Select the definition of diffusion. The gradual dispersal of one substance through another substance. The movement of a gas particle from one area to another. The escape of a gas through a small opening in a barrier into a region of lower pressure. The random motion of gas particles. Select the definition of effusion. The random motion of gas particles. The gradual dispersal of one substance through another substance. The escape of a gas through a small opening in a barrier into a region of lower pressure. The movement of a gas particle from one area to another.

Answer:

Nitrogen

Explanation: nitrogen is around 70% of the air hope this helps god bless

Answer:

3.8 is the van't Hoff factor for iron(III) chloride in X.

Explanation:

[tex]\Delta T_f=i\times K_f\times m[/tex]

where,

[tex]\Delta T_f[/tex] =depression in freezing point =

[tex]K_f[/tex] = freezing point constant

m = molality =[tex]\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in kg}}[/tex]

i = van't Hoff factor

we have :

Mass of glycine = 63.4 g

Molar mass of glycine = 71 g/mol

Mass of solvent X = 700. g = 0.7 kg

1 g = 0.001 kg

[tex]K_f[/tex] of solvent X= ?

i = 1  (non electrolyte)

Depression in freezing point= [tex]7.9^oC[/tex]

[tex]7.9^oC=1\times K_f \times \frac{63.4 g}{71 g/mol\times 0.7 kg}[/tex]

[tex]K_f=6.19 ^oC/m[/tex]

When iron(III) chloride is dissolved in 0.7 kg of solvent X

Mass of  iron(III) chloride = 63.4 g

Molar mass of  iron(III) chloride= 162.5 g/mol

Mass of solvent X = 700. g = 0.7 kg

1 g = 0.001 kg

[tex]K_f[/tex] of solvent X= [tex]6.19 ^oC/m[/tex]

i = ?

Depression in freezing point:[tex]13.3^oC[/tex]

[tex]13.3^oC=i\times 6.19^oC\times \frac{63.4 g}{162.5 g/mol\times 0.7 kg}[/tex]

Solving for i:

i = 3.85 ≈ 3.8

3.8 is the van't Hoff factor for iron(III) chloride in X.

Answer: The molar concentration of sulfuric acid in the original sample is 1.943 M

Explanation:

To calculate the molarity of acid, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]H_2SO_4[/tex]

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.

We are given:

[tex]n_1=2\\M_1=?\\V_1=56.5mL\\n_2=1\\M_2=0.5824M\\V_2=43.37mL[/tex]

Putting values in above equation, we get:

[tex]2\times M_1\times 56.5=1\times 0.5824\times 43.37[/tex]

[tex]M_1=0.2235[/tex]

Now to calculate the molarity of original solution:

[tex]M_1\times 6.5=0.2235\times 56.5[/tex]

[tex]M_1=1.943[/tex]

Thus the molar concentration of sulfuric acid in the original sample is 1.943 M

Final answer:

The molar concentration of sulfuric acid in the original 6.5 mL sample is calculated to be approximately 1.9436 M, based on titration with a 0.5824 M NaOH solution.

Explanation:

Determining the Concentration of Sulfuric Acid in a Sample Using Titration:

To determine the molar concentration of sulfuric acid, H2SO4, we will use the data that 43.37 mL of a 0.5824 M NaOH solution was needed to titrate a 6.5 mL sample of the acid.

Firstly, we calculate the moles of NaOH used in the titration:

Moles of NaOH = Volume (in Liters)  imes Molarity

Moles of NaOH = 0.04337 L  imes 0.5824 mol/L

Moles of NaOH = 0.02526888 mol

According to the balanced chemical equation, H2SO4 + 2NaOH
ightarrow Na2SO4 + 2H2O, the stoichiometry of the reaction is 1:2. This means one mole of sulfuric acid reacts with two moles of sodium hydroxide. Thus, the moles of H2SO4 will be half of the moles of NaOH used.

Moles of H2SO4 = 0.02526888 mol / 2

Moles of H2SO4 = 0.01263444 mol

Now, we calculate the molarity of the sulfuric acid in the original sample:

Molarity of H2SO4 = Moles of H2SO4 / Volume of sample in liters

Molarity of H2SO4 = 0.01263444 mol / 0.0065 L

Molarity of H2SO4 = 1.9436 M

Therefore, the molar concentration of the sulfuric acid in the original sample is approximately 1.9436 M.

Answer:

Volume = 5.71mL

Explanation:

Applying Boyle's law

P1V1= P2V2

P1= 80kPa, V1= 5mL,P2= 70kPa, V2=?

Substitute into above formula

80×5= 70×V2

V2= (80×5)/70 = 5.71mL

Answer:

6.941

Explanation:

Step 1:

Representation.

Let Lithium−6 be isotope A

Let Lithium−7 be isotope B

Let the abundance of isotope A (Lithium-6) be A%

Let the abundance of isotope B (Lithium-6) be B%

Step 2:

Data obtained from the question. This includes:

Mass of isotope A (Lithium−6) =

6.01512

Abundance of isotope A (Lithium−6) = A% = 7.47%

Mass of isotope B (Lithium−7) = 7.016

Abundance of isotope B (Lithium−7) = B% = 92.53%

Atomic weight of lithium =?

Step 3:

Determination of the atomic weight of lithium. This is illustrated below:

Atomic weight = [(Mass of AxA%)/100] + [(Mass of BxB%)/100]

Atomic weight = [(6.01512x7.47)/100] + [(7.016x92.53)/100]

Atomic weight = 0.449 + 6.492

Atomic weight of lithium is 6.941

Final answer:

The atomic weight of lithium is approximately 6.9423 u, which is the weighted average of the masses of its isotopes, lithium-6 and lithium-7, with respect to their natural abundances.

Explanation:

To calculate the atomic weight of lithium, we consider the relative abundances and masses of its naturally occurring isotopes, lithium-6 and lithium-7. The atomic weight is the weighted average of the atomic masses of an element's isotopes, based on their natural abundance. Using the provided information:

Lithium-6 has a mass of 6.01512 and an abundance of 7.47%.

Lithium-7 has a mass of 7.016 and an abundance of 92.53%.

The calculation is as follows:

(6.01512 × 0.0747) + (7.016 × 0.9253) = 0.4487 + 6.4936 = 6.9423

Therefore, the atomic weight of lithium is approximately 6.9423 u (atomic mass units).

Answer:

Overall reaction

H2(g) + 2ICI(g) -----> I2(g) +2HCl(g)

Overall Rate = k1[H2] [ICl]

Explanation:

Overall reaction

H2(g) + 2ICI(g) -----> I2(g) +2HCl(g)

The overall reaction is the sum of the two two reactions shown in the question. After the two reactions are summed up properly, this overall reaction equation his obtained.

Since K1<<K2 it means that step 1 is slower than step 2. Recall that the rate if reaction depends on the slowest step of the reaction. Hence

Overall Rate = k1[H2] [ICl]

The balanced chemical equation for the overall reactions is H2 (g) + 2ICl (g) → I2 (g) + 2HCl (g). The rate of the reaction is determined by the slowest step, in this case, the first one. So, the experimentally-observable rate law for the overall reaction would be Rate = k1[H2][ICl].

The overall reaction is obtained by adding up the given elementary reactions. Adding step 1 and 2, we have:

H2 (g) + 2ICl (g) → 2HI (g) + HCl (g) → H2 (g) + I2 (g) + HCl (g)

After cancelling like terms the balanced chemical equation is:

H2 (g) + 2ICl (g) → I2 (g) + 2HCl (g)

Since the first step is much slower than the second, it's the rate-determining step. The experimentally-observable rate law for the overall reaction will depend on the rate-determining step, hence, Rate = k1[H2][ICl]

https://brainly.com/question/34346242

#SPJ3

1. The concentration of[tex]\( \text{Q}^- \)[/tex] at equilibrium is[tex]\( 2.10 \times 10^{-3} \)[/tex] M.

2. The[tex]\( K_{sp} \)[/tex] value for [tex]\( \text{XQ}_3 \) is \( 6.48 \times 10^{-12} \).[/tex]

To find the concentration of [tex]\( \text{Q} \)[/tex] at equilibrium and the solubility product constant [tex](\( K_{sp} \))[/tex]for the ionic compound [tex]\( \text{XQ}_3 \)[/tex], follow these steps:

The compound [tex]\( \text{XQ}_3 \)[/tex] dissociates in water as follows:

[tex]\[ \text{XQ}_3(s) \rightleftharpoons \text{X}^{3+}(aq) + 3\text{Q}^-(aq) \][/tex]

Let ( s ) be the solubility of [tex]\( \text{XQ}_3 \)[/tex] in mol/L, which is given as \[tex]7.00 \times 10^{-4} \) M.[/tex]

When [tex]\( \text{XQ}_3 \)[/tex] dissociates:

- The concentration of[tex]\( \text{X}^{3+} \)[/tex] ions will be ( s ).

- The concentration of [tex]\( \text{Q}^- \)[/tex] ions will be ( 3s ) (since three[tex]\( \text{Q}^- \)[/tex] ions are produced for each formula unit of[tex]\( \text{XQ}_3 \)).[/tex]

Given [tex]\( s = 7.00 \times 10^{-4} \) M:[/tex]

[tex]{Q}^-] = 3s = 3 \times 7.00 \times 10^{-4} = 2.10 \times 10^{-3} \text{ M} \][/tex]

Thus, the concentration of [tex]\( \text{Q}^- \)[/tex] at equilibrium is[tex]\( 2.10 \times 10^{-3} \)[/tex].

The solubility product constant [tex]\( K_{sp} \) for \( \text{XQ}_3 \)[/tex]   can be calculated using the equilibrium concentrations of the ions:

[tex][ K_{sp} = [\text{X}^{3+}][\text{Q}^-]^3 \][/tex]

Substitute the equilibrium concentrations:

[tex][ [\text{X}^{3+}] = s = 7.00 \times 10^{-4} \text{ M} \][/tex]

[tex][ [\text{Q}^-] = 3s = 2.10 \times 10^{-3} \text{ M} \][/tex]

Now, calculate[tex]K_{sp} \):[/tex]

[tex]\[ K_{sp} = (7.00 \times 10^{-4}) \times (2.10 \times 10^{-3})^3 \[/tex]

First, compute[tex]( (2.10 \times 10^{-3})^3 \)[/tex]

[tex]\[ (2.10 \times 10^{-3})^3 = 2.10^3 \times (10^{-3})^3 = 9.261 \times 10^{-9} \][/tex]

Then, multiply by[tex]7.00 \times 10^{-4} \):[/tex]

[tex][ K_{sp} = (7.00 \times 10^{-4}) \times (9.261 \times 10^{-9}) \][/tex]

[tex][ K_{sp} = 6.4827 \times 10^{-12} \][/tex]

1. The concentration of[tex]{Q}^- \)[/tex] at equilibrium is [tex]\( 2.10 \times 10^{-3} \) M.[/tex]

2. The [tex]K_{sp} \) value for {XQ}_3 \) is \( 6.4827 \times 10^{-12} \).[/tex]

Final answer:

Internal standards and standard addition are techniques used in analytical chemistry to ensure accuracy. The incorrect statement is about the use of internal standards.

Explanation:

Internal standards and standard addition are both techniques used in analytical chemistry to ensure the accuracy and reliability of quantitative measurements. In internal standard method, a known amount of a compound is added to all samples and standards, which allows for compensation of errors that may occur during sample preparation and analysis.

Standard addition, on the other hand, involves adding known amounts of a standard solution to an unknown solution to determine the concentration of the analyte of interest. It can be a single standard addition or multiple standard additions, depending on the requirements of the analysis.

Based on the given options, the incorrect description is:

d) Internal standard is used when instrument response varies from run to run.

The correct answer is a) Internal standard is useful when the matrix in the unknown is complicated.

Let's analyze each option to understand why option (a) is incorrect:

a) Internal standard is useful when the matrix in the unknown is complicated.

This statement is incorrect because it confuses the roles of internal standards and standard addition.

An internal standard is a substance that is added in a constant amount to all samples, including the calibration standards and the unknowns. It is used to correct for any variations in the analytical procedure, such as changes in instrument response, sample preparation, and matrix effects that affect the analyte's signal.

However, it is not specifically used because the matrix is complicated; rather, it is used to account for variations in the analytical process. The complexity of the matrix is typically addressed by the standard addition method, which involves adding known quantities of the analyte to the sample to overcome matrix effects.

b) Standard addition could be a single standard addition or multiple standard additions to an unknown solution.

This statement is correct. Standard addition can involve adding a known amount of analyte to the sample once (single standard addition) or several times (multiple standard additions) to construct a calibration curve.

This method is used to compensate for matrix effects that might not be accounted for by using external calibration alone.

c) Standard addition is useful when the matrix in the unknown is complicated.

This statement is correct. The standard addition method is particularly useful for samples with complex matrices that can interfere with the analysis.

By adding known amounts of the analyte directly to the sample, the method allows for the determination of the analyte's concentration while accounting for the matrix effects.

d) Internal standard is used when instrument response varies from run to run.

This statement is correct. An internal standard is used to normalize the response of the analyte and to correct for any variations in the analytical procedure, including changes in instrument response over time.

It helps to ensure the accuracy and precision of the analytical results, regardless of the variations that occur between different runs of the analysis.

Therefore, the statement in option (a) is the one that is not correct, as it misrepresents the use of internal standards in the context of complex matrices.

Answer:

The daughter nuclide is selenium; 76:34 Se

Explanation:

The complete question is as follows;

is desired to determine the concentration of arsenic in a lake sediment sample by means of neutron activation analysis. The nuclide 75:33 As captures a neutron to form 76:33 As, which in turn undergoes beta decay. The daughter nuclide produces the characteristic gamma rays used for the analysis. What is the daughter nuclide?

solution:

Please check attachment for decay equations and explanations

Answer:

c = 4

Explanation:

In general, for the reaction

       a A     +    b B     ⇒ c C   +    d D

the rate is given by:

rate = - 1/a ΔA/Δt = - 1/b ΔB/Δt = + 1/c ΔC/Δt = + 1/d ΔD/Δt

this is done so as to express the rate in a standarized way which is the same to all the reactants and products irrespective of their stoichiometric coefficients.

For this question in particular we know the coefficient of A and need to determine the coefficient c.

- 1/2 ΔA/Δt = + 1/c ΔC/Δt

- 1/2 (-0.0080 ) = + 1/c ( 0.0160 mol L⁻¹s⁻¹ )

0.0040 mol L⁻¹s⁻¹  c = 0.0160 mol L⁻¹s⁻¹

∴ c = 0.0160 / 0.0040 = 4

Answer:

When atoms other than hydrogen form covalent bonds, an octet is accomplished by sharing. The octet rule can be used to explain the number of covalent bonds an atom forms. This number normally equals the number of electrons that atom needs to have a total of eight electrons (an octet) in its outer shell

Explanation:

chemistry, the octet rule explains how atoms of different elements combine to form molecules. ... In a chemical formula, the octet rule strongly governs the number of atoms for each element in a molecule; for example, calcium fluoride is CaF2 because two fluorine atoms and one calcium satisfy the rule.

octet rule: Atoms lose, gain, or share electrons in order to have a full valence shell of eight electrons. Hydrogen is an exception because it can hold a maximum of two electrons in its valence level.

There is another rule, called the duplet rule, that states that some elements can be stable with two electrons in their shell. Hydrogen and helium are special cases that do not follow the octet rule but the duplet rule. ... They are stable in a duplet state instead of an octet state.

To calculate the atoms of an element in a given molecule, we need to multiply stoichiometry by the number that is written on the foot of that element. Therefore, the balanced equation is

1B+4St+1Y+3Ic[tex]\rightarrow[/tex]1 BSt[tex]_4[/tex]YIc[tex]_3[/tex]

Balanced equation is the one in which the total number of atoms of a species on reactant side is equal to the total number of atoms on product side. The mass of the overall reaction should be conserved. There are so many types of chemical reaction reaction like combination reaction, displacement reaction.

The other characteristic of balanced reaction is that physical state should be written with each compound or molecule on reactant and product side. Physical state should be written in brackets. s means solid, l means liquid, g means gas.

The word equation is

1 banana+ 4 strawberries+1 container of yogurt+3 ice cubes[tex]\rightarrow[/tex] 1 smoothie

The balanced chemical equation is

1B+4St+1Y+3Ic[tex]\rightarrow[/tex]1 BSt[tex]_4[/tex]YIc[tex]_3[/tex]

Therefore, the balanced equation is

1B+4St+1Y+3Ic[tex]\rightarrow[/tex]1 BSt[tex]_4[/tex]YIc[tex]_3[/tex]

Learn more about the balanced equation, here:

https://brainly.com/question/7181548

#SPJ6

To determine the volume of ethanolamine needed, the student should divide the desired mass (10.0g) by the density of ethanolamine (1.02g/cm³), which gives approximately 9.8 cm³. A trusted source, such as the CRC Handbook of Physics and Chemistry was referenced.

To calculate the volume of ethanolamine needed for the experiment, we need to use the formula for density, which is mass/volume. By rearranging the formula to solve for volume gives us volume = mass/density. So, the volume of ethanolamine will be 10.0g / 1.02g/cm3. This calculation gives us approximately 9.80 cm3. Therefore, the student needs to pour out about 9.8 cm3 of ethanolamine for the experiment. The student was able to easily determine this by consulting a trusted resource, the CRC Handbook of Physics and Chemistry.

https://brainly.com/question/32822827

#SPJ3

Final answer:

To find the volume of ethanolamine needed, divide the mass needed (10.0 g) by the density (1.02 g/cm³) to obtain approximately 9.80 cm³.

Explanation:

To calculate the volume of ethanolamine the student should pour out, we can use the density formula, which is density (d) equals mass (m) divided by volume (V), rearranged to solve for volume. Given that the density of ethanolamine is 1.02 g/cm³ and the student needs 10.0 g of ethanolamine, the equation would be:

V = m / d

Substituting the given values, we get:

V = 10.0 g / 1.02 g/cm³

The calculation results in a volume of approximately 9.80 cm³ of ethanolamine the student should pour out.

Answer:

A --- (E)-oct-2-en-1-o1

B ----(E)-oct-2-enal

Explanation:

See the attached file for the structure.

Answer:

Molar mass= 297.1gmol-1

Explanation:

BaBr2 = 137.3+ (79.9*2)= 297.1gmol-1

Answer:

1. 3.29mol

2. 0.125molH2o/mol

3. 52.5'C

Explanation: The step by step explanation are attached to the answer.

Answer:

Mole fraction of propane = 0.74 mol C₃H₈/mol

Mole fraction of water = 0.29 mol H₂0 /mol

Dew point temperature = 52.5°C

Explanation:

See the attached file for the calculation.

Answer:

copper is being reduced

Explanation: I got 100% on the quiz good luck this class is hard lol

Answer:

The answer is the second option

Copper is being reduced

Explanation:

Have a good day just took test

Answer:

17.5 g

Explanation:

Given data

The concentration by mass of NaCl in the solution is 35.0%, that is, there are 35.0 grams of sodium chloride per 100 grams of solution. We will use this ratio to find the mass of sodium chloride required to prepare 50.0 grams of a 35.0% salt solution.

[tex]50.0gSolution \times \frac{35.0gNaCl}{100gSolution} = 17.5gNaCl[/tex]

To prepare a 50 gram 35% salt solution, you need 17.5 grams of NaCl.

To find out how many grams of NaCl are needed to prepare 50.0 grams of a 35.0% salt solution, you use the definition of percent concentration by mass: (mass of solute/mass of solution) x 100%. The mass of the solution is the total mass, which is the sum of the mass of the solute (NaCl in this case) and the solvent (usually water). For a 35% solution, this means that 35 grams of NaCl are in every 100 grams of solution.

However, we want to prepare 50 grams of solution. So, you set up a ratio: (35 g NaCl/100 g solution) = (x g NaCl/50 g solution). Solving this equation, you find that x = 17.5. Therefore, 17.5 grams of NaCl are needed to prepare a 50 gram 35% salt solution.

https://brainly.com/question/35910758

#SPJ3

Answer:

-to see if food went bad

-to test the safety of water

-to make sure conditions are safe

Explanation:

Answer:

A, B, C

Explanation:

Answer:

The total heat required is 35807 J

Explanation:

Step 1: Data given

Mass of water = 15.0 grams

Initial temperature of water = 87.0 °C

Temperature of steam = 135.0 °C

ΔHfus = 334 J/g

ΔHVap = 2260 J/g

Step 2: Calculate heat required to heat water from 85 to 100 °C

Q = m*c*ΔT

⇒Q = the heat required = TO BE DETERMINED

⇒m = the mass of water = 15.0 grams

⇒c= the specific heat of water = 4.18 J/g°C

⇒ΔT = the change of temperature = T2 - T1 = 100 - 87 = 13°C

Q = 15.0 * 4.18 J/g°C * 13 °C

Q = 815.1 J

Step 3: Calculate heat required to change water at 100 °C to steam

Q = m * ΔHVap

Q = 15.0 grams * 2260 J/g

Q = 33900 J

Step 4: Calculate heat required to heat steam from 100 °C to 135 °C

Q = m*c*ΔT

⇒Q = the heat required = TO BE DETERMINED

⇒m = the mass of water = 15.0 grams

⇒c= the specific heat of steam = 2.09 J/g°C

⇒ΔT = the change of temperature = T2 - T1 = 135 - 100 = 35°C

Q = 15.0 * 2.09 * 35 °C

Q = 1097.25 J

Step 5: Calculate the total heat required

Q = 35807 J

The total heat required is 35807 J

T

Answer:

the ionic radius of the anion [tex]r^- = 312.52 \ pm[/tex]

Explanation:

From the diagram shown below :

The anion [tex]Cl^-[/tex] is located at the corners

The cation [tex]Cs^+[/tex] is located at the body center

The Body diagonal length =  [tex]\sqrt{3 \ a }[/tex]

∴ [tex]2 \ r^+ \ + 2r^- \ = \sqrt{3 \ a} \\ \\ r^+ +r^- = \frac{\sqrt{3}}{2} a[/tex]

Given that :

[tex]\frac{r^+}{r^-} =0.84[/tex]   (i.e the  ratio of the ionic radius of the cation to the ionic radius of

                 the anion )

[tex]0.84r^- \ + r^- \ = \frac{\sqrt{3}}{2}a \\ \\ 1.84 r^- = \frac{3}{2}a \\ \\ r^- = \frac{\sqrt{3}}{2*1.84}a[/tex]

Also ; a =  664 pm

Then :

[tex]r^- = \frac{\sqrt{3} }{2*1.84}*664 \ pm\\ \\ r^- = 312.52 \ pm[/tex]

Therefore,  the ionic radius of the anion [tex]r^- = 312.52 \ pm[/tex]

The ionic radius of the anion  [tex]r^-=312.52pm[/tex]

The anion is placed on the corners and the cation  is placed on the frame center.

The Body diagonal length =  [tex]\sqrt{3a}[/tex]

[tex]2r^++2r^-=\sqrt{3a} \\\\r^++r^-=\sqrt{3}/2a }[/tex]

Given:

Ratio=  0.840

[tex]\frac{r^+}{r^-}=0.840[/tex]

[tex]0.84r^-+r^-=\sqrt{3}/2a \\\\1.84r^-=3/2a\\\\r^-=\sqrt{3}/2*1.84a[/tex]

Also ; a =  664 pm

Then : [tex]r^-[/tex] =312.52 pm

Therefore,  the ionic radius of the anion = 312.52 pm

Find more information about Cubic structure here:

brainly.com/question/7959646

Answer:

I think it is D.

Explanation:

Answer: i know it d but i think it is c to

Explanation:

Answer:

0.0676 moles of oxygen gas is produced

Explanation:

Step 1: Data given

Volume of the flask = 730 mL = 0.730 L

The pressure of the gas in the flask is 2.5 atm

The temperature is recorded to be 329 K

Step 2: The balanced equation

2NO3- → 2NO2- + O2

Step 3: Calculate moles oxygen gas (O2)

p*V =n * R*T

⇒with p = the pressure of oxygen gas = 2.5 atm

⇒with V = the volume of the flask = 0.730 L

⇒with n = the number of moles O2 gas

⇒with R = the gas constant = 0.0821 L*atm/mol*K

⇒with T ⇒ the temperature = 329 K

n = (p*V) / (R*T)

n = (2.5 * 0.730 ) / (0.0821*329)

n = 0.0676 moles

0.0676 moles of oxygen gas is produced

Answer:

523°C

Explanation:

Step 1:

Data obtained from the question.

Diameter (d) = 31.0 cm

Height (h) = 37.2 cm

Pressure (P) = 5.90 MPa

Mass of ClF5 = 3.27 kg

Temperature (T) =?

Step 2:

Determination of the volume of the stainless-steel cylinder.

Volume of cylinder = πr^2h = π/4d^2h

V = π/4 x (31)^2 x 37.2

V = 28077.36 cm3

Converting the volume to L, we have:

1 cm3 = 0.001 L

Therefore, 28077.36 cm3 = 28077.36 x 0.001 = 28.07736 L

Step 3:

Conversion of 5.90 MPa to atm.

1 MPa = 9.869 atm

Therefore, 5.90 MPa = 5.90 x 9.869

= 58.2271 atm

Step 4:

Determination of the number of mole of chiorine pentafluoride gas (ClF5). This is illustrated below:

Molar Mass of ClF5 = 35.5 + (19x5) = 35.5 + 95 = 130.5g/mol

Mass of ClF5 = 3.27kg =3.27x10000

Mass of ClF5 = 3270g

Number of mole = Mass /Molar Mass

Number of mole of ClF5 = 3270/130.5

Number of mole of ClF5 = 25.057 moles

Step 5:

Determination of the temperature.

Applying the ideal gas equation

PV = nRT, the temperature T, can be obtained as follow:

P = 58.2271 atm

V = 28.07736 L

n = 25.057 moles

R (gas constant) = 0.082atm.L/Kmol

T=?

PV = nRT

Divide both side by nR

T = PV /nR

T = (58.2271x28.07736)/(25.057x0.082)

T = 795.68 K

Step 6:

Conversion of Kelvin temperature to celsius temperature.

°C = K - 273

K = 795.68 K

°C = 795.68 - 273

°C = 523°C

Therefore, the maximum safe operating temperature the engineer should recommend is 523°C

The reaction between solutions of silver nitrate and sodium chloride results in a precipitate due to a double-replacement reaction. The solid precipitate formed is silver chloride (AgCl). The net ionic equation becomes: Ag+ (aq) + Cl- (aq) → AgCl (s).

When you mix solutions of silver nitrate (AgNO3) and sodium chloride (NaCl), a precipitate forms. This is because of a double-replacement reaction that occurs between the two compounds. You start with AgNO3 (silver nitrate) and NaCl (sodium chloride), and end up with NaNO3 (sodium nitrate) and AgCl (silver chloride). The silver chloride precipitates, or becomes solid, leaving the sodium nitrate in solution. This can be shown by the net ionic equation:Ag+ (aq) + Cl- (aq) → AgCl (s). This equation represents the formation of the silver chloride precipitate. The ions Na+ and NO3- do not take part in the reaction and so are not included in the net ionic equation.

https://brainly.com/question/35304253

#SPJ12

Answer:

See the answer below.

Explanation:

Fire has three major components:

If the victim had died as a result of the fire, he/he would have inhaled smoke and hot gases from the fire. These components would have resulted in traces of burns and soot deposition in the trachea and lungs as well as traces of CO in the blood of the victim.

If the analysis of the victim's corpse does not reflect some of the results above, it can be effectively concluded that the victim has been dead before the fire.

The single most important indicator of death by the fire would be the presence of CO in the blood of the victim's corpse. All others might be to a less significant degrees.

Final answer:

To determine if the victim died before the fire, forensic anthropologists analyze perimortem injuries, signs of decomposition, and toxicology for smoke inhalation indicators.

Explanation:

To determine whether a victim died before a fire, a forensic anthropologist would analyze the remains for signs of perimortem trauma. This includes examining cut marks or injuries sustained around the time of death which could indicate homicide. Disarticulation of joints and the degree of skeletal preservation can also help establish the time since death, negating the possibility that the fire caused the death if the individual was already deceased and decomposing. Moreover, the presence of smoke inhalation in the lungs can suggest if the person was alive during the fire. Toxicology reports may provide evidence of substances such as carbon monoxide or soot. The presence or absence of such substances can indicate whether death occurred before or during the fire.

Answer:

See explanation

Explanation:

the compound that gives the fastest SN2 reaction with sodium methoxide- 1-bromohexane

the compound that gives the fastest SN1 reaction- 3-bromo-3-methylpentane

the compound(s) that undergo an SN1 reaction to give rearranged products- 1-bromo-2,2-dimethylbutane

the compound that is least reactive to sodium methoxide in methanol -

3-bromo-3-methylpentane

the compound(s) that can exist as diastereomers - 3-bromo-3-methylpentane

the compound(s) that can exist as enantiomers- 3-bromo-2-methylpentane

Explanation:

2 ClO2 (aq) + 2OH- (aq)→ ClO3- (aq) + ClO2- + H2O (l)

The data is given as;

Experiment [ClO2] (M) [OH-] (M) Initial Rate (M/s)

1 0.060 0.030 0.0248

2 0.020 0.030 0.00276

3 0.020 0.090 0.00828

a) Rate law is given as;

Rate = k [ClO2]^x [OH-]^y

From  experiments 2 and 3, tripling the concentration of  [OH-] also triples the rate of the reaction. This means the reaction  is first order with respect to  [OH-]

From experiments 1 and 2, when the [ClO2] decreases by a factor of 3, the rate decreases by a factor of 9. This means the reaction is second order with respect to [ClO2]

Rate =  k [ClO2]² [OH-]

b. Calculate the value of the rate constant with the proper units.

Taking experiment 1,

0.0248 = k (0.060)²(0.030)

k = 0.0248 / 0.000108

k = 229.63 M-2 s-1

c. Calculate the rate when [ClO2] = 0.100 M and [OH-] = 0.050 M.

Rate =  229.63  [ClO2]² [OH-]

Rate = 229.63 (0.100)²(0.050)

Rate = 0.1148 M/s

Final answer:

The rate law is rate = k[ClO2]^2[OH-], with k ≈ 22.2222 M^-2s^-1. The calculated rate with [ClO2] = 0.100 M and [OH-] = 0.050 M is approximately 0.111 M/s.

Explanation:

To determine the rate law for the reaction given, we should look at the changes in concentration and the effect on the initial rate.

Comparing experiment 1 and 2:

Comparing experiment 2 and 3:

Therefore, the rate law is: rate = k[ClO2]^2[OH-].

Using experiment 1 data to calculate the rate constant (k):

To calculate the rate when [ClO2] = 0.100 M and [OH-] = 0.050 M:

Answer:

6

Explanation:

This atom is sulfur (if the electrons are equal to the protons/not an ion). You can tell the number of valence electrons by looking at the individual shell. The first shell (1s) can only hold 2 electrons. The second shell (2s and 2p) can hold 8 electrons. The third shell (3s and 3p), which is the valence shell, only has 6 out of its possible 8 electrons, so this atom has 6 valence electrons.

The number of valence electron that the atom with the electronic configuration of 1s²2s²2p⁶3s²3p⁴ has is 6

Valence electron(s) are the electrons located on the outermost shell of an atom.

With the above information in mind, we shall determine the number of  valence electron that the atom has. This is illustrated as follow:

Electronic configuration => 1s²2s²2p⁶3s²3p⁴

Valence shell => 3s²3p⁴

Valence electron = 2 + 4

Therefore, the atom has 6 valence electrons.

See attachment image

Learn more: https://brainly.com/question/18214726

Answer : The rate constant at 525 K is, [tex]0.0606M^{-1}s^{-1}[/tex]

Explanation :

According to the Arrhenius equation,

[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]

or,

[tex]\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]

where,

[tex]K_1[/tex] = rate constant at [tex]701K[/tex] = [tex]2.57M^{-1}s^{-1}[/tex]

[tex]K_2[/tex] = rate constant at [tex]525K[/tex] = ?

[tex]Ea[/tex] = activation energy for the reaction = [tex]1.5\times 10^2kJ/mol=1.5\times 10^5J/mol[/tex]

R = gas constant = 8.314 J/mole.K

[tex]T_1[/tex] = initial temperature = 701 K

[tex]T_2[/tex] = final temperature = 525 K

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

[tex]\log (\frac{K_2}{2.57M^{-1}s^{-1}})=\frac{1.5\times 10^5J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{701K}-\frac{1}{525K}][/tex]

[tex]K_2=0.0606M^{-1}s^{-1}[/tex]

Therefore, the rate constant at 525 K is, [tex]0.0606M^{-1}s^{-1}[/tex]