The temperature versus time graph of a solid substance absorbing heat is shown. A graph is shown with Temperature followed by degree Celsius in parentheses labeled on the y axis and Time followed by minutes in parentheses labeled on the x axis. An upwards arrow is shown between y axis and the label and a right pointing arrow is shown between the x axis and the label. A slanting graph line starting at a point A near the intersection of the two axis is shown. The slanting graph after a point B starts running parallel to the x axis till point C. The line after point C slopes upwards till point D and then runs parallel to the x axis till point E after which it again slopes upwards to finally terminate at point F. What best describes the change taking place in section CD of the graph? The intermolecular bonds of the solid state are being broken as particles vibrate faster. The intermolecular bonds of the liquid state are being broken as particles flow faster. The particles of the solid vibrate faster as the kinetic energy of the particles increases. The particles of the liquid slide around faster as the kinetic energy of the particles increases.

Answer :

Oxidation number or oxidation state : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.

Rules for Oxidation Numbers are :

Now we have to determine the oxidation state of the elements in the compound.

(a) [tex]H_2SO_4[/tex]

Let the oxidation state of 'S' be, 'x'

[tex]2(+1)+x+4(-2)=0\\\\x=+6[/tex]

Hence, the oxidation state of 'S' is, (+6)

(b) [tex]Ca(OH)_2[/tex]

Let the oxidation state of 'Ca' be, 'x'

[tex]x+2(-2+1)=0\\\\x=+2[/tex]

Hence, the oxidation state of 'Ca' is, (+2)

(c) [tex]BrOH[/tex]

Let the oxidation state of 'Br' be, 'x'

[tex]x+(-2)+1=0\\\\x=+1[/tex]

Hence, the oxidation state of 'Br' is, (+1)

(d) [tex]ClNO_2[/tex]

Let the oxidation state of 'N' be, 'x'

[tex]-1+x+2(-2)=0\\\\x=+5[/tex]

Hence, the oxidation state of 'N' is, (+5)

(e) [tex]TiCl_4[/tex]

Let the oxidation state of 'Ti' be, 'x'

[tex]x+4(-1)=0\\\\x=+4[/tex]

Hence, the oxidation state of 'Ti' is, (+4)

(f) [tex]NaH[/tex]

Let the oxidation state of 'Na' be, 'x'

[tex]x+(-1)=0\\\\x=+1[/tex]

Hence, the oxidation state of 'Na' is, (+1)

Answer:

Concentration of hydrogen ion, [tex][H^+]=5.0118*10^{-6} M[/tex]

Explanation:

pH is defined as the negative logarithm of hydrogen ion's concentration.

The lower the value of pH, the higher the acidic the solution is.

The formula for pH can be written as:

[tex]pH=-log[H^+][/tex]

Given,

pH of the saliva of Marco = 5.3

To calculate: Hydrogen ion concentration in the saliva

Thus, applying in the formula as:

[tex]pH=-log[H^+][/tex]

[tex]5.3=-log[H^+][/tex]

So,

[tex]log[H^+]=-5.3[/tex]

[tex][H^+]=10^{(-5.3)}[/tex]

[tex][H^+]=5.0118*10^{-6} M[/tex]

The hydrogen ion concentration of Marco's saliva is approximately 5.01 x 10^-6 M.

The pH scale ranges from 0 to 14, where anything below 7 is acidic and above 7 is alkaline. The hydrogen ion (H+) concentration of a solution can be determined using the pH value. A change of one unit on the pH scale represents a ten-fold change in the concentration of H+ ions. Given that Marco's saliva has a pH of 5.3, we can calculate the hydrogen ion concentration as follows:

Therefore, the hydrogen ion concentration of Marco's saliva is approximately 5.01 x 10⁻⁶M.

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Answer:

so the reaction rate increases by a factor 6.

Explanation:

For the given equation the reaction is first order with respect to both ester and sodium hydroxide

So we can say that the rate law is

[tex]Rate(initial)=K[NaOH][CH_{3}COOC_{2}H_{5}][/tex]

now as per given conditions the concentration of ester is increased by half it means that the new concentration is 1.5 times of old concentration

The concentration of NaOH is quadrupled means the new concentration is 4 times of old concentration.

The new rate law is

[tex]Rate(final)=K[1.5XNaOH][4XCH_{3}COOC_{2}H_{5}][/tex]

the final rate = 6 X initial rate

so the reaction rate increases by a factor 6.

The reaction rate of ethyl acetate with sodium hydroxide would increase by a factor of 6 if the concentration of ethyl acetate is increased by half and the concentration of NaOH is quadrupled, as it is first order in both reactants.

The reaction of ethyl acetate with sodium hydroxide is:

This reaction is first order concerning both CH₃COOC₂H₅ and NaOH. The rate law for this reaction can be written as:

If the concentration of CH₃COOC₂H₅ is increased by half, its new concentration becomes 1.5 times its initial concentration. If the concentration of NaOH is quadrupled, its new concentration becomes 4 times its initial concentration. Therefore, the rate of the reaction increases by a factor of:

So, the reaction rate would increase by a factor of 6.

Answer:

Here's what I get.

Explanation:

The MO diagrams of KrBr, XeCl, and XeBr are shown below.

They are similar, except for the numbering of the valence shell orbitals.

Also, I have drawn the s and p orbitals at the same energy levels for both atoms in the compounds. That is obviously not the case.

However, the MO diagrams are approximately correct.

The ground state electron configuration of KrF is

[tex](1\sigma_{g})^{2}\, (1\sigma_{u}^{*})^{2} \, (2\sigma_{g})^{2} \, (2\sigma_{u}^{*})^{2} \, (3\sigma_{g})^{2} \,  (1\pi_{u})^{4} \, (1\pi_{g}^{*})^{4} \, (3\sigma_{g}^{*})^{1}[/tex]

KrF⁺ will have one less electron than KrF.

You remove the antibonding electron from the highest energy orbital, so the bond order increases.

The KrF bond will be stronger.

Final answer:

The molecular orbital diagram for noble gas compounds like KrF can be explained by the ground-state electron configuration and the expected bond strength of its cationic form.

Explanation:

Noble Gas Compounds Molecular Orbital Diagram:

The ground-state electron configuration of KrF is [Kr] 5s² 5p⁶. The cationic analog of KrF is likely to have a stronger bond due to the removal of an electron, leading to a decrease in repulsion between the nuclei and shared electrons.

Comparing Bond Strength in KrF vs. KrF⁺:

In KrF, the valence electrons fill the σ bonding and σ non-bonding orbitals, and the π bonding orbitals are empty.

When KrF loses an electron to become KrF⁺, it removes an electron from the σ non-bonding orbital (lone pair).

Losing this electron strengthens the bond because it removes electron density that opposes the bonding interaction in the σ bonding orbital. This is similar to losing a lone pair in other molecules.

Therefore, KrF⁺ is likely to have a stronger bond than KrF.

Answer:

100.71 cm H2O

Explanation:

In a fluid column barometer, the height of the fluid column is proportional to the pressure. The pressure is by definition:

[tex]P=\frac{F}{A}[/tex], where F is a force and A is the area.

In a column barometer the force is given by the weight of the fluid:

[tex]F=m*g[/tex], and the mass may be expressed as [tex]m=p*V[/tex], where [tex]p[/tex] is the density and V is the volume.

Replacing this in the pressure definition:

[tex]P=\frac{pVg}{A}[/tex]

In a constant cross section area column, the volume may be calculated as:

[tex]V=A*h[/tex] where A is the area and h the height. Replacing this in the previous equation:

[tex]P=\frac{pAhg}{A}=pgh[/tex]

Different columns may be over the same pressure, so:

[tex]P_{w}=P_{Hg}\\ p_{w}h_{w}g=p_{Hg}h_{Hg}g\\[/tex]

Dividing each part for gravity constant:

[tex]p_{w}h_{w}=p_{Hg}h_{Hg}[/tex]

And isolating hw:

[tex]h_{w}=\frac{p_{Hg}h_{Hg}}{p_{w} } \\h_{w}=\frac{13.5*746}{1.00}=10071[/tex] mm

It is equal to 1007,1 cm.

Try the suggested option; answer is marked with red colour (18.4953 °C).

All the details are in the attached picture.

The equilibrium temperature of the system is required.

The equilibrium temperature of the mixture is [tex]18.48^{\circ}\text{C}[/tex].

[tex]m_i[/tex] = Mass of ice = 25 kg

[tex]m_s[/tex] = Mass of steam = 4 kg

[tex]T_i[/tex] = Temperature of ice = [tex]0^{\circ}\text{C}[/tex]

[tex]T_s[/tex] = Temperature of steam = [tex]100^{\circ}\text{C}[/tex]

[tex]L_f[/tex] = Latent heat of fusion = [tex]3.34\times 10^5\ \text{J/kg}[/tex]

[tex]L_v[/tex] = Latent heat of vaporization = [tex]2.23\times 10^6\ \text{J/kg}[/tex]

[tex]c_w[/tex] = Specific heat of water = [tex]4180\ \text{J/kg}^{\circ}\text{C}[/tex]

The heat balance of the system will be

[tex]m_iL_f+m_ic_w(T-T_i)=m_sL_v+m_sc_w(T_s-T)\\\Rightarrow m_iL_f+m_ic_wT-m_ic_wT_i=m_sL_v+m_sc_wT_s-m_sc_wT\\\Rightarrow T=\dfrac{m_ic_wT_i+m_sL_v+m_sc_wT_s-m_iL_f}{m_ic_w+m_sc_w}\\\Rightarrow T=\dfrac{25\times 4180\times 0+4\times 2.23\times 10^6+4\times 4180\times 100-25\times 3.34\times 10^5}{25\times 4180+4\times 4180}\\\Rightarrow T=18.49^{\circ}\text{C}[/tex]

The equilibrium temperature of the mixture is [tex]18.49^{\circ}\text{C}[/tex].

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Answer : The heat of combustion is, 15642 J

Solution :

Formula used :

[tex]q=m\times c\times \Delta T[/tex]

where,

[tex]q[/tex] = heat of combustion = ?

[tex]m[/tex] = mass of fructose = 1.0 g

[tex]c[/tex] = heat capacity of the calorimteter  = [tex]9.90KJ/g^oC=9900J/g^oC[/tex]

conversion used : 1 KJ = 1000 J

[tex]\Delta T[/tex] = change in temperature = [tex]1.58^oC[/tex]

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

[tex]q=1.0g\times 9900J/g^oC\times 1.58^oC[/tex]

[tex]q=15642J[/tex]

Therefore, the heat of combustion is, 15642 J

Answer:

q combustion = -15.6 kJ (exothermic)

Explanation:

Bomb calorimeter questions are interesting because we are usually given the heat capacity of the bomb calorimeter not the specific heat capacity (which includes grams in the unit). Therefore we don't actually need to include the mass of the fructose as long as we know how much the temperature of the calorimeter changed (ΔT) and the heat capacity of the calorimeter.

We know that:

-q (combustion) = q (calorimeter) and we have enough information to calculate q (calorimeter):

   q (calorimeter) = (Heat Capacity)(Change in Temp.)

⇒ q(cal) = (9.90 kJ/°C)(1.58 °C) = 15.642 kJ

                                                  = 15.6 kJ (3 sig figs)

Since -q (combustion) = q (calorimeter), then:  

q (combustion) = -15.6 kJ (negative sign simply means heat released)

Answer:

The molecular formula of mono sodium glutamate is [tex]C_5H_8O_4N_1Na_1[/tex]

Explanation:

Molar mass of sodium glutamate,M = 169 g/mol

let the molecular formula be [tex]C_aH_bO_cN_dNa_e[/tex]

Percentage of carbon in the M.S.G. =35.52 %

[tex]35.51\%=\frac{a\times 12 g/mol}{169 g/mol}[/tex]

a = 5

Percentage of Hydrogen in the M.S.G. = 4.77 %

[tex]4.77\%=\frac{b\times 1 g/mol}{169 g/mol}[/tex]

b = 8

Percentage of oxygen in the M.S.G. =37.85 %

[tex]8.29\%=\frac{c\times 16 g/mol}{169 g/mol}[/tex]

c = 3.99 ≈ 4

Percentage of nitrogen in the M.S.G. = 8.29 %

[tex]4.77\%=\frac{d\times 14 g/mol}{169 g/mol}[/tex]

d = 1

Percentage of sodium in the M.S.G. =13.60 %

[tex]13.60\%=\frac{c\times 23g/mol}{169 g/mol}[/tex]

e = 0.99 ≈ 1

The molecular formula be :[tex]C_aH_bO_cN_dNa_e=C_5H_8O_4N_1Na_1[/tex]

The molecular formula of Monosodium glutamate is C2H4O4N2Na2

To determine the molecular formula of monosodium glutamate based on its elemental composition, we'll first find the empirical formula, and then calculate the molecular formula.

Find the moles of each element:

Carbon (C): 35.51%

Hydrogen (H): 4.77%

Oxygen (O): 37.85%

Nitrogen (N): 8.29%

Sodium (Na): 13.60%

Calculate the moles of each element using their molar masses:

Moles of C = (35.51/100) * 169 g / (12.01 g/mol) ≈ 5.97 moles

Moles of H = (4.77/100) * 169 g / (1.01 g/mol) ≈ 7.90 moles

Moles of O = (37.85/100) * 169 g / (16.00 g/mol) ≈ 8.37 moles

Moles of N = (8.29/100) * 169 g / (14.01 g/mol) ≈ 9.99 moles

Moles of Na = (13.60/100) * 169 g / (22.99 g/mol) ≈ 9.43 moles

Find the smallest whole number ratio of moles.

Divide all moles by the smallest number of moles (approximately 5.97).

Empirical Formula:

C1H1.32O1.4N1.68Na1.58

Round the subscripts to whole numbers (since you can't have fractions of atoms):

CH2O2N2Na2

The empirical formula of Monosodium glutamate is CH2O2N2Na2.

To find the molecular formula, you need to determine the molar mass of the empirical formula and compare it to the given molar mass (169 g/mol).

The empirical formula mass is 85 g/mol (approximately), which is half of the molar mass.

Therefore, the molecular formula is twice the empirical formula:

Molecular Formula: C2H4O4N2Na2

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Answer:

(a) forward direction

(b) forward direction

(c) backward direction.

Explanation:

Given , the chemical reaction in equilibrium is,

SO₂(g)  + Cl₂(g)  ⇄  SO₂Cl₂ (g)

The direction of the reaction by changing the concentration can be determined by Le Chatelier's principle,

It states that ,

When a reaction is at equlibrium , Changing the concentration , pressure,  temperature disturbs the equilibrium , and the reaction again tries to attain equilibrium by counteracting the changes.

(a)

For the reaction , Cl₂ is added to the system , i.e. , increasing the concentration of Cl₂ ,Now, according to Le Chatelier , The reaction will move in forward direction , to reduce the increased amount of Cl₂.

Hence, reaction will go in forward direction.

(b)

Removing SO₂Cl₂ from the system ,i.e. , decreasing the concentration of SO₂Cl₂  , according to Le Chatelier , the reaction will move in forward direction , to increase the amount of reduced SO₂Cl₂.

Hence, reaction will go in forward direction.

(c)

Removing SO₂ from the system , i.e. decreasing the concentration of SO₂ ,  according to Le Chatelier , the reaction will move in backward direction , to increase the amount of reduced SO₂.

Hence, reaction will go in backward direction.

Answer:So we can say that the rate of reaction increases by factor of 6.

Explanation:

The rate law for any given reaction  

A+B⇄C+D

Rate law for the above reaction is:

R=K[A]ᵃ[B]ᵇ

a and b are the order of reaction and it is an experimentally determined quantity.

K is the rate constant and it is constant for a given reaction

[A] and [B] are the concentrations of the reactants.

R is the rate of reaction

For the given reaction :

H₂O₂(aq.)+I₂(aq.)⇆OH⁻(aq.)+HIO(aq.)

Also it is given for the reaction that order with respect to H₂O₂ is 1 and order with respect to I₂ is also 1

The rate law can be written as :

R=K[H₂O₂]¹[I₂]¹

k=rate constant

When we increase the  concentration of H₂O₂ by half which meansthat new concentration of H₂O₂ will be= 3/2[H₂O₂].

When we increase the  concentration of I₂ by 4 which means that new concentration of I₂ will be= 4[I₂]

Putting the values of our new concentration in the rate law:

R=K3/2[H₂O₂]¹4[I₂]¹

R=6K[H₂O₂]¹[I₂]¹=New rate

So as we can see that the new rate of the reaction using the new concentration is 6 times the older rate of reaction.

So we can say that the rate of reaction increases by factor of 6.

Answer:

1.08 atm is the total pressure in the flask at equilibrium.

The value of the [tex]K_p=1.44[/tex].

Explanation:

Partial pressure of A at equilibrium = [tex]p_a=0.37 atm[/tex]

Partial pressure of B at equilibrium = [tex]p_b=2x[/tex]

                           A(g)      ⇄    2B(g)

at t=0                 0.73 atm           0

At equilibrium  (0.73- x)         2x

[tex]p_a=0.37 atm=(0.73- x)[/tex]

x = 0.36 atm

[tex]p_b=2x=2\times 036 atm=0.72 atm[/tex]

Total pressure in the flask at equilibrium = P

[tex]P=p_a+p_b=0.36 atm+0.72 atm = 1.08 atm[/tex]

The expression of equilibrium constant will be given as:

[tex]K_p=\frac{p_{b}^2}{p_a}=\frac{(0.72 atm)^2}{0.36 atm}=1.44[/tex]

The partial pressure is the pressure exerted by individual gases in the volume it is present. 1.08 atm is the total pressure and 1.44 is the equilibrium constant.

The total static and velocity pressure of a system is called total pressure. While the equilibrium constant is the proportion of the partial pressure of products and reactants.

Given,

From the reaction,

[tex]\begin{aligned}p_{a} &= 0.37 \rm\; atm \\\\\\&= (0.73 - x)\\\\\rm x &= 0.36 \;\rm atm\end{aligned}[/tex]

Solving the partial pressure of B:

[tex]\begin{aligned}p_{b} &= 2\rm x\\\\&= 2 \times 0.36\\\\&= 0.72\;\rm atm\end{aligned}[/tex]

Total pressure (P) in the flask will be:

[tex]\begin{aligned} \rm P &= p_{a}+ p_{b}\\\\&=0.36 + 0.72\\\\&= 1.08 \;\rm atm\end{aligned}[/tex]

Also, the equilibrium constant  will be calculated as:

[tex]\begin{aligned} K_{p} &= \dfrac{p_{b}^{2}}{p_{a}}\\\\&=\dfrac{(0.72)^{2}}{0.3.6}\\\\&=1.44 \end{aligned}[/tex]

Therefore, A. 1.08 atm is the total pressure and B. 1.44 is the equilibrium constant.

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To find the molar mass of a gas collected over water, apply the ideal gas law to determine the moles of gas. Then, divide the mass of the gas by the number of moles.

To determine the molar mass of a gas from an experiment in a general chemistry laboratory, you first need to use the ideal gas law, which is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant and T is the temperature in Kelvin.

The student collected a gas with a volume of 265 mL, which is equivalent to 0.265 L, at a pressure of 753 torr. Since 1 atm is equal to 760 torr, the pressure in atmospheres is 753 torr / 760 torr/atm = 0.9911 atm. The temperature must be converted from Celsius to Kelvin; thus, 27 °C is equal to 300.15 K (27 + 273.15 = 300.15 K).

To solve for n (the number of moles), you rearrange the ideal gas law to n = PV / RT. With the previously mentioned values and the gas constant R as 0.0821 L·atm/K·mol, n = (0.9911 atm × 0.265 L) / (0.0821 L·atm/K·mol × 300.15 K). After calculation, the number of moles of gas n is found.

Once n is calculated, the molar mass (M) can be found using the formula M = mass of gas (g) / number of moles (mol). Therefore, with the mass of the gas being 0.472 g, we calculate M = 0.472 g / n moles. By plugging in the value of n from our previous calculation, we can determine the molar mass.

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Final answer:

To calculate the molar mass of a gas, the ideal gas law is used with the given pressure, volume, and temperature. The molar mass is found to be approximately 44.22 g/mol after doing the conversions and calculations.

Explanation:

The student's question asks about determining the molar mass of a collected gas sample. To find out the molar mass of the gas, we use the ideal gas law formula (PV = nRT), where P is pressure, V is volume, n is number of moles, R is the gas constant, and T is temperature. Here's the step-by-step solution:

Convert the given pressure from torr to atm (753 torr / 760 torr/atm).

Convert the volume from mL to L (265 mL / 1000 mL/L).

Convert the temperature from Celsius to Kelvin (27 °C + 273.15).

Solve the ideal gas law equation for the number of moles (n).

Use the mass of the gas and the number of moles calculated to find the molar mass (molar mass = mass/n).

The detailed calculations show:

Solve for n: PV = nRT → n = PV / RT = (0.991 atm × 0.265 L) / (0.0821 L·atm/(mol·K) × 300.15 K) ≈ 0.01067 mol.

Finally, determine the molar mass: molar mass = mass/n = 0.472 g / 0.01067 mol ≈ 44.22 g/mol.

Hence, the molar mass of the gas is approximately 44.22 g/mol.

Answer:The reaction would shift to the left on adding Br2(g)

             The reaction would shift to right on adding BrNO(g)

Explanation:

The concept can be explained on the basis of LeChateliers principle.

LeChateliers principle explain the effect on equilibrium when a reaction system at equilibrium is subjected to any disturbance or change in conditions then the reaction shifts in such a way so as to reduce the effect of that change or disturbance and again establish the equilibrium.

For example if we add more reactants in a given reaction at equilibrium, the reaction would shift in such a way so that it can reduce the effect of increasing the concentration of reactants and hence the reaction would favor that direction in which it can reduce the concentration of reactants.So when we increase the concentration of reactants  the reaction would move towards the formation of more products and so the concentration of reactants would be less in the reaction.

Likewise if we increase the concentration of products so the reaction would shift in such a way so that it can oppose the increased concentration of products so the reaction moves towards more formation of reactants that is the products decompose to form reactants and reaction moves backwards.

In this reaction:

2BrNO(g)⇄2NO(g)+Br₂(g)

When we add more amount of Br₂(g) the reaction would proceed in such a ways so that oppose the increased concentration of Br₂(g).Hence the reaction would move towards left that is backwards.

When we add more amount of BrNO(g)the reaction would proceed in such a ways so that oppose the increased concentration of BrNO(g).Hence the reaction would move towards right that is forward.

Answer : The partial pressure of helium in this mixture is, 12.4 atm.

Explanation : Given,

Partial pressure of oxygen = 0.10 atm

Total partial pressure = 12.5 atm

Now we have to calculate the partial pressure of helium.

According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gasses.

[tex]P_T=p_{He}+p_{O_2}[/tex]

where,

[tex]P_T[/tex] = total partial pressure = 12.5 atm

[tex]P_{O_2}[/tex] = partial pressure of oxygen = 0.10 atm

[tex]P_{He}[/tex] = partial pressure of hydrogen = ?

Now put all the given values is expression, we get the partial pressure of the helium gas.

[tex]12.5atm=p_{He}+0.10atm[/tex]

[tex]p_{He}=12.4atm[/tex]

Therefore, the partial pressure of helium in this mixture is, 12.4 atm.

The partial pressure of helium in a deep-sea diving mixture of heliox, given a total pressure of 12.5 atm and an oxygen partial pressure of 0.10 atm, is 12.4 atm.

The subject of this question pertains to the principles of gas laws and partial pressures in a gas mixture, specifically applicable to scuba diving environments where deep-sea divers use a unique mixture of gases like heliox (mixture of helium and oxygen).

When the total pressure is 12.5 atm and the oxygen has a partial pressure of 0.10 atm, the partial pressure of helium in the mixture can be found via subtraction. Hence, the partial pressure of helium in this mixture can be found by subtracting the partial pressure of oxygen from the total pressure. That is, 12.5 atm - 0.10 atm = 12.4 atm. In other words, helium, as part of this heliox mixture, offers a partial pressure of 12.4 atm.

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Answer:

-4.15*[tex]10^{-5}mol/(L.s)[/tex]

Explanation:

Average rate            = [tex]\frac{final concentration -initial concentration}{change in time} = \frac{0.1076mol/L-0.1103mol/L}{65s}=-4.15.10^{-5}mol/(L.s)[/tex]

The rate is negative because it is a decomposition and our focus is the reactant which is depleting.

Answer:

False

Explanation:

Glucose is a monosachharide carbohydrate, with the molecular formula C₆H₁₂O₆.

Glucose molecule can exist in two forms-

1. Open chain form

2. cyclic form  

The open chain form of the glucose is an unbranched 6 carbon atom  chain. The carbon 1 of the molecule is an aldehyde group and the rest of the five carbon atoms have one hydroxyl group each.

The cyclic form of the glucose can be-

a. Pyranose: The pyranose form is a 6-membered cyclic ring, which consists of 5 carbon atoms and 1 oxygen atom in the ring.

b. Furanose: The furanose form is a 5- membered cyclic ring, which consists of 4 carbon atoms and 1 oxygen atom in the ring.

In an aqueous solution, 99% glucose molecule exists in the cyclic pyranose form as it is energetically more stable.

Therefore, in aqueous solution, the glucose molecule does not prefer the open-chain conformation.

Therefore, the statement is false.

Answer:

For a: The mass of one unit cell of copper is [tex]1.0553\times 10^{-22}g[/tex]

For b: The volume of copper unit cell is [tex]4.726\times 10^{-23}cm^3[/tex]

For c: The edge length of the unit cell is [tex]3.615\times 10^{-8}cm[/tex]

For d: The radius of a copper atom 127.82 pm.

Explanation:

We know that:

Mass of copper atom = 63.55 g/mol

According to mole concept:

1 mole of an atom contains [tex]6.022\times 10^{23}[/tex] number of atoms.

If, [tex]6.022\times 10^{23}[/tex] number of atoms occupies 63.55 grams.

So, 1 atom will occupy = [tex]\frac{63.55g}{6.022\times 10^{23}atom}\times 1 atom=1.0553\times 10^{-22}g[/tex]

Hence, the mass of one unit cell of copper is [tex]1.0553\times 10^{-22}g[/tex]

Copper crystallizes with a face-centered cubic lattice. This means that 4 number of copper atoms are present in 1 units cell.

Mass of 4 atoms of copper atom = [tex]1.0553\times 10^{-22}g/atom \times 4atoms=4.2212\times 10^{-22}g[/tex]

We are given:

Density of copper = [tex]8.93g/cm^3[/tex]

To find the volume of copper, we use the equation:

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

Putting values in above equation, we get:

[tex]8.93g/cm^3=\frac{4.2212\times 10^{-22}}{\text{Volume of copper}}\\\\\text{Volume of copper}=4.726\times 10^{-23}cm^3[/tex]

Hence, the volume of copper unit cell is [tex]4.726\times 10^{-23}cm^3[/tex]

Edge length of the unit cell is taken as 'a'

Volume of cube = [tex]a^3[/tex]

Putting the value of volume of unit in above equation, we get:

[tex]\sqrt[3]{4.726\times 10^{-23}}cm^3=3.615\times 10^{-8}cm[/tex]

Hence, the edge length of the unit cell is [tex]3.615\times 10^{-8}cm[/tex]

The relation of radius and edge length for a face-centered lattice follows:

[tex]a=r\sqrt{8}[/tex]

Putting values in above equation, we get:

[tex]3.615\times 10^{-8}=r\sqrt{8}\\\\r=1.2782\times 10^{-8}cm[/tex]

Converting cm to pm, we get:

[tex]1cm=10^{10}pm[/tex]

So, [tex]1.2782\times 10^{-8}cm=127.82pm[/tex]

Hence, the radius of a copper atom 127.82 pm.

hey there!:

Electron density on Cl atom depends on electronegativity difference between Cl and other bonded atom. If the electrnegativity difference is more then Cl has greater electron density, that menas if the bonded atom has less electronegativity then bonded electrons are more attracted by Cl and it has greater electron density.  

Among the five atoms which are bonded to Cl atom H has low electronegativity. So in H-Cl two bonded electrons are closer to Cl atom as it has greater electronegativity than H. This results more electron density on Cl atom.

Hence in H-Cl bond Cl atom have the highest electron density

Hope this helps!

The bond in which the chlorine atom has the highest electron density is H − C l.

The polarity of a bond depends on the magnitude of electronegativity difference between the atoms in the bond. The greater the electronegativity difference between the atoms in a bond the more the polarity of the bond.

The magnitude of electron density on each atom in a bond depends on its electronegativity. The more electronegative an atom is, the more it is able to accommodate larger electron density. Looking at the options, hydrogen is far less electronegative than chlorine so a large magnitude of electron density resides on the chlorine atom in HCl.

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Answer:

A. Mafic; iron and/or magnesium

Explanation:

Let's find the answer by naming some minerals and their chemistry.

Mafic minerals are dark-colored whereas felsic minerals are light-colored, thats way mafic rocks are dark-colored because they are mainly composed by mafic minerals and the other way around for felsic rocks.

But remember that mafic minerals as amphiboles, pyroxenes or biotites, involve in their chemical structure iron and/or magnesium. Although calcium and sodium can be incorporated in amphiboles and clinopyroxenes, they are not involved in orthopyroxenes and biotites. On the other hand, although potassium is involved in biotite and in some extent in amphiboles, this element is not involved in pyroxenes.

So in conclusion, mafic minerals are usually dark-colored because they involve iron and/or magnesium in their chemical structures.  

Answer: The mass of sodium carbonate reacted is 0.205 g.

Explanation:

To calculate the moles of a solute, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]

We are given:

Volume of nitric acid = 25 mL = 0.025 L   (Conversion factor: 1 L = 1000 mL)

Molarity of the solution = 0.155 moles/ L

Putting values in above equation, we get:

[tex]0.155mol/L=\frac{\text{Moles of nitric acid}}{0.025L}\\\\\text{Moles of nitric acid}=0.003875mol[/tex]

For the given chemical reaction:

[tex]Na_2CO_3+2HNO_3\rightarrow H_2CO_3+2NaNO_3[/tex]

By Stoichiometry of the reaction:

If 2 moles of nitric acid reacts with 1 mole of sodium carbonate.

So, 0.003875 moles of nitric acid will react with [tex]\frac{1}{2}\times 0.003875=0.0019375moles[/tex] of sodium carbonate.

To calculate the mass of sodium carbonate, we use the equation:

Molar mass of sodium carbonate = 105.98 g/mol

Moles of sodium carbonate = 0.0019375 moles

Putting values in above equation, we get:

[tex]0.0019375mol=\frac{\text{Mass of sodium carbonate}}{105.98g/mol}\\\\\text{Mass of sodium carbonate}=0.205g[/tex]

Hence, the mass of sodium carbonate reacted is 0.205 g.

To find out how many grams of sodium carbonate are required to completely react with 25.0 mL of 0.155 M nitric acid, we used stoichiometry. First, we calculated the number of moles of nitric acid, then used the balanced equation to find out the moles of sodium carbonate, which is equal to the moles of nitric acid. Finally, we multiplied this by the molar mass of sodium carbonate to get the weight in grams, which is 0.411 grams.

The subject of this question is related to a chemical reaction between sodium carbonate and nitric acid. In order to determine the amount of sodium carbonate required to completely react with the given amount of nitric acid, we need to use stoichiometry.

Here are the steps:

So, we need 0.411 grams of sodium carbonate to completely react with 25.0 mL of 0.155 M nitric acid.

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Explanation:

Gram is unit which is used to expressed mass of a substance. Many other units can be used to express the mass of a substance. All the units are inter changeable.

Conversion factors used to convert the units are:

(a) Gram to mega gram

[tex]1g=1\times 10^{-6}Mg[/tex]

(b) Gram to kilo gram

[tex]1g=1\times 10^{-3}kg[/tex]

(c) Gram to deci gram

[tex]1g=1\times 10^{1}1dg[/tex]

(d) Gram to centi gram

[tex]1g=1\times 10^{2}cg[/tex]

(e) Gram to mili gram

[tex]1g=1\times 10^{3}mg[/tex]

(f) Gram to micro gram

[tex]1g=1\times 10^{6}\mu g[/tex]

(g) Gram to nano gram

[tex]1g=1\times 10^{9}ng[/tex]

(h) Gram to pico gram

[tex]1g=1\times 10^{12}pg[/tex]

Hence, the conversion factors for gram equivalents are given above.

To convert 1 gram to different metric prefixes in scientific notation, we use the powers of ten specific to each prefix, ensuring the numeric value is between 1 and 1000. The result is megagrams (1e-6), kilograms (1e-3), decigrams (1e1), centigrams (1e2), milligrams (1e3), micrograms (1e6), nanograms (1e9), and picograms (1e12).

The task involves converting the mass of 1 gram into various metric prefixes and expressing them in scientific notation, where the numeric value is greater than one but less than 1000. The prefixes include mega, kilo, deci, centi, milli, micro, nano, and pico. Below are the conversions using the appropriate powers of ten to express 1 gram in the specified units:

Answer:

Roughly C100 H140 N3 O

Explanation:

Gilsonite is a bituminous product that resembles shiny black obsidian.

It contains more than 100 elements.

Its mass composition varies but is approximately 84 % C, 10 % H, 3 % N, and 1 % O.

Its empirical formula is roughly C100 H140 N3 O.

Answer:

-3.7555 M/s is the rate of change of B.

Explanation:

3A + 5B  → 4C

Given that rate of the reaction ,R= 0.7511 M/s

Rate of the reaction is defined as change in concentration of any one of the reactant or product with respect to time.

[tex]R=\frac{-1}{3}\frac{dA}{dt}=\frac{-1}{5}\frac{dB}{dt}=\frac{1}{4}\frac{dC}{dt}[/tex]

[tex]R=0.7511 M/s=\frac{-1}{5}\frac{dB}{dt}[/tex]

[tex]\frac{dB}{dt}=5\times 0.7511 M/s=-3.7555 M/s[/tex]

The negative sign indicates the concentration of reactant B is decreasing with progress in time. This mean reactant B is disappearing.

-3.7555 M/s is the rate of change of B.

Answer: Cis-1-bromo-4-tert-butylcyclohexane would undergo faster elimination reaction.

Explanation:

The two primary requirements for an E-2 elimination reaction are:

1.There must be availability of β-hydrogens that is presence of hydrogen on the carbon next to the leaving group.

2.The hydrogen and leaving group must have a anti-periplanar position .

Any substrate which would follow the above two requirements can give elimination reactions.

For the structure of trans-1-bromo-4-tert-butylcyclohexane and cis-1-bromo-4-tert-butylcyclohexane  to be stable it  must have the tert-butyl group in the equatorial position as it is a bulky group and at equatorial position it would not repel other groups. If it is kept on the axial position it would undergo 1,3-diaxial interaction and would destabilize the system and that structure would be unstable.

Kindly find the structures of trans-1-bromo-4-tert-butylcyclohexane and cis-1-bromo-4-tert-butylcyclohexane in attachment.

The cis- 1-bromo-4-tert-butylcyclohexane has the leaving group and β hydrogens in anti-periplanar position so they can give the E2 elimination reactions easily.

The trans-1-bromo-4-tert-butylcyclohexane  does not have the leaving group and βhydrogen in anti periplanar position so they would not give elimination reaction easily.

so only the cis-1-bromo-4-tert butyl cyclohexane would give elimination reaction.

Trans-1-bromo-4-tert-butylcyclohexane is expected to undergo E2 elimination faster than cis-1-bromo-4-tert-butylcyclohexane due to less steric hindrance.

In determining the rate of E2 elimination, the trans-1-bromo-4-tert-butylcyclohexane would undergo E2 elimination faster than the cis-1-bromo-4-tert-butylcyclohexane. This is due to the larger degree of steric hindrance in the case of the cis isomer.

In trans-1-bromo-4-tert-butylcyclohexane, the bromine is at the equatorial position while the tert-butyl group is axial. It forms a structure that allows the compound to experience less steric hindrance with bromine in a more favorable position for leaving.

In comparison, cis-1-bromo-4-tert-butylcyclohexane has a bromine and tert-butyl group both at equatorial positions. This causes steric hindrance, and in turn, slows down the E2 elimination rate. Despite the more stable conformation, the bromine is not well-oriented for a leaving group in E2 elimination.

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Answer:

Kgoal = 8.15 X 10⁻²⁸

Explanation:

The goal reaction is

4PCl₅(g)⇌P₄(s)+10Cl₂(g)   Kgoal = ?

The given reactions are

P₄(s)+6Cl₂(g)⇌4PCl₃(g), K₁ = 2.00×10¹⁹

PCl₅(g)⇌PCl₃(g)+Cl₂(g), K₂ = 1.13×10⁻²

We can obtain the goal equation by

i) multiplying the second equation with four

ii) subtracting the equation one from above equation

We know that

i) If we multiply an equation with a number the equilbirium constant increases that times (we have to raise the power of equilibrium constant by that number)

ii) if we subtract two equations the equilibrium constants are divided

Kgoal = (K₂)⁴ / K₁

Kgoal = 8.15 X 10⁻²⁸

To determine the equilibrium constant, Kgoal, for the reaction 4PCl5(g)⇌P4(s)+10Cl2(g), we can use the given equilibrium constants for the two reactions involving the same species. By multiplying the equations for the given reactions, we can obtain the equation for the desired reaction and calculate the value of Kgoal. The value of Kgoal is found to be 2.26×1017.

To determine the equilibrium constant, Kgoal, for the reaction 4PCl5(g)↔P4(s)+10Cl2(g), we can use the given equilibrium constants for the two reactions involving the same species. We start by writing the equation for the desired reaction as the sum of the two given reactions:

By multiplying these two equations together, we can obtain the equation for the desired reaction and calculate the value of Kgoal. Multiplying the equations gives:
4PCl5(g)+24Cl2(g)↔16PCl3(g)+4P(s)+(4)(10)Cl2(g)

Since Kgoal is the product of the equilibrium constants for the forward and reverse reactions, we can calculate it as:

Kgoal = K1 × K2 = (2.00×1019)(1.13×10-2) = 2.26×1017

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When an electron jumps from higher energy level to a lower energy level it radiates or gives out energy in the form of radiation.

Electrons present in an atom revolve in different orbits which are stationary states and are also called as energy levels. The energy levels are numbered as integers which are also called  as principal quantum numbers.

Energy of the stationary state  is given as E= -R[tex]_h[/tex] 1/n² where R[tex]_h[/tex] is  the Rydberg's constant. When an electron is excited, and it moves from lower to higher energy levels there is absorption of energy, while when it moves from higher energy level to lower energy level it radiates or gives out energy in the form of radiation.

They can also be defined as the distances  between electron and nucleus of an atom . Electrons present in K energy level have least energy .Energy level diagrams are studied to understand nature of bonding , placement of electrons in orbits and  and elemental behavior under certain conditions.

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Hey there!:

Total volume of wine = 750ml

volume ℅ of ethanol = 12 %

volume of ethanol = (12ml/100ml)*750ml = 90ml

Density of Ethanol = 0.789 g/ml

Mass of Ethanol = 0.789 g/ml × 90ml = 71.01 g

Molar mass of ethanol = 46 g/mol  Nº of mole of ethanol = Mass/molar mass

=>  71.01 g /46(g/mol)= 1.5437 moles

Hope this helps!

1.541 moles of ethanol are present in a 750 mL bottle of wine.

Number of moles = [tex]\frac{\text{Mass}}{\text{Molar mass}}[/tex]

The substance per unit volume is called Density. SI unit of density is kg/m.

It is expressed as:

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

Volume of ethanol = 12%

                               = [tex]\frac{12}{100}[/tex]

                               = 0.12

Volume of ethanol = 0.12 × 750

                               = 90

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

0.789 g/mL = [tex]\frac{\text{Mass of ethanol}}{90}[/tex]

Mass of ethanol = 0.789 × 90

                           = 71.01 g

Now put the value in above formula we get

Number of moles = [tex]\frac{\text{Mass}}{\text{Molar mass}}[/tex]

                             = [tex]\frac{71.01\ g}{46.06\ \text{g/mol}}[/tex]

                             = 1.541 mol

Thus from the above conclusion we can say that 1.541 moles of ethanol are present in a 750 mL bottle of wine.

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Answer:

1. Multiply the equivalence point volume by the NaOH concentration.

2. Divide that result by the volume taken from the formic acid solution for titration.

Explanation:

The reaction between NaOH and formic acid is one to one, it means that a NaOH mol reacts for each mole of formic acid in the solution. So the number of moles who react in the NaOH titrator is the same number of moles that were in the volume taken of the formic acid solution for titration. So, the number of formic acid moles can be calculated as:

[tex]n_{CHOOH}=V_{NaOH}*C_{NaOH}[/tex]

Then, divide that number of moles by the volume taken from the formic acid solution:

[tex]C_{CHOOH}=\frac{V_{NaOH}*C_{NaOH}}{V_{CHOOH}}[/tex]

The concentrations must be in Molar units (Mol/Liter) and the volume in liters, so:

[tex]C_{CHOOH}=\frac{0.02666*C_{NaOH}}{V_{CHOOH}}[/tex]

The concentration of the formic acid solution can be determined using the titration curve provided by the student.

To determine the concentration of the formic acid solution using the student's titration curve, follow these steps:

1. Identify the equivalence point on the titration curve. This is the point where the pH changes rapidly and corresponds to the addition of 26.66 mL of NaOH.

2. At the equivalence point, the moles of NaOH added are equal to the moles of formic acid in the initial solution.

3. Calculate the moles of NaOH added using the concentration of the NaOH solution[tex](C_NaOH)[/tex] and the volume of NaOH added [tex](V_NaOH)[/tex] at the equivalence point:

[tex]\[ \text{moles of NaOH} = C_{\text{NaOH}} \times V_{\text{NaOH}} \][/tex]

4. Since the stoichiometry of the reaction between formic acid (HCOOH) and NaOH is 1:1, the moles of NaOH added will be equal to the moles of formic acid (HCOOH) in the initial solution.

5. Calculate the concentration of the formic acid solution using the moles of formic acid and the initial volume of the formic acid solution [tex](V_HCOOH):[/tex]

[tex]\[ \text{Concentration of formic acid} = \frac{\text{moles of HCOOH}}{V_{\text{HCOOH}}} \][/tex]

6. The concentration of formic acid can be expressed as:

[tex]\[ C_{\text{HCOOH}} = \frac{C_{\text{NaOH}} \times V_{\text{NaOH}}}{V_{\text{HCOOH}}} \][/tex]

7. Plug in the known values to find the concentration of formic acid:

[tex]\[ C_{\text{HCOOH}} = \frac{C_{\text{NaOH}} \times 26.66 \text{ mL}}{25.00 \text{ mL}} \][/tex]

8. If the concentration of NaOH is not given, it can be determined from the pH of the solution before the equivalence point where the pH starts to increase sharply. At this point, the concentration of [tex]OH^-[/tex] ions can be calculated, and since[tex][OH^-] = \(\frac{K_w}{[\text{H}_3\text{O}^+]}\),[/tex] the concentration of NaOH can be found.

9. Once the concentration of NaOH is known, use it in the equation from step 7 to find the concentration of formic acid.

Answer:

The correct option is : b) noncompetitive

Explanation:

There are three main types of inhibition:

1. Competitive: In this inhibition, the inhibitor molecule competes with the substrate to bind on the active site of the enzyme.

2. Uncompetitive: In this inhibition, the inhibitor molecule binds to the enzyme- substrate activated complex and thus, does not compete with the substrate to bind on the active site of the enzyme.

3. Non-competitive: In this inhibition, the inhibitor molecule can bind to both the enzyme molecule or to the enzyme-substrate activated complex.

Therefore, In non- competitive inhibition, the inhibitor molecule binds to the enzyme regardless of whether the substrate molecule is bound to the enzyme active site or not.