A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 MPa(50 ksi) is exposed to a stress of 1205 MPa(174800psi) . (a) If the largest surface crack is 0.8 mm (0.03150 in.) long, determine the critical stress ?c. (b) Will this specimen experience fracture? Assume that the parameter Y has a value of 0.99.

Answer:

0.00642mole

Explanation:

Molar Mass of Cu = 63.5g/mol

Mass of Cu from the question = 0.4076g

Number of mole =?

Number of mole = Mass /Molar Mass

Number of mole of Cu = 0.4076/63.5 = 0.00642mole

The question is incomplete, here is the complete question:

At 45°C, Kc = 0.619 for the reaction N₂O₄(g) ⇌ 2 NO₂(g).

If 50.0 g of N₂O₄ is introduced into an empty 2.12 L container, what are the partial pressures of NO₂ and N₂O₄ after equilibrium has been achieved at 45°C?

Answer: The equilibrium partial pressure of [tex]NO_2\text{ and }N_2O_4[/tex] is 7.12 atm and 3.133 atm respectively.

Explanation:

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

PV = nRT

Or,

[tex]PV=\frac{w}{M}RT[/tex]

where,

P = Pressure of the gas  = ?

V = Volume of the gas  = 2.12 L

w = Weight of the gas  = 50.0 g

M = Molar mass of gas  = 92 g/mol

R = Gas constant = [tex]0.0821\text{ L atm }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]P\times 2.12L=\frac{50.0g}{92g/mol}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 318\\\\P=\frac{50.0\times 0.0821\times 318}{2.12\times 92}=6.693atm[/tex]

Relation of [tex]K_p[/tex] with [tex]K_c[/tex] is given by the formula:

[tex]K_p=K_c(RT)^{\Delta ng}[/tex]

where,

[tex]K_p[/tex] = equilibrium constant in terms of partial pressure

[tex]K_c[/tex] = equilibrium constant in terms of concentration = 0.619

R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]

T = temperature  = [tex]45^oC=[45+273]K=318K[/tex]

[tex]\Delta n_g[/tex] = change in number of moles of gas particles = [tex]n_{products}-n_{reactants}=(2-1)=1[/tex]

Putting values in above equation, we get:

[tex]K_p=0.619\times (0.0821\times 318)^{1}\\\\K_p=16.16[/tex]

For the given chemical equation:

                 [tex]N_2O_4(g)\rightleftharpoons 2NO_2(g)[/tex]

Initial:            6.693

At eqllm:    6.693-x           2x

The expression of [tex]K_p[/tex] for above equation follows:

[tex]K_p=\frac{(p_{NO_2})^2}{p_{N_2O_4}}[/tex]

Putting values in above expression, we get:

[tex]16.16=\frac{(2x)^2}{6.693-x}\\\\x=-7.59,3.56[/tex]

So, equilibrium partial pressure of [tex]NO_2=2x=(2\times 3.56)=7.12atm[/tex]

Equilibrium partial pressure of [tex]N_2O_4=(6.693-x)=(6.693-3.56)=3.133atm[/tex]

Hence, the equilibrium partial pressure of [tex]NO_2\text{ and }N_2O_4[/tex] is 7.12 atm and 3.133 atm respectively.

Answer:

1. HNO3

2. HNO2

Explanation:

Nitrogen acid may refer to any of:

1. Nitric acid, HNO3 or

2. Nitrous acid, HNO2

Acid rain is caused by a chemical reaction that begins when compound like nitrogen oxides are released into the air. These substances can rise very high into the atmosphere, where they mix and react with water, oxygen, and other chemicals to form more acidic pollutants, known as acid rain. nitrogen oxides dissolve very easily in water and can be carried very far by the wind. As a result, the compounds can travel long distances where they become part of the rain, sleet, snow, and fog as acids.

Equation of reaction:

NO2 + H2O ==> HNO3

NO + H2O ==> HNO2

Final answer:

Nitrogen oxides (NO and NO2) from automobile exhaust contribute to acid rain primarily through the formation of nitric acid and nitrous acid.

Explanation:

High temperatures in the automobile engine cause nitrogen and oxygen gases from the air to combine to form nitrogen oxides (NO and NO2). These nitrogen oxides in automobile exhaust contribute to acid rain by forming nitric acid and nitrous acid. Nitric acid is formed when nitrogen dioxide (NO2), a highly reactive gas and a major component of nitrogen oxides, reacts with atmospheric water. Nitrous acid forms under similar conditions but is less prevalent. The reaction involving nitrogen oxides, particularly with the presence of water vapor, contributes significantly to the phenomenon of acid rain, which has adverse effects on natural water bodies, soil, and vegetation by altering their chemical composition.

Answer:

K = 0,00000135 = 1.35 * 10^-6

Explanation:

Step 1: Data given

The equilibrium constant, K, for any reaction is defined as the concentrations of the products raised by their coefficients divided by the concentrations of the reactants raised by their coefficients. In this case, the concentrations are given as partial pressures.

The partial pressures of H2O = 0.0500 atm

The partial pressures of H2 = 0.00150 atm

The partial pressures of O2 = 0.00150 atm

Step 2: The balanced equation

2H2O(g) ⇆ 2H2(g) + O2(g)

Step 3: Calculate K

K = [O2][H2]² / [H2O]²

K = 0.00150 * 0.00150² /  0.0500²

K = 0,00000135 = 1.35 * 10^-6

The mass of ammonium phosphate produced by the reaction of 4.9 g of phosphoric acid is 7.45 g.

The question asks for the mass of ammonium phosphate produced by the reaction of 4.9 g of phosphoric acid. To determine the mass of ammonium phosphate produced, we need to balance the chemical equation and calculate the molar mass of both reactants and products.

The balanced equation for the reaction is:3H3PO4 + (NH4)OH → (NH4)3PO4 + 3H2O

The molar mass of phosphoric acid (H3PO4) is 97.99 g/mol. The molar mass of ammonium phosphate ((NH4)3PO4) is 149.0 g/mol.

Using the molar mass of phosphoric acid and the ratio of the reactants and products in the balanced equation, we can calculate the mass of ammonium phosphate produced.First, calculate the moles of phosphoric acid:

moles of H3PO4 = mass (g) / molar mass (g/mol)

moles of H3PO4 = 4.9 g / 97.99 g/mol = 0.050 moles

Since the stoichiometry of the reaction is 1:1 between phosphoric acid and ammonium phosphate, the moles of ammonium phosphate produced is also 0.050 moles.

Finally, calculate the mass of ammonium phosphate:mass of (NH4)3PO4 = moles of (NH4)3PO4 × molar mass of (NH4)3PO4

mass of (NH4)3PO4 = 0.050 moles × 149.0 g/mol = 7.45 g

Therefore, 7.45 g of ammonium phosphate is produced by the reaction of 4.9 g of phosphoric acid.

Answer:

a) 0.000112 M/s is the average reaction rate between 0.0 seconds and 1500.0 seconds.

b) 0.00011 M/s is the average reaction rate between 200.0 seconds and 1200.0 seconds.

c) Instantaneous rate of the reaction at t=800 s :

Instantaneous rate  : [tex]\frac{0.031 M}{800.0 s}=3.875\times 10^{-5} M/s[/tex]

Explanation:

Average rate of the reaction is given as;

[tex]R_{avg}=-\frac{\Delta A}{\Delta t}=\frac{A_2-A_1}{t_2-t_1}[/tex]

a.) The average reaction rate between 0.0 s and 1500.0 s:

At 0.0 seconds the concentration was  = [tex]A_1=0.184 M[/tex]

[tex]t_1=0.0s[/tex]

At 1500.0 seconds the concentration was  = [tex]A_2=0.016 M[/tex]

[tex]t_2=1500 s[/tex]

[tex]R_{avg]=-\frac{0.016 M-0.184 M}{1500.0 s-0.0 s}=0.000112 M/s[/tex]

0.000112 M/s is the average reaction rate between 0.0 seconds and 1500.0 seconds.

b.) The average reaction rate between 200.0 s and 1200.0 s:

At 0.0 seconds the concentration was  = [tex]A_1=0.129 M[/tex]

[tex]t_1 =200.0 s[/tex]

At 1500.0 seconds the concentration was  = [tex]A_2=0.019M[/tex]

[tex]t_2=1200 s[/tex]

[tex]R_{avg]=-\frac{0.019 M-0.129M}{1200.0s-200.0s}=0.00011 M/s[/tex]

0.00011 M/s is the average reaction rate between 200.0 seconds and 1200.0 seconds.

c.) Instantaneous rate of the reaction at t=800 s :

At 800 seconds the concentration was  = [tex]A=0.031 M[/tex]

[tex]t =800.0 s[/tex]

Instantaneous rate  : [tex]\frac{0.031 M}{800.0 s}=3.875\times 10^{-5} M/s[/tex]

Explanation:

It is given that 100 kg feed solution  contains 30 wt% solution NaOH.  Hence, feed contains 70 kg of water and 30 kg of NaOH.

Therefore, mixer outlet shows 5 wt% NaOH.

Let us assume that the mixture outlet be F kg  so, by applying mass balance of NaOH we get the value of force as follows.

          [tex]F \times 0.05[/tex] = 30

                 F = 600 Kg

In that 30 kg is NaOH and 570 kg is water  which also means that initially water present is 70 kg. And, additional water added is 500 kg .

Thus, water feed rate is 500 kg/hr.

Answer:

The concentration of the monoprotic acid is about 0.114 mol/L.

The correct answer is E none of the above

Explanation:

Step 1: Data given

Volume of a monoprotic acid = 25.0 mL = 0.025 L

Molarity of the NaOH solution = 0.115 M

The end point was obtained at about 24.8 mL.

Step 2: Calculate the concentration of the monoprotic acid

b*Ca*Va = a*Cb*Vb

⇒with B = the coefficient of NaOH = 1

⇒with Ca = the concentration of the monoprotic acid = ?

⇒with Va = the volume of the monoprotic acid = 0.025 L

⇒with a = the coefficient of the monoprotic acid = 1

⇒with Cb = the concentration of NaOH = 0.115M

⇒with Vb = the volume of NaOH= 0.0248 L

1*Ca*0.025 = 1*0.115*0.0248

Ca = (0.115*0.0248)/0.025

Ca = 0.114 M

The concentration of the monoprotic acid is about 0.114 mol/L.

The correct answer is E none of the above

Answer:

[tex]\large \boxed{\text{88.1 $^{\circ}$C}}[/tex]

Explanation:

There are two heat transfers involved: the heat gained by the spoon and the heat lost by the coffee.

According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.

Let the spoon be Component 1 and the coffee be Component 2.

Data:  

For the spoon:

[tex]m_{1} =\text{100.0 g; }T_{i} = 10.0 ^{\circ}\text{C; }\\C_{1} = 0.235 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}[/tex]

For the coffee:

[tex]m_{2} =\text{240.0 g; }T_{i} = 90.0 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}[/tex]

Calculations

1. The relative temperature changes

[tex]\begin{array}{rcl}\text{Heat gained by spoon + heat lost by coffee} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{100.0 g}\times 0.235 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{240. g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\23.50\Delta T_{1} + 1004\Delta T_{2} & = & 0\\\end{array}[/tex]

[tex]\begin{array}{rcl}1004\Delta T_{2} & = & -23.50\Delta T_{1}\\\Delta T_{2} & = & -0.02340\Delta T_{1}\\\end{array}\\\text{(The temperature change for the coffee is about 1/40 that of the spoon.)}[/tex]

2. Final temperature of coffee

[tex]\Delta T_{1} = T_{\text{f}} - 10.0 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 90.0 ^{\circ}\text{C}[/tex]

[tex]\begin{array}{rcl}\Delta T_{2} & = & -0.02340\Delta T_{1}\\T_{\text{f}} - 90.0 \, ^{\circ}\text{C} & = & -0.02340 (T_{\text{f}} - 10.0 \, ^{\circ}\text{C})\\& = & -0.02340T_{\text{f}} + 0.2340 \, ^{\circ}\text{C}\\T_{\text{f}} & = & -0.02340T_{\text{f}}+ 90.23 \, ^{\circ}\text{C}\\1.02430T_{\text{f}}& = & 90.23 \, ^{\circ}\text{C}\\\end{array}\\[/tex]

[tex]\begin{array}{rcl}T_{\text{f}} & = & \dfrac{ 90.23 \, ^{\circ}\text{C}}{1.02430}\\\\ & = & \mathbf{88.1 \,^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature of the coffee is $\large \boxed{\textbf{88.1 $\,^{\circ}$C}}$}\\\text{This is hot enough to cause third-degree burns in less than 1 s.}[/tex]

Using the principles of physics and the formula for heat transfer, the final temperature after introducing a cold spoon to a cup of hot coffee ends up being around 89.6°C. It's further noted that spilling coffee at this temperature could potentially cause 3rd-degree burns, although the specific outcome may vary.

This question can be answered through the principles of energy conservation. In this situation, the chilled spoon will absorb heat from the hot coffee until they both reach a thermal equilibrium. This means they will eventually have the same temperature. The formula for this heat transfer can be expressed as Q (heat energy transferred) = mcΔT, where 'm' is the mass, 'c' is the specific heat, and 'ΔT' is the change in temperature.

Applying this to the spoon and coffee respectively, we will end up with an equation: (mass of coffee * specific heat of coffee * (initial coffee temp - final temp)) = -(mass of spoon * specific heat of spoon * (final temp - initial spoon temp)). Considering the specific heat of water (essentially coffee without impurities) is approximately 4.186 J/g °C, the initial math leads us to a final temperature of around 89.6°C.

Mentioning the last thought in your question, it's likely that spilling coffee at this temperature could potentially cause 3rd-degree burns. The exact effect can vary depending on various factors such as exposure duration and skin sensitivity, but typically, exposure to liquid at a temperature above 70°C can cause severe burns in a matter of seconds.

https://brainly.com/question/13433948

#SPJ11

Answer:

Number of atoms, N = 17,259,259

Explanation:

Given:

Radius of Iridium = 135 pm

distance = 2.33 mm

To determine number of iridium atom that would be laid side by side to span a distance of 2.33 mm, we say let the number of the atom = N

[tex]N =\frac{Distance}{Iridium. Radius} = \frac{2.33X10^{-3}}{135 X10^{-12}} \\\\N = 17259259.26[/tex]

Therefore, If a iridium atom has a radius of  135 pm, then 17,259,259 atoms of Iridium would be laid side by side to span a distance of 2.33 mm.

Number of atoms, N = 17,259,259

17,259,259 atoms of iridium are present in a distance of 2.33 mm

First, to calculate the amount of iridium atoms in a distance of 2.33 mm it is necessary to divide the two values:

                                       [tex]N = \frac{Distance}{Radius} [/tex]

As the element radius value is in pm, it is necessary to transform this unit to suit the other:

                                         [tex]135pm = 135\times 10^{-9} mm[/tex]

Now, we can apply the values in the expression:

                                             [tex]N = \frac{2.33}{135\times10^{-9}}[/tex]

                                        [tex]N = 17,259,259 [/tex] atoms

So, 17,259,259 atoms of iridium are present in a distance of 2.33 mm.

Learn more about calculation of atoms in: brainly.com/question/3320707

Answer: The value of equilibrium constant for new reaction is [tex]1.92\times 10^{-25}[/tex]

Explanation:

The given chemical equation follows:

[tex]O_2(g)\rightarrow \frac{2}{3}O_3(g)+\frac{1}{2}O_2(g)[/tex]  

The equilibrium constant for the above equation is [tex]5.77\times 10^{-9}[/tex]

We need to calculate the equilibrium constant for the equation of 3 times of the above chemical equation, which is:

[tex]3O_2(g)\rightarrow 2O_3(g)[/tex]

The equilibrium constant for this reaction will be the cube of the initial reaction.

If the equation is multiplied by a factor of '3', the equilibrium constant of the new reaction will be the cube of the equilibrium constant of initial reaction.

The value of equilibrium constant for reverse reaction is:

[tex]K_{eq}'=(5.77\times 10^{-9})^3=1.92\times 10^{-25}[/tex]

Hence, the value of equilibrium constant for new reaction is [tex]1.92\times 10^{-25}[/tex]

The equilibrium constant for equation (b) is 1.73x10^8.

Therefore, the equilibrium constant (Kc) for equation (b) is 1 / (5.77x10^-9) = 1.73x10^8.

https://brainly.com/question/32442283

#SPJ11

Answer:

A. m and n are independent from the molar coefficients of the reactants in the balanced chemical equation.

B. m and n must be determined by experiment.

Explanation:

rate = k[H2O2]^m × [I-]^n

The Order of Reaction refers to the power dependence of the rate on the concentration of each reactant.

Either the differential rate law or the integrated rate law can be used to determine the reaction order of reactants from experimental data.

The question is incomplete. The complete question is:

The half-life for the decay of carbon-14 is 5.73x10^3 years. Suppose the activity due to the radioactive decay of the carbon-14 in a tiny sample of an artifact made of woodfrom an archeological dig is measured to be 2.8x10^3 Bq. The activity in a similiar-sized sample of fresh wood is measured to be 3.0x10^3 Bq. Calculate the age of the artifact. Round your answer to 2 significant digits.

Answer:

570 years

Explanation:

The activity of the fresh sample is taken as the initial activity of the wood sample while the activity measured at a time t is the present activity of the wood artifact. The time taken for the wood to attain its current activity can be calculated from the formula shown in the image attached. The activity at a time t must always be less than the activity of a fresh wood sample. Detailed solution is found in the image attached.

Answer:

H+  +  OH−  --> H2O

Explanation:

Hydrochloric acid is represented by the chemical formular; HCl. This is an ionic substance so in water it breaks apart into hydrohrn ions; H+ and chloride ions; Cl−. It is a strong acid, hence it completely dissociates.

Potassium Hydroxide is also an ionic substance it also breaks apart in water into potassium ions; K+ and hydroxide ions; OH−. It is a strong base, hence it completely dissociates.

The complete ionic equation for the reaction is given as;

H+  +  Cl−  +  K+  +  OH−  -->  K+  +  Cl−  +  H2O

The Hydrogen ion and the Hydroxide ions combine to form water.

The net ionic equation is given as;

H+  +  OH−  --> H2O

Cl- and K+ ions were cancelled out because they do not undergo any changes therefore are not part of the net ionic equation. They are referred to as spectator ions.

Answer: The final temperature of the coffee is 43.9°C

Explanation:

To calculate the final temperature, we use the equation:

[tex]q=mC(T_2-T_1)[/tex]

where,

q = heat released = [tex]1.0\times 10^3J=1000J[/tex]

m = mass of water = 10.0 grams

C = specific heat capacity of water = 4.184 J/g°C

[tex]T_2[/tex] = final temperature = ?

[tex]T_1[/tex] = initial temperature = 20°C

Putting values in above equation, we get:

[tex]1000J=10.0g\times 4.184J/g^oC\times (T_2-20)\\\\T_2=43.9^oC[/tex]

Hence, the final temperature of the coffee is 43.9°C

Final answer:

The final temperature of the coffee, after adding 1.0 x 10° J of heat to 10.0 g of water initially at 20°C, is approximately 44°C, calculated using the heat transfer formula in thermodynamics.

Explanation:

The subject of this question is Physics, specifically the concept of energy transfer through heating in thermodynamics. To calculate the final temperature of the prepared coffee, we can use the formula q = mcΔT, where q is the heat added, m is the mass of the water, c is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature. By rearranging the formula to solve for ΔT, we get ΔT = q / (mc).

Given that 1.0 x 103 J of heat is added to 10.0 g of water initially at 20°C, we can calculate ΔT as follows:

ΔT = (1.0 x 103 J) / (10.0 g * 4.18 J/g°C) = 23.923°C (approximately)

Therefore, the final temperature of the water (or coffee) is 20°C + 23.923°C = 43.923°C, which when rounded off, gives us approximately 44°C.

The equilibrium expression (Ka) for the ionization of bicarbonate ion is given by the formula [H3O+][CO32-] / [HCO3-], where A, B, and C are the equilibrium concentrations of H3O+, CO32-, and HCO3- respectively.

To find the equilibrium expression (Ka) for the ionization reaction of the bicarbonate ion (HCO3-(aq)), we must consider the reaction, HCO3-(aq) + H2O(l) ⇌ H3O+(aq) + CO32-(aq). The expression for Ka, known as the acid-ionization constant, follows the general form:

Ka = [H3O+][CO32-] / [HCO3-]

For the reaction given:

The reaction quotient will be Ka = [A][B] / [C]. Utilizing the values provided, assuming that the concentrations of the products [A] and [B] are equivalent as they form in a 1:1 ratio, and given that we're focusing on the ionization equilibrium of bicarbonate ion as a weak acid, the Ka value can be derived.

Equilibrium expression [tex]\( K_a = \frac{[H_3O^+][CO_3^{2-}]}{[HCO_3^-]} \)[/tex]. Concentrations at equilibrium: [tex]\([A] = [HCO_3^-]\), \([B] = [H_3O^+]\), \([C] = [CO_3^{2-}]\)[/tex].

The equilibrium expression (Ka) for the ionization reaction of [tex]HCO_3^-[/tex](aq) in water is given by the following general formula for an acid dissociation constant:

[tex]\[ K_a = \frac{[H_3O^+][CO_3^{2-}]}{[HCO_3^-]} \][/tex]

Here, the concentration of water ([tex]H_2O[/tex]) is not included in the expression because it is the solvent and its concentration remains relatively constant during the reaction.

For the given reaction:

[tex]\[ HCO_3^-(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + CO_3^{2-}(aq) \][/tex]

The expressions for the concentrations of A ([tex]HCO_3^-[/tex]), B ([tex]H_3O^+[/tex]), and C [tex](CO_3^{2-})[/tex] at equilibrium would be:

[tex]\[ [A] = [HCO_3^-] \] \[ [B] = [H_3O^+] \] \[ [C] = [CO_3^{2-}] \][/tex]

These expressions represent the molar concentrations of the species A, B, and C at equilibrium.

The equilibrium constant expression [tex]\( K_a \)[/tex] is then derived from these concentrations, with the product of the concentrations of B and C divided by the concentration of A, as shown in the initial equation.

Explanation:

Expression to calculate the value of for the given reaction is as follows.

And, it is given that

        [CO] = [tex][H_{2}][/tex] = 0.13 mol

        = 0.43 mol

Putting the given values into the above formula as follows.

         [tex]K_{c} = \frac{[CO][H_{2}]}{[H_{2}O]}[/tex]

                  = [tex]\frac{0.13 \times 0.13}{0.43}[/tex]

                   = 0.04

When additional amount of is added then all of has reacted.

So, new = 0 mols

       new = 0.43 + 0.13 = 0.56 mols

The reaction equation is as follows.

                  [tex]C + H_{2}O \rightleftharpoons CO + H_{2}[/tex]

Initial:    -     0.56     0.13     0    

Change:  -      -x         +x         +x

Equilibm.: -    0.56 - x   0.13 + x    x

So,

            0.04 = [tex]\frac{(0.13 + x)(x)}{(0.56-x)}[/tex]

            [tex]0.0224 - 0.04x = x^{2} + 0.13x[/tex]

            [tex]x^{2} + 0.17x - 0.0224[/tex] = 0

                  x = 0.087 mols

Therefore, the amount of [CO] at equilibrium is as follows.

             0.13 + 0.087

            = 0.217 mols

thus, we can conclude that the amount of CO in the flask when the system returns to equilibrium is 0.217 moles.

The question in incomplete, complete question is;

Determine the theoretical yield:

Excess aqueous copper(II) nitrate reacts with aqueous sodium sulfide to produce aqueous sodium nitrate and copper(II) sulfide as a precipitate. In this reaction 469 grams of copper(II) nitrate were combined with 156 grams of sodium sulfide to produce 272 grams of sodium nitrate.

Answer:

The theoretical yield of sodium nitrate is 340 grams.

Explanation:

[tex]Cu(NO_3)_2(aq)+Na_2S(aq)\rightarrow 2NaNO_3(aq)+CuS(s)[/tex]

Moles of copper(II) nitrate = [tex]\frac{469 g}{187.5 g/mol}=2.5013 mol[/tex]

Moles of sodium sulfide = [tex]\frac{156 g}{78 g/mol}=2 mol[/tex]

According to reaction, 1 mole of copper (II) nitrate reacts with 1 mole of sodium sulfide.

Then 2 moles of sodium sulfide will react with:

[tex]\frac{1}{1}\times 2mol= 2 mol[/tex] of copper (II) nitrate

As we can see from this sodium sulfide is present in limiting amount, so the amount of sodium nitrate will depend upon moles of sodium sulfide.

According to reaction, 1 mole of sodium sulfide gives 2 mole of sodium nitrate, then 2 mole of sodium sulfide will give:

[tex]\frac{2}{1}\times 2mol=4 mol[/tex] sodium nitrate

Mass of 4 moles of sodium nitrate :

85 g/mol × 4 mol = 340 g

Theoretical yield of sodium nitrate = 340 g

The theoretical yield of sodium nitrate is 340 grams.

Explanation:

Below is an attachment containing the solution.

Answer:

[tex]\large \boxed{0.0106}[/tex]

Explanation:

We have three equations:

1. M(g) ⇌ Z(g);                      Kc₁ =    3.15

2. 6R(g) ⇌ 2N(g) + 4Z(g); Kc₂ =    0.509

3. 3X(g) + 3Q(g) ⇌ 9R(g); Kc₃ = 12.5

From these, we must devise the target equation:

4. Q(g) + X(g) ⇌ 2M(g) + N(g); Kc = ?

The target equation has Q(g) on the left, so you divide Equation 1 by 3.

When you divide an equation by 3, you take the cube root of its Kc.

5. X(g) + Q(g) ⇌ 3R(g): K₅ = ∛(Kc₃)

Equation 5 has 3R on the right, and that is not in the target equation.

You need an equation with 3R on the left, so you divide Equation 2 by 2.  

When you divide an equation by 2, you take the square root of its Kc.

6. 3R(g) ⇌ N(g) + 2Z(g); K₆ = √ (Kc₂)

Equation 6 has 2Z on the right, and that is not in the target equation.

You need an equation with 2Z on the left, so you reverse Equation 2 by and double it.

When you reverse an equation, you take the reciprocal of its K.

When you double an equation, you square its K.

7. 2Z(g) ⇌ 2M(g); K₇ = (1/Kc₁)²

Now, you add equations 5, 6, and 7, cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you multiply their K values.

You get the target equation 4:

5. X(g) + Q(g) ⇌ 3R(g);              K₅ = ∛(Kc₃)

6. 3R(g) ⇌ N(g) + 2Z(g);             K₆ = √(Kc₂)

7. 2Z(g) ⇌ 2M(g);                        K₇ = (1/Kc₁)²

4. Q(g) + X(g) ⇌ 2M(g) + N(g); Kc = K₅K₆K₇ =  [∛(Kc₃)√(Kc₂)]/(Kc₁)²

Kc =  [∛(12.5)√(0.509)]/(12.5)² = (2.321 × 0.7120)/156.2 = 0.0106

[tex]K_{c} \text{ for the reaction is $\large \boxed{\mathbf{0.0106}}$}[/tex]

Answer:

The value of the equilibrium constant is 0.167

Explanation:

Step 1: The target equation

Q(g) + X(g) ⇔ 2M(g) + N(g)

Given is:

(1) M(g)⇔Z(g)   c1=3.15

(2) 6R(g) ⇔ 2N(g) + 4Z(g)   c2=0.509

(3) 3X(g) +3Q(g) ⇔ 9R(g)    c3=12.5

Step 2: Rearange the equation

We have to rearange the equation to come to the final result

This is Hess' Law

In the target equation we have Q(g) + X(g)

In (3)  we have 3X(g) +3Q(g) ⇔ 9R(g)

To get the target of  Q(g) + X(g) we have to divide (3) by 3. This will give us:

X(g) +Q(g) ⇔ 3R(g)   Kc = ∛12.5 = 2.32  (Note: to get Kc of the target equation we use cube root)

The target equation has as product  2M(g) + N(g)

To get M(g) we will use the (1) equation

Since M(g) is a product and not a reactant, we have to reverse the equation. Next to that we also have to double the equation because we need 2M(g) and not M(g)

2Z(g) ⇔ 2M(g)        Kc = 1/(3.15)²  = 0.101   (Note: to get Kc' after reversing the equation we calculate 1/Kc.   To get Kc'' after doubling and reversing the equation we calculate 1/(Kc²)

To get N(g) we will use (2) 6R(g) ⇔ 2N(g) + 4Z(g)

Since we only need N(g) we will divide this equation by 2. This will get us:

3R(g) ⇔ N(g) + 2Z(g)    Kc = √0.509   = 0.713   (Note: if we divide the equation by 2, to calculate Kc' we use square root)

Now we have all the components we will add the 3 equations:

X(g) +Q(g) + 2Z(g)  + 3R(g)⇔ 3R(g) + 2M(g) + N(g) + 2Z(g)

We will simplify this equation:

X(g) +Q(g) ⇔  2M(g) + N(g)  this is our target equation

The value of the  equilibrium constant, Kc is:

Kc = 2.32 * 0.101*0.713

Kc = 0.167

Note: to calculate Kc after adding several equations,we'll multiply Kc1* Kc2 * Kc3 etc...

The value of the equilibrium constant is 0.167  

Answer:

7.85 g H₃BO₃

2.92 g NaOH

Explanation:

The strategy for solving this question is to first utilize the Henderson- Hasselbach equation to calculate the ratio of conjugate base concentration to weak acid:

pH = pKa + log [A⁻]/ [HA]

In this case:

pH = pKa + log [H₂BO₃⁻]/[H₃BO₃]

We know pH and indirectly pKa ( = - log Ka ).

9.00 = -log(5.8 x 10⁻¹⁰) + log [H₂BO₃⁻]/[H₃BO₃]

9.00 = 9.24 + log [H₂BO₃⁻]/[H₃BO₃]

log [H₂BO₃⁻]/[H₃BO₃] = - 0.24

taking inverse log function to both sides of the equation:

[H₂BO₃⁻]/[H₃BO₃]  = 10^-0.24 = 0.58

We are also told we want to have a total concentration of boron of 0.200 mol/L, and if we call x the concentration of  H₂BO₃⁻ and y the concentration of H₃BO₃, it follows that:

x + y = 0.200 ( since we have 1 Boron atom per formula of each compound)

and from the Henderson Hasselbach calculation, we have that

x / y = 0.58

So we have a system of 2 equations with two unknowns, which when solved give us that

x = 0.073  and y = 0.127

Because we are told the volume is one liter it follows that the number of moles of boric acid and the salt are the same numbers 0.073 and 0.127

gram boric acid = 0.127 mol x molar mass HBO₃ = 0127 mol x 61.83 g/mol

                          = 7.85 g boric acid

grams NaOH = 0.073 mol x molar mass NaOH = 0.073 x 40 g/mol

                          = 2.92 g NaOH

Answer:

e.It would decrease by a factor of 4.

Explanation:

first order in [NO2]

first order in [F2]

The rate is then given as;

Rate = k [NO2][F2]

where k = rate constant =  And is constant for a reaction.

Let's insert some dummy values (Any values work, just be consistent);

[NO2] = 2

[F2] = 2

K = 3

Rate = 3*2*2

Rate = 12

What would happen to the reaction rate if the concentrations of both reactants were halved with everything else held constant?

[NO2] = 2 / 2 = 1

[F2] = 2 / 2 = 1

K = 3

Rate = 3*1*1

Rate = 3

Comparing both rates (12 and 3); the correct option is;

e.It would decrease by a factor of 4.

The question is incomplete, here is the complete question:

While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene with water vapor at elevated temperatures.

A chemical engineer studying this reaction fills a 1.5 L flask at 12°C with 1.8 atm of ethylene gas and 4.7 atm of water vapor. When the mixture has come to equilibrium she determines that it contains 1.16 atm of ethylene gas and 4.06 atm of water vapor.

The engineer then adds another 1.2 atm of ethylene, and allows the mixture to come to equilibrium again. Calculate the pressure of ethanol after equilibrium is reached the second time. Round your answer to 2 significant digits.

Answer: The partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

Explanation:

We are given:

Initial partial pressure of ethylene gas = 1.8 atm

Initial partial pressure of water vapor = 4.7 atm

Equilibrium partial pressure of ethylene gas = 1.16 atm

Equilibrium partial pressure of water vapor = 4.06 atm

The chemical equation for the reaction of ethylene gas and water vapor follows:

                     [tex]CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)[/tex]

Initial:                  1.8                4.7

At eqllm:           1.8-x             4.7-x

Evaluating the value of 'x'

[tex]\Rightarrow (1.8-x)=1.16\\\\x=0.64[/tex]

The expression of [tex]K_p[/tex] for above equation follows:

[tex]K_p=\frac{p_{CH_3CH_2OH}}{p_{CH_2CH_2}\times p_{H_2O}}[/tex]

[tex]p_{CH_2CH_2}=1.16atm\\p_{H_2O}=4.06atm\\p_{CH_3CH_2OH}=0.64atm[/tex]

Putting values in above expression, we get:

[tex]K_p=\frac{0.64}{1.16\times 4.06}\\\\K_p=0.136[/tex]

When more ethylene is added, the equilibrium gets re-established.

Partial pressure of ethylene added = 1.2 atm

                     [tex]CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)[/tex]

Initial:                2.36             4.06               0.64

At eqllm:           2.36-x        4.06-x             0.64+x

Putting value in the equilibrium constant expression, we get:

[tex]0.136=\frac{(0.64+x)}{(2.36-x)\times (4.06-x)}\\\\x=0.363,13.41[/tex]

Neglecting the value of x = 13.41 because equilibrium partial pressure of ethylene and water vapor will become negative, which is not possible.

So, equilibrium partial pressure of ethanol = (0.64 + x) = (0.64 + 0.363) = 1.003 atm

Hence, the partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

Answer:

[HCl] = 0.035 M

Explanation:

COMPLETE QUESTION:

What is the concentration (M) of HCl in a solution that is prepared by dissolving 25.5 g of HCl in 20.0 L  of water? (Assume volume does not change after adding HCl.)

Molarity → moles of solute in 1L of solution

As we assume volume does not change after adding  HCl, solution's volume is 20L

We convert solute's mass to moles → 25.5 g. 1mol / 36.45 g = 0.699 moles

Molarity (mol/L) → 0.699 mol/20L = 0.035 M

To find the concentration of HCl, divide the mass of HCl (25.5 g) by its molar mass to get moles, then divide by the volume of the solution in liters, assuming a volume of 1L for simplicity.

Firstly, the molar mass of HCl must be determined, which is approximately 36.46 g/mol. To find the concentration (Molarity, M) of HCl in the solution, the mass of HCl (25.5 g) must be converted into moles by dividing by the molar mass (moles of HCl = 25.5 g / 36.46 g/mol = 0.699 moles).

Then, the volume of water needs to be converted into liters (assuming it to be 1L for this example, as the exact volume was not provided). Finally, the concentration can be calculated by dividing the moles of HCl by the volume of the solution in liters (Concentration, M = 0.699 moles / 1L = 0.699 M)

Answer:

1.6mM

Explanation:

NaCl(aq) -----------> Na+ (aq) + Cl- (aq)

There are two ions present and the total concentration of both ions is 3.2mM. Since the total concentration of both sodium ions and chloride ions is 3.2mM, then one of the ions will have a concentration which is half of the given value. That is, concentration of chloride ion will be 1/2× 3.2 = 1.6mM. Since the two ions are present in a ratio of one is to one, when you add the concentrations of both ions, you obtain the total concentration of ions in solution.

The pH of the given solution is approximately 12.1.

Let's go through the calculations step by step:

Step 1: Moles of Hydrogen Ions in Sulfuric Acid Solution (a)

a = Molarity × Volume

a = 0.100 M × 0.050 L

a = 0.005 mol

Step 2: Moles of Hydrogen Ions in HOCl Solution (b)

Dissociation: HOCl ↦ H+ + OCl-

Ka = [H+][OCl-]/[HOCl] = 3.5 × 10^-8

At equilibrium: (0.1133 - x) M (x) M

x^2/(0.1133 - x) = 3.5 × 10^-8

Solving for x gives x ≈ 1.87 × 10^-4 mol

b = moles of H+ in HOCl solution = x

Step 3: Total Moles of Hydrogen Ions

Total moles of H+ = a + b

Step 4: Moles of Hydroxide Ions in NaOH Solution (c)

c = Molarity × Volume

c = 0.200 M × 0.025 L

c = 0.005 mol

Step 5: Moles of Hydroxide Ions in Ba(OH)2 Solution (d)

d = Molarity × Volume

d = 0.100 M × 0.025 L

d = 0.0025 mol

Step 6: Moles of Hydroxide Ions in KOH Solution (e)

e = Molarity × Volume

e = 0.170 M × 0.010 L

e = 0.0017 mol

Step 7: Total Moles of Hydroxide Ions

Total moles of hydroxide ions = c + d + e

Step 8: Determine Excess Ions (Hydrogen or Hydroxide)

Since Total moles of H+ < Total moles of hydroxide ions, there is excess hydroxide ions.

Step 9: Moles of Excess Hydroxide Ions

Moles of hydroxide left after neutralization = Total moles of hydroxide ions - Total moles of H+

Step 10: Concentration of Hydroxide Ions

Concentration of hydroxide ions left in the solution = Moles of hydroxide left/Total volume of solution

Step 11: Calculate pOH

pOH = -log(OH- concentration)

Step 12: Calculate pH

pH = 14 - pOH

Let's substitute the values and calculate:

pOH = -log(0.001698/0.140) ≈ 1.9

pH = 14 - 1.9 ≈ 12.1

So, the pH of the solution is approximately 12.1.

The calculated pH of the solution is approximately 12.09.

Let's start with the calculations:

1. Calculate Moles of Each Component

2. Determine Neutralization

Strong acids (H₂SO₄) and bases (NaOH, KOH, Ba(OH)₂) will neutralize each other:

3. Calculate Final pH

Since 0.0017 moles of OH- remain:

The resulting pH of the mixed solution is approximately 12.09.

Answer:

5

Explanation:

We can obtain the value of x by doing the following:

Mass of hydrated salt (CuSO4.xH2O) = 1.50g

Mass of anhydrous salt (CuSO4) = 0.96g

Mass of water molecule(xH2O) = 1.50 — 0.96 = 0.54g

Molar Mass of CuSO4.xH2O = 63.5 + 32 + (16x4) + x(2 +16) = 63.5 + 32 + 64 + 18x = 159.5 + 18x

Mass of water(xH2O) molecules in the hydrate salt is given by:

xH2O/CuSO4.xH2O = 0.54/1.5

18x/(159.5 + 18x) = 0.36

Cross multiply to express in linear form

18x = 0.36 (159.5 + 18x)

18x = 57.42 + 6.48x

Collect like terms

18x — 6.48x = 57.42

11.52x = 57.42

Divide both side by 11.52

x = 57.42/11.52

x = 5

Therefore, the unknown integer x is 5 and the formula for the hydrated salt is CuSO4.5H2O

Answer: Ecell = -0.110volt

Explanation:

Zn--->Zn^+2 + 2e^-.........(1) oxidation

Cu^2+ 2e^- --->Cu........(2)reduction

Zn + Cu^2+ ----> Cu + Zn^+2 (overall

For an electrochemical cell, the reduction potential set up is given by

E(cell) = E(cathode) - E(anode)

E(cell) = E(oxidation) - E(reduction)

E(cathode) = E(oxidation)

E(anode) = E(reduction)

Given that

E(oxidation) = -0.763v

E(reduction) = +0.337v

E(cell) = -0.763 - (+0.337)

E(cell) = -0.763- 0.337

E(cell) = -0.110volt

Answer:

A. AN INCREASE IN BLOOD ACIDITY NEAR THE TISSUES

B. AN INCREASE IN BLOOD TEMPERATURE NEAR THE TISSUES.

C. THE PRESENCE OF A PRESSURE GRADIENT FOR OXYGEN.

Explanation:

Metabolically active tissues need more oxygen to carry out theirs functions. They are involved during excercise and other active phsiological conditions.

There is the reduction in the amount of oxygen reaching these tissues resulting in carbon IV oxide build up, lactic acid formation and temperature increases.

The acidity of the blood near the tissues is increased due to the accumulation of carbon IV oxide in the tissues resulting into a decreased pH. This reduces the affinity of heamoglobin to oxygen in the blood near the metabollically active tissues.

There is also the increase in temperature causing rapid offload of oxygen from oxy-heamoglobin molecules.

The partial pressure of oxygen gradient also affects the rate of oxygen offload by the blood. In metabollically active tissues, the partial pressure of oxygen is reduced in the tissues causing a direct offloading of oxygen to the tissues.

Explanation:

It is known that [tex]Fe(OH)_{3}[/tex] is insoluble in water. As a result, plants are not able to absorb [tex]Fe^{3+}[/tex] readily through osmosis.

Therefore, the [tex]Fe^{3+}[/tex] in [tex]Fe(OH)_{3}[/tex] would be released in acidic environments, using neutralization, Iron(III) ions can be released.

Hence, the easiest way is to add low concentrations of [tex]H_{2}SO_{4}[/tex] to the soil is as follows.

         [tex]Fe(OH)_{3} + H_{2}SO_{4} \rightarrow H_{2}O + Fe_{2}(SO_{4})_{3}[/tex]

Thus, we can conclude that [tex]Fe_{2}(SO_{4})_{3}[/tex] is soluble and is good for plants too.

Plants can be iron deficient even when grown in soils with high iron content because the form of iron present, Fe(OH)3, is insoluble and not easily accessible by plants. Therefore, they may not be able to absorb enough iron to meet their requirements.

The fact that plants can still be iron deficient even when grown in soils with high iron content can be explained by the form of iron present. In this case, the iron is in the form of Fe(OH)3. While plants require iron for proper growth and development, they can only absorb it in a specific form, called Fe²+. Fe(OH)3 is insoluble and cannot be easily accessed by plants. Therefore, even with high iron content, the plants may still not be able to absorb enough iron to meet their requirements, resulting in iron deficiency.

https://brainly.com/question/36255477

#SPJ3

Answer:

Nickel

Explanation:

This element and its alloys are used in pumps, valves, and other components that are in contact with acid and petroleum solutions

Final answer:

Nickel and its alloys are used in components like pumps and valves that are exposed to acid and petroleum solutions due to their corrosion resistance.

Explanation:

The element and its alloys mentioned in the question that are used in pumps, valves, and other components in contact with acid and petroleum solutions is likely to be Nickel. This is because nickel and its alloys are known for their corrosion resistance and are used in environments that would rapidly degrade other metals. For instance, nickel is used in the desalination of seawater and nickel steel is used for manufacturing armor plates and burglar-proof vaults. Alloys such as brass (copper and zinc) and bronze (copper, tin, and sometimes zinc) are also important but are not specific to the context given in the question, which highlights usage in acid and petroleum solutions where nickel's properties are most relevant.