute to the mass of the solution, multiplied by one million.

Parts per million (ppm) = (Mass of solute / Mass of solution) × 10^6

In this case, the mass of solute is given as 1.5 g/L (since sulfate concentration is 1.5 g/L). The mass of the solution is assumed to be equal to 1 liter.

Parts per million sulfate = (1.5 g/L / 1 g/L) × 10^6 = 1,500,000 ppm

Therefore, the concentration of sulfate in the estuarine water sample is approximately 1,500,000 ppm.

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Answer:

0 m/s

Explanation:

An object at rest stays at rest unless acted upon therefore it can only be 0.

Unlike the leaf cells, the root cells have no chloroplasts because root cells do not need to produce or use energy. Option A

We have to note that the cell is the smallest part of an organism that can have independent existence. We know that the microscope is the device that we can be able to use to view the cells.

Now, we know that the cell is composed of the organelles of the cell. The organelles of the cell include the chloroplast of the cell as well as the cell wall of the cell.

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Missing parts;

The image below shows two types of plant cells under a microscope.

root cells and leaf cells

Which statement explains the difference in chloroplasts between the two types of plant cells?

A.

Unlike the leaf cells, the root cells have no chloroplasts because root cells do not need to produce or use energy.

B.

Unlike the root cells, the leaf cells have many chloroplasts because leaf cells produce sugars for the plant.

C.

The root cells have smaller chloroplasts because root cells do not need to store food and water.

D.

The leaf cells have larger chloroplasts because leaf cells need more support and a stronger structure.

Answer:

there are no valence electrons left over, so the molecule has four bond pairs and no lone pairs.

Explanation:

Answer:

to accelerate the reaction

Explanation:

In the context, a reaction is used at an extra reagent that is needed to push the copper through its cycle in the reaction because using an extra amount of copper or excess amount speed up the reaction and the products are proportionately formed. The extra copper acts as the catalyst and boost the reaction speed accelerating to from its products.

Answer:

outer or non - luminous

Explanation:

complete combustion of wax takes place in non-luminous zone of a candle flame.

Answer:

Explanation:

The rate enhancement that could be accomplished by the enzyme forming one low barrier hydrogen bond with transition state at 25 °C is 10⁷.

The decrease is about 5.7 kJ/mol that is observed in the free energy of the activation of the reaction when the 10 fold increase will occurs in the rate of the reaction at 25ºC.

The hydrogen bond  free energy = 40 kJ/mol.

Now, for the hydrogen bond, the times of the  10 fold increase

= (40 kJ/mol) / (5.7 kJ/mol)

= 7 times.

Hence, the rate that show the 10 fold increase 7 times. Therefore, the enhancement in the rate will be 10⁷.

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This question is incomplete, the complete question is :

calculate the rate enhancement that could be accomplished by an enzyme forming one low barrier hydrogen bond with transition state at 25 °C.

Answer:

c

Explanation:

Answer:

Boron and Carbon

Explanation:

They both are in Group 14 in the Periodic Table of Elements.

Kw of water is the product concentration of H+ and concentration OH- ions . The OH - concentration for a Kw of 1.995 ×10⁻¹³ is 4.46 × 10⁻⁷. Hence, the pOH is 6.35.

pOH of a solution is the measure of its OH- ion concentration as pH for the H+ ion concentration. pOH is the negative logarithm of the molar concentration of OH- ions.

Like pH, pOH also indicates the acidity or basicity of the solution.

The equilibrium constant Kw of water = [OH-] [H+]

for neutral water = [H+] = [OH-]

Then [OH-]² = kw = 1.995 ×10⁻¹³

[OH-] = √(1.995 ×10⁻¹³) = 4.46 × 10⁻⁷

pOH = - log [OH-]

        = - log (4.46 × 10⁻⁷)

        = 6.35.

Therefore, pOH of water at 75 degree celsius is 6.35.

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A sound wave generated by a musical note has the characteristics presented in the table. The missing value is 85.Therefore, option B is correct.

An example of an energy type that propagates through a medium using adiabatic loading and unloading is acoustic waves. Acoustic pressure, particle velocity, particle displacement, and acoustic intensity are crucial parameters for defining acoustic waves.

A sound source, such as the vibrating speaker's diaphragm in a stereo, produces the sound waves. Vibrations are produced in the medium surrounding the sound source. The medium continues to vibrate as the source does, and as they do, the vibrations move away from the source at the speed of sound, resulting in the formation of the sound wave.

A vibration that moves across a medium, such as air or water, is what we refer to as a sound in terms of energy. Even a gaseous medium can carry sound. There is something audible about it.

Thus, option B is correct.

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The statement which best describes the relationship between matter, pure substances, and mixtures among the following is ''Matter can be divided into two groups: pure substances and mixtures.'' Hence, option (2) is correct

A pure substance is a single kind of matter that cannot be separated into other kinds of matter by any physical means.

Matter can be divided into two groups that is, pure substances and mixtures.

Hence, option (2) is correct which best describes the relationship between matter, pure substances, and mixtures among the given statements.

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