1. A smoothie contains 1 banana (B), 4 strawberries (St), 1 container of yogurt (Y), and 3 ice cubes (Ic). Write a balanced equation to describe the relationship. *

Write a conversion factor to show the relationship between the number of ice cubes and the number of smoothies produced. *

How many strawberries would you need to make 12 smoothies? *

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:

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.

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Answer:

Mix 11.73 mL of CHCl₃ and 8.27 mL of CHBr₃

Explanation:

To begin, it seems your question lacks the name of the pure samples that will be mixed to prepare the liquid mixture. The names are not necessary, as the numerical values are there. However, an internet search tells me they are CHCl₃ (d=1.492g/mL) and CHBr₃ (d=2.890 g/mL).

The mixture has a density of 2.07 g/mL, so 20 mL of the mixture would weigh:

20 mL * 2.07 g/mL = 41.4 g

Let X be the volume of CHCl₃ and Y the volume of CHBr₃:

X + Y = 20 mL

The mass of CHCl₃ and CHBr₃ combined have to be equal to the mass of the mixture. We can write that equation using the volume of the samples and their density:

X * 1.492 + Y * 2.890 = 41.4 g

So now we have a system of two equations and two unknowns, we use algebra to solve it:

   1. Express Y in terms of X:

X + Y = 20

   2. Replace Y in the second equation:

X * 1.492 + Y * 2.890 = 41.4

   3. Solve for X:

   4. Using the now known value of X, solve for Y:

X + Y = 20

So, to prepare the liquid mixture we would mix 8.27 mL of CHBr₃ and 11.73 mL of  CHCl₃.

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:

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.

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Answer:

Molar mass= 297.1gmol-1

Explanation:

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

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.

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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.

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]

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]

Answer:

A,C,F/Air,water,orginisams in water

Explanation:

The resources and organisms are affected by oil spillages are Air, Water, and Organisms in water

Oil spillage can cause to the general environment and the organisms in the environment.

Based on the passage, the resources and organisms are affected by oil spillages include the following;

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Answer:

see explanation below

Explanation:

First, let's remember the principle of a friedel crafts alkylation. This kind of reaction are often used to alkylate an aromatic ring like bencene, or a mono substitued bencene. This reactions often involves reactions with alkyl halides or alkenes and aluminum or iron halides.

Now, in this case we have Iodobenzene, bromobenzene, fluorobenzene and benzoic acid.

First, the reactivity order always raise with the polarization of the C - X bond. Therefore, the one with the most polar bond, will react first. In this case, with the halides, the reactivity order would be RI > RBr > RCl >> RF

In the case of the benzoic acid, the COOH is a very strong deactivating group by resonance, therefore, the benzene it's really weak to do a reaction of alkylation, so this will most likely to not react.

In conclude, we have the following order:

Compound A: Iodobencene (1)

Compound B: Bromobencene (2)

Compound C: Fluorobencene (3)

Compound D: Benzoic acid (non)

In Friedel-Crafts alkylation, compounds are ordered by reactivity based on the strength of their carbon-halogen bonds and ability to form stable carbocations. The ranking is Bromobenzene, Iodobenzene, Fluorobenzene, and Benzoic Acid, with the latter being 'non' reactive due to the presence of a deactivating carboxylic acid group.

The reactivity of these compounds in a Friedel-Crafts alkylation scenario depends on their ability to form stable carbocation intermediates and the strength of the carbon-halogen bond. The stronger the bond, the less likely it is to break and form the carbocation necessary for the reaction. Therefore, the reactivity order is as follows:

It's worth noting that Benzoic Acid (Compound D) would be not considered in Friedel-Crafts alkylation due to the presence of the deactivating carboxylic acid group, which hinders this reaction. Therefore, in effect, it would be ranked as 'non' reactive for this specific reaction.

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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:

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:

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

Question: The question is incomplete and can't be comprehended. See the complete question below and the answer.

Consider a galvanic cell based upon the following half reactions: Ag+ + e- → Ag 0.80 V Cu2+ + 2 e- → Cu 0.34 V

How would the following changes alter the potential of the cell?

a) Adding Cu2+ ions to the copper half reaction (assuming no volume change).

b) Adding equal amounts of water to both half reactions.

c) Removing Cu2+ ions from solution by precipitating them out of the copper half reaction (assume no volume change).

d) Adding Ag+ ions to the silver half reaction (assume no volume change)

Explanation:

Nernst equation relates the reduction potential of an electrochemical reaction (half-cell or full cell reaction) to the standard electrode potential, temperature, and activities (often approximated by concentrations) of the chemical species undergoing reduction and oxidation.

Reaction under consideration:

Ag+ + e- → Ag 0.80 V

Cu+2 + 2 e- → Cu 0.34 V

Clearly, Ag reduction potential is high and this indicates that it gets reduced readily which leaves Cu to oxidize. Cu+2 ions are products of reaction and Ag+ ions are reactant ions.

Nernst equation : Ecell = E°cell­ – (2.303 RT / n F) log Q    

where                            

             Ecell = actual cell potential

             E°cell­ ­​ = standard cell potential

             R = the universal gas constant = 8.314472(15) J K−1 mol−1

             T = the temperature in kelvins

              n = the number of moles of electrons transferred                                    

 F = the Faraday constant, the number of coulombs per mole of electrons:

  (F = 9.64853399(24)×104 C mol−1)

 Q = [product ion]y / [reactant ion] x

Accordingly when applied to above reaction one will get the following

= E°cell­ – (2.303 x RT / 6 F) log [Cu+2] / [Ag+]

Now the given variables can be studied according to Le Chatelier's principle which states when any system at equilibrium is subjected to change in its concentration, temperature, volume, or pressure, then the system readjusts itself to (partially) counteract the effect of the applied change and a new equilibrium is established.

a)        Adding Cu2+ ions to the copper half reaction (assuming no volume change).

Addition of Cu+2 ions increases its concentration and consequently increases the Q value which results in reduction of Ecell. In other words the addition of Cu+2 ions favors the backward reaction to maintain the equilibrium of reaction and hence the forward reaction rate decreases.

b)       Adding equal amounts of water to both half reactions.

Addition of water increases the dilution of the electrochemical cell. For weak electrolytes such as Ag+/ Cu+2 with increase in dilution, the degree of dissociation increases and as a result molar conductance increases.

c)        Removing Cu2+ ions from solution by precipitating them out of the copper half reaction (assume no volume change).

Based on Le Chatelier's principle when Cu+2 ions amount is decreased by its continuous removal from  the system the forward reaction is favored. As the Cu+2 ions is removed the system attempts to generate more Cu+2 ions to counter the affect of its removal.

d)       Adding Ag+ ions to the silver half reaction (assume no volume change)

Addition of reactant ions, i.e. Ag+ ions, will favour the forward reaction, which results in more product formation.

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.

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Answer:

Nitrogen

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

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: 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:

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

Answer:

During a phase change, the temperature of a substance remains constant

Explanation:

Step 1:

During phase change, the energy that is required, will be used to separate the molecules ( to change phase).

This energy is not used to change the temperature.

Since the kinetic energy of the molecules remains the same, the temperature of the substance remains constant.

Answer:

fraction of the compound will be in the unprotonated, amine (amino) form is evaluated to be 83.37%

Explanation:

check the attachment for explicit explanations.

Using the Henderson-Hasselbalch equation, we determine that approximately 86% of the compound will be in the unprotonated, amine form after the pH change from 11 to 10.6.

We can use the Henderson-Hasselbalch equation to find the ratio of the protonated form to the unprotonated form:

Henderson-Hasselbalch equation:

Insert the known values:

Rearrange to solve for the ratio of [tex][R-NH_2][/tex] to [tex][R-NH_3^+][/tex]:

Raise 10 to the power of both sides to remove the logarithm: [tex]10^{0.8} \approx 6.31[/tex]

To find the fraction of the compound in the unprotonated form, we can use:

Convert this fraction to a percentage:

Therefore, approximately 86% of the compound will be in the unprotonated, amine form.

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:

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:

13

Explanation:

1M MOH solution dissociates into 0.1M OH- ions.

pOH= -log(OH-)

       =    1

pH=14-1

    =13

The pH of this solution is 13.

We can solve this question knowing that a hydroxide as MOH, dissociates in water as follows:

MOH(aq) ⇄ M⁺(aq) + OH⁻(aq)

Based on the reaction, 1 mole of MOH produces 1 mole of OH⁻

If 10% of the 1M MOH dissociates, a 0,1M of M⁺ and a 0.1M of OH⁻ are produced.

Now, with molarity of OH⁻ we can find pOH (pOH = -log[OH⁻]) and pH (pH = 14-pOH) as follows:

pOH:

pOH = -log [OH⁻] = -log [0.1] =1

And pH:

pH:

pH = 14 - pOH

pH = 13

You can learn more about pH and pOH in:

https://brainly.com/question/17144456

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 Keq value for the reaction is 0.13

The reactants are favored in the reaction.

Explanation:

The equilibrium expression:  N[tex]^{2}[/tex]O4(g) ⇌ 2 [NO[tex]^{2}[/tex]](g)  

Keq = squared of  [NO[tex]^{2}[/tex]] divided by [N[tex]^{2}[/tex]O4]

Keq = squared of [0.10] divided by [0.075]

Keq = 0.13

Equilibrium constant (Keq) indicates what will be favoured at equilibrium, as it is a ratio that quantifies the position of a chemical equilibrium at a given temperature. It measures the extent to which reactants are converted to products.

If equilibrium constant is less than 1, the mixture contains mostly reactants. Which indicates more reactants are present at equilibrium. If equilibrium constant is equal to 1, then, the amount of products is roughly equal to the amount of reactants at equilibrium. If equilibrium constant is much greater than 1, more products are present at equilibrium.

In the question treated above, the reactants are N[tex]^{2}[/tex]O4(g). While, the products are: 2 [NO[tex]^{2}[/tex]](g).

The value of [tex]\( K_{eq} \)[/tex] is 0.1333. Since [tex]\( K_{eq} \)[/tex] is less than 1, the reactants are favored at equilibrium is 0.1333.

The value of [tex]\( K_{eq} \)[/tex] for the decomposition of dinitrogen tetroxide [tex](\( N_2O_4 \))[/tex] into nitrogen dioxide [tex](\( NO_2 \))[/tex] is calculated by the following equilibrium expression:

[tex]\[ K_{eq} = \frac{[NO_2]^2}{[N_2O_4]} \][/tex]

Given that the initial concentration of [tex]\( N_2O_4 \)[/tex] is 0.125 mol/L and the equilibrium concentration is 0.0750 mol/L, we can determine the concentration of [tex]\( NO_2 \)[/tex] at equilibrium by using the stoichiometry of the reaction:

[tex]\[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \][/tex]

For every mole of [tex]\( N_2O_4 \)[/tex] that decomposes, 2 moles of [tex]\( NO_2 \)[/tex] are produced. The change in concentration of [tex]\( N_2O_4 \) (\( \Delta[N_2O_4] \))[/tex] is the initial concentration minus the equilibrium concentration:

[tex]\[ \Delta[N_2O_4] = [N_2O_4]_{initial} - [N_2O_4]_{equilibrium} \][/tex]

[tex]\[ \Delta[N_2O_4] = 0.125 \text{ mol/L} - 0.0750 \text{ mol/L} \][/tex]

[tex]\[ \Delta[N_2O_4] = 0.0500 \text{ mol/L} \][/tex]

Since 2 moles of [tex]\( NO_2 \)[/tex] are produced for every mole of [tex]\( N_2O_4 \)[/tex] that decomposes, the change in concentration of [tex]\( NO_2 \) (\( \Delta[NO_2] \))[/tex] is twice the change in concentration of [tex]\( N_2O_4 \)[/tex]:

[tex]\[ \Delta[NO_2] = 2 \times \Delta[N_2O_4] \][/tex]

[tex]\[ \Delta[NO_2] = 2 \times 0.0500 \text{ mol/L} \][/tex]

[tex]\[ \Delta[NO_2] = 0.1000 \text{ mol/L} \][/tex]

The equilibrium concentration of [tex]\( NO_2 \) (\( [NO_2]_{equilibrium} \))[/tex] is the initial concentration of [tex]\( NO_2 \)[/tex] (which is 0, since none was present initially) plus the change in concentration:

[tex]\[ [NO_2]_{equilibrium} = [NO_2]_{initial} + \Delta[NO_2] \][/tex]

[tex]\[ [NO_2]_{equilibrium} = 0 + 0.1000 \text{ mol/L} \][/tex]

[tex]\[ [NO_2]_{equilibrium} = 0.1000 \text{ mol/L} \][/tex]

Now we can calculate [tex]\( K_{eq} \)[/tex]:

[tex]\[ K_{eq} = \frac{[NO_2]_{equilibrium}^2}{[N_2O_4]_{equilibrium}} \][/tex]

[tex]\[ K_{eq} = \frac{(0.1000)^2}{0.0750} \][/tex]

[tex]\[ K_{eq} = \frac{0.01000}{0.0750} \][/tex]

[tex]\[ K_{eq} = 0.1333 \][/tex]

Answer:

I think it is D.

Explanation:

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

Explanation:

The enthalpy of combustion of ethanol is approximately -118.45 kJ/mol.

To determine the enthalpy of combustion (ΔH) of ethanol, we can use the heat transfer equation:

q = mcΔT

where q is the heat absorbed by the water, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.

First, calculate the heat absorbed by the water:

q = (730 g)(4.184 J/g ⋅ K)(20.5 °C - 16.8 °C)

q = 730 × 4.184 × 3.7 J

q = 11,300.984 J

Next, convert the mass of ethanol burned (4.40 g) to moles using the molar mass of ethanol (C2H5OH):

Molar mass of ethanol = 2 × molar mass of C + 6 × molar mass of H + molar mass of O

Molar mass of ethanol = 2 × 12.01 + 6 × 1.01 + 16.00 g/mol

Molar mass of ethanol = 46.07 g/mol

Moles of ethanol = 4.40 g / 46.08 g/mol

Moles of ethanol ≈ 0.0954 mol

Now, calculate ΔH using the formula:

ΔH = -q / moles of ethanol

ΔH = -11,300.984 J / 0.0954 mol

ΔH ≈ -118,458.952 J/mol

Convert to kJ/mol:

ΔH ≈ -118.45 kJ/mol

Answer:

Explanation:

Find attached the solution

Complete Question (in order)

The growth of baker's yeast (S. cerevisiae) on glucose may be simply described by following equation:

C₆H₁₂0₆ +  30₂ + 0.48 NH₃ => 0.48C₆H₁₂NO₃ + 4.32H₂O + 3.12CO₂ yeast

In a batch reactor of volume 10 1, the final desired yeast concentration is 50 gdw/1.

a. Determine the concentration and total amount of glucose and (NH4)₂SO4 in the nutrient medium.

b. Determine the yield coefficients [tex]Y_{x/s}[/tex] (biomass/glucose) and [tex]Y_{x/o}[/tex] (biomass/oxygen).

c. Determine the total amount of oxygen required.

d. If the rate of growth at exponential phase is [tex]r_x[/tex] = 0.7 gdw/l-h, determine the rate of oxygen consumption (g O₂/1-h)

e. Calculate the heat-removal requirements for the reactor (recall equation 6.26).

Answer:

a. 2.3 × 10⁴ kg

b. 0.384 and 0.72

c. 6.94 × 10³ kg

d. 0.973

Explanation:

The given equation that describes the growth of baker's yeast on glucose

C₆H₁₂0₆ +  30₂ + 0.48 NH₃ => 0.48C₆H₁₂NO₃ + 4.32H₂O + 3.12CO₂ yeast

The volume of batch reactor is given as 10⁵ Liters

The final desired yeast concentration is 50 gdw/l

The following attached documents gives additional solution for all other answers