Antifreeze is used in car engines. It is produced by a local manufacturer, but they’ve run out of an ingredient: ethane (C2H6). As a junior scientist, you know you can produce ethane by reacting sulfuric acid with diethyl zinc gas. The local manufacturer will pay $1.25 per gram of ethane you produce. If you have 45.00 mL of 8.00 M H2SO4 and 30.00 mL of Zn(CH3CH2)2, what is the maximum amount of money you can make?

Answer:

The new pressure at constant volume is 1066.56 kPa

Explanation:

Assuming constant volume, the pressure is diectly proportional to the temperature of a gas.

Mathematically, P1/T1 = P2 /T2

P1 = 880 kPA= 880 *10^3 Pa

T1 = 250 K

T2 = 303 K

P2 =?

Substituting for P2

P2 = P1 T2/ T1

P2 = 880 kPa * 303 / 250

P2 = 266,640 kPa/ 250

P2 = 1066.56 kPa.

The new pressure of the gas is 1066.56 kPa

At constant volume, if the temperature of the container is heated to the given value, the pressure of the gas increases to 1066.5kPa.

Gay-Lussac's law states that the pressure exerted by a given quantity of gas varies directly with the absolute temperature of the gas.

It is expressed as;

P₁/T₁ = P₂/T₂

Given the data in the question;

P₁/T₁ = P₂/T₂

P₁T₂ = P₂T₁

P₂ = P₁T₂ / T₁

P₂ = (8.68492atm × 303K) / 250K

P₂ = (2631.53atmK) / 250K

P₂ = 10.526atm

P₂ = (10.526 × 101.325)kPa

P₂ = 1066.5kPa

Therefore, at constant volume, if the temperature of the container is heated to the given value, the pressure of the gas increases to 1066.5kPa.

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Answer:

32.6 Jmol-1

Explanation:

Now we have to use the formula

∆G= ∆H-T∆S where

∆G= change in free energy (the unknown)

∆H= enthalpy change = 91.8KJ

∆S= entropy change= 0.1987 kJ/K

T= temperature= 298K

∆G= 91.8 - (298× 0.1987)

∆G= 91.8 - 59.2126

∆G= 32.6 Jmol-1

Answer:

the reaction is spontaneous

Explanation:

Answer:

Find the given attachment.

Answer:

0.333 M

Explanation:

Step 1: Write the balanced equation

2 HCl(aq) + Mg(OH)₂(aq) = MgCl₂(aq) + 2 H₂O(l)

Step 2: Calculate the reacting moles of Mg(OH)₂

25.00 mL of a 0.200 M magnesium hydroxide react. The reacting moles of Mg(OH)₂ are:

[tex]25.00 \times 10^{-3} L \times \frac{0.200mol}{L} = 5.00 \times 10^{-3} mol[/tex]

Step 3: Calculate the reacting moles of HCl

The molar ratio of HCl(aq) to Mg(OH)₂ is 2:1. The reacting moles of HCl are:

[tex]5.00 \times 10^{-3} molMg(OH)_2 \times \frac{2mol}{1molMg(OH)_2} =1.00 \times 10^{-2}molHCl[/tex]

Step 4: Calculate the molarity of HCl

[tex]\frac{1.00 \times 10^{-2}mol}{30.00 \times 10^{-3}L} = 0.333 M[/tex]

Answer:

a) ΔH°rxn = -9.2kJ/mol

b) ΔH°rxn = -9.2kJ/mol

Explanation:

Using Hess's law, you can find ΔH of a reaction from ΔH of formation of the substances involved in the reaction, thus:

ΔH°rxn = ∑(BE(reactants)) − ∑(BE(products))

Or:

ΔH°rxn = ∑(nΔH°f (products)) − ∑(mΔH°f (reactants))

For the reaction:

H₂(g) + I₂(g) → 2HI(g)

a) Using the first equation:

ΔH°rxn = ΔH (H-H) + ΔH (I-I) - 2ΔHBE (H-I)

ΔH°rxn = 436.4kJ + 151kJ - 2×298.3kJ

ΔH°rxn = -9.2kJ/mol

b) Using the second equation:

ΔH°rxn = 2Δ°f (HI) − ΔH°f (H₂) - ΔH°f (I₂)

ΔH°rxn = 2×25.9kJ - 0kJ - 61.0kJ

ΔH°rxn = -9.2kJ/mol

The reaction H₂(g) + I₂(g) → 2HI(g) has a ΔHo value of -9.2 kJ/mol, which is calculated using both bond dissociation energies and enthalpies of formation. This indicates the reaction is exothermic.

To calculate ΔHo for the reaction H₂(g) + I₂(g) → 2HI(g),first we will use the bond dissociation energies (BE), and second the enthalpies of formation (ΔHof).

(a) Using the equation ΔHorxn = ∑BE(reactants) − ∑BE(products), we get ΔHorxn = [1(436.4 kJ/mol) + 1(151 kJ/mol)] - 2(298.3 kJ/mol) = -9.2 kJ/mol. This asserts that the reaction is exothermic and energy is released in the process. The sum of bond energies of reactants is smaller than that of the products.

 (b) Using the equation ΔHorxn = ∑nΔHof(products) − ∑mΔHof(reactants), we get ΔHorxn = 2(25.9 kJ/mol) - [1(0 kJ/mol) + 1(61.0 kJ/mol)] = -9.2 kJ/mol. This value is same as obtained by the first method and thus validates it.

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Answer:

2H2 + O2 = 2H2O

Explanation:

From the original equation, you first need to write each component separately.

Left side: H = 2 ; O = 2 (number based on the subscript)

Right side: H = 2 ; O = 1 (number based on the subscripts; no subscript means that the element is just 1)

Notice that the number of H is already balance but the number of O is not. In order to balance the O, you need to multiply the element by 2, but you CANNOT do this by simply changing the subscript.

Left side: H = 2 ; O = 2

Right side: H = 2 x 2 = 4 ; O = 1 x 2 = 2

Now, notice that the number of O is now balance (both are 2) but the number of H is not (since  

You also multiply the left side H by 2. Hence,

Left side: H = 2 x 2 = 4 ; O = 2

Right side: H = 2 x 2 = 4 ; O = 1 x 2 = 2

The equation is now balanced.

Answer:

HNO₂(aq) + H₂O(l) ⇄ NO₂⁻(aq) + H₃O⁺

NO₂⁻(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq)

Explanation:

According to Brönsted-Lowry acid-base theory, nitrous acid is an acid because it transfers an H⁺ to another compound. The corresponding reaction is:

HNO₂(aq) + H₂O(l) ⇄ NO₂⁻(aq) + H₃O⁺

According to Brönsted-Lowry acid-base theory, nitrite ion is a base because it accepts an H⁺ from another compound. The corresponding reaction is:

NO₂⁻(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq)

The chemical reaction for nitrous acid in water is HNO₂(aq) + H₂O(l) ⇄ NO₂⁻(aq) + H₃O⁺

And, the chemical reaction for the nitrite ion in water, is  

NO₂⁻(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq)

Since nitrous acid should be an acid due to this it transferred an H⁺ to another compound so here the reaction should be HNO₂(aq) + H₂O(l) ⇄ NO₂⁻(aq) + H₃O⁺

Here nitrite ion should have a base due to this,  it accepts an H⁺ from another compound. So here the reaction is NO₂⁻(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq)

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Answer:

The volume of the air is 0.662 L

Explanation:

Charles's Law is a gas law that relates the volume and temperature of a certain amount of gas at constant pressure. This law says that for a given sum of gas at a constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases because the temperature is directly related to the energy of the movement they have. the gas molecules. This is represented by the quotient that exists between volume and temperature will always have the same value:

[tex]\frac{V}{T}=k[/tex]

If you have a certain volume of gas V1 that is at a temperature T1 at the beginning of the experiment and several the volume of gas to a new value V2, then the temperature will change to T2, and it will be true:

[tex]\frac{V1}{T1}=\frac{V2}{T2}[/tex]

In this case:

Replacing:

[tex]\frac{0.730 L}{301K}=\frac{V2}{273K}[/tex]

Solving:

[tex]V2=273K*\frac{0.730L}{301K}[/tex]

V2=0.662 L

The volume of the air is 0.662 L

Answer:

M = 0.3077 M

Explanation:

As I said in the comments, you are missing the required volume of the base to react with the KHP. I found this on another site, and the volume it used was 21.84 mL.

Now, KHP is a compound often used to standarize NaOH or KOH solutions. This is because it contains a mole ratio of 1:1 with the base, so it's pretty easy to use and standarize any base.

Now, as we are using an acid base titration, the general expression to use when a acid base titration reach the equivalence point would be:

n₁ = n₂   (1)

This, of course, if the mole ratio is 1:1. In the case of KHP and NaOH it is.

Now, we also know that moles can be expressed like this:

n = M * V   (2)

And according to this, we are given the volume of base and the required mass of KHP. So, if we want to know the concentration of the base, we need to get the moles of the KHP, because in the equivalence point, these moles are the same moles of base.

The reported molar mass of KHP is 204.22 g/mol, so the moles are:

n = 1.372 / 204.22 = 6.72x10⁻³ moles

Now, we will use expression (2) to get the concentration of the diluted base:

n = M * V

M = n / V

M = 6.72x10⁻³ / 0.02184

M NaOH = 0.3077 M

This is the concentration of the dilute solution of NaOH

The molar concentration of sodium hydroxide ( NaOH ) is ; 0.3077 M

Given data :

Volume of base = 21.84 mL = 0.02184 L  ( missing data )

Molar mass of KHP = 204.22 g/mol

mass of KHP = 1.372 g

At equivalence point

n1 = n2   ( ratio of KHP to NaOH )

note : n = M * V  ---- ( 1 )

First step : calculate the number of moles of KHP

n = mass / molar mass

n = 1.372 / 204.22 =   6.72 * 10⁻³ moles

Determine the molar concentration of NaOH

From equation ( 2 )

M ( molar concentration ) = n / V

                                          = 6.72 * 10⁻³ moles / 0.02184 L

                                          = 0.3077 M

Hence we can conclude that The molar concentration of sodium hydroxide ( NaOH ) is  0.3077 M

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Answer:

A. Reject Upper H 0. There is sufficient evidence to conclude that a higher proportion of subjects in group 1 experienced drowsiness as a side effect than subjects in group 2 at the alphaequals0.10 level of significance.

Explanation:

H0: p1 = p2

Ha: p1 > p2

pcap = (101 + 72)/(489 + 625) = 0.1553

SE = sqrt(0.1553*(1-0.1553)*(1/489 + 1/625)) = 0.0219

Test statistic,

z = (101/489 - 72/625)/0.0219 = 4.1710

p-value = 0.0000

Reject H0

A. Reject Upper H 0. There is sufficient evidence to conclude that a higher proportion of subjects in group 1 experienced drowsiness as a side effect than subjects in group 2 at the alphaequals0.10 level of significance.

Answer:

see explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem

The heat capacity of the hydrocarbon is 0.389 kJ/°C.

To find the heat capacity of the hydrocarbon, we can use the formula:

Heat Capacity = q / ∆T

where q is the heat added and ∆T is the change in temperature.

Given that q = 9.54 kJ and ∆T = 45.0 °C - 20.5 °C = 24.5 °C, we can plug these values into the formula:

Heat Capacity = 9.54 kJ / 24.5 °C = 0.389 kJ/°C

Therefore, the heat capacity of the hydrocarbon is 0.389 kJ/°C.

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Answer:

Theoretical yield is 20%.

Explanation:

Percent yield = [(actual yield)/(theoretical yield)][tex]\times 100[/tex] %

Theoretical yield is calculated amount of product by using stoichiometric ratio between reactant and product.

Actual yield is the amount of product which is experimentally obtained.

Here theoretical yield is 20.0 g and actual yield is 4.0 g.

So, percent yield = [tex]\frac{4.0g}{20.0}\times 100[/tex] % = 20%

Answer:

There will be produced 30.64 grams of nitrogen dioxide

The statement is false

Explanation:

Step 1: Data given

MAss of nitrogen produced = 20.0 grams

Molar mass of N2 = 28.0 g/mol

Mass of NO = 20.0 grams

Molar mass of NO = 30.01 g/mol

Volume of O2 at STP = 29.8 L

Step 2: the balanced equation

2NO + O2 → 2N02

Step 3: Calculate moles NO

Moles NO = mass NO / molar mas NO

Moles NO = 20.0 grams / 30.01 g/mol

Moles NO = 0.666 moles

Step 4: Calculate moles O2

1 mol O2 at STP = 22.4 L

29.8 L = 29.8/22.4 = 1.33 moles

Step 5: Calculate the limiting reactant

For 2 moles NO we need 1 mol O2 to produce 2 moles NO2

NO is the limiting reactant. IT will completely be consumed (0.666 moles).

O2 is in excess. There will react 0.666/2 = 0.333 moles O2

There will remain 1.333 - 0.333 = 0.997 moles O2

Step 6: Calculate moles NO2

For 2 moles NO we need 1 mol O2 to produce 2 moles NO2

For 0.666 moles NO we'll have 0.666 moles NO2

Step 7: Calculate mass NO2

Mass NO2 = moles NO2 * molar mass NO2

Mass NO2 = 0.666 moles * 46.0 g/mol

Mass NO2 = 30.64 grams

There will be produced 30.64 grams of nitrogen dioxide

The statement is false

Answer:

CO2 contains sp hybridized carbon and sp2 oxygen atoms (linear shape)

Explanation:

There are two sigma bonds and two pi binds in the CO2 molecule. Carbon in its ground state contains one outer 2s orbital filled with two electrons and two outer 2p orbitals which are singly filled. Oxygen contains in its ground state contains an outer 2s orbital and three 2p orbitals filled with four electrons.

When these orbitals on oxygen are sp2 hybridized, one orbital is left unhybridized in each oxygen atom. Recall that two hybrized sp2 orbitals on oxygen atom accommodate the two lone pairs and one sp2 hybridized orbital in each oxygen atom forms a sigma bond to carbon. The remaining unhybridized orbital on oxygen is used by each oxygen atom to overlap with each unhybridized p orbital on the sp hybridized carbon.

Carbon forms two sigma bonds to oxygen via the two hybridized sp orbitals on carbon. The two unhybridized orbitals overlap sideways and firm pi bonds with oxygen p orbitals.

Triatomic molecules must be either linear or bent. In CO2, the bond angle must be 180° giving a linear molecule.

The carbon atom in CO₂ has sp hybridization due to its two regions of high electron density. Each of these forms a sigma bond with an oxygen atom, resulting in a linear geometry. Each C-O bond consists of one sigma bond and one pi bond, forming double bonds.

The question asks for a hybridization and bonding scheme for the molecule CO₂ based on the valence bond theory. In CO₂, the central atom is Carbon (C). This atom is surrounded by two regions of high electron density, each represented by the double bonds with oxygen atoms.

According to the valence bond theory, a central atom with two regions of electron density (Lone pairs or bonds) is associated with sp hybridization. Therefore, in the case of CO₂, the Carbon atom will have sp hybridization. Since sp hybridization generates two hybrid orbitals, each of these will form a σ (sigma) bond with one of the Oxygen atoms maintaining a linear geometry (180° angle).

Moving onto the molecule's bond character, each C-O bond in CO₂ is a double bond, consisting of a σ bond and a π bond. The σ bonds are formed by the overlapping of sp hybrid orbitals of carbon with p orbitals of oxygen, while the π bonds are formed by the side-by-side overlapping of unhybridized p orbitals from both Carbon and Oxygen atoms.

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Answer:

1) 25.44 moles H

2) 0.444 moles H

3) 792 moles H

Explanation:

Step 1: Number of moles H in 8.48 moles NH3

In 1 mol NH3 we have 3 moles H

For 8.48 moles NH3 we have 3*8.48 = 25.44 moles H

Step 2: Number of moles H in 0.111  moles N2H4

For 1 mol N2H4 we have 4 moles H

For 0.111 moles N2H4 we have 4*0.111 = 0.444 moles H

Step 3: Number of moles H in 36.0 moles C1OH22

For 1 mol C10H22 we have 22 moles of H

For 36.0 moles C1OH22 we have 22*36.0 = 792 moles H

Answer:

See explaination

Explanation:

moles of benzamide = mass / Molar mass of it = 70.4g / ( 121.14g/mol) = 0.58 mol

Molality = moles of solute ( benzamide) / ( solvent mass in kg) = 0.58 mol / ( 0.85kg) = 0.6837

we have formula dT = i x Kf x m , where dT = change in freezing point = 2.7C , i = vantoff factor = 1 for non dissociable solutes , Kf = freezing oint constant of solvent , m = 0.6837

hence 2.7C = 1 x Kf x 0.6837m

Kf = 3.949 C/m

we use this Kf value for calculating i for NH4Cl , where moles of NH4Cl = ( 70.4g/53.491g/mol) =1.316 mol

molality = ( 1.316mol) / ( 0.85kg) = 1.5484 , dT = 9.9

hence 9.9 = i x 3.949C/m x 1.5484 m

i = 1.62

Answer:

The reaction reaches equilibrium at the fourth panel.

Explanation:

Chemical Equilibrium is achieved when the overall properties the system seem to be constant, that is, stop changing.

Although, for chemical equilibrium, the right term for this equilibrium is dynamic equilibrium; the rate of forward reaction balances the rate of backward reaction, but concentrations can keep changing.

The point where equilibrium is achieved is when exactly when we reach the panel where the spheres that make up this panel is the same as the next panel and the next, that is, the specific colour and number of spheres start to become unchanged.

And from the description given in the question,

- In the first panel, there are ten large red spheres.

- In the second panel, there are 8 large red spheres and two small blue spheres.

- In the third panel, there are six large red spheres and four small blue spheres.

- In the fourth panel, there are four large red spheres and six small blue spheres.

- In the fifth panel, there are four large red spheres and six small blue spheres.

It is evident that the make-up of the spheres have become the same as at the fourth and fifth panel. This means that the first point where this final configuration of spheres first appeared is the fourth panel.

Hence, equilibrium is first reached at the fourth panel.

Hope this Helps!!!

The reaction reaches equilibrium at the fourth panel. A reaction is at equilibrium when the amounts of reactants or products no longer change. Chemical equilibrium is a dynamic process

It is the rate of formation of products by the forward reaction is equal to the rate at which the products re-form reactants by the reverse reaction.

The point where equilibrium is achieved is when exactly when we reach the panel where the spheres that make up this panel is the same as the next panel and the next, that is, the specific color and number of spheres start to become unchanged.

As per the descriptions of panel given in question:

It is evident that the make-up of the spheres have become the same as at the fourth and fifth panel. This means that the first point where this final configuration of spheres first appeared is the fourth panel.

Hence, equilibrium is first reached at the fourth panel.

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Answer:

B) increase by 0.18 V

Explanation:

The given chemical spontaneous reaction is :

[tex]2 AgCl_{(s)} + Zn _{(s)} \to 2Ag (s) + 2Cl^- _{(aq)} + Zn ^{2+} _{(aq)}[/tex]

By applying Nernst Equation:

[tex]E_{cell} = E^0 - \frac{0.059}{n} log [\frac{product}{reactant}][/tex]

here ;

n = number of electrons transferred in the reaction

n =2

[tex]E^0 = E^0_{cathode } - E^0_{anode}[/tex]

[tex]E^0 = E^0_{Ag^+/Ag } - E^0_{Zn^{2+}/Zn}[/tex]

[tex]E^0 =+0.80 \ V -(-0.76 \ V)[/tex]

[tex]E^0 =1.56 \ V[/tex]

When it happens to occur that the concentration of chlorine (aq) and Zn²⁺ (aq) is 1 M ;

[tex]E^0_{cell} = E^0[/tex] is  as follows:

[tex]E_{cell} = E^0 - \frac{0.059}{n} log [\frac{product}{reactant}][/tex]

[tex]E_{cell} = 1.56 - \frac{0.059}{n} log [\frac{[Zn^{2+}]}{[Cl^-]^2}][/tex]

[tex]E_{cell} = 1.56 - \frac{0.059}{n} log (1) \ \ \ \ \ \ \ ( where \ log (1) = 0)[/tex]

[tex]E_{cell} = 1.56 \ V[/tex]

Now; the [tex]E_{cell}[/tex] value in the decreased concentration of chlorine (aq) ion is calculated as:

[tex]E_{cell} = E^0 - \frac{0.059}{n} log [\frac{product}{reactant}][/tex]

[tex]E_{cell} = 1.56 - \frac{0.059}{n} log [\frac{[Zn^{2+}]}{[Cl^-]^2}][/tex]

[tex]E_{cell} = 1.56 - \frac{0.059}{n} log (1*0.001^2)[/tex]

[tex]E_{cell} = +1.737 \ V[/tex]

Hence; the change in voltage = [tex]E_{cell} - E^0[/tex]

= 1.737 - 1.56

= 0.177 V

≅ 0.18 V

We therefore conclude that: since the [tex]E^0_{cell}[/tex] value after the decreased concentration of Chlorine is greater than the [tex]E^0[/tex] before the change; then there is increase in the value by 0.18 V

To calculate αY4-, we need to find the fraction of EDTA in its protonated forms at the given pH and divide the concentration of the completely unprotonated form by the total concentration of EDTA.

To calculate the fraction of EDTA that is in the completely unprotonated form, Y4- (designated as αY4-), at a given pH value, we need to find the values of αY4- at two different pH values.

To calculate αY4-, we first need to find the fraction of EDTA in its protonated forms at the given pH.

For example, at pH = 2, we can calculate the fractional composition of each protonated form by dividing the concentration of the protonated form by the total concentration of all protonation states of EDTA. We can repeat this calculation for another pH value to find αY4- at that pH.

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The fraction of EDTA in the completely unprotonated form at pH3 is approximately [tex]\( 9.28 \times 10^{-4} \)[/tex], and at pH8 is approximately [tex]\( 8.83 \times 10^{-7} \)[/tex].

To calculate the fraction of EDTA that is in the completely unprotonated form, [tex]\( \alpha_{Y^{4-}} \)[/tex], at a given pH, we use the following equation:

[tex]\[ \alpha_{Y^{4-}} = \frac{1}{1 + \sum_{i=1}^{6} 10^{(pH - pKa_i)}} \][/tex]

Here, [tex]\( pKa_i \)[/tex] represents the [tex]\( i \)-th[/tex] acid dissociation constant for EDTA. Since EDTA is a hexaprotic acid, it has six [tex]\( pKa \)[/tex] values, which are given as:

[tex]\( pKa_1 = 0.00 \), \( pKa_2 = 1.50 \), \( pKa_3 = 2.00 \), \( pKa_4 = 2.69 \), \( pKa_5 = 6.13 \), and \( pKa_6 = 10.37 \).[/tex]

Calculate [tex]\( \alpha_{Y^{4-}} \)[/tex] for two pH values, say pH = 3 and pH = 8.

For pH = 3:

[tex]\[ \alpha_{Y^{4-}} = \frac{1}{1 + 10^{(3 - 0.00)} + 10^{(3 - 1.50)} + 10^{(3 - 2.00)} + 10^{(3 - 2.69)} + 10^{(3 - 6.13)} + 10^{(3 - 10.37)}} \][/tex]

[tex]\[ \alpha_{Y^{4-}} = \frac{1}{1 + 10^{3} + 10^{1.5} + 10^{1} + 10^{0.31} + 10^{-3.13} + 10^{-7.37}} \][/tex]

[tex]\[ \alpha_{Y^{4-}} = \frac{1}{1 + 1000 + 31.62 + 10 + 2.08 + 0.00074 + 4.17 \times 10^{-8}} \][/tex]

[tex]\[ \alpha_{Y^{4-}} \approx \frac{1}{1 + 1000 + 31.62 + 10 + 2.08} \][/tex]

[tex]\[ \alpha_{Y^{4-}} \approx \frac{1}{1033.62 + 31.62 + 10 + 2.08} \][/tex]

[tex]\[ \alpha_{Y^{4-}} \approx \frac{1}{1077.32} \][/tex]

[tex]\[ \alpha_{Y^{4-}} \approx 9.28 \times 10^{-4} \][/tex]

For pH = 8:

[tex]\[ \alpha_{Y^{4-}} = \frac{1}{1 + 10^{(8 - 0.00)} + 10^{(8 - 1.50)} + 10^{(8 - 2.00)} + 10^{(8 - 2.69)} + 10^{(8 - 6.13)} + 10^{(8 - 10.37)}} \][/tex]

[tex]\[ \alpha_{Y^{4-}} = \frac{1}{1 + 10^{8} + 10^{6.5} + 10^{6} + 10^{5.31} + 10^{1.87} + 10^{-2.37}} \][/tex]

[tex]\[ \alpha_{Y^{4-}} = \frac{1}{1 + 10000000 + 316227.76 + 1000000 + 20794.42 + 74.12 + 0.00417} \][/tex]

[tex]\[ \alpha_{Y^{4-}} \approx \frac{1}{10000000 + 316227.76 + 1000000 + 20794.42 + 74.12} \][/tex]

[tex]\[ \alpha_{Y^{4-}} \approx \frac{1}{1131992.3} \][/tex]

[tex]\[ \alpha_{Y^{4-}} \approx 8.83 \times 10^{-7} \][/tex]

Therefore, the fraction of EDTA in the completely unprotonated form at pH 3 is approximately [tex]\( 9.28 \times 10^{-4} \)[/tex], and at pH 8 is approximately [tex]\( 8.83 \times 10^{-7} \)[/tex].

The complete question is:

EDTA is a hexaprotic system with the PK, values:

PKa1 = 0.00, pKa2 = 1.50, pKa3 2.00, pKa4 = 2.69,

pKa5 6.13, and pKa6 10.37.

The distribution of the various protonated forms of EDTA will therefore vary with pH. For equilibrium calculations involving metal complexes with EDTA, it is convenient to calculate the fraction of EDTA that is in the completely unprotonated form, Y4-. This fraction is designated αY4−. Calculate αY4−  at two pH values.

pH = 3.10

pH = 10.55

Answer:

50 mL of 12M HCl should be added into 100 mL of water to produce a 4M solution.

Explanation:

According to law of dilution-

                                      [tex]C_{1}V_{1}=C_{2}V_{2}[/tex]

where [tex]C_{1},C_{2},V_{1}[/tex] and [tex]V_{2}[/tex] are initial concentration, final concentration, initial volume and final volume respectively.

Here, [tex]C_{1}=12M[/tex] , [tex]C_{2}=4M[/tex] , [tex]V_{1}=50mL[/tex]

So, [tex]V_{2}=\frac{C_{1}V_{1}}{C_{2}}[/tex] = [tex]\frac{(12M)\times (50mL)}{(4M)}[/tex] = 150 mL

Hence, final volume of HCl solution should be 150 mL.

So, volume of water needed = (150-50) mL = 100 mL

Therefore 50 mL of 12M HCl should be added into 100 mL of water to produce a 4M solution.

You need to add [tex]\( {150} \)[/tex] mL of water to 50 mL of 12 M hydrochloric acid to produce a 4 M solution.

To determine how much water should be added to 50 mL of 12 M hydrochloric acid (HCl) to produce a 4 M solution, we can use the dilution formula which states:

[tex]\[ C_1V_1 = C_2V_2 \][/tex]

where:

- [tex]\( C_1 \)[/tex] and [tex]\( V_1 \)[/tex] are the concentration and volume of the initial solution (12 M and 50 mL, respectively),

- [tex]\( C_2 \)[/tex] and [tex]\( V_2 \)[/tex] are the concentration and volume of the final solution (4 M and the unknown volume of water, respectively).

Let's solve for [tex]\( V_2 \)[/tex]:

[tex]\[ C_1V_1 = C_2V_2 \][/tex]

Substitute the given values:

[tex]\[ 12 \times 50 = 4 \times V_2 \][/tex]

Solve for [tex]\( V_2 \)[/tex]:

[tex]\[ 600 = 4 \times V_2 \][/tex]

[tex]\[ V_2 = \frac{600}{4} \][/tex]

[tex]\[ V_2 = 150 \text{ mL} \][/tex]

Answer:

The amount of Chlorodecane in the unknown is 0.105nmols

Explanation:

a) Since the GC is in an isothermal state, Chlorohexane C6H13Cl (1.69 nmols) because of its lower boiling point will elute first and Chlorodecane C12H21Cl will elute second.

The area of the first peak corresponding to Chlorohexane is 32434 units.

The area of the second peak corresponding to chlorodecane is 2022 units.

Since the response factor of the compound is not given in question and considering the response factor is same for both the compounds, the answer will be as follow:

1.69 nmols of Chlorohexane gives 32434 units

How much of chlorodecane gives 2022 units

By cross multiplication;

Moles of Chlorodecane = 2022*1.69/32434

                                         =0.105nmols

Final answer:

After calculating T1, we will have our answer in Kelvin, which can be converted back to degrees Celsius by subtracting 273.15.

Explanation:

To determine at what temperature the NFL footballs were inflated before the pressure dropped to 10.5 psi from 12.5 psi, we can use the ideal gas law and assume that the volume of the footballs remains constant (since they're made of materials that don't expand or contract much with temperature). The ideal gas law implies a direct relationship between pressure and temperature, meaning that if the temperature drops, the pressure will also drop, provided the number of moles of gas and the volume remain constant.

Starting with the initial condition, where P1 = 12.5 psi and the final condition where P2 = 10.5 psi at T2 = 5°C, we can find the initial temperature (T1) using the following relation from Charles's Law:

P1/T1 = P2/T2

However, to use this formula, we need to convert the temperatures into Kelvin:

T1 = (12.5 psi × 278.15 K) / 10.5 psi

After calculating T1, we will have our answer in Kelvin, which can be converted back to degrees Celsius by subtracting 273.15.

Final answer:

The formula unit for an ionic compound represents the simplest ratio of positive and negative ions that result in a neutral compound, typically involving metals and nonmetals. A molecular formula for a molecule indicates the actual number of atoms of each element in a molecule, formed by covalent bonds between nonmetals. The periodic table helps distinguish between the two.

Explanation:

The difference between a formula unit for an ionic compound and a molecular formula for a molecule lies in the types of elements involved and the nature of their bonding. A formula unit refers to the simplest, neutral collection of ions in an ionic compound, which is comprised of metals and nonmetals. The periodic table can be used to determine which elements fall into these categories. On the other hand, a molecular formula represents the actual number and type of atoms in a molecule, which are bonded covalently and generally involve nonmetal elements.

For ionic compounds, the formula mass is used, which is the sum of atomic masses of all the elements in the empirical formula (simplest ratio of ions), with each multiplied by its corresponding subscript. This contrasts with a molecular compound's molecular mass, which is computed using the molecular formula, representing the total mass of all atoms in the actual molecule.

To summarize, the key distinction is that molecular formulas represent covalently-bonded nonmetals and give the specific numbers and types of atoms, whereas formula units represent ionic compounds composed of metals and nonmetals and indicate the simplest ratio of ions that results in neutrality.

Answer:

Hydrogen

Explanation:

It gains oxygen from carbon dioxide to form water

Oxidation is the addition of oxygen to an element in a chemical reaction

Answer:

Its D

Explanation:

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Answer:

The calorimeter constant would be 567.62 J/C

Explanation:

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Answer:

Collagen is a "fibrous proteins" present in the extracellular matrix.

Explanation:

Protein are made up of hundreds or thousands of smaller units called amino acids and is made of carbon, hydrogen, oxygen, nitrogen and sometimes sulfur and is found in many foods. Proteins are organic molecules found in living organisms.

There are three types of proteins; fibrous, globular, and membrane.

Collagen which is a fibrous proteins, form muscle fiber, tendons, connective tissue and bone.

Collagen are naturally occurring proteins that consist of single molecules made up of amino-acids, which are in turn built of carbon, oxygen and hydrogen. It is mostly found in fibrous tissues such as tendons, ligaments, and skin.

The synthesis of collagen occurs in two stages: intracellular and extracellular.

Collagen is an abundant connective tissues such as cartilage, tendons, bones, and ligaments. Collagen is a contributing factor to variation in meat tenderness and texture.

Answer: The decay energy is the energy released by a radioactive decay. Radioactive decay is the process in which an unstable atomic nucleus loses energy by emitting ionizing particles and radiation.

Answer:

Explanation:

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Answer:

Yes

Explanation: