Formaldehyde is a carcinogenic volatile organic compound with a permissible exposure level of 0.75 ppm. At this level, how many grams of formaldehyde are permissible in a 6.0-L breath of air having a density of 1.2 kg/m3?

Answer: (a) 2 (b) 6 (c) 14

Explanation:

In the Azimuthal quantum number(l) electrons in a particular subshell (such as s, p, d, or f) are defined by values of l (0, 1, 2, or 3).

s is l=0, p is l=1, d is l=2, f is l=3.

The magnetic quantum number (ml)  The value of ml can range from -l to +l, including zero. Thus the s, p, d, and f subshells contain 1, 3, 5, and 7 orbitals each, with values of m within the ranges 0, ±1, ±2, ±3 respectively. Each shell can have 2 x l + 1 sublevels, and each of these sublevel can accommodate up to two electrons.

(a) n=2, l=1, ml=0. If l=1 then 2 x 1+ 1=3 sublevels, 3*2=6 electrons. When l=1, ml =-1,0,+1, ml=0  accommodate two(2)electrons

(b) 5p. p is l=1  If l=1 then 2 x 1+ 1=3 sublevels, 3*2= electrons. This means in the 5 shell, the p orbital has 3 subshell and accommodate 6 electrons.

(c) n = 4, l = 3 if l=3 then 2 x 3+ 1=7 sublevel 7*2=14 electrons. This means the in the 4 shell, the f orbital has 7 subshell and accomdate 14 elections.

(a) The maximum number of electrons an atom can have with the sublevel designation is 2 electrons.

(b) The maximum number of electrons for p-orbital is 6 electrons.

(c) The maximum number of electrons for f-orbital is 14 electrons.

The energy level of each of an atom with the given sublevel designations determines the maximum number of electrons the atom can occupy.

(a) For n = 2, I = 1, ml = 0,

The energy level is calculated as;

Energy level = 2(1) + 1 = 3 sub-orbtials

= -1  0  1  : ( 2 electrons each)

Thus, the maximum number of electrons an atom can have with the sublevel designation is 2 electrons.

(b) 5p

p-orbital has 3 sub-orbitals and maximum of 6 electrons

(c) n= 4, I = 3

energy level = 2(3) + 1 = 7 sub-orbitals

7 sub-orbitals corresponds f-orbital and it has maximum 14 electrons.

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Answer: option A. volume

Explanation:

Answer: the molecular formula is C10H20O

Explanation:Please see attachment for explanation

Answer:

To prepare 24 mL of 0.050 M NaOH solution, you would add 4 mL of 0.300 M NaOH stock solution and 20 mL of 0.300 M NaCl solution.

Explanation:

Molarity of the NaOH solution = [tex]M_1=0.300 M[/tex]

Volume of the NaOH solution = [tex]V_1=?[/tex]

Molarity of the NaOH solution after dilution= [tex]M_2=0.050 M[/tex]

Volume of the NaOH solution after dilution= [tex]V_2=24 mmL[/tex]

[tex]M_1V_1=M_2V_2[/tex] (Dilution )

[tex]V_1=\frac{M_2V_2}{M_1}=\frac{0.050 M\times 24 mL}{0.300 M}=4 mL[/tex]

Volume of NaCl solution of 0.300 M = 24 mL - 4 mL = 20 mL

To prepare 24 mL of 0.050 M NaOH solution, you would add 4 mL of 0.300 M NaOH stock solution and 20 mL of 0.300 M NaCl solution.

To prepare 24 mL of 0.050 M NaOH solution, you would add 4 mL of 0.300 M NaOH stock solution and 20 mL of 0.300 M NaCl solution

•Molarity of stock solution (M₁) = 0.3 M

•Molarity of diluted solution (M₂) = 0.05 M

•Volume of diluted solution (V₂) = 24 mL

•Volume of stock solution needed (V₁) =?

Using the dilution formula, the volume of the stock solution needed can be obtained as follow:

M₁V₁ = M₂V₂

0.3 × V₁ = 0.05 × 24

0.3 × V₁ = 1.2

Divide both side by 0.3

V₁ = 1.2 / 0.3

V₁ = 4 mL

•Volume of NaOH needed = 4 mL

•Volume of diluted solution of NaOH = 24 mL

•Volume of NaCl needed =?

Volume of NaCl needed = (Volume of diluted solution of Na) – (Volume of NaOH needed)

Volume of NaCl needed = 24 – 4

Volume of NaCl needed = 20 mL

Therefore, we can conclude as follow:

To prepare 24 mL of 0.050 M NaOH solution, you would add 4 mL of 0.300 M NaOH stock solution and 20 mL of 0.300 M NaCl solution.

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Answer:

56.06 g/mol is the molar mass

Explanation:

Vapor pressure lowering → P° - P' = P° . Xm

Where P° is vapor pressure of pure solvent

P' is vapor pressure of solution

Xm is the mole fraction (moles of solute / total moles)

Total moles = moles of solute + moles of solvent

Let's replace the data.

100 Torr - 87.7 Torr = 100 Torr . Xm

12.3 Torr = 100 Torr . Xm

0.123 = Xm

We know the moles of solvent because we know the molar mass from benzene and its mass in the solution. (mass / molar mass)

100 g / 78 g/mol = 1.28 moles

Let's build the equation where the unknown is the moles of solute

0.123 = moles of solute / moles of solute + 1.28 moles

0.123 (moles of solute + 1.28 moles) = moles of solute

0.123 moles of solute + 0.158 moles = moles of solute

0.158 = 1moles of solute - 0.123moles of solute

0.158 moles = 0.877 moles of solute

0.158 / 0.877 = moles of solute → 0.180

These moles corresponds to 10.1 g of the unknown, non volatile and non electrolyte X compound so:

molar mass (g/mol) → 10.1 g / 0.180 mol = 56.06 g/mol

To calculate the molar mass of the solute, we can use the formula: molar mass = (mass of solute / moles of solute). First, find the moles of solute using the relationship between the freezing point depression and the moles of solute. Next, calculate the molality of the solution using the given freezing point depression constant and mass of the solute, and use that to calculate the moles of solute. Finally, divide the mass of solute by the moles of solute to find the molar mass.

To calculate the molar mass of the solute, we can use the formula:

molar mass = (mass of solute / moles of solute)

We first need to find the moles of solute using the relationship between the freezing point depression and the moles of solute:

ΔTf = Kf * m

where ΔTf is the freezing point depression, Kf is the freezing point depression constant, and m is the molality of the solution.

In this case, we are given that the freezing point depression is 0.40°C, the freezing point depression constant of benzene is 5.12 K kg/mol, and the mass of the solute is 2 grams. We can use these values to calculate the molality of the solution:

m = (ΔTf / Kf)

m = (0.40°C / 5.12 K kg/mol)

m = (0.078125 mol/kg)

Now we can calculate the moles of solute:

moles of solute = (m * mass of solvent)

moles of solute = (0.078125 mol/kg * 0.1 kg)

moles of solute = 0.0078125 mol

Finally, we can calculate the molar mass:

molar mass = (mass of solute / moles of solute)

molar mass = (2 grams / 0.0078125 mol)

molar mass ≈ 256 g/mol

Therefore, the molar mass of the solute, X, is approximately 256 g/mol.

Answer: The solubility of ethylene gas in water is [tex]9.16\times 10^{-2}g/L[/tex]

Explanation:

To calculate the molar solubility, we use the equation given by Henry's law, which is:

[tex]C_{C_2H_4}=K_H\times p_{C_2H_4}[/tex]

where,

[tex]K_H[/tex] = Henry's constant = [tex]4.78\times 10^{-3}mol/L.atm[/tex]

[tex]C_{C_2H_4}[/tex] = molar solubility of ethylene gas = ?

[tex]p_{C_2H_4}[/tex] = partial pressure of ethylene gas = 0.684 atm

Putting values in above equation, we get:

[tex]C_{C_2H_4}=4.78\times 10^{-3}mol/L.atm\times 0.684atm\\\\C_{C_2H_4}=3.27\times 10^{-3}mol/L[/tex]

Converting this into grams per liter, by multiplying with the molar mass of ethylene:

Molar mass of ethylene gas = 28 g/mol

So, [tex]C_{C_2H_6}=3.27\times 10^{-3}mol/L\times 28g/mol=9.16\times 10^{-2}g/L[/tex]

Hence, the solubility of ethylene gas in water is [tex]9.16\times 10^{-2}g/L[/tex]

The solubility of ethylene in water at 25 °C with a partial pressure of 0.684 atm is approximately 0.0919 grams per liter.

To calculate the solubility of ethylene in water, we can use Henry's Law. Henry's Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. The equation is given by:

S = kH * P

Where:

First, we can calculate the solubility of ethylene in units of moles per liter:

S = (4.78×10-3 mol/L·atm) * (0.684 atm) = 0.00327 mol/L

Then, we can convert the solubility to grams per liter using the molar mass of ethylene:

Molar mass of C2H4 = 2(12.01 g/mol) + 4(1.01 g/mol) = 28.05 g/mol

Grams/Liter = (0.00327 mol/L) * (28.05 g/mol) = 0.0919 g/L

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Answer:

Wavelength = 13492242 nm

Wavelength = 134922420 Å

Explanation:

The relation between frequency and wavelength is shown below as:

[tex]c=frequency\times Wavelength [/tex]

c is the speed of light having value [tex]3\times 10^8\ m/s[/tex]

Given, Frequency = [tex]22.235\ GHz=22.235\times 10^{9}\ Hz[/tex]

Thus, Wavelength is:

[tex]Wavelength=\frac{c}{Frequency}[/tex]

[tex]Wavelength=\frac{3\times 10^8}{22.235\times 10^{9}}\ m[/tex]

[tex]Wavelength=0.013492242\ m=13492242\times 10^{-9}\ m[/tex]

Also, 1 m = [tex]10^{-9}[/tex] nm

So,  

Wavelength = 13492242 nm

Also, 1 m = [tex]10^{-10}[/tex] Å

Wavelength = 134922420 Å

Answer:

a.A pure substance is made up of only 1 type of particle

Explanation:

When a substance is pure, it has only one type of particle. These particles maybe molecules, ions or atoms linked in a definite way throughout the substance. If a substance contains different particles, it cannot be regarded as a pure substance because its properties will be observed as a compromise of the individual properties of its different components.

The statement that accurately describes a pure substance is that it is made up of only one type of particle. Both elements and compounds can be considered as pure substances.

The correct statement to describe a pure substance is:

a. A pure substance is made up of only 1 type of particle.

This means that a pure substance is only made up of identical atoms if it is an element, or identical molecules if it is a compound. Therefore, both elements and compounds can be considered as pure substances, contradicting options c and d. It also contradicts option b because a pure substance is not made up of more than one type of particle.

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Answer:Hydrogen is placed such because it exhibits some similar characteristics of both group1 and group VII elements.

Explanation:

The reason why hydrogen is similar to group 1 metals:

#It has same valence electron and inorder achieve octet state it can lose that electron and forms H+ ion

#It acts as a good reducing agent similar to group1 metals

#It can also halides

Similarity to halogens:

#hydrogen can also gain one electron to gain noble gas configuration. It can combine with other non metals to form molecules with covalent bonding.

#It exists as diatomin molecule,H2

#Have the same electronegativity nature

#its reaction with other metal

Answer:

Calculate the atomic radii of two touching or overlapping atoms.

Explanation:

No doubt, we can't find the atomic boundary of a single atom, but when atoms are in the form of pairs it becomes very easy to measure the atomic radii of two and then dividing it by 2 to get an estimate of atomic radius of a single atom.

It is also called as covalent radius which is half of the total inter-nuclear distance between two same bonded atoms (Homo-nuclear).

If two adjacent mettalic ions are joined by such pairing then the same half of the distance between the nucleus is termed as metallic radii.

Explanation:

Case 1.  When circumference of cylinder is 27.94 cm and its height is 21.59 cm. Therefore, we will calculate the radius of cylinder as follows.

            Circumference = [tex]2 \pi r[/tex]

                        27.94 cm = [tex]2 \times 3.14 \times r[/tex]

                         r = 4.45 cm

Now, we will calculate the volume of cylinder as follows.

              Volume = [tex]\pi r^{2}h[/tex]

                            = [tex]3.14 \times (4.45 cm)^{2} \times 21.59 cm[/tex]

                            = [tex]1342.46 cm^{3}[/tex]

Case 2.  When circumference of cylinder is 21.59 cm and its height is 27.94 cm. Therefore, we will calculate the radius of cylinder as follows.

            Circumference = [tex]2 \pi r[/tex]

                        21.59 cm = [tex]2 \times 3.14 \times r[/tex]

                         r = 3.44 cm

Now, we will calculate the volume of cylinder as follows.

              Volume = [tex]\pi r^{2}h[/tex]

                            = [tex]3.14 \times (3.44 cm)^{2} \times 27.94 cm[/tex]

                            = [tex]1038.18 cm^{3}[/tex]

Answer: The empirical formula for the given compound is [tex]Fe_2O_3[/tex]

Explanation:

We are given:

Mass of Fe = 4.166 g

Mass of O = 1.803 g

To formulate the empirical formula, we need to follow some steps:

Moles of Iron =[tex]\frac{\text{Given mass of Iron}}{\text{Molar mass of Iron}}=\frac{4.166g}{55.85g/mole}=0.0746moles[/tex]

Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.803g}{16g/mole}=0.113moles[/tex]

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0746 moles.

For Iron = [tex]\frac{0.0746}{0.0746}=1[/tex]

For Oxygen = [tex]\frac{0.113}{0.0746}=1.5[/tex]

Converting the mole ratio into whole number by multiplying with '2'

Mole ratio of Fe = (1 × 2) = 2

Mole ratio of O = (1.5 × 2) = 3

The ratio of Fe : O = 2 : 3

Hence, the empirical formula for the given compound is [tex]Fe_2O_3[/tex]

Answer:

12.96 seconds

Explanation:

Assuming the reaction follows a first order

Rate = K[NOBr] = change in concentration of NOBr/time

K = 0.556 L mol^-1 s^-1

Change in concentration of NOBr = 0.32M - 0.039M = 0.281M

0.281/t = 0.556×0.039

t = 0.281/0.021684 = 12.96 seconds

It will take approximately 40.48 seconds for the concentration of NOBr to decrease from 0.32 M to 0.039 M.

To determine how long it will take for the concentration of NOBr to decrease from 0.32 M to 0.039 M, we need to use the integrated rate law for a second-order reaction:

For a second-order reaction:

1 / [A]₁ = kt + 1 / [A]₀

Where:

Now, let's rearrange and solve for time (t):

(1 / [A]₁) - (1 / [A]₀) = kt

[1 / 0.039 M] - [1 / 0.32 M] = (0.556 L mol⁻¹ s⁻¹) * t

25.64 - 3.125 = 0.556t

22.515 = 0.556t

t = 22.515 / 0.556

t ≈ 40.48 seconds

Answer:

.

Explanation:

Answer:

Hund's rule: states that electrons always enter an empty orbital before they pair up.

In this exercise, we notice that orbital p has three suborbitals, then,

we must start filling each suborbitals with one electrons and after that we start to pairing them up.

The result must be,

1st suborbital with 2 electrons

2nd suborbital with 1 electron

3 rd suborbital with 1 electron

Answer:

74.344 kJ.

Explanation:

Below is an attachment containing the solution.

Answer : The formic acid pressure is, 99 torr

Explanation :

The given chemical reaction is:

                            [tex]HCOOH(g)\rightarrow CO_2(g)+H_2(g)[/tex]

Initial pressure        a                   0              0

At time 't'                (a-x)                 x              x

According to the Dalton's law,

[tex]P_{Total}=P_{HCOOH}+P_{CO_2}+P_{H_2}[/tex]

[tex]P_{Total}=(a-x)+x+x=a+x[/tex]    .........(1)

As we are given that:

Initial pressure = a = 220 torr

[tex]P_{Total}=319torr[/tex]

Now put the value of 'a' in equation 1, we get:

[tex]P_{Total}=a+x[/tex]

[tex]319torr=220torr+x[/tex]

[tex]x=99torr[/tex]

Thus, the formic acid pressure is, 99 torr

Answer: The vapor pressure of solution is 459.17 mmHg

Explanation:

To calculate the number of moles, we use the equation:

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

Given mass of testosterone = 7.752 g

Molar mass of testosterone = 288.4 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of testosterone}=\frac{7.752g}{288.4g/mol}=0.027mol[/tex]

Given mass of diethyl ether = 208.0 g

Molar mass of diethyl ether = 74.12 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of diethyl ether}=\frac{208.0g}{74.12g/mol}=2.81mol[/tex]

Mole fraction of a substance is calculated by using the equation:

[tex]\chi_A=\frac{n_A}{n_A+n_B}[/tex]

[tex]\chi_{\text{testosterone}}=\frac{n_{\text{testosterone}}}{n_{\text{testosterone}}+n_{\text{diethyl ether}}}[/tex]

[tex]\chi_{\text{testosterone}}=\frac{0.027}{0.027+2.81}\\\\\chi_{\text{testosterone}}=0.0095[/tex]

The formula for relative lowering of vapor pressure will be:

[tex]\frac{p^o-p_s}{p^o}=i\times \chi_{\text{solute}}[/tex]

where,

[tex]p^o[/tex] = vapor pressure of solvent (diethyl ether) = 463.57 mmHg

[tex]p^s[/tex] = vapor pressure of the solution = ?

i = Van't Hoff factor = 1 (for non electrolytes)

[tex]\chi_{\text{solute}}[/tex] = mole fraction of solute (testosterone) = 0.0095

Putting values in above equation, we get:

[tex]\frac{463.57-p^s}{463.57}=1\times 0.0095\\\\p^s=459.17mmHg[/tex]

Hence, the vapor pressure of solution is 459.17 mmHg

Answer : The percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.

Explanation :

To calculate the percentage composition of element in sample, we use the equation:

[tex]\%\text{ composition of element}=\frac{\text{Mass of element}}{\text{Mass of sample}}\times 100[/tex]

Given:

Mass of carbon = 1.94 g

Mass of hydrogen = 0.48 g

Mass of sulfur = 2.58 g

First we have to calculate the mass of sample.

Mass of sample = Mass of carbon + Mass of hydrogen + Mass of sulfur

Mass of sample = 1.94 + 0.48 + 2.58 = 5.0 g

Now we have to calculate the percentage composition of a compound.

[tex]\%\text{ composition of carbon}=\frac{1.94g}{5.0g}\times 100=38.8\%[/tex]

[tex]\%\text{ composition of hydrogen}=\frac{0.48g}{5.0g}\times 100=9.6\%[/tex]

[tex]\%\text{ composition of sulfur}=\frac{2.58g}{5.0g}\times 100=51.6\%[/tex]

Hence, the percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.

Answer: beta

Explanation: the beta carbon is the second after the alpha carbon( carbon bonding to the functional group)

Answer:

Explanation:

The period law state that when elements are listed in order of their atomic numbers, the elements fall into recurring groups, so that there is a recurrence of similar properties at regular intervals.

Na and K in the periodic table fall into the same group, this is because they both have one electrons in their outermost shell.

Na 11 -1s2 2s2 2p6 3s1

K 19 - 1s2 2s2 2p6 3s2 3p6 4s1

They share similar chemical and physical properties. Na and K are very reactive metals, they can loose/donate their outermost electron to non metals in other to attain stable octet state.

The form ionic compound when they react with non metals.

Answer:

Explanation:

In the picture you have the answer.

Now, let's analize the structure, so you can know why the structure in the picture is the correct structure.

The aniline is the name that receives the benzene with a NH2 group as one of it's substituent. Now, This group is a really strong activating group and in the nomenclature priority, it has more order priority than any halide.

Now, it says that the chloro it's on the para position. The "para" position in a aromatic ring, in this case, the benzene, refers to the position of this substituent to the first substitued position. In this case, the NH2 it's on the position 1 or carbon 1, the para position, means that it's on position 4 of the ring. The ortho position is carbon 2, and meta position is carbon 3 of the benzene. So, according to this, the p-chloroaniline it's on picture attached.

Answer:

(a) Ga (b) Si (d) Te

Explanation:

Paramagnetic are those which has unpaired electrons and diamagnetic are those in which all electrons are paired.

(a) Ga

The electronic configuration is -  

[tex]1s^22s^22p^63s^23p^63d^{10}4s^24p^1[/tex]

The electrons in 4p orbital = 1 (Unpaired)

Thus, the element is paramagnetic as the electrons are unpaired.

(b) Si

The electronic configuration is -  

[tex]1s^22s^22p^63s^23p^2[/tex]

The electrons in 3p orbital = 2 (Unpaired)

Thus, the element is paramagnetic as the electrons are unpaired.

(c) Be

The electronic configuration is -  

[tex]1s^22s^2[/tex]

The electrons in 2s orbital = 2 (paired)

Thus, the element is diamagnetic as the electrons are paired.

(d) Te

The electronic configuration is -  

[tex]1s^22s^22p^63s^23p^63d^{10}4s^24p^64d^{10}5s^25p^4[/tex]

The electrons in 5p orbital = 4 (1 pair and 2 Unpaired)

Thus, the element is paramagnetic as the electrons are unpaired.

Answer:

a. Zero unpaired electron

b. 3 unpaired electrons

c. Zero unpaired electron

d. 1 unpaired electron

Explanation:

a. 2A(2) has configuration => 1s2. Since the s-orbital is completely filled, Therefore it has zero unpaired electrons

b. 5A(15) has configuration =>

1s2 2s2 2p6 3s2 3p3

Since the p-orbital is not completely filled, It has 3 unpaired electrons

c. 8A(18) has configuration =>

1s2 2s2 2p6 3s2 3p6

Since the p-orbital is completely filled, therefore it has zero unpaired electrons

d. 3A(13) has configuration =>

1s2 2s2 2p6 3s2 2p1

Since the p-orbital is not completely filled, therefore it has 1 unpaired

Answer:

(a) 2A(2) - it has 2 valence electrons

(b) 5A(15) -

Explanation:

A)To determine the number of unpaired electrons for atoms in group 2A (2)

Using beryllium (it belongs to group 2A) as an example

The atomic number of Be is 4

The electronic configuration is 1s²2s²

The highest principal quantum number is 2, therefore all electrons with n=2 are valence/unpaired electron

Beryllium has 2 valence/unpaired electrons, this applies to all other elements in this group

Therefore group 2A atoms have 2 unpaired electrons  

B) To determine the number of unpaired electrons for atoms in group 5A(15)

Using Nitrogen as an example

The atomic number of Nitrogen is 7

The electronic configuration  is 1s²2s²2p³

The highest principal quantum number for nitrogen is 2, therefore all electrons with n=2 are valence/unpaired electrons

Nitrogen has 2+3= 5 valence/unpaired electrons, this applies to all other elements in this group

Therefore, group 5A atoms have 5 unpaired electrons

C) To determine the number of unpaired electrons for atoms in group 8A(18)

Using Neon as an example

The atomic number of Neon is 2

The electronic configuration of Neon is 1s²2s²2p⁶3s²3p⁶

The highest principal quantum number for 3, therefore all electrons with n=3 are valence/unpaired electrons

Neon has 2+6 = 8 valence/unpaired electrons,this applies to all other elements in this group except Helium whose number of unpaired electrons is 2

Therefore, group 8A atoms have 8 unpaired electrons

D) To determine the number of unpaired electrons for atoms in group 3A(13)

Using Aluminium as an example

The atomic number of Aluminium is 13

The electronic configuration of Neon is 1s²2s²2p⁶3s²3p¹

The highest principal quantum number for 3, therefore all electrons with n=3 are valence/unpaired electrons

Neon has 2+1 = 3 valence/unpaired electrons,this applies to all other elements in this group

Therefore, group 3A atoms have 3 unpaired electrons

Note: The number of valence electrons of atoms in a group is the same as the group number that the atom belongs to

Explanation:

Lithium, sodium and potassium are all group 1A elements and when we move down a group then there occurs an increase in atomic size of the elements. As lithium is the smallest and potassium being the largest so, when each of them will lose an electron and obtain a positive charge then size of lithium will further decrease.

Therefore, ions are ranked according to their increase in size as follows.

                      [tex]Li^{+} < Na^{+} < K^{+}[/tex]  

When an atom tends to gain electrons then it acquires a negative charge. This means that size of the atom increases.  

So, more is the negative charge present on an atom more will be its atomic size. Therefore, correct order of increasing size for [tex]Se^{2-}, Rb^{+}, Br[/tex] is as follows.

                      [tex]Br < Rb^{+} < Se^{2-}[/tex]

Similarly, order of increasing size of [tex]O^{2-}, F^{-}, N^{3-}[/tex] is as follows.

                    [tex]F^{-} < O^{2-} < N^{3-}[/tex]

The root-mean-square speed for diatomic oxygen at 50°C can be calculated using the formula Urms =

oot(3RT/M), where R is the gas constant, T is the temperature in Kelvin, and M is the molar mass of oxygen. Multiplication options presented in the question are incorrect and should not be used.

To calculate the root-mean-square speed (Urms) for diatomic oxygen at a given temperature, we use the formula derived from the kinetic theory of gases:

Urms =

oot(3RT/M)

Where:

The student provides information that diatomic hydrogen has a particular Urms value at 50°C (which needs to be converted to Kelvin by adding 273.15, resulting in 323.15K). However, to find the Urms for oxygen directly, we will use the molar mass of oxygen (32.00 g/mol or 0.032 kg/mol) and the same temperature in Kelvins:

Urms =

oot(3  imes 8.314  imes 323.15 / 0.032)

Calculating the above will give us the correct Urms for oxygen, the presented multiplication options (16)(2000 m/s), etc., are misleading and should not be used for this calculation.

Final answer:

The rate law and the Arrhenius equation help calculate the ratio of rate constants for reactions with similar frequency factors. In this case, with identical frequency factors, the ratio of the rate constants is 1.

Explanation:

The rate law for a reaction:

rate = k[A] [B]

The Arrhenius equation:

k = Ae(-Ea/RT)

For the given question:

Given k1 and k2 are the same, k1/k2 = 1.

The ratio of rate constants depends on the exponential term in the Arrhenius equation, specifically the difference in activation energies. If the activation energies are very close, the ratio [tex]\( \frac{k_1}{k_2} \)[/tex] will be close to 1, indicating similar rate constants for the two reactions at 25 °C.

The ratio of reaction rate constants [tex](\(k_1/k_2\))[/tex] for two reactions with similar frequency factors can be expressed using the Arrhenius equation, which relates the rate constant (k) to temperature. The Arrhenius equation is given by:

[tex]\[ k = A \cdot e^{-\frac{E_a}{RT}} \][/tex]

Where:

- ( k ) is the rate constant,

- ( A) is the frequency factor (pre-exponential factor),

- ( E_a ) is the activation energy,

- ( R ) is the gas constant (8.314 J/(mol·K)),

- ( T ) is the temperature in Kelvin.

Assuming the frequency factors[tex](\( A_1 \) and \( A_2 \))[/tex] are the same, the ratio of rate constants [tex](\( \frac{k_1}{k_2} \))[/tex] can be simplified to the ratio of the exponential term:

[tex]\[ \frac{k_1}{k_2} = e^{-\frac{E_{a1} - E_{a2}}{RT}} \][/tex]

Given that the reactions are at 25 °C (298 K), the ratio [tex]\( \frac{k_1}{k_2} \)[/tex]will depend on the difference in activation energies [tex](\( E_{a1} - E_{a2} \)).[/tex]

Answer:

The complete answer is in the diagram.

Answer:

For the first question the products elute slower. The answer to the second question is it would not be a good option as an eluent

Explanation:

Solvents are classified into polar and nonpolar. While in polar solvents, the distribution of the electronic cloud is asymmetric; therefore, the molecule has a positive and a negative pole. Low molecular weight alcohols such as methanol belong to this type.

While in apolar solvents, the distribution of the electronic cloud is symmetric; Therefore, these substances lack a positive and negative pole in their molecules. Some solvents such as hexane.

The miscibility of methanol (polar solvent) in hexane (apolar solvent) is low, miscibility is the main reason for not using this mixture as eluent.

Answer:

(a)  7.11 x 10⁻³⁷ m

(b)  1.11 x 10⁻³⁵ m

Explanation:

(a)  The de Broglie wavelength is given by the expression:

λ = h/p = h/mv

where h is plancks constant, p is momentum which is equal to mass times velocity.

We have all the data required to calculate the wavelength, but first we will have to convert the velocity to m/s, and the mass to kilograms to work in metric system.

v = 19.8 mi/h x ( 1609.34 m/s ) x ( 1 h / 3600 s ) = 8.85 m/s

m = 232 lb x ( 0.454 kg/ lb ) = 105.33 kg

λ = h/ mv = 6.626 x 10⁻³⁴ J·s / ( 105.33 kg x 8.85 m/s ) = 7.11 x 10⁻³⁷ m

(b) For this part we have to use the uncertainty principle associated with wave-matter:

ΔpΔx > = h/4π

mΔvΔx > = h/4π

Δx = h/ (4π m Δv )

Again to utilize this equation we will have to convert the uncertainty in velocity to m/s for unit consistency.

Δv = 0.1 mi/h x ( 1609.34 m/mi ) x ( 1 h/ 3600 s )  

     = 0.045 m/s

Δx = h/ (4π m Δv ) = 6.626 x 10⁻³⁴ J·s / (4π x 105.33 kg x 0.045 m/s )

     = 1.11 x 10⁻³⁵ m

This calculation shows us why we should not be talking of wavelengths associatiated with everyday macroscopic objects for we are obtaining an uncertainty of 1.11 x 10⁻³⁵ m for the position of the fullback.

This question discusses the fundamentals quantum mechanics. It asks about the de Broglie wavelength and uncertainty principle of a football player treating as a particle. The de Broglie wavelength can be calculated using Planck's constant and momentum, and the uncertainty can be determine using the uncertainty principal formula.

In this question, we're asked about two core principles in quantum mechanics: the de Broglie wavelength and the uncertainty principle. As such, we're essentially treating the football player as both a particle and a wave, which is a fundamental concept of quantum mechanics.

Let's address the points one by one.

(a) The de Broglie wavelength of a particle is given by λ=h/p. Where h is Planck's constant (6.62607015 × 10-34 m2 kg / s) and p is momentum. The momentum of a player can be calculated as mass x velocity. Convert the player's mass from lbs to kg and speed from mi/h to m/s. Substitute these values into the equation to get the de Broglie wavelength.

(b) The uncertainty principle states that the more precisely the momentum of a particle is known, the less precisely its position can be known, and vice versa. In this case, we're given the uncertainty in the player's speed (which contributes to an uncertainty in his momentum, and thus in his position). By using the uncertainty principal formula Δx=Δp*h/4π, where Δx represents the position uncertainty and Δp is the momentum uncertainty. Using the figures provided in the question, you can calculate the position uncertainty.

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The question provides us with the chemical formula for an unknown compound, and values for moles of oxygen and calcium. However, to solve this problem, we need to have the number of moles for phosphorus, not calcium. Without this, it's impossible to proceed further and derive a concrete answer.

The task requires us to find the accurate value of x in the chemical formula P2Ox. To do this, we consider that the unknown compound is composed entirely of phosphorus (P) and oxygen (O), but not calcium (Ca). Calcium seems to be irrelevant for this particular problem.

In the formula P2Ox, '2' is the stoichiometric index for phosphorus and 'x' for oxygen. However, we are not given the moles of phosphorus, which makes the problem unsolvable with the information provided.

Ideally, we would divide the number of moles of oxygen by the number of moles of phosphorus to find the value of 'x', but since the moles of phosphorus are not provided, we are unable to proceed further with the provided information.

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