Calculate the number of grams of xenon in 4.658 g of the compound xenon tetrafluoride.

Final answer:

The problem cannot be solved as presented because the initial energy flux of the black body at 255 K is not given. Without it, we cannot apply the Stefan-Boltzmann law to find the new temperature after adding 150 W/m^2.

Explanation:

The student is asking how to find the new temperature of a black body that was initially at 255 K after adding an energy flux of 150 W/m2. To solve this, we can apply the Stefan-Boltzmann law, which states that the energy radiated by a black body per unit area is directly proportional to the fourth power of the black body's temperature (E = σT4). The constant of proportionality, σ, is known as the Stefan-Boltzmann constant (σ = 5.67 x 10-8 W/m2K4).

where

q= heat flux = 155 W/m²+150 W/m² = 255 W/m²

σ= Stephan-Boltzmann constant = 5.67*10⁻⁸ W/m²K⁴

T= absolute temperature

T₀= absolute initial temperature = 255 K

solving for T

q = σ*(T⁴-T₀⁴)

T = (q/σ + T₀⁴)^(1/4)

replacing values

T = (q/σ + T₀⁴)^(1/4) = (255 W/m²/(5.67*10⁻⁸ W/m²K⁴) + (255 K)⁴)^(1/4) = 305.63 K

T=305.63 K

thus the final temperature is T=305.63 K

Answer:

CH3CH2CH(OH)CH3

Explanation:

This reaction follows the Markonikovs rule which states that the addition of an acid yo an assymetric alkene, the proton given off by the acid goes to the Carbon attached/with more hydrogen substituents and the rest of the acid ion goes to the Carbon with more alkyl substituents.

The reaction under goes 2 steps,

The first step called Oxymercuration involves the dissociation of Hg(OAc)2 into HgOAc+ and OAc-. The double bond in 1-butene (high electron site) bonds to an electron deficient HgOAc+ thereby forming an unstable complex, the lone pairs of oxygen in the water molecule (H2O) is attracted and a conplex is further formed with a proton on the now bonded H2O molecule.

To stabilise this, the proton(H+) is removed and then Deoxymercuration occurs. Deoxymercuration is simply removing the HgOAc+ molecule from the compound. This is done by the further reaction with the mixture of NaBH4 in alkaline medium; this is a string reducing reagent (reducting reagent - addition of excess Hydrogen) to form a secondary alcohol(Butan-2-ol)

Below us the mechanism in the attachment, i hope this was helpful.

Answer:

              Ethanol is completely miscible due to presence of Hydrogen bonding.

              Ethanethiol is partially miscible due to absence of Hydrogen Bonding.

Explanation:

                     The miscibility of liquids depend upon the intermolecular interactions between the two liquids. The stronger the intermolecular interactions the more miscible will be the liquids.

Among the two given examples, Ethanol is more miscible in water because it exhibits hydrogen bonding which is considered the strongest intermolecular interaction. Hydrogen bonding occurs when the hydrogen atom is bonded to more electronegative atoms like Fluorine, Oxygen and Nitrogen. In this way the hydrogen atom gets partial positive charge and the electronegative atom gets partial negative charge. Hence, these partial charges results in attracting the opposite charges on other surrounding atoms.

While, in case of Ethanethiol the hydrogen atom is not bonded to any high electronegative atom hence, there will be no hydrogen bonding and therefore, there will be less interactions between the neighbour atoms.

The difference in solubility between ethanol and ethanethiol in water can be explained by the difference in polarity and the ability to form hydrogen bonds with water molecules.

Ethanol (C2H5OH) has a hydroxyl (-OH) group, which is polar and can form hydrogen bonds with water molecules. The oxygen atom in the hydroxyl group has a partial negative charge due to its higher electronegativity, and the hydrogen atoms in water have a partial positive charge. This allows for the formation of hydrogen bonds between the ethanol molecules and water molecules, which overcomes the energy required to break the hydrogen bonding network in water. As a result, ethanol can mix with water in any proportion, making it completely miscible.

On the other hand, ethanethiol (C2H5SH) contains a thiol (-SH) group instead of a hydroxyl group. Although the thiol group can still form hydrogen bonds, the sulfur atom is less electronegative than the oxygen atom in ethanol. This results in a weaker hydrogen bond with water molecules. Additionally, the larger size of the sulfur atom compared to the oxygen atom leads to a greater Van der Waals radius, which can disrupt the hydrogen bonding network in water more than the smaller oxygen atom. As a result, ethanethiol is not as soluble in water as ethanol, and its solubility is limited to 1.5 grams per 100 milliliters of water.

In summary, the greater polarity and hydrogen bonding capability of ethanol with water molecules make it completely miscible, while the weaker hydrogen bonding and disruption of water's hydrogen bonding network by ethanethiol limit its solubility in water.

Answer: 54.5%

Explanation:Please see attachment for explanation

Answer: The concentration of calcium, magnesium, sodium and potassium are [tex]1.25\times 10^{-4}mol/L[/tex], [tex]4.20\times 10^{-5}mol/L[/tex], [tex]5.91\times 10^{-3}mol/L[/tex] and [tex]3.08\times 10^{-5}mol/L[/tex] respectively

Explanation:

To convert the mass from milligrams to grams, we use the conversion factor:

1 g = 1000 mg

To convert the given concentration fro grams to moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]    .....(1)

Concentration given = 5.0 mg/L = [tex]5.0\times 10^{-3}g/L[/tex]

We know that:

Molar mass of calcium element = 40 g/mol

Putting values in equation 1, we get:

[tex]\text{Concentration in mol/L}=\frac{5.0\times 10^{-3}g/L}{40g/mol}\\\\\text{Concentration in mol/L}=1.25\times 10^{-4}mol/L[/tex]

Concentration given = 1.0 mg/L = [tex]1.0\times 10^{-3}g/L[/tex]

We know that:

Molar mass of magnesium element = 24 g/mol

Putting values in equation 1, we get:

[tex]\text{Concentration in mol/L}=\frac{1.0\times 10^{-3}g/L}{24g/mol}\\\\\text{Concentration in mol/L}=4.20\times 10^{-5}mol/L[/tex]

Concentration given = 136 mg/L = [tex]136\times 10^{-3}g/L[/tex]

We know that:

Molar mass of sodium element = 23 g/mol

Putting values in equation 1, we get:

[tex]\text{Concentration in mol/L}=\frac{136\times 10^{-3}g/L}{23g/mol}\\\\\text{Concentration in mol/L}=5.91\times 10^{-3}mol/L[/tex]

Concentration given = 1.2 mg/L = [tex]1.2\times 10^{-3}g/L[/tex]

We know that:

Molar mass of potassium element = 39 g/mol

Putting values in equation 1, we get:

[tex]\text{Concentration in mol/L}=\frac{1.2\times 10^{-3}g/L}{39g/mol}\\\\\text{Concentration in mol/L}=3.08\times 10^{-5}mol/L[/tex]

Hence, the concentration of calcium, magnesium, sodium and potassium are [tex]1.25\times 10^{-4}mol/L[/tex], [tex]4.20\times 10^{-5}mol/L[/tex], [tex]5.91\times 10^{-3}mol/L[/tex] and [tex]3.08\times 10^{-5}mol/L[/tex] respectively

Final answer:

The effects on the equilibrium of converting carbon to methane through the reaction C(s) + 2H2(g) ⇌ CH4(g) depend on changes in the system. Adding more carbon or hydrogen, or lowering volume, favors the formation of methane, while raising the temperature favors reactants. Adding a catalyst speeds up equilibrium but doesn't affect its position.

Explanation:

Effects of Changes on a Chemical Equilibrium

The conversion of carbon to methane in the presence of hydrogen gas is described by the exothermic reaction C(s) + 2H2(g) ⇌ CH4(g). This process is subjected to changes that can affect its equilibrium according to Le Châtelier's principle:

Regarding which options favor CH4 at equilibrium:

Thus, both adding more H2 to the reaction mixture and lowering the volume of the reaction mixture would favor the formation of CH4 at equilibrium.

Adding more H₂ or lowering the volume of the reaction mixture will shift the equilibrium to favor CH₄ formation. Raising the temperature will decrease CH₄ production and adding a catalyst or more C will not change the equilibrium position of the reaction.

The reaction given, C(s) + 2H₂(g) ⇌ CH₄(g), is governed by the principles of chemical equilibrium. Let’s analyze each part:

Answer: 0.83mol/m3

Explanation:

Number of mole = 2.5 mmol = 0.0025mol

Recall

1L = 0.001m3

Therefore 3L = 3x0.001 = 0.003m3

0.003m3 contains 0.0025mol

Therefore, 1m3 will contain = 0.0025/0.003 = 0.83mol

The concentration in mol/m3 is 0.83mol/m3

The concentration of this solution in number of molecules per cubic meter is equal to 5 x 10^23 molecules.

To calculate the amount of molecules present in a given substance, one must use Avogadro's constant, which corresponds to:

                                     [tex]6 \times 10 ^{23}molecules =1mol[/tex]

For this question, we must first calculate the molarity, so that:

                                               [tex]\frac{2.5 \times 10^{-3}mol}{xmol} =\frac{3L}{1L}[/tex]

                                            [tex]x = 0.83 \times 10^{-3} mol/L[/tex]

After that, you must convert the unit of volume to cubic centimeters, that is, milliliters, so that:

                                   [tex]0.83 \times 10^{-3} mol/L = 0.83 mol/ml[/tex]

Finally, just multiply the value obtained by the avogadro constant:

                                [tex]0.83 \times 6\times 10^{23}= 5 \times 10^{23} molecules[/tex]

So, the concentration of this solution in number of molecules per cubic meter is equal to 5 x 10^23 molecules.

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Answer:

26.0 g/mol is the molar mass of the gas

Explanation:

We have to combine density data with the Ideal Gases Law equation to solve this:

P . V = n . R .T

Let's convert the pressure mmHg to atm by a rule of three:

760 mmHg ____ 1 atm

752 mmHg ____ (752 . 1)/760 =  0.989 atm

In density we know that 1 L, occupies 1.053 grams of gas, but we don't know the moles.

Moles = Mass / molar mass.

We can replace density data as this in the equation:

0.989 atm . 1L = (1.053 g / x ) . 0.082 L.atm/mol.K . 298K

(0.989 atm . 1L) / (0.082 L.atm/mol.K . 298K) = 1.053 g / x

0.0405 mol = 1.053 g / x

x =  1.053 g / 0.0405  mol = 26 g/mol

Answer:

The precautionary principle would prevent the implementation of technologies that possess a risk to humans, animals, and the environment. The strengths are that it will definately save the population and planet from a new technology that could cause long-term harm.

The weakness is that this principle may inhibit new technologies that are needed to help under-developed countries from preventing diseases. The precautionary principle states that technologies should entirely risk-free.

Explanation:

Final answer:

Filtration devices used in different situations include membrane filtration, gravity filtration, microfiltration, vacuum filtration, activated carbon filtration, and simple filtration.

Explanation:

For A, to remove powdered decolorizing charcoal from 20mL of solution, you can use membrane filtration with a syringe filter. It allows you to push the solution through by depressing the syringe's plunger.

For B, to collect crystals obtained from crystallizing a substance from about 1 mL of solution, you can use gravity filtration. It involves pouring the solution through a funnel with filter paper to separate the crystals.

For C, to remove a very small amount of dirt from 1 mL of liquid, you can use microfiltration. It uses a membrane filter with a pore size small enough to capture the dirt particles.

For D, to isolate 2.0 g of crystals from about 50 mL of solution after performing a crystallization, you can use vacuum filtration. It involves connecting a filtration unit to a vacuum to draw the solution through the filter.

For E, to remove dissolved colored impurities from about 3 mL of solution, you can use activated carbon filtration. It utilizes activated carbon to adsorb the impurities.

For F, to remove solid impurities from 5 mL of liquid at room temperature, you can use simple filtration. It involves pouring the liquid through a filter to separate the solid impurities.

Answer:

We need 4.06 J to heat of 406.0 mg of cyclohexane

Explanation:

Step 1: Data given

Mass of cyclohexane = 406.0 mg = 0.406 grams

The initial temperature = 33.5 °C

The final temperature = 38.9 °C

The specific heat of cyclohexane = 1.85 J/g*K

Step 2: Calculate the energy required to heat

Q = m*c*ΔT

⇒ with m = the mass of cyclohexane = 0.406 grams

⇒ with c = the specific heat of cyclohexane = 1.85J/g*K

⇒ with ΔT = The change of temperature = T2 - T1 = 38.9 - 33.5 = 5.4

Q = 0.406 g * 1.85J/g*K * 5.40

Q = 4.06 J

We need 4.06 J to heat of 406.0 mg of cyclohexane

To obtain accurate melting points on a Mel-Temp, grind the sample into a fine powder, put the smallest visible amount in a capillary tube, fill the tube with about 3mm of powder, set the voltage to rise quickly to 20 degrees below expected temperature and slow down to 2 degrees per minute, and record the temperature at which liquid first appears and the last crystal disappears.

The four important steps needed to obtain accurate melting points on a Mel-Temp are as follows:

By following these steps, you can ensure that you are determining the melting point accurately.

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Explanation:

Expression for rate of the given reaction is as follows.

             Rate = k[HgCl_{2}]x [C_{2}O^{2-}_{4}]y[/tex]

Therefore, the reaction equations by putting the given values will be as follows.

       [tex]1.8 \times 10^{-5} = k[0.105]x [0.15]y[/tex] ............. (1)

       [tex]7.2 \times 10^{-5} = k [0.105]x [0.30]y[/tex] ........... (2)

       [tex]3.6 \times 10^{-5} = k [0.0525]x [0.30]y[/tex] ............ (3)

Now, solving equations (1) and (2) we get the value of y = 2. Therefore, by solving equation (2) and (3)  we get the value of x = 1.

Therefore, expression for rate of the reaction is as follows.

     Rate = [tex]k[HgCl_{2}]x [C_{2}O^{2-}_{4}]y[/tex]

          Rate = [tex]k [HgCl2]1 [C_{2}O^{-2}_{4}]2[/tex]

Hence, total order = 1 + 2 = 3

According to equation (1),

               [tex]1.8 \times 10^{-5} = k[0.105]x [0.15]y[/tex]  

            [tex]1.8 \times 10^{-5} = k [0.105]1 [0.15]2[/tex]  

                      k = [tex]7.6 \times 10^{-3} M^{-2} min^{-1}[/tex]  

Thus, we can conclude that rate constant for the given reaction is [tex]7.6 \times 10^{-3} M^{-2} min^{-1}[/tex].

Final answer:

The kinetics of the reaction is first-order with respect to HgCl₂, second-order with respect to C₂O4²⁻, and third-order overall. The rate constant (k) is 7.6x10⁻³ M−²min−1, which can be calculated using the initial rates method from the given data.

Explanation:

The kinetics of the reaction 2HgCl₂(aq) + C₂O4²⁻(aq) → 2Cl–(aq) + 2CO₂(g) + HgCl₂(s) can be determined by using the method of initial rates. From the given data, we observe that when the concentration of HgCl₂ is halved (0.105 M to 0.0525 M), keeping the concentration of C₂O4²⁻ constant (0.30 M), the initial rate is also halved, indicating that the reaction is first-order with respect to HgCl₂. When the concentration of C₂O4²⁻ is doubled (0.15 M to 0.30 M), keeping the concentration of HgCl₂ constant (0.10⁵ M), the initial rate increases by a factor of four (1.8x10⁻⁵ to 7.2x10⁻⁵), suggesting the reaction is second-order with respect to C₂O4²⁻. Thus, the reaction is first-order with respect to HgCl₂ and second-order with respect to C₂O4²⁻, making it third-order overall.

To find the rate constant (k), we can choose any of the given rate data. For example, taking the data from the first experiment:

rate = k[HgCl₂][C₂O4²⁻]²
1.8x10⁻⁵ M/min = k(0.105 M)(0.15 M)²
k = 7.6x10⁻³ M−²min−1

Explanation:

The given data is as follows.

        Density of gasoline = 0.77 g/ml

         Volume of gasoline = 35 L = 35000 ml     (as 1 L = 1000 ml)

As we know that density of a substance is equal to its mass divided by its volume.

Mathematically,    Density = [tex]\frac{mass}{volume}[/tex]

Hence, calculate the mass of given gasoline as follows.

                  Density = [tex]\frac{mass}{volume}[/tex]

                         0.77 g/ml = [tex]\frac{mass}{35000 ml}[/tex]

                       mass = 26950 g

Also, it is given that 1 g of gasoline on combustion produces 45.0 kJ of energy.

Therefore, energy produced by 26950 g of combustion of gasoline will be as follows.

                 [tex]45.0 \times 26950[/tex]

                  = 1212750 kJ

Thus, we can conclude that the amount of energy produced by burning 35 L of gasoline is 1212750 kJ.

Answer:The result indicates that the polar component of the mixture fail to separate. In a case like this a third liquid like acid, base or complexing agent is often added to retain more water in the organic solvent.

Explanation:

Answer:

Seems to me that the sample had not been properly adsorbed, did not establish a true equilibrium, and therefore washed through the column rather that separated. The sample should be allowed to stand a short time before elution rather than putting the solvent in immediately.

Explanation:

Answer:

1. 4C₃H₅(NO₃)₃ (l) → 12CO₂(g) + 10H₂O(g) + O₂(g) + 6N₂(g)

2. 238 g of C₃H₅(NO₃)₃ has been reacted.

Explanation:

This is the chemical reaction:

4C₃H₅(NO₃)₃ (l) → 12CO₂(g) + 10H₂O(g) + O₂(g) + 6N₂(g)

For the second part, let's apply the Ideal Gases Law to find out the moles of CO₂ that were produced.

P . V = n . R . T

1 atm . 69L = n . 0.082 L.atm/mol.K . 268K

(1 atm . 69L) / (0.082 L.atm/mol.K . 268K) = n  → 3.14 moles

In the equation, ratio between nitroglycerin and CO₂ is 12:4.

12 moles of CO₂ were produced by 4 moles of C₃H₅(NO₃)₃

Then, 3.14 moles of CO₂ would have been produced by (3.14  .4) / 12 = 1.04 moles of C₃H₅(NO₃)₃

Let's convert the moles to mass, to find out the mass of nitroglycerin that must have reacted (mol . molar mass)

1.04 mol . 227.08 g/mol = 238 g

The molecularity of each reaction is determined by the number of molecules or atoms participating as reactants. The first is bimolecular, the second is termolecular, and the third is unimolecular.

Certainly, here's a more detailed explanation for each reaction:

1. The reaction F(g) + H2(g) → HF(g) + H(g) is bimolecular in nature. Bimolecular reactions involve two molecules or atoms as reactants that come together to form products. In this case, one molecule of fluorine (F2) is reacting with one molecule of hydrogen gas (H2) to produce one molecule of hydrogen fluoride (HF) and one hydrogen atom (H). The balanced equation reflects this stoichiometry: 1 F2 + 1 H2 → 2 HF + 1 H.

2. The reaction H2(g) + 2I(g) → 2HI(g) is termolecular. Termolecular reactions are relatively rare and involve three molecules or atoms as reactants. In this reaction, one molecule of hydrogen gas (H2) is reacting with two molecules of iodine gas (I2) to form two molecules of hydrogen iodide (HI). The balanced equation represents this as 1 H2 + 2 I2 → 2 HI. The presence of three molecules as reactants in this equation makes it termolecular.

3. NO2Cl(g) → NO2(g) + Cl(g) is a unimolecular reaction. Unimolecular reactions involve only one molecule or species as a reactant that decomposes or rearranges to form products. In this reaction, a single molecule of nitrogen dioxide chloride (NO2Cl) is decomposing into one molecule of nitrogen dioxide (NO2) and one chlorine atom (Cl). The equation 1 NO2Cl → 1 NO2 + 1 Cl illustrates the unimolecular nature of this reaction.

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Answer:

The alkene has Z configuration as shown in the figure.

The products of bromination of this compound are shown in the figure too with they stereochemistry.

Explanation:

The cis/trans nomenclature system in alkenes is insufficient when there are three or more different substituents in the double bond. In these cases, the Z/E nomenclature system, adopted by IUPAC, is used for all alkenes. Z comes from the German word zusammen which means together and E from the German word entgegen which means opposite. They would be equivalent to the terms cis and trans respectively.

If a molecular configuration is Z or E is determined by the priority rules of Cahn, Ingold and Prelog. For each of the two carbon atoms of the double bond, it is determined individually which of the two substituents has the highest priority. If both substituents of higher priority are on the same side, the arrangement is Z. On the other hand, if they are on opposite sides, the arrangement is E.

Halogenation of alkenes

Takes place with the addition of halogen atoms to the double bond to give a neighborhood dihaloalkane.

Halogenation mechanism

Halogenations are carried out at room temperature and in inert solvents such as carbon tetrachloride. In the mechanism it is observed that the opening of the bromonium ion occurs on the opposite side to the positive bromine that is the leaving group, this causes the halogens to be anti in the final product.

Answer : The molality of NaCl is, 1.95 mol/kg

Explanation :

Formula used for Elevation in boiling point :

[tex]\Delta T_b=i\times k_b\times m[/tex]

where,

[tex]\Delta T_b[/tex] = change in boiling point = [tex]2^oC[/tex]

[tex]k_b[/tex] = boiling point constant  for water = [tex]0.512^oC/m[/tex]

m = molality

i = Van't Hoff factor = 2 (for electrolyte)

The dissociation [tex]NaCl[/tex] will be,

[tex]NaCl\rightarrow Na^++Cl^{-}[/tex]

So, Van't Hoff factor = Number of solute particles = [tex]Na^++Cl^{-}[/tex] = 1 + 1 = 2

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

[tex]2^oC=2\times (0.512^oC/m)\times m[/tex]

[tex]m=1.95mol/kg[/tex]

Therefore, the molality of NaCl is, 1.95 mol/kg

The molality of NaCl is, 1.95 mol/kg

[tex]\Delta T _ b = i \times k_b \times m\\\\2^{\circ}C = 2\times (0.512^{\circ} C/m) \times m[/tex]

m = 1.95 mol/kg

Here m represent the molarity

[tex]\Delta T_b[/tex] represents a change in boiling point i.e. [tex]2^{\circ}C[/tex]

[tex]k_b[/tex] represents the constant boiling point for water  

i = Van't Hoff factor = 2 (for electrolyte)

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Answer:

This is due to the physical properties of the sample, since it affects the volume dispensed.

Explanation:

For example, in the case of very dense samples, selected samples to adhere to the surface of the tip, dispensing more slowly. In contrast, ethanol samples are less viscous and more volatile and are dispensed more rapidly. Some of the ways to minimize these inconveniences are the use of ultra low retention pipette tips, since they have a hydrophobic plastic additive that prevents the liquid from adhering to the inside of the tip.

Another way is to use the reverse pipetting.

Answer: v = 2π2 Kme2 Z / nh

Explanation:

The formula for velocity of an electron in the nth orbit is given as,

v = 2π2 Kme2 Z / nh

v = velocity

K = 1/(4πε0)

m= mass of an electron

e = Charge on an electron

Z= atomic number

h= Planck’s constant

n is a positive integer.

Answer:

18.9

Explanation:

Given data

We can find the moles (n) of glucose using the following expression.

n = C × V

n = 0.137 mol/L × 0.7649 L

n = 0.105 mol

The molar mass of glucose is 180.16 g/mol. The mass corresponding to 0.105 moles is:

m = 0.105 mol × 180.16 g/mol = 18.9 g

Answer:

A

Explanation:

ethane weaker than ethyne because it is not as unsaturated as ethyne

Ethene is longer because it need to bond with two more hydrogen atom

Answer: The new water level of the cylinder is 24.16 mL

Explanation:

To calculate the volume of water displaced by silver, we use the equation:

[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]

Density of silver = 10.49 g/mL

Mass of silver = 35.2 g

Putting values in above equation, we get:

[tex]10.49g/mL=\frac{35.2g}{\text{Volume of silver}}\\\\\text{Volume of silver}=\frac{35.2g}{10.49g/mL}=3.36mL[/tex]

We are given:

Volume of graduated cylinder = 20.8 mL

New water level of the cylinder = Volume of graduated cylinder + Volume of water displaced by silver

New water level of the cylinder = (20.8 + 3.36) mL = 24.16 mL

Hence, the new water level of the cylinder is 24.16 mL

Answer:

0.370 micromoles of magnesium fluoride the chemist has added to the flask.

Explanation:

[tex]Molarity=\frac{moles}{\text{Volume of solution(L)}}[/tex]

Moles of  magnesium fluoride  = n

Volume of the solution = 380.0 mL = 0.380 L (1 mL = 0.001 L)

Molarity of the solution = [tex]9.75\times 10^{-4} mM=9.75\times 10^{-7} M[/tex]

(1 mM = 0.001 M)

[tex]9.75\times 10^{-7} M=\frac{n}{0.380 L}[/tex]

[tex]n=3.705\times 10^{-7} mol[/tex]

1 mole = [tex]10^6[/tex] micro mole

[tex]n=3.705\times 10^{-7} \times 10^6 \mu mol=0.3705 \mu mol[/tex]

0.370 micromoles of magnesium fluoride the chemist has added to the flask.

Explanation:

A volatile substance is defined as the substance which can easily evaporate into the atmosphere due to weak intermolecular forces present within its molecules.

Whereas a flammable substance is defined as a substance which is able to catch fire easily when it comes in contact with flame.

Hence, when we heat a flammable or volatile solvent for a recrystallization then it should be kept in mind that should heat the solvent in a stoppered flask to keep vapor away from any open flames so that it won't catch fire.

And, you should ensure that no one else is using an open flame near your experiment.

Thus, we can conclude that following statements are correct:

When heating a flammable or volatile solvent for recrystallization, one should not do the heating using an open flame or heat near any open flame in the laboratory.

They are liquids that can catch flame easily either directly or through their vapors.

This means that such liquids should not be heated using an open flame or near any open flame to prevent fire accidents.

Heating near an open flame may cause the vapor to catch fire. Heating in a stoppered flask is not ideal because the main aim of heating is to allow the vapor to escape.

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Answer:

b.  CI (1255 kJ/mol), Ge (784 kJ/mol), and K (418 kJ-mol)

Explanation:

In general, the first ionization energy for a given period increases as we go from left to right in the periodic table (there are some exeptions as with every rule), and the first ionization energy decreases a we go down in the periodic table. The reason for this are:

1. As we move from left to right in a group, the effective nuclear charge increases which makes it harder to remove the electron.

2. As we increase the period in going top to bottom in the periodic table we are adding another shell farther away from the nucleus, making it easier to remove the electron.

Given the three values for the first ionization energy in kJ/mol : 418, 784 and 1255, we expect the highest value to correspond to Cl which belongs to period 3 (K and Ge belong to period 4).

Now comparing K and Ge which belong to period 4, Ge will have a higher effective nuclear charge than K .

So the match will be Cl (1255 kJ/mol), Ge (784) and K(418)

Correct answer is b.

Note: there could be some confusion since the value of 784 was misplaced in the question statement, but we can deduce that in this question we are asked to match the values for the atoms.

Answer:

[NaOH] = 0.045 M

[red dye] = 0.031 M

Explanation:

The NaOH and the red dye will not react between them, so, the process that it's occurring is only a dilution. The final volume of the solution will be the volume of the flask, which is 100 mL.

The number of moles (n) of each substance will not vary, and it's calculated by the multiplication of the concentration (C) by the volume (V). If 1 is the initial solution, and 2 the diluted:

n1 = n2

C1*V1 = C2*V2

For NaOH:

C1 = 0.150 M

V1 = 30.0 mL

V2 = 100 mL

0.150*30 = C2*100

100C2 = 4.5

C2 = 0.045 M

For the red dye:

C1 = 0.125 M

V1 = 25.0 mL

V2 = 100 mL

0.125*25 = C2*100

100C2 = 3.125

C2 = 0.031 M

Final answer:

The final concentration of NaOH is 0.045 M and the final concentration of the red dye is 0.03125 M.

Explanation:

To calculate the final concentrations of NaOH and the red dye, we can use the formula:

11 = 22

Let's calculate the final concentration of NaOH first:

1 (concentration of NaOH) = 0.150 M

1 (volume of NaOH) = 30.0 mL = 0.030 L

2 (final concentration of NaOH) = ?

2 (final volume of solution) = 100 mL = 0.100 L

Plugging in the values into the formula gives us:

(0.150 M)(0.030 L) = 2 (0.100 L)

Solving for 2, we find that the final concentration of NaOH is 0.045 M.

Now, let's calculate the final concentration of the red dye:

1 (concentration of red dye) = 0.125 M

1 (volume of red dye) = 25.0 mL = 0.025 L

2 (final concentration of red dye) = ?

2 (final volume of solution) = 100 mL = 0.100 L

Using the formula:

(0.125 M)(0.025 L) = 2 (0.100 L)

We find that the final concentration of the red dye is 0.03125 M.

The pH of the solution is 3,76

Why?

To solve this problem we have to apply the Henderson-Hasselbach equation, which is used whenever we need to calculate the pH of a solution of an acid HA (Lactic Acid) and its conjugate base A⁻ (Sodium Lactate)

We can use either moles or concentrations for this equation. In this case, we are going to use the moles, and we are going to take the pKa of lactic acid as 3,86:

[tex]pH=pKa+log(\frac{nA^- }{nHA} )=3,86+log(\frac{(0,04moles)}{(0,05mL)*(1.0M)} )\\\\pH=3,76[/tex]

Have a nice day!

#LearnwithBrainly

Final answer:

To calculate the pH of a buffer solution containing lactic acid and its conjugate base sodium lactate, the Henderson-Hasselbalch equation is used. The calculated pH for the buffer solution is approximately 3.76.

Explanation:

The question involves calculating the pH of a solution containing a weak acid and its conjugate base, which is a typical buffer solution scenario. We recognize it as a buffer solution because it contains both the weak acid (lactic acid, HC3H5O3) and its conjugate base (sodium lactate, NaC3H5O3). The Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]), can be used to calculate the pH of a buffer solution where [A-] is the concentration of the conjugate base and [HA] is the concentration of the acid.

Firstly, we need to determine the concentrations of the acid and the base. The concentration of the lactic acid remains at 1.0 M since the added volume is negligible, while the amount of sodium lactate added is given as 0.040 moles. Since the volume is 50.0 mL (or 0.050 L), the concentration of sodium lactate will be 0.040 moles/0.050 L = 0.80 M.

Using this information and the known pKa of lactic acid, which is approximately 3.86, we can now use the Henderson-Hasselbalch equation to calculate the pH of the buffer solution:

pH = pKa + log([A-]/[HA])
pH = 3.86 + log(0.80/1.0)
pH = 3.86 + log(0.80)
pH ≈ 3.86 + (-0.09691)
pH ≈ 3.76

Therefore, the pH of the buffer solution is approximately 3.76.

Answer:

a)  the system is closed

b) the system is open

Explanation:

a) the system can be characterised as closed system. If we define the boundaries of this system as an envelope that contains the bathtub, then there will not be any mass transfer to the environment through the boundary → then the system is closed ( the mass of the system remains constant)

b) on the other hand , taking the same system as in the first case, if the hot water is continually added to the tub , then there is water flow through the boundary (from the environment to the system) → then the system is open ( the mass of the system can be altered)