How many moles of disulfur decafluoride are present in 2.45 grams of this compound? moles

Answer : The volume of [tex]CO_2[/tex] gas is 1.00 L

Explanation :

First we have to determine the moles of [tex]C_3H_4PO_7^{3-}[/tex].

Molar mass of [tex]C_3H_4PO_7^{3-}[/tex] = 182.9 g/mole

[tex]\text{ Moles of }C_3H_4PO_7^{3-}=\frac{\text{ Mass of }C_3H_4PO_7^{3-}}{\text{ Molar mass of }C_3H_4PO_7^{3-}}=\frac{15.0g}{182.9g/mole}=0.0820moles[/tex]

Now we have to calculate the moles of [tex]CO_2[/tex].

The given balanced chemical reaction is:

[tex]C_5H_8P_2O_{11}^{4-}(aq)+H_2O(aq)+CO_2(g)\rightarrow 2C_3H_4PO_7^{3-}(aq)+2H^+(aq)[/tex]

From the reaction we conclude that,

As, 2 moles of [tex]C_3H_4PO_7^{3-}[/tex] produce from 1 mole of [tex]CO_2[/tex]

So, 0.0820 moles of [tex]C_3H_4PO_7^{3-}[/tex] produce from [tex]\frac{0.0820}{2}=0.041moles[/tex] of [tex]CO_2[/tex]

Now we have to calculate the volume of [tex]CO_2[/tex] gas.

Using ideal gas equation:

[tex]PV=nRT[/tex]

where,

P = pressure of gas = 1.00 atm

V = volume of gas = ?

T = temperature of gas = 298 K

n = number of moles of gas = 0.041 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get:

[tex](1.00atm)\times V=(0.041mole)\times (0.0821L.atmK^{-1}mol^{-1})\times (298K)[/tex]

[tex]V=1.00L[/tex]

Therefore, the volume of [tex]CO_2[/tex] gas is 1.00 L

Answer:

mass flow = 64.9086 Lbm/min

Explanation:

SG = 0.700 = ρ gas / ρ H2O

∴ ρ H2O = 1000 Kg/m³

⇒ ρ gas = 700 Kg/m³

∴ F auto tank = 20.0 gal / 1.8 min = 11.11 gal/min * ( m³/264.172 gal)

⇒ F auto = 0.042 m³/min

⇒ mass flow = 0.042 m³/min * (700 Kg/m³) * ( 2.20462 Lbm/ Kg )

⇒ mass flow =  64.9086 Lbm/min

Answer:

big question

Explanation:

Answer: The density of substance is [tex]3.38g/cm^3[/tex]

Explanation:

To calculate density of a substance, we use the equation:

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

We are given:

Mass of substance = 61.6 g

Volume of substance = [tex]18.2cm^3[/tex]

Putting values in above equation, we get:

[tex]\text{Density of substance}=\frac{61.6g}{18.2cm^3}\\\\\text{Density of substance}=3.38g/cm^3[/tex]

Hence, the density of substance is [tex]3.38g/cm^3[/tex]

Final answer:

The density of the substance with a mass of 61.6 g and a volume of 18.2 [tex]cm^3[/tex] is 3.385 g/[tex]cm^3[/tex], calculated by dividing mass by volume using the density formula.

Explanation:

The density of a substance is calculated using the formula d = m/V, where d is the density, m is the mass, and V is the volume.

To find the density of a substance with a given mass and volume, you simply divide the mass by the volume. For a substance with a mass of m = 61.6 g and a volume of V = 18.2 [tex]cm^3[/tex], the density d would be calculated as follows:

d = m/V
d = 61.6 g / 18.2 [tex]cm^3[/tex]
d \3.385 g/[tex]cm^3[/tex]

This means that the density measurement of the substance is 3.385 grams per cubic centimeter, expressed as g/[tex]cm^3[/tex].

Answer:

The answer is c. Express Scripts Company

Explanation:

DrugDigest is a website tool that helps you to find information about medicaments  you consume or you see for example you introduce the first three letters of the drug in the website and you will have a detailed description of brands, supplied information eg. if the medicament comes by pills or by syrup and also the concentration of the medicament and recommendations ( under which case you can take the medicine, effects, side effects)  

Answer : The correct answer is, [tex]6.054\times 10^5[/tex]

Explanation :

Scientific notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.

For example :

5000 is written as [tex]5.0\times 10^3[/tex]

889.9 is written as [tex]8.899\times 10^{-2}[/tex]

In this examples, 5000 and 889.9 are written in the standard notation and [tex]5.0\times 10^3[/tex]  and [tex]8.899\times 10^{-2}[/tex]  are written in the scientific notation.

[tex]8.89\times 10^{-2}[/tex]  this is written in the scientific notation and the standard notation of this number will be, 0.00889.

If the decimal is shifting to right side, the power of 10 is negative and if the decimal is shifting to left side, the power of 10 is positive.

As we are given the 605,400.34 in standard notation.

Now converting this into scientific notation, we get:

[tex]\Rightarrow 605,400.34=6.054\times 10^5[/tex]

As, the decimal point is shifting to left side, thus the power of 10 is positive.

Hence, the correct answer is, [tex]6.054\times 10^5[/tex]

To find the molar mass of the protein subunit, use the formula for osmotic pressure. Rearrange the formula to solve for M and plug in the given values. Convert the mass of the protein to moles using the molar mass.

To find the molar mass of the protein subunit, we can use the formula for osmotic pressure, II = MRT. Given that the osmotic pressure is 0.138 atm and the temperature is 28 °C (which is 310 K), we can rearrange the formula to solve for M. The value of R is 0.08206 L atm/mol K. Plugging in the values, we get:

0.138 atm = M * (0.08206 L atm/mol K) * (310 K)

Solving for M gives:

M = 0.138 atm / (0.08206 L atm/mol K * 310 K)

Calculating this expression will give us the molarity of the solution. Since we dissolved 0.150 g of the protein subunit in enough water to make 2.00 mL of solution, we can convert this to moles using the molar mass of the protein. The molar mass is given by:

Molar mass = Mass / Moles

Therefore, we can rearrange the equation to solve for the molar mass:

Molar mass = Mass / Moles = (0.150 g / M) * (1 mol / 1000 g)

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Answer : The mass of helium gas added must be 12.48 grams.

Explanation : Given,

Mass of helium (He) gas = 6.24 g

Molar mass of helium = 4 g/mole

First we have to calculate the moles of helium gas.

[tex]\text{Moles of }He=\frac{\text{Mass of }He}{\text{Molar mass of }He}=\frac{6.24g}{4g/mole}=1.56moles[/tex]

Now we have to calculate the moles of helium gas at doubled volume.

According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,

[tex]V\propto n[/tex]

or,

[tex]\frac{V_1}{V_2}=\frac{n_1}{n_2}[/tex]

where,

[tex]V_1[/tex] = initial volume of gas  = V

[tex]V_2[/tex] = final volume of gas = 2V

[tex]n_1[/tex] = initial moles of gas  = 1.56 mole

[tex]n_2[/tex] = final moles of gas  = ?

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

[tex]\frac{V}{2V}=\frac{1.56mole}{n_2}[/tex]

[tex]n_2=3.12mole[/tex]

Now we have to calculate the mass of helium gas at doubled volume.

[tex]\text{Mass of }He=\text{Moles of }He\times \text{Molar mass of }He[/tex]

[tex]\text{Mass of }He=3.12mole\times 4g/mole=12.48g[/tex]

Therefore, the mass of helium gas added must be 12.48 grams.

Final answer:

To double the volume of a helium balloon at constant temperature and pressure, an additional 6.24 grams of helium must be added, resulting in a total of 12.48 grams of helium in the balloon.

Explanation:

The question involves the concept of the molar volume of a gas, which is a central principle in chemistry, particularly in the context of gases under the ideal gas law. To double the volume of helium in a balloon at constant temperature and pressure, the amount of helium in moles must be doubled as well. If initially there are 6.24 grams of helium in the balloon, we first convert this mass to moles using the molar mass of helium (4.00 g/mol) and then calculate the additional moles (and hence grams) of helium needed to double the volume.

Initial moles of helium (n1): 6.24 g / 4.00 g/mol = 1.56 moles

To double the volume, we need another 1.56 moles of helium. The mass of additional helium required is:

Additional mass (Δm): 1.56 moles x 4.00 g/mol = 6.24 g

Therefore, to double the volume of the helium balloon, an additional 6.24 grams of helium gas must be added.

Explanation:

It is known that number of moles of a substance equal to mass divided by its molar mass.

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

It is given that no. of oles present are 27.2 mol and molar mass of N is 14 g/mol.

Therefore, calculate mass of N as follows.

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

                       27.2 mol = [tex]\frac{mass}{14 g/mol}[/tex]

                      mass = 380.8 g

Thus, we can conclude that mass of N is 380.8 g.

Explanation:

It is known that pinch point temperature difference is defined as the difference between the temperature of exhaust existing the evaporator and the temperature of water evaporation.

     [tex]\Delta T_{p.p}[/tex] = Exhaust existing the evaporator temperature - Temperature of water evaporation

Since, it is given that exhaust existing the evaporator temperature = [tex]1000^{o}F[/tex]

           Temperature of water evaporation = [tex]800^{o}F[/tex]

Hence, calculate [tex]\Delta T_{p.p}[/tex] using the above formula as follows.

                   [tex]\Delta T_{p.p}[/tex] = Exhaust existing the evaporator temperature - Temperature of water evaporation

                                  = [tex]1000^{o}F[/tex] - [tex]800^{o}F[/tex]

                                  = [tex]200^{o}F[/tex]

Thus, we can conclude that the temperature difference at the pinch point in the generator is [tex]200^{o}F[/tex].

Answer: [tex]4.4\times 10^{2}[/tex]

Explanation:

The chemical reaction follows the equation:

     [tex]2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)[/tex]

t = 0 [tex]5.000\times 10^{-3}[/tex] [tex]2.50\times 10^{-3}[/tex]    0

At eqm [tex](5.000\times 10^{-3}-2x)[/tex] [tex](2.50\times 10^{-3}-x)[/tex]   (2x)      

The expression for [tex]K_c[/tex] for the given reaction follows:

[tex]K_c=\frac{[SO_3]^2}{[SO_2]^2[O_2]}[/tex]

[tex]K_c=\frac{[2x]^2}{[5.000\times 10^{-3}-2x]^2[2.50\times 10^{-3}-x]}[/tex]

Given : [tex][SO_2]_{eqm}=2.80\times 10^{-3}[/tex]

[tex]5.000\times 10^{-3}-2x=2.80\times 10^{-3}[/tex]

[tex]x=1.1\times 10^{-3}[/tex]

Putting the values we get:

[tex]K_c=\frac{[2\times 1.1\times 10^{-3}]^2}{[5.000\times 10^{-3}-2\times 1.1\times 10^{-3}]^2[2.50\times 10^{-3}-1.1\times 10^{-3}]}[/tex]

[tex]K_c=4.4\times 10^2[/tex]

Therefore, the equilibrium concentration [tex]4.4\times 10^{2}[/tex]

Answer: 15.850

Explanation:

The conversion used from liters to gallons is:

1 L = 0.264172 gallon

The conversion used from sec to min is:

60 sec = 1 min

1 sec =[tex]\frac{1}{60}\times 1=0.017min[/tex]

We are asked: liters/sec = gallons/min

[tex]liters/sec=\frac{0.264172}{0.017}=15.850gallons/min[/tex]

Therefore, to convert from liters/second to gallons/minute, multiply the number of liters/second by 15.850.

Answer:  The nature of an amphoteric substance is amphoprotic; it has ability to donate proton or gain proton. So the amphotric substances are those which can react as both acid or base in a reaction. Many oxides of metals when subjected to a reaction mixture can acts as both an acid or either a base, they are called amphoteric oxides. some of the examples of these amphoteric oxide is zinc oxide and lead oxide.

Hence, the correct option here is (c )

Answer:

a) pH = 2,88

b) pH = 4,58

c) pH = 5,36

d) pH = 8,79

e) pH = 12,10

Explanation:

In a titration of a strong base (KOH) with a weak acid (CH₃COOH) the reaction is:

CH₃COOH + KOH → CH₃COOK + H₂O

a) Here you have just CH₃COOH, thus:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ where ka =1,74x10⁻⁵ and pka = 4,76

When this reaction is in equilibrium:

[CH₃COOH] = 0,100 -x

[CH₃COO⁻] = x

[H⁺] = x

Thus, equilibrium equation is:

1,74x10⁻⁵ = [tex]\frac{[x][x] }{[0,100-x]}[/tex]

The equation you will obtain is:

x² + 1,74x10⁻⁵x - 1,74x10⁻⁶ = 0

Solving:

x = -0,0013278193 ⇒ No physical sense. There are not negative concentrations

x = 0,0013104193

As x = [H⁺] and pH = - log [H⁺]

pH = 2,88

b) Here, it is possible to use:

CH₃COOH + KOH → CH₃COOK + H₂O

With adition of 5,0 mL of 0,200M KOH solution the initial moles are:

CH₃COOH = [tex]0,025 L.\frac{0,100 mol}{L} =[/tex] = 2,5x10⁻³ mol

KOH = [tex]0,005 L.\frac{0,200 mol}{L} =[/tex] = 1,0x10⁻³ mol

CH₃COOK = 0.

In equilibrium:

CH₃COOH = 2,5x10⁻³ mol - 1,0x10⁻³ mol = 1,5x10⁻³ mol

KOH = 0 mol

CH₃COOK = 1,0x10⁻³ mol

Now, you can use Henderson–Hasselbalch equation:

pH = 4,76 + log [tex]\frac{1,0x10^{-3} }{1,5x10^{-3} }[/tex]

pH = 4,58

c) With adition of 10,0 mL of 0,200M KOH solution the initial moles are:

CH₃COOH = [tex]0,025 L.\frac{0,100 mol}{L} =[/tex] = 2,5x10⁻³ mol

KOH = [tex]0,010 L.\frac{0,200 mol}{L} =[/tex] = 2,0x10⁻³ mol

CH₃COOK = 0.

In equilibrium:

CH₃COOH = 2,5x10⁻³ mol - 2,0x10⁻³ mol = 0,5x10⁻³ mol

KOH = 0 mol

CH₃COOK = 2,0x10⁻³ mol

Now, you can use Henderson–Hasselbalch equation:

pH = 4,76 + log [tex]\frac{2,0x10^{-3} }{0,5x10^{-3} }[/tex]

pH = 5,36

d) With adition of 12,5 mL of 0,200M KOH solution the initial moles are:

CH₃COOH = [tex]0,025 L.\frac{0,100 mol}{L} =[/tex] = 2,5x10⁻³ mol

KOH = [tex]0,0125 L.\frac{0,200 mol}{L} =[/tex] = 2,5x10⁻³ mol

CH₃COOK = 0.

Here we have the equivalence point of the titration, thus, the equilibrium is:

CH₃COO⁻ + H₂O ⇄ CH₃COOH + OH⁻ kb = kw/ka where kw is equilibrium constant of water = 1,0x10⁻¹⁴; kb = 5,75x10⁻¹⁰

Concentrations is equilibrium are:

[CH₃COOH] = x

[CH₃COO⁻] = 0,06667-x

[OH⁻] = x

Thus, equilibrium equation is:

5,75x10⁻¹⁰ = [tex]\frac{[x][x] }{[0,06667-x]}[/tex]

The equation you will obtain is:

x² + 5,75x10⁻¹⁰x - 3,83x10⁻¹¹ = 0

Solving:

x = -0.000006188987⇒ No physical sense. There are not negative concentrations

x = 0.000006188

As x = [OH⁻] and pOH = - log [OH⁻]; pH = 14 - pOH

pOH = 5,21

pH = 8,79

e) The excess volume of KOH will determine pH:

With 12,5mL is equivalence point, the excess volume is 15,0 -12,5 = 2,5 mL

2,5x10⁻³ L × [tex]\frac{0,200 mol}{1L}[/tex] ÷ 0,040 L = 0,0125 = [OH⁻]

pOH = - log [OH⁻]; pH = 14 - pOH

pOH = 1,90

pH = 12,10

I hope it helps!

Answer:

The pH of solution on addition of 0.0, 5.0, 10.0 12.5 and 15 ml of KOH will be 2.87, 4.56, 5.34, 4.74,  and 12.09 respectively.

Explanation:

pH can be calculated by the evaluation of Hydronium ions in the solution.

[tex]\rm K_a[/tex] of [tex]\rm CH_3COOH[/tex] is 1.8 [tex]\rm \times10^-^5[/tex]

(a) Hydrogen ion concentration = [tex]\rm \sqrt{K_a\;\times\;CH_3COOH\;concentartion}[/tex]

Hydrogen ion concentration = [tex]\rm \sqrt{1.8\;\times\;10^-^5\;\times\;0.1}[/tex]

Hydrogen ion concentration = 1.34 [tex]\times\;10^-^3[/tex] M

pH of solution = log [Hydrogen ion concentration]

pH = log [[tex]\rm 1.34\;\times10^-^5[/tex]]

pH = 2.87

(b) On addition of 5 ml of KOH of 0.200 M, the moles of KOH added are:

moles of KOH = [tex]\rm \frac{volume\;(ml)}{1000}\;\times\;\frac{molarity}{L}[/tex]

moles of KOH = [tex]\rm \frac{5}{1000}\;\times\;\frac{0.200}{L}[/tex]

moles of KOH = [tex]\rm 1\;\times\;10^-^3[/tex] M

The initial moles of [tex]\rm CH_3COOH[/tex] are:

moles of [tex]\rm CH_3COOH[/tex] = [tex]\rm \frac{25}{1000}\;\times\;\frac{0.100}{L}[/tex]

moles of [tex]\rm CH_3COOH[/tex] = [tex]\rm 2.5\;\times\;10^-^3[/tex]

At equilibrium, the concentration of [tex]\rm CH_3COOH[/tex] = [tex]\rm 2.5\;\times\;10^-^3[/tex] - [tex]\rm 1\;\times\;10^-^3[/tex] mol

concentration of [tex]\rm CH_3COOH[/tex] = [tex]\rm 1.5\;\times\;10^-^3[/tex] moles.

The concentration of [tex]\rm CH_3COOK[/tex] = [tex]\rm 1\;\times\;10^-^3[/tex] moles

pH = [tex]\rm pK_a\;+\;log\;\frac{salt}{acid}[/tex]

pH = 4.76 + log [tex]\rm \frac{1\;\times\;10^-^3}{1.5\;\times\;10^-^3}[/tex]

pH = 4.58

(c) On addition of 10 ml of KOH,

moles of KOH = [tex]\rm \frac{10}{1000}\;\times\;\frac{0.200}{L}[/tex]

moles of KOH = [tex]\rm 2\;\times\;10^-^3[/tex] moles

moles of [tex]\rm CH_3COOH[/tex] = [tex]\rm 2.5\;\times\;10^-^3[/tex] moles

moles of [tex]\rm CH_3COOK[/tex] = [tex]\rm 2\;\times\;10^-^3[/tex] moles

pH = 4.76 + log [tex]\rm \frac{2\;\times\;10^-^3}{0.5\;\times\;10^-^3}[/tex]

pH = 5.36

(d) On addition of 12.5 ml of KOH,

moles of KOH = [tex]\rm \frac{12.5}{1000}\;\times\;\frac{0.200}{L}[/tex]

moles of KOH = [tex]\rm 2.5\;\times\;10^-^3[/tex]

moles of  [tex]\rm CH_3COOH[/tex] = [tex]\rm 2.5\;\times\;10^-^3[/tex] moles

moles of [tex]\rm CH_3COOK[/tex] = [tex]\rm 2.5\;\times\;10^-^3[/tex] moles

pH = 4.76 + log [tex]\rm \frac{2\;\times\;10^-^3}{0}[/tex]

pH = 4.74

(e) On addition of 15.0 ml of KOH,

The excess point is reached after the addition of 12.5 ml of KOH. After further addition of KOH, the pOH will be there.

The OH concentration on addition of KOH = [tex]\rm \frac{(\frac{15}{1000}\;\times0.200\;moles)\;-\;(\frac{25}{1000}\;\times0.100\;moles) }{\frac{25}{1000}\;+\;\frac{15}{1000} }[/tex]

= 0.0125 M

pOH = -log [OH concentration]

pOH = -log [0.0125]

pOH = 1.903

pH = 14 - pOH

pH = 14 - 1.903

pH = 12.097

The pH of solution on addtion of KOH will be :

O ml KOH = 2.87

5 ml KOH= 4.56

10 ml KOH = 5.36

12.5 ml KOH = 4.74

15 ml KOH = 12.907.

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Answer:

6298702 potassium atoms would have to be laid side by side to span a distance of 2,91 mm

Explanation:

If the radius of a potassium atom is 231 pm, then the diameter would be:

d = 2r = 2*(231) = 462 pm

So each potassium atom occupies a space of 462 pm, we can express this relationship as follows:

[tex]\frac{1 potassium atom}{462 pm}[/tex]

To solve this problem, we'll use the following conversion factors:

1 pm = 1×E−12 m

1000 mm = 1 m

We begin to accommodate all our relationships starting from that numerical expression that is not written as a relationship (usually the one in the question), and in such a way that the units are eliminated between them.

[tex]2, 91 mm * \frac{1 m}{1000 mm}*\frac{ 1 pm}{1*10^{-12}m }*\frac{1 potassium atom}{462 pm}= 6298701.3[/tex] potassium atoms

So, we'll need 6298702 potassium atoms to span a distance of 2,91 mm

Explanation:

Fischer Projections allow to represent the three dimensional molecular structures in two dimensional environment without the change in the properties or the structural integrity of the compound. It consists of horizontal as well as vertical lines both, where horizontal lines represent atoms which are pointed toward viewer while vertical line represents atoms which are pointed away from viewer. The point of the intersection between horizontal and vertical lines represents central carbon.

Answer:

Part A : n = 1.77 moles

Part B : λ = [tex]2.01*10^{6} m^{-1}[/tex]

Part C : n = 0.154 moles

Explanation:

Part A

The problem gives you the equation for molarity M:

[tex]M=\frac{n}{V}[/tex]

n is the number of moles of solute and V is the volume

Then the problem gives you the molarity of a substance [tex]M=2.73\frac{mol}{L}[/tex] and the volume V = 0.650L, so you need to solve the equation for n:

[tex]M=\frac{n}{V}[/tex]

as the V is dividing it passes to multiply the M:

n = M*V

and you should replace the values:

[tex]n = 2.73\frac{mol}{L}*0.650L[/tex]

n = 1.77 moles

Part B

This time you have to solve the equation E = hcλ for λ that is the unknown information, so you have:

E = hcλ

h and c are multiplying so they pass to divide the E:

λ = [tex]\frac{E}{hc}[/tex]

and replacing the values:

λ = [tex]\frac{3.98*10^{-19}J}{(6.626*10^{-34}J.s)(2.99*10^{8}\frac{m}{s})}[/tex]

λ = [tex]2.01*10^{6} m^{-1}[/tex]

PartC

In this part the problem gives you the equation PV=nRT and the first thing you should do is to verify that all the quantities are in consistent units so:

[tex]R=8.206*10^{-2} \frac{L.atm}{K.mol}[/tex] so you need to convert the pressure to atmospheres and convert the volume to liters.

- Convert the pressure to atmospheres:

[tex]P=899torr*\frac{0.00131579atm}{1torr}[/tex]

P = 1.18 atm

- Convert the volume to liters:

[tex]V=3280mL*\frac{1L}{1000mL}[/tex]

V = 3.28L

To find the number of moles n, you should solve the equation for n:

Pv = nRT

As R and T are multiplying the n, they pass to divide to the other side of the equation:

[tex]n=\frac{PV}{RT}[/tex]

And finally you should replace the values:

[tex]n=\frac{(1.18atm)(3.28L)}{(8.206*10^{-2}\frac{L.atm}{K.mol})(307K)}[/tex]

n = 0.154 moles

Answer:

PART A 1st order in A and 0th order in B

Part B The reaction rate increases

Explanation:

PART A

The rate law of the arbitrary chemical reaction is given by

[tex]-r_A=k\times\left[A\right]^\alpha\times\left[B\right]^\beta\bigm[/tex]

Replacing for the data

Expression 1 [tex]0.00639=k\times{0.12}^\alpha\times{0.22}^\beta[/tex]

Expression 2 [tex]0.01280=k\times{0.24}^\alpha\times{0.22}^\beta[/tex]

Expression 3 [tex]0.00639=k\times{0.12}^\alpha\times{0.11}^\beta[/tex]

Making the quotient between the fist two expressions

[tex]\frac{0.00639}{0.01280}=\left(\frac{0.12}{0.24}\right)^\alpha[/tex]

Then the expression for [tex]\alpha[/tex]

[tex]\alpha=\frac{ln\frac{0.00639}{0.01280}}{ln\frac{0.12}{0.24}}=1\bigm[/tex]

Doing the same between the expressions 1 and 3  

[tex]\frac{0.00639}{0.00639}=\left(\frac{0.22}{0.11}\right)^\beta[/tex]

Then

[tex]\beta=\frac{ln\frac{0.00639}{0.00639}}{ln\frac{0.22}{0.11}}=0\bigm[/tex]

This means that the reaction is 1st order respect to A and 0th order respect to B .

PART B

By the molecular kinetics theory, if an increment in the temperature occurs, the molecules will have greater kinetic energy and, consequently, will move faster. Thus, the possibility of colliding with another molecule increases. These collisions are necessary for the reaction. Therefore, an increase in temperature necessarily produces an increase in the reaction rate.

Answer: The correct answer is Option 4.

Explanation:

There are three sub-atomic particles present in an atom. They are: electrons, protons and neutrons.

Protons constitute in each and every atom.

The charge on proton is of equal magnitude as that of electron but having opposite sign. Proton carry a positive charge and electron carry a negative charge.

Protons and neutrons, both determine the mass of an atom.

Mass of 1 proton = 1.007276 u

Mass of 1 neutron = 1.008664 u

Mass of 1 electron = 0.00054858 u

Mass of proton is almost same as that of neutron but is more than the mass of electron.

Hence, the correct answer is Option 4.

Final answer:

The false statement about protons is that they have about the same mass as electrons. In reality, a proton's mass is approximately 1836 times greater than an electron's mass, making this statement incorrect.

Explanation:

The statement about protons that is false is: Protons have about the same mass as electrons.

Let's review the statements to identify the false one:

Understanding the basic properties of subatomic particles, such as their mass and charge, is essential in chemistry.

Answer: The moles of silver nitrate reacted is 5.72 moles, moles of iron (II) nitrate produced is 2.86 moles and moles of silver chloride produced is 5.72 moles

Explanation:

We are given:

Moles of iron (II) chloride = 2.86 moles

For the given chemical equation:

[tex]FeCl_2(aq.)+2AgNO_3(aq.)\rightarrow Fe(NO_3)_2(aq.)+2AgCl(s)[/tex]

By Stoichiometry of the reaction:

1 mole of iron (II) chloride reacts with 2 moles of silver nitrate.

So, 2.86 moles of iron (II) chloride will react with = [tex]\frac{2}{1}\times 2.86=5.72mol[/tex] of silver nitrate

Moles of silver nitrate reacted = 5.72 moles

By Stoichiometry of the reaction:

1 mole of iron (II) chloride produces 1 mole of iron (II) nitrate

So, 2.86 moles of iron (II) chloride will produce = [tex]\frac{1}{1}\times 2.86=2.86mol[/tex] of iron (II) nitrate

Moles of iron (II) nitrate produced = 2.86 moles

By Stoichiometry of the reaction:

1 mole of iron (II) chloride produces 2 moles of silver chloride

So, 2.86 moles of iron (II) chloride will produce = [tex]\frac{2}{1}\times 2.86=5.72mol[/tex] of silver chloride

Moles of silver chloride produced = 5.72 moles

Hence, the moles of silver nitrate reacted is 5.72 moles, moles of iron (II) nitrate produced is 2.86 moles and moles of silver chloride produced is 5.72 moles

When 2.86 moles of FeCl2 react, they consume 5.72 moles of AgNO3, produce 2.86 moles of Fe(NO3)2, and produce 5.72 moles of AgCl.

In the balanced equation FeCl2 (aq) + 2AgNO3(aq) --> Fe(NO3)2(aq) + 2AgCl(s), each FeCl2 molecule reacts with 2 AgNO3 molecules. This means that for every mole of FeCl2, 2 moles of AgNO3 are consumed. Therefore, if we have 2.86 moles of FeCl2, it will consume 2 x 2.86 = 5.72 moles of AgNO3.

In this reaction, for every mole of FeCl2 consumed, 1 mole of Fe(NO3)2 is produced. Therefore, if we have 2.86 moles of FeCl2, 2.86 moles of Fe(NO3)2 will be produced.

In addition, for every mole of FeCl2 consumed, 2 moles of AgCl are produced. Therefore, if we have 2.86 moles of FeCl2, 2 x 2.86 = 5.72 moles of AgCl will be produced.

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Answer: The number of moles of gas present is 0.276 moles

Explanation:

To calculate the number of moles of gas, we use the equation given by ideal gas:

PV = nRT

where,

P = Pressure of the gas = 725 mm Hg

V = Volume of the gas = 7.55 L

n = number of moles of gas = ?

R = Gas constant = [tex]62.3637\text{ L.mmHg }mol^{-1}K^{-1}[/tex]

T = Temperature of the gas = [tex]45^oC=(45+273)K=318K[/tex]

Putting values in above equation, we get:

[tex]725mmHg\times 7.55L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 318K\\\\n=0.276mol[/tex]

Hence, the number of moles of gas present is 0.276 moles

Answer:

b) Thermally conductive

Explanation:

Few properties of metals are:

Hence, correct options is B. The rest are not properties of metals.

Answer:

30 m/v %

Explanation:

First, we need to understand the definition of mass/volume percentage (m/v %). This is a way of expressing concentration of a substance and it means grams of a determined compound per 100 mL of solution:

m/v% = g of X substance/100 mL of solution

Having said that, if we have 0.600 g of NaOH per 2.00 mL of solution, then in 100 mL of solution we will have:

2.00 mL solution ---- 0.600 g of NaOH

100 mL solution---- x=(100 mL × 0.600 g NaOH)/ 2.00 mL = 30 g  = 30 %m/v

So, the concentration of a solution containing 0.600 g of NaOH per 2.00 mL is a 30 m/v% solution.-

Answer:

The correct option is: B. 33%

Explanation:

Orbital hybridisation refers to the mixing of atomic orbitals of the atoms in order to form new hybrid orbitals. The concept of orbital hybridization is used to explain the structure of a molecule.

The sp² hybrid orbitals are formed by the hybridization of one 2s orbital and two 2p orbitals. The three sp² hybrid orbitals formed have 33% s character and 67% p character.

Answer:

It's B. 33%

Explanation:

Because it refers to the mixing of atomic orbitals of the atoms in order to form new hybrid orbitals. The concept of orbital hybridization is used to explain the structure of a molecule.

Also, the sp² hybrid orbitals are formed by the hybridization of one 2s orbital and two 2p orbitals. Then the three sp² hybrid orbitals formed have 33% s character and 67% p character.

Answer:

option D - coking or fouling

Explanation:

Coking (not cocking) is the process involving the deposition of small carbon particles (created by simply putting carbon atoms) on a catalyst's accessible surface, leading in a reduction in the area accessible for catalytic activity. This is also sometimes related to fouling catalysts or merely fouling them.

Answer:

a) The concentration in ppm (mg/L) is 5.3 downstream the release point.

b) Per day pass 137.6 pounds of pollutant.  

Explanation:

The first step is to convert Million Gallons per Day (MGD) to Liters per day (L/d). In that sense, it is possible to calculate with data given previously in the problem.  

Million Gallons per day [tex]1 MGD = 3785411.8 litre/day = 3785411.8 L/d[/tex]

[tex]F_1 = 5 MGD (\frac{3785411.8 L/d}{1MGD} ) = 18927059 L/d\\F_2 =10 MGD (\frac{3785411.8 L/d}{1MGD} )= 37854118 L/d [/tex]

We have one flow of wastewater released into a stream.  

First flow is F1 =5 MGD with a concentration of C1 =10.0 mg/L.

Second flow is F2 =10 MGD with a concentration of C2 =3.0 mg/L.  

After both of them are mixed, the final concentration will be between 3.0 and 10.0 mg/L. To calculate the final concentration, we can calculate the mass of pollutant in total, adding first and Second flow pollutant, and dividing in total flow. Total flow is the sum of first and second flow. It is shown in the following expression:  

[tex]C_f = \frac{F1*C1 +F2*C2}{F1 +F2}[/tex]

Replacing every value in L/d and mg/L

[tex]C_f = \frac{18927059 L/d*10.0 mg/L +37854118 L/d*10.0 mg/L}{18927059 L/d +37854118 L/d}\\C_f = \frac{302832944 mg/d}{56781177 L/d} \\C_f = 5.3 mg/L[/tex]

a) So, the concentration just downstream of the release point will be 5.3 mg/L it means 5.3 ppm.

Finally, we have to calculate the pounds of substance per day (Mp).  

We have the total flow F3 = F1 + F2 and the final concentration [tex]C_f[/tex]. It is required to calculate per day, let's take a time of t = 1 day.  

[tex]F3 = F2 +F1 = 56781177 L/d \\M_p = F3 * t * C_f\\M_p = 56781177 \frac{L}{d} * 1 d * 5.3 \frac{mg}{L}\\M_p = 302832944 mg[/tex]

After that, mg are converted to pounds.  

[tex]M_p = 302832944 mg (\frac{1g}{1000 mg} ) (\frac{1Kg}{1000 g} ) (\frac{2.2 lb}{1 Kg} )\\M_p = 137.6 lb[/tex]

b) A total of 137.6 pounds pass a given spot downstream per day.

Answer: The volume of liquid X is 246 mL

Explanation:

To calculate volume of a substance, we use the equation:

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

We are given:

Mass of liquid X = 205 g

Density of liquid X = 0.834 g/mL

Putting values in above equation, we get:

[tex]0.834g/mL=\frac{205g}{\text{Volume of liquid X}}\\\\\text{Volume of liquid X}=246mL[/tex]

Hence, the volume of liquid X is 246 mL

Answer:

2 Pb(OH)2 + 2H2SO4 => 2 PbSO4 + 4 H20

Explanation:

Since there's no "?" shown in the equation, let's balance it and solve it entirely.

Pb(OH)2 + 2H2SO4 => PbSO4 + 2H20

1Pb + 10O + 6H + 2S ≠ 1Pb + 6O + 4H + 1S → it needs to be balanced.

To do this, let's start by looking at the elements that are only presnet once on each side:

On the products half, S is only present in PbSO4 → if we look at the reagents half, we can see it needs a "2" → then Pb is multiplied by 2 too → so Pb(OH)2 on the reagents half will also need a "2" → final count on O and H on the reagents side: 12O and 8H  → to balance it, you need 4 water molecules on the products side.

Answer:

Explanation:

I hope it helps!

Answer:

[tex]H_{3}O^{+}[/tex] concentration is 10000 times higher in diet cola than milk

Explanation:

Final answer:

The concentration of hydronium ions (H3O+) is 10,000 times higher in diet cola with a pH of 3.0 than in milk with a pH of 7.0, due to the logarithmic nature of the pH scale where each pH unit represents a tenfold change in H3O+ concentration.

Explanation:

The pH scale is logarithmic, meaning that each whole number on the pH scale represents a tenfold difference in hydronium ion concentration. Therefore, a change in pH unit corresponds to a change of a factor of 10 in the H3O+ concentration. Diet cola has a pH of about 3.0, and milk has a pH of about 7.0, a difference of 4 pH units. This implies that the hydronium ion concentration in diet cola is 104 times greater than in milk, as each unit difference accounts for a tenfold increase. Thus, the H3O+ concentration is 10,000 times higher in diet cola than in milk.