Green plants use light from the Sun to drive photosynthesis, a chemical reaction in which liquid water and carbon dioxide gas from aqueous glucose and oxygen gas. Calculate the moles of glucose produced by the reaction of 2.40 moles of water. Be sure your answer has a unit symbol, if necessary, and round it to significant digits.

Explanation:

Isotope are the species which have same atomic number (number of the protons) but number of neutrons different. which corresponds to different mass of isotopes.

For example,

[tex]^6_{12}C, ^6_{13}C, ^6_{14}C[/tex] are the three isotopes of carbon.

Isoelectronic are the species which have the same number of electrons.

For example,

Ne is isoelectronic with [tex]Na^+[/tex]

Isotopes of [tex]^{18}_{40}Ar[/tex] are:

[tex]^{18}_{36}Ar, ^{18}_{38}Ar[/tex]

The number of the electrons in [tex]^{18}_{40}Ar[/tex] = 18

Number of the electrons in [tex]K^+[/tex] = 18

Thus, [tex]K^+[/tex] is isoelectronic to Argon-40

Answer:

28.52 L

Explanation:

First, let's calculate the density of the ocean, which is the mass divided by the volume:

d = m/V

d = 35.06/1

d = 35.06 g/L

So, for a mass of 1.00 kg = 1000.00 g

d = m/V

35.06 = 1000.00/V

V = 1000.00/35.06

V = 28.52 L

How all the data are expressed with two significant figures, the volume must also be expressed with two.

Answer:

66622.653 mi/hr

Explanation:

The average speed at which the Earth rotates around the Sun = 29.783 km/s

Also, The conversion of km to miles is shown below:

1 km = 0.621371 miles

The conversion of s to hr is shown below:

1 s = 1 / 3600 hr

So,

[tex]29.783\ km/s=\frac {29.783\times 0.621371\ miles}{\frac {1}{3600}\ hr}[/tex]

Thus,

The average speed at which the Earth rotates around the Sun in miles per hour =  66622.653 mi/hr

Answer:

specific volume O2 = 0.5 Kg/m³

molar volume O2 = 8 E-3 m³/mol

Explanation:

specific volume (Sv):

∴ Sv = 1 / ρ

∴ ρ = mass / volume

∴ V = 500 L * ( m³/1000 L) = 0.5 m³

∴ ρ = 1 Kg / 0.5m³ = 2 Kg/m³

⇒ Sv = 1 / 2 Kg/m³ = 0.5 m³/Kg

molar volume ( Vm ):

∴ Vm = volume/mol

∴ Mw O2 = 1000g O2 * ( mol/16g O2) = 62.5 mol O2

⇒ Vm = 0.5 m³ / 62.5 mol

⇒ Vm = 8 E-3 m³/mol

Answer:

The atomic mass of the given metal M is 24.3 g/mol.

Therefore, the given metal M is magnesium (Mg)

Explanation:

Reaction involved: MSO₄ + BaCl₂ → BaSO₄ + MCl₂

Atomic mass (g/mol): oxygen (O)=16, sulphur (S)=32,

Molar mass of SO₄²⁻ = 96 g/mol and molar mass of BaSO₄ = 233.38 g/mol

Let the atomic mass of M be m g/mol.

Therefore, molar mass of MSO₄= (m + 96) g/mol

If 0.1131 g of MSO₄ gives 0.2193 g of BaSO₄ on reaction

Then, (m + 96) g/mol of MSO₄ gives 233.38 g/mol of BaSO₄

Therefore, (m + 96) g/mol of MSO₄ = [(233.38 g/mol) × (0.1131 g)] ÷ (0.2193 g)

⇒(m + 96) g/mol = [26.395] ÷ (0.2193)

⇒(m + 96) g/mol = 120.36 g/mol

⇒m = 120.36 g/mol - 96 g/mol = 24.36 g/mol ≈ 24.3 g/mol

Since, the atomic mass of the given metal M is 24.3 g/mol.

Therefore, the given metal M is magnesium (Mg)

To determine the number of photons emitted per second by a 45 kW FM radio transmitter at 98.4 MHz, the energy of one photon is calculated using Planck's constant and the frequency, and then the power output is divided by this energy. The result is approximately 6.90 × 1028 photons per second.

To calculate the number of photons emitted per second by an FM radio transmitter, we need to know the energy of one photon and the total energy emitted per second (power output of the transmitter).

The energy of a photon, E, can be calculated using the Planck's equation: E = h * f, where h is Planck's constant (6.626 × 10-34 J·s), and f is the frequency of the photon in hertz. In this case, the frequency (f) is 98.4 MHz, or 98.4 × 106 Hz.

The total power output in joules per second is given by the transmitter's power, which is 45 kW, or 45,000 J/s. By dividing the total energy emitted per second by the energy of one photon, we obtain the number of photons emitted per second.

Number of Photons per Second = Power Output (J/s) / Energy per Photon (J)

Now, let's perform the calculations:

Energy per Photon, E = (6.626 × 10-34 J·s) × (98.4 × 106 Hz) = 6.522 × 10-26 J

Photons per Second = 45,000 J/s / 6.522 × 10-26 J = 6.90 × 1028 photons/s

Answer:

fluid pressure = 132590 pascal

Explanation:

Fluid pressure can be  calculated by using following relation:

[tex]P_o - P_g = g\time p\times (Z_b - Z_a )[/tex]

Where

Pb = fluid pressure at a distance of 3.2 m below from surface

Pa = fluid pressure at surface = atmospheric pressure = 101325 Pascal

g = gravitational constant = 9.8 m/sec2

[tex]\rho = density of fluid = 997 kg/m³[/tex]

Z = location depth from surface of lake

Therefore , [tex]Z_a = 0 meter[/tex]

[tex]Z_b = 3.2 meter[/tex]

[tex]P_b - P_a =g\times\rhop\times (Z_a - Z_b)[/tex]

               [tex]= 9.8 \times 997\times (3.2-0)[/tex]

               [tex]= 31265 kg/m-sec^2 [/tex]

1 Pascal = 1 Kg/(m.Sec)

[tex]P_b = P_a +31265[/tex]

       = 101325 + 132590 Pascal

       =  132590 pascal

Final answer:

The balanced equation for the decomposition of barium carbonate is BaCO3(s) → BaO(s) + CO2(g). Heating 71.0 g of barium carbonate will produce 16.06 g of carbon dioxide.

Explanation:

The balanced equation for the decomposition of barium carbonate (BaCO3) is:

   BaCO3(s) → BaO(s) + CO2(g)

From the equation, it can be observed that one mole of barium carbonate produces one mole of carbon dioxide.

To calculate the number of grams of carbon dioxide produced by heating 71.0 g of barium carbonate, we need to use the molar mass of barium carbonate (197.34 g/mol). Since one mole of barium carbonate produces one mole of carbon dioxide, we can use the molar mass of carbon dioxide (44.01 g/mol) to calculate the mass of carbon dioxide produced.

   71.0 g BaCO3 x (1 mol CO2/197.34 g BaCO3) x (44.01 g CO2/1 mol CO2) = 16.06 g CO2

Therefore, heating 71.0 g of barium carbonate will produce 16.06 g of carbon dioxide.

Answer:

[tex]\large \boxed{\text{250 lb}}[/tex]

Explanation:

1. Weight of oranges

An average box contains 102 oranges, so

Five boxes contain 510 oranges

[tex]\text{Weight of oranges} = \text{510 oranges} \times \dfrac{\text{7.2 oz}}{\text{1 orange}} = \text{3672 oz}\\\\\text{Weight} = \text{3672 oz} \times \dfrac{\text{1 lb}}{\text{16 oz}} = \text{230 lb}[/tex]

2. Weight of boxes

[tex]\text{Weight} = \text{5 boxes} \times \dfrac{\text{3.2 lb}}{\text{1 box}} = \text{16 lb}[/tex]

3. Total weight

[tex]\begin{array}{rcr}\text{Oranges} & = & \text{230 lb}\\\text{Boxes} & = & \text{16 lb}\\\text{TOTAL} & = & \textbf{250 lb}\\\end{array}\\\text{On average, the total weight is $\large \boxed{\textbf{250 lb}}$}[/tex]

Note: The answer can have only two significant figures because that is all you gave for the average weight of an orange.

Answer:

30.73 Doses and 39.42 g

Explanation:

It is necessary to know the conversion factors in that manner change the initial 500 mL to the number of doses as follow:

[tex]500 ml * \frac{1tsp}{4.93mL} * \frac{1TBS}{3tsp} * \frac{1 Dose }{1.1 TBS} = 30.7 doses\\[/tex]

And the for b. it is such necessary to take in account that 1 mL = 1 [tex]cm^{3}[/tex]

35.2 mL * 1.12 [tex]\frac{g}{cm^{3} }[/tex] = 39.42 g

Answer:

The molar composition of the equilibrium mixture is

NO: 0.338 = 33.8%

Br2: 0.169 = 16.9%

NOBr: 0.493 = 49.3%

Explanation:

The reaction of nitrosyl bromide formation can be written as

[tex]NO+0.5\,Br_2\rightarrow NOBr[/tex]

To form 1 mol of NOBr, we need 1 mol of NO and 0.5 mol of Br2.

A sample of 0.0524 mol NO with 0.0262 mol Br2 gives an equilibrium mixture containing 0.0311 mol NOBr.

Then, to form 0.0311 mol NOBr, were needed 0.0311 mol NO and 0.01555 mol of Br2.

The amount of NO that stay in the same form is (0.0524-0.0311)=0.0213 mol NO.

The amount of Br2 that stay in the same form is (0.0262-0.01555)=0.01065 mol Br2.

The total amount of mol is (0.0213 mol NO + 0.01065 mol Br2 + 0.0311 mol NOBr) = 0.06305.

The molar composition is

NO: 0.0213/0.06305 = 0.338 = 33.8%

Br2: 0.01065/0.06305 = 0.169 = 16.9%

NOBr: 0.0311/0.06305 = 0.493 = 49.3%

Answer:

The answer is 916.67 g

Explanation:

48.0 wt% NaOH means that there are 48 g of NaOH in 100 g of solution. With this information and the molecular weight of NaOH (40 g/mol), we can calculate the number of mol there are in 100 g of this solution:

[tex]\frac{48 g NaOH}{100 g solution}[/tex] x [tex]\frac{1 mol NaOH}{40 g}[/tex] = 0.012 mol NaOH/100 g solution

Finally, we need 0.11 mol in 1 liter of solution to obtain a 0.11 M NaOH solution.

0.012 mol NaOH ------------ 100 g solution

0.11 mol NaOH------------------------- X

X= 0.11 mol NaOH x 100 g/ 0.012 mol NaOH= 916.67 g

We have to weigh 916.67 g of 48.0%wt NaOH and dilute it in a final volume of 1 L of water to obtain a 0.11 M NaOH solution.

Answer:

The hydrogen cannot participate in hydrogen bonding with other molecules.

Explanation:

Hydrogen bonding is a special type of the dipole-dipole interaction and it occurs between hydrogen atom that is bonded to highly electronegative atom which is either fluorine, oxygen or nitrogen atom.

Thus, hydrogen bonding is present in alcohols as the oxygen atom is linked to hydrogen

Partially positive end of the hydrogen atom is attracted to partially negative end of the oxygen atom. It is strong force of attraction between the molecules. Thus, hydrogen acquires slight positive charge and become electron deficient or acidic.

Hence, option C is incorrect.

Answer:

0.002 %

Explanation:

Given that:

[tex]K_{a}=4.0\times 10^{-10}[/tex]

Concentration = 1.0 M

Consider the ICE take for the dissociation of Hydrocyanic acid as:

                                      HCN    ⇄     H⁺ +        CN⁻

At t=0                            1.0                -              -

At t =equilibrium        (1.0-x)                x           x            

The expression for dissociation constant of Hydrocyanic acid is:

[tex]K_{a}=\frac {\left [ H^{+} \right ]\left [ {CN}^- \right ]}{[HCN]}[/tex]

[tex]4.0\times 10^{-10}=\frac {x^2}{1.0-x}[/tex]

x is very small, so (1.0 - x) ≅ 1.0

Solving for x, we get:

x = 2×10⁻⁵  M

Percentage ionization = [tex]\frac {2\times 10^{-5}}{1.0}\times 100=0.002 \%[/tex]

Answer: [tex]2.47\times 10^{6}[/tex]

Explanation:

Engineering notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form times 10 raise to the power.  It is similar to the scientific notation but in engineering notation, the powers of ten are always multiples of 3.

The engineering notation written in the form:

[tex]a\times 10^b[/tex]

where,

a = the number which is greater than 0 and less than 999

b = an integer multiple of 3

Now converting the given value of 2,469,100 into engineering notation, we get [tex]2.47\times 10^{6}[/tex]

Hence, the correct answer is, [tex]2.47\times 10^{6}[/tex]

Answer:

a) p = 1.10 * 10⁻²⁷ kg·m/s

b) p = 9.46 * 10⁻²⁴ kg·m/s

c) p = 3.31 * 10⁻³⁶ kg·m/s

Explanation:

To solve this problem we use the de Broglie's equation, which describes the wavelenght of a photon with its momentum:

λ=h/p

Where λ is the wavelength, h is Planck's constant (6.626 * 10⁻³⁴ J·s), and p is the linear momentum of the photon.

Rearrange the equation in order to solve for p:

p=h/λ

And now we proceed to calculate, keeping in mind the SI units:

a) 600 nm= 600 * 10⁻⁹ m

p=(6.626 * 10⁻³⁴ J·s) / (600*10⁻⁹m) = 1.10 * 10⁻²⁷ kg·m/s

b) 70 pm= 70 * 10⁻¹² m

p=(6.626 * 10⁻³⁴ J·s) / (70*10⁻¹²m) = 9.46 * 10⁻²⁴ kg·m/s

c) 200 m

p=(6.626 * 10⁻³⁴ J·s) / (200m) = 3.31 * 10⁻³⁶ kg·m/s

Answer:

[tex]-r_{O_2}=1\times{10}^{-5}\frac{M}{s}[/tex]

Explanation:

The decomposition of [tex]NO_2[/tex] follows the equation  

[tex]2NO_2\rightarrow2NO+O_2[/tex]

By definition, the rate of a chemical reaction can be expressed by  

[tex]-r_{NO_2}=\frac{d\left[NO_2\right]}{dt}=\frac{0.010-0.008\ M}{100\ s}=2\times{10}^{-5}\frac{M}{s}[/tex]

The rate of appearance of [tex]O_2[/tex] is related to the rate of disappearance of [tex]NO_2[/tex] by the stoichiometry. This means that, for each mole of [tex]O_2[/tex] that appears 2 moles of [tex]NO_2[/tex]  are consumed. So  

[tex]-r_{O_2}=-r_{NO_2}\times\frac{1\ mole\ \ O_2}{2\ mole\ \ {\rm NO}_2}=1\times{10}^{-5}\frac{M}{s}[/tex]

The rate of appearance of O2 for this period is -1.0x10^-5 M/s.

The rate of appearance of O2 for a given period can be determined using the change in concentration of NO2 over that period. In this case, the concentration of NO2 drops from 0.0100 to 0.00800 M in 100 s. First, calculate the change in concentration of NO2:

Δ[NO2] = [NO2]final - [NO2]initial = 0.00800 M - 0.0100 M = -0.00200 M

Since the reaction is 2NO2 → 2NO + O2, the stoichiometric ratio between NO2 and O2 is 2:1. Therefore, the change in concentration of O2 is half of the change in concentration of NO2:

Δ[O2] = -0.00200 M ÷ 2 = -0.00100 M

Finally, divide the change in concentration of O2 by the time period to find the rate of appearance of O2:

Rate = Δ[O2] ÷ time = -0.00100 M ÷ 100 s = -1.0x10-5 M/s

Answer:

There will be 13107200 number of bacteria after 12 hours

Explanation:

1 hour = 60 minutes

So, 12 hours = ([tex]12\times 60[/tex]) minutes = 720 minutes

Initially there are 50 bacteria.

As number of bacteria doubles in every 40 minutes therefore rate of increase in number of bacteria will be similar to a first order reaction.

Hence, number of bacteria after 720 minutes = [tex]50\times (2)^{\frac{720}{40}}[/tex] = 13107200

Answer:

option D= temperature and volume

Explanation:

Definition:

Charle's law stated that ,

" At constant pressure, the volume of given amount of gas is directly proportional to its absolute temperature"

V∝ T

V= kT

OR

V/T = k     (k is proportionality constant)

if the temperature is changed from T1 to T2 then the volume of gas is also changed from V1 to V2. Then expression will be:

V1/T1= k  and   V2/T2=k

V1/T1=V2/T2

suppose a cylinder is filled with a gas having volume V1 at temperature T1. When the gas is heated its temperature raises from T1 to T2 and its volume also increased with increase of temperature from V1 TO V2.

In Charles' law, the two changing properties are temperature and volume, with pressure maintained constant. The law indicates a direct proportionality between temperature and volume for a given amount of gas.

The two changing properties in Charles' law are temperature and volume. Charles' law states that for a given amount of gas at constant pressure, the volume of the gas is directly proportional to its absolute temperature. This means if the temperature of the gas increases, its volume increases as well, provided the pressure and the amount of gas remain constant. In other words, doubling the absolute temperature at constant pressure will double the volume. It is important to note that in gas laws, temperatures must always be expressed in kelvins.

Answer: The area of square is [tex]4.84cm^2[/tex]

Explanation:

To calculate the area of square, we use the formula:

[tex]A=l^2[/tex]

where,

A = area of square = ?

l = edge length of square  = 2.2 cm

Putting values in above equation, we get:

[tex]\text{Area of square}=(2.2)^2=4.84cm^2[/tex]

Hence, the area of square is [tex]4.84cm^2[/tex]

Answer:

4687.5 mL

Explanation:

Given that the density of the substance = 8.96 g/mL

Mass = 42.0 kg

The conversion of kg into g is shown below as:

1 kg = 1000 g

So, mass = 42.0 ×  1000 g = 42000 g

Volume = ?

So, volume:  

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

[tex]Volume=\frac {42000\ g}{8.96\ g/mL}[/tex]  

The volume of the 42.0 kg of the substance = 4687.5 mL

Answer:

The mole fraction composition of the liquid is :

Mole fraction of butane, pentane and hexane are 0.3638,0.3908 and 0.2454 respectively.

Explanation:

Mass of the liquid mixture = 200 g

Percentage of butane = 30%

Mass of butane = [tex]\frac{30}{100}\times 200 g=60 g[/tex]

Moles of butane = [tex]n_1=\frac{60 g}{58 g/mol}=1.0345 mol[/tex]

Percentage of pentane= 40%

Mass of pentane= [tex]\frac{40}{100}\times 200 g=80 g[/tex]

Moles of pentane= [tex]n_2=\frac{80 g}{58 g/mol}=1.1111 mol[/tex]

Percentage of hexane = 100% - 30% - 40% = 30%

Mass of hexane = [tex]\frac{30}{100}\times 200 g=60 g[/tex]

Moles of hexane = [tex]n_2=\frac{60 g}{86 g/mol}=0.6977 mol[/tex]

Mole fraction of butane, pentane and hexane : [tex]\chi_1, \chi_2 \& \chi_3[/tex]

[tex]\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{1.0345 mol}{1.0345 mol+1.1111 mol+0.6977 mol}=0.3638[/tex]

[tex]\chi_2=\frac{n_2}{n_1+n_2+n_3}=\frac{1.1111 mol}{1.0345 mol+1.1111 mol+0.6977 mol}=0.3908[/tex]

[tex]\chi_3=\frac{n_1}{n_1+n_2+n_3}=\frac{0.6977 mol}{1.0345 mol+1.1111 mol+0.6977 mol}=0.2454[/tex]

Answer:

c- Hot water remains almost same but cold water temperature increases drastically.

Explanation:

In a process of heat exchange where a hot and a cold fluid are present, the amount of heat that is transfered from one to another, the heat lost by the hot fluid, is gained by the cold fluid. That can be calculated by the equation : Q = m * Cp * dT.

Where:

m= mass flowrate

Cp=specific-heat-capacity

dT= difference between the inlet and outlet temperatures

We can state the equality of this equation for both fluids:

(mass-flowrate * specific-heat-capacity * (temperature-in – temperature-out) ) for hot medium = (mass-flowrate * specific-heat-capacity * (temperature-out – temperature-in) ) for cold medium

Now if we increase the value of the mass flow rate of the hot fluid to its maximum and decrease the mass flowrate of the cold fluid to its minimum. That will mean that the difference between the inlet and outlet temperatures for the cold fluid must increase so as to compensate the increase on the other side of the equation and guarantee the equality. And taking in account that the inlet temperature of the cold fluid is known and what can change is the outlet one, the correct answer is c that says that the outlet temperature of the cold fluid increases drasically.

Explanation:

Atomic number of P = 15

Common oxidation state of P = -3, +3 and +5

Electronic configuration of P^3+: [tex]1s^2 2s^22p^63s^2[/tex]

Electronic configuration of P^3-: [tex]1s^2 2s^22p^63s^23p^6[/tex]

Electronic configuration of P^5+: [tex]1s^2 2s^22p^6[/tex]

Atomic number of Ar = 18

Argon has stable octet, so it generally does not exist as ions.

Electronic configuration of Ar: [tex]1s^2 2s^22p^63s^23p^6[/tex]

Atomic number of Te = 52

Common oxidation states of Te = -2, +2, +4, +6

Electronic configuration of Te^2-: [tex][Kr]4d^10 5s^2 5p^6[/tex]

Electronic configuration of Te^2+: [tex][Kr]4d^10 5s^2 5p^2[/tex]

Electronic configuration of Te^4+: [tex][Kr]4d^10 5s^2[/tex]

Electronic configuration of Te^6+: [tex][Kr]4d^10[/tex]

Boiling point of HF is more as compared to HCl

Reason:

Molecular weight of HF is low, but is boiling point is high because of presence of hydrogen bonding. Whereas in case of HCl, its molecular weight is high but has only weak van der Waals intercations. As hydrogen bonding is stronger than van der Waals interactions, therefore, boiling point of HF is more.

Answer:

Explained

Explanation:

Erucamide is plastic additive used as anti slip agent and used as scratch resistant. It is used as scratch resistance because

- It reduces coefficient of friction between the polymer layer.

- Slip additives in Erucamide form coalesce along the surface.

-These additives move from the bulk to the part surface to form closed pack layers.

Answer: Option (c) is the correct answer.

Explanation:

Entropy is defined as the degree of randomness that is present within the particles of a substance.

As [tex]NH_{4}NO_{3}[/tex] is ionic in nature. Hence, when it is added to water then it will readily dissociate into ammonium ions ([tex]NH^}{+}_{4}[/tex]) and nitrate ions ([tex]NO^{-}_{3}[/tex]).

Therefore, it means that ions of ammonium nitrate will be free to move from one place to another. Hence, there will occur an increase in entropy.

Thus, we can conclude that ammonium nitrate ([tex]NH_{4}NO_{3}[/tex]) dissolve readily in water even though the dissolution process is endothermic by 26.4 kJ/mol because the overall entropy of the system increases upon dissolution of this strong electrolyte.

Answer:

2.1 °C/m

Explanation:

Hello, for this exercise, consider the formula:

[tex]T_{solution}-T{solvent}=K_bm_solute[/tex]

Considering that the difference in the temperature is 0.284°C, and the given molality by:

[tex]m_{solute}=\frac{10.6g\frac{1mol}{106g}}{740g*\frac{1kg}{1000g} } \\m_{solute}=0.135m[/tex]

Now, solving for [tex]K_b[/tex], we get:

[tex]K_b=\frac{0.284C}{0.135m}\\K_b=2.1 C/m[/tex]

Best regards.

Answer:

A. Oxidation refers to loss of electron or increase in oxidation state. Reduction refers to gain of electrons or decrease in oxidation state.

B. The oxidation number of S in S₂O₃²⁻ is +2.

C. The oxidation number of S in S₄O₆²⁻ is +5/2.

Explanation:

A. Oxidation refers to loss of electron or increase in oxidation state. Reduction refers to gain of electrons or decrease in oxidation state.

The overall charge on a molecule is equal to the sum of the oxidation states of all the atoms in a molecule.

B. S₂O₃²⁻ : The oxidation state of oxygen (O) is -2 and overall charge on molecule is -2.

Therefore, the oxidation state of sulfur (x) can be calculated by

2x + 3 (-2) = -2

2x + (-6) = -2

2x  = -2 + 6 = 4

x = 4 ÷ 2= +2

Therefore, the oxidation number of S is +2.

C. S₄O₆²⁻ : The oxidation state of oxygen (O) is -2 and overall charge on molecule is -2.

Therefore, the oxidation state of sulfur (x) can be calculated by

4x + 6 (-2) = -2

4x + (-12) = -2

4x  = -2 + 12 = 10

x = 10 ÷ 4 = +5/2

Therefore, the oxidation number of S is +5/2.

Explanation:

The given data is as follows.

       P = 28.3 mm = [tex]\frac{28.3}{760}[/tex],            V = 600 mL = [tex]600 ml \frac{0.001 L}{1 ml}[/tex] = 0.6 L

      R = 0.082 L atm/mol K,                 T = (28 + 273) K = 301 K

Therefore, according to ideal gas law PV = nRT. Hence, putting the values into this equation calculate the number of moles as follows.

                                    PV = nRT

                 [tex]\frac{28.3}{760} \times 0.6 L[/tex] = [tex]n \times 0.082 L atm/mol K \times 301 K[/tex]            

                       n = [tex]9.03 \times 10^{-4}[/tex] mol  

As it is known that number of moles equal to mass divided by molar mass. Hence, mass of water vapor present will be calculated as follows. (molar mass of water is 18 g/mol)

               No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]

        [tex]9.03 \times 10^{-4}[/tex] mol = [tex]\frac{mass}{18 g/mol}[/tex]

                            mass = [tex]162.5 \times 10^{-4}g[/tex]

                                      = [tex]163 \times 10^{-4}g[/tex] (approx)

Since, 1 g = 1000 mg. Therefore, [tex]163 \times 10^{-4}g[/tex] will be equal to  [tex]163 \times 10^{-4}g \times \frac{10^{3}mg}{1 g}[/tex]

                          = 16.3 mg

Thus, we can conclude that mass of water vapor present is 16.3 mg.

To find the mass of water vapor in a given vapor volume at a certain temperature, you can use the ideal gas law. Convert the temperature from Celsius to Kelvin, calculate the number of moles using the ideal gas law equation, and then convert moles to grams using the molar mass of water. Finally, convert grams to milligrams. The mass of water vapor in a vapor volume of 600 mL at 28 °C is 5097 mg.

To find the mass of water vapor, we can use the ideal gas law equation: PV = nRT, where P represents pressure, V represents volume, n represents the number of moles, R represents the ideal gas constant, and T represents temperature in Kelvin. First, we need to convert the temperature from Celsius to Kelvin by adding 273.15. Then, we can rearrange the equation to solve for n, the number of moles. Once we have the moles, we can use the molar mass of water to convert it to mass in grams. Finally, we can convert grams to milligrams by multiplying by 1000.

Given:

Vapor pressure of water = 28.3 mm Hg
Temperature = 28 °C
Vapor volume = 600 mL

Converting the temperature to Kelvin:

T = 28 °C + 273.15 = 301.15 K

Using the ideal gas law equation to find the number of moles (n):

n = PV / RT

Substituting the values:

n = (28.3 mm Hg * 600 mL) / (62.36 L mm/mol K * 301.15 K)

n = 0.2829 mol

Using the molar mass of water (18.015 g/mol) to find the mass:

mass = n * molar mass

mass = 0.2829 mol * 18.015 g/mol

mass = 5.097 g

Converting grams to milligrams:

mass = 5.097 g * 1000 mg/g

mass = 5097 mg

Therefore, the mass of water vapor in a vapor volume of 600 mL at 28 °C is 5097 mg.

Answer:

Part A

[tex]rate= 4.82*10^{-3}s^{-1} * [N2O5][/tex]

Part B

[tex]rate= 1.35*10^{-4}Ms^{-1}[/tex]

Explanation:

Part A

The rate law is the equation that relates the rate of the reaction, the kinetic constant and the concentration of the reactant or reactants.

For the given chemical reaction we can write a general expression for the rate law as follows:

[tex]rate= k * [N2O5]^{x}[/tex]

where k is the rate constant and x is the order of the reaction with respect of N2O5 concentration. Particularly, a first order reaction kinetics indicate that the rate of the reaction is directly proportional to the concentration of only one reactant. Then x must be 1.

Replacing the value of the rate constant given in the text we can arrive to the following expression for the rate law:

[tex]rate= 4.82*10^{-3}s^{-1} * [N2O5][/tex]

Part B

Replacing the value of the concentration of N2O5 given, we can get the rate of reaction:

[tex]rate= 4.82*10^{-3}s^{-1} *2.80*10^{-2}M[/tex]

[tex]rate= 1.35*10^{-4}Ms^{-1}[/tex]