A cannonball made of iron has a volume of 121 cm3. What is the mass, in kilograms, of the cannonball? Express your answer to three significant figures and include the appropriate units.

Answer:

[HI] = 0.264M

Explanation:

Based on the equilibrium:

2HI(g) ⇄ H₂(g) + I₂(g)

It is possible to define Kc of the reaction as the ratio between concentration of products and reactants using coefficients of each compound, thus:

Kc = 0.0156 = [H₂] [I₂] / [HI]²

As initial concentration of HI is 0.660mol / 2.00L = 0.330M, the equlibrium concentrations will be:

[HI] = 0.330M - 2X

[H₂] = X

[I₂] = X

Where X is reaction coefficient.

Replacing in Kc:

0.0156 = [X] [X] / [0.330M - 2X]²

0.0156 = X² / [0.1089 - 1.32X + 4X² ]

0.00169884 - 0.020592 X + 0.0624 X² = X²

0.00169884 - 0.020592 X - 0.9376 X² = 0

Solving for X:

X = - 0.055 → False solution, there is no negative concentrations

X = 0.0330 → Right solution.

Replacing in HI formula:

[HI] = 0.330M - 2×0.033M

Answer:

[tex]\large \boxed{5.1}[/tex]

Explanation:

1. Initial concentrations of reactants

[SO₂] = (3.8 mol)/(75 L) = 0.0507 mol·L⁻¹

 [O₂] = (7.0 mol)/(75 L) = 0.0933 mol·L⁻¹

2. Equilibrium concentration of SO₃

[SO₃] = (1.5 mol)/(75 L) = 0.0200 mol·L⁻¹

3. Set up an ICE table

                      2SO₂     +      O₂     ⇌     2SO₃

I/mol·L⁻¹:     0.0507         0.0933             0

C/mol·L⁻¹:        -2x                -x                +2x

E/mol·L⁻¹:   0.0507-2x    0.0933-x           2x

4. Calculate x

We know the final concentration of SO₃ is 0.0200 mol·L⁻¹, so

2x = 0.0200

x = 0.0100

5. Find the final concentrations of the reactants

Insert the numbers into the ICE table.

                      2SO₂     +       O₂     ⇌     2SO₃

I/mol·L⁻¹:     0.0507         0.0933             0

C/mol·L⁻¹:    -0.0200       -0.0100      +0.0200

E/mol·L⁻¹:     0.0307         0.0833        0.0200

6. Calculate K

[tex]K_{\text{eq}} = \dfrac{\text{[SO$_{3}$]}^{2}}{\text{[SO}_{2}]^{2}\text{[O$_{2}$]}} = \dfrac{0.0200^{2}}{0.0307^{2}\times0.0833} =\mathbf{5.1}\\\\\text{The value of the equilibrium constant is $\large \boxed{\mathbf{5.1}}$}[/tex]

Answer:

1. The coefficients are: 1, 3, 2

2. From the balanced equation, we obtained the following:

3 moles oxygen, O2 reacted.

2 moles of Hydrogen sulfide, H2S reacted.

2 moles of water were produced.

2 moles of sulphur dioxide, SO2 were produced.

Explanation:

1. Determination of the coefficients of the equation.

This is illustrated below:

P2O5(s) + H2O(l) <==> H3PO4(aq)

There are 2 atoms of P on the left side and 1 atom on the right side. It can be balance by putting 2 in front of H3PO4 as shown below:

P2O5(s) + H2O(l) <==> 2H3PO4(aq)

There are 2 atoms of H on the left side and 6 atoms on the right side. It can be balance by putting 3 in front of H2O as shown below:

P2O5(s) + 3H2O(l) <==> 2H3PO4(aq)

Now the equation is balanced.

The coefficients are: 1, 3, 2.

2. We'll begin by writing the balanced equation for the reaction. This is given below:

2H2S(g) + 3O2(g) => 2H2O(l) + 2SO2(g)

From the balanced equation above,

3 moles of oxygen, O2 reacted with 2 moles of Hydrogen sulfide, H2S to produce 2 moles of water, H2O and 2 moles of sulphur dioxide, SO2.

Answer:

24.15%

Explanation:

According to the given situation the computation of the percent yield of the reaction is shown below:-

PV = NRT = N = [tex]\frac{PV}{RT}[/tex]

Mole of [tex]N_2[/tex] = [tex]\frac{PV}{RT}[/tex]

= [tex]\frac{1\times 0.450}{0.0821\times 295}[/tex]

= [tex]\frac{0.450}{24.2195}[/tex]

= 0.0186

Mole of [tex]N_2H_4 = \frac{2.45}{32}[/tex]

= 0.077

Now, the percentage of yield is

= [tex]\frac{Practical\ yield}{Theoretical\ yield}\times 100[/tex]

= [tex]\frac{0.0186}{0.077}\times 100[/tex]

= 24.15%

Therefore for computing the percentage of yield we simply divide the practical yield by theoretical yield and multiply with 100 so that we can get the result into the percentage form.

Answer:

PhCH2CH2COOH

Explanation:

This is a reaction of PhCH2CH2Br with KCN in the presence of H3O^+. The reaction first leads to the formation of PhCH2CH2CN.

We must recall that part of the properties of nitriles is that they can be converted to carboxylic acids in the presence of H3O^+. This is a common synthetic route for carboxylic acids.

Therefore, when the PhCH2CH2CN is now further reacted with H3O^+, the carboxylic acid PhCH2CH2COOH is formed as the major organic product of the reaction, hence the answer given above.

Answer:

0.460 g  

Explanation:

Mass of flask + acid = 98.778 g

Mass of flask            = 98.318 g

Mass of acid             =  0.460 g

Answer:

-62.12kJ/mol is heat of neutralization

Explanation:

The neutralization reaction of HCl and NaOH is:

HCl + NaOH → NaCl + H₂O + HEAT

An acid that reacts with a base producing a salt and water

You can find the released heat of the reaction  -heat of neutralization- (Released heat per mole of reaction) using the formula:

Q = C×m×ΔT

Where Q is heat, C specific heat of the solution (4.06J/gºC), m its mass of the solution and ΔT change in temperature (27.5ºC-20.0ºC = 7.5ºC).

The mass of the solution can be found with the volume of the solution (50.0mL of HCl solution + 50.0mL of NaOH solution = 100.0mL) and its density (1.02g/mL), as follows:

100.0mL × (1.02g / mL) = 102g of solution.

Replacing, heat produced in the reaction was:

Q = C×m×ΔT

Q = 4.06J/gºC×102g×7.5ºC

Q = 3106J = 3.106kJ of heat are released.

There are 50.0mL ×1M = 50.0mmoles = 0.0500 moles of HCl and NaOH that reacts releasing 3.106kJ of heat. That means heat of neutralization is:

3.106kJ / 0.0500mol of reaction =

The - is because heat is released, absorbed heat has a + sign

Answer:

[tex][OH^-]=3.33x10^{-7}M[/tex]

Explanation:

Hello,

In this case, for the given concentration of hydronium, we can compute the pH as shown below:

[tex]pH=-log([H^+])=-log(3.0x10^{-8})=7.52[/tex]

Now, given the relationship between pH and pOH we can compute the pOH which is directly related with the concentration of hydroxyl in the solution:

[tex]pOH=14-pH=14-7.52=6.48[/tex]

Then, the concentration of hydroxyl turns out:

[tex][OH^-]=10^{-pOH}=10^{-6.48}[/tex]

[tex][OH^-]=3.33x10^{-7}M[/tex]

Best regards.

A process that is not a reversible reaction is frying an egg.

Reversible reactions are those reactions in which product will again change into the reactant.

Hence option (D) is correct.

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Answer:

0.00295M

Explanation:

Mass Concentration = mass/vol

= 0.203 g/ 0.02414 L = 8.409 g/L

But molarity = Mass conc / molar mass

∴ Molar mass(mol/L) = mass conc / molarity

= .84909 / 0.0295

= 285.06 g/mol

If 1 mol of green stock - 285.06g

? mol - 0.203 g

= 0.00071213 g

= 0.00071213 g / .2414L = 0.0095 mol/L.

Answer:

0.56M

Explanation:

Molar concentration is defined as the ratio between moles of solute and volume in liters of solution.

In a 0.28M H₂ there are 0.28moles of H₂ per liter of solution.

Now, in 1 molecule of H₂ there are 2 atoms of H. Following this idea, in 0.28 moles of H₂ there are 0.28*2 = 0.56 moles of H atoms.

Thus, molar concentration of H atoms in a 0.28M H₂ is 0.56M

Answer:

0.394 atm

Explanation:

Mathematically, when we increase the pressure of a gas, we are increasing its concentration and when we decrease the pressure, we are decreasing its concentration.l at same temperature

What this means is that pressure and concentration are directly proportional.

Representing concentration by c and pressure by p, we have;

P1/C1 = P2/C2

From the question;

P1 = 0.719 atm

P2 = ?

C1 = 429.7 ppm

C2 = 235.3 ppm

Now, we can rewrite the equation to be;

P1C2/C1 = P2

Substituting the values we have;

0.719 * 235.3/429.7 = 0.394 atm

Answer:

D. Poop Butt.

Explanation:

Based on the given data, we can propose a possible structure that fits the criteria: a. carbonyl group (C=O) and an aromatic ring b. there are seven unique carbon environments. c. Presence of a functional group like a carboxylic acid or phenol .

a. The IR peaks at 1603 [tex]cm^{-1}[/tex] and 1495[tex]cm^{-1}[/tex]suggest the presence of both a carbonyl group (C=O) and an aromatic ring.

b. The 13C NMR having a total of 7 signals indicates that there are seven unique carbon environments in the molecule.

c. Considering the presence of an acidic proton, it suggests the presence of a functional group like a carboxylic acid (COOH) or phenol ([tex]C_6H_5OH[/tex]).

Putting all this information together, a possible structure that fits the data could be benzoic acid ([tex]C_6H_5COOH[/tex]). It contains a benzene ring (giving 6 unique carbon environments), a carbonyl group (giving 1 unique carbon environment), and an acidic proton in the carboxylic acid group. This structure satisfies all the given data.

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Answer:

a) [tex]\Delta G=2.6kJ[/tex]

b) [tex]\Delta G=-979.57kJ[/tex]

c) [tex]\Delta G=264.21kJ[/tex]

Explanation:

Hello,

In this case, in each reaction we must subtract the Gibbs free energy of formation the reactants to the Gibbs free energy of formation of the products considering each species stoichiometric coefficients. In such a way, the Gibbs free energy of formations are:

[tex]\Delta _fG_{H_2}=\Delta _fG_{I_2}=0kJ/mol\\\Delta _fG_{HI}=1.3kJ/mol\\\Delta _fG_{CO_2}=-394.4kJ/mol\\\Delta _fG_{CO}=-137.3 kJ/mol\\\Delta _fG_{NH_3}=16.7 kJ/mol\\\Delta _fG_{HCl}=-95.3kJ/mol\\\Delta _fG_{MnO_2}=465.37kJ/mol\\\Delta _fG_{Mn}=0kJ/mol\\\Delta _fG_{NH_4Cl}=-342.81kJ/mol[/tex]

So we proceed as follows:

a)

[tex]\Delta G=2\Delta _fG_{HI}-\Delta _fG_{H_2}-\Delta _fG_{I_2}\\\\\Delta G=2*1.3\\\\\Delta G=2.6kJ[/tex]

b)

[tex]\Delta G=\Delta _fG_{Mn}+2*\Delta _fG_{CO_2}-\Delta _fG_{MnO_2}-2*\Delta _fG_{CO}\\\\\Delta G=0+2*-394.4-465.37-2*-137.3\\\\\Delta G=-979.57kJ[/tex]

c)

[tex]\Delta G=\Delta _fG_{NH_3}+\Delta _fG_{HCl}-\Delta _fG_{NH_4Cl}\\\\\Delta G=16.7-95.3-(-342.81)\\\\\Delta G=264.21kJ[/tex]

Regards.

Answer:

New pH = 3.84

Explanation:

First of all we may think that if the buffer has pH 3.98 and we're adding H⁺, pH's buffer will be lower, as the [H⁺] is been increased.

Let's determine the moles of each compound:

0.23 M . 1.3L = 0.299 moles of NaBz

0.38 M . 1.3L = 0.494 moles of HBz

We add 0.058 of HCl, which is the same as 0.058 moles of H⁻

HCl →  H⁺  +  Cl⁻

As we add the moles of protons, these are going to react to the Bz⁻

In the buffer system we have these dissociations:

NaBz  →  Na⁺  +  Bz⁻

HBz →  H⁺  + Bz⁻

So, as we add protons, we have a new equilibrium:

          Bz⁻      +      H⁺       ⇄   HBz    

In    0.299         0.058          0.494

Eq   0.241               -              0.552

Protons are substracted to benzoate, so the [HBz] is now higher than before. We calculate the new pH, with the Henderson Hasselbach equation

pH = pKa + log (Bz⁻/HBz)

pH = 4.20 + log (0.241 / 0.552) → 3.84

Answer:

ye, it will be optically active

Explanation:

a compound is said to be optically active if it can optically rotate.

the removal of carboxyl group and release of cabon dioxide from carboxylic acid in acetoacetate synthesis which will result in production of ketone as given the attachment below.

Answer:

Following are the answer to this question:

Explanation:

The value of pH solution is =5.17 So, the p^{OH}:

[tex]p^{OH}[/tex]=14-56.17

      =8.823

The volume of the [tex]NH_{3}[/tex] = 40.00 ml  

convert into the liter= 0.040L

The value of the concentrated [tex]NH_{3}[/tex] =0.10 M

The volume of the [tex]NH_{4}Cl[/tex]= 50.00 ml

convert into the liter= 0.050L

 The value of concentrated [tex]NH_{4}Cl[/tex]= 0.10 M

The volume of the [tex]H_{2}So_{4}[/tex]= 30 ml

convert into the liter= 0.030L  

The value of concentrated [tex]H_2So_4[/tex]=0.05 M

Calculating total volume=(0.40+0.050+0.030)

                                       =0.120 L

calculating the new concentrated value of [tex]NH_3[/tex] = [tex]\frac{0.10\times 0.040}{0.120}= 0.33 \ M[/tex]

calculating the new concentrated value of [tex]NH_4Cl[/tex]= [tex]\frac{0.050\times 0.10}{0.120}= 0.04166 \ M[/tex]calculating the new concentrated value of [tex]H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M[/tex] when 1 mol [tex]H_2So_4[/tex] produced 2 mols [tex]H^{+}[/tex] so, 0.0125 in [tex]H_2So_4[/tex]produced:

[tex]=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}[/tex]

create the ICE table:    

[tex]NH_3 \ \ \ \ \ \ \ \ + H^{+} \ \ \ \ \ \ \longrightarrow NH_4^{+}[/tex]                    

I (m)       0.033(m)            0.025                       0.04166

C            -0.025                 -0.025                       + 0.025  

E            8.3\times 10^{-3}     0                    0.0667

now calculating pH:

when ph= 8.83:

[tex]P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769[/tex]

Answer:

The drawing that represents the result of placing all three liquids into the same graduated cylinder will have the liquid arranged one on top of the other from top to bottom in the order of A, C, B.

Explanation:

The image with the options is not provided in this question, but I can answer this fairly so that you can pick from the question, the correct drawing.

We know that two or more immiscible liquids contained together in a container will always separate in the order of their density from top to bottom, with the densest at the bottom, and the least densest at the top. In this case, liquid A is the least densest, and liquid B is the densest. Liquid A will stay on top, and liquid B will be at the bottom. Liquid C will be in between liquid A and liquid B.

Answer:

The initial temperature was  [tex]36.4^\circ \:C[/tex]

Explanation:

[tex]\Delta t=\frac{q}{m\cdot C_s}=\frac{5.83\times10^5}{2190\times 4.184}\\\\=63.6^\circ\:C[/tex]

The temperature difference [tex]=100-63.6=36.4^\circ\:C[/tex]

Best Regards!

Answer:

Explanation:

2NOCl(g)    ⇄      2NO(g) + Cl 2(g)

C ( 1 - .28 )            .28 C         .14 C

Kc = [ NO ]²  x [ Cl₂ ] / [ NOCl ]²

= (.28 C )² x .14 C / C² ( 1 - .28 )²

= .021173 x C

C = concentration of reactant

= 2.5 / 2/5 = 1 M

Kc = .021173 x 1

= 211.73 x 10⁻⁴ M .

Given:

At equilibrium,

Now,

The concentration of NOCl will be:

= [tex]\frac{Number \ of \ moles}{Volume \ of \ solution}[/tex]

= [tex]\frac{2.5}{2.5}[/tex]

= [tex]1 \ M[/tex]

At equilibrium,

The concentration of NOCl will be:

= [tex]1-0.28[/tex]

= [tex]0.72 \ M[/tex]

hence,

The equilibrium constant,

→ [tex]K_c =\frac{ [NO]^2 [Cl_2]}{[NOCl]^2}[/tex]

By substituting the values, we get

        [tex]= \frac{(0.28)^2\times (0.14)}{(0.72)^2}[/tex]

        [tex]= 2.117\times 10^{-2}[/tex]

Thus the above answer is right.

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Answer:

Coordinate or r = (7,9).

Explanation:

Data obtained from the question include the following:

Mid point = (8,7)

Coordinate of S = (9,5)

Coordinate of r =...?

We shall determine the coordinate of r as follow:

Let the coordinate of r be (x2, y2)

Mid point = (x1 + x2)/2 , (y1 + y2)/2

Mid point = (8,7)

Coordinate of S = (9,5)

x1 = 9

y1 = 5

x2 =?

y2 =?

The value of x2 can be obtained as follow:

8 = (x1 + x2)/2

8 = (9 + x2)/2

Cross multiply

9 + x2 = 2 × 8

9 + x2 = 16

Collect like terms

x2 = 16 – 9

x2 = 7

The value of y2 can be obtained as follow:

5 = (y1 + y2)/2

7 = (5 + y2)/2

Cross multiply

5 + y2 = 2 × 7

5 + y2 = 14

Collect like terms

y2 = 14 – 5

y2 = 9

Coordinate of r = (x2, y2)

Coordinate or r = (7,9)

Answer: When a solute is added to water, the vapor pressure will decrease and the boiling point will increase.

Explanation:

Answer:

Alternative C would be the correct choice.

Explanation:

The other three choices are not related to the given circumstances. So that option C would be the appropriate choice.

Answer:

0.44 moles

Explanation:

Given that :

A mixture of water and graphite is heated to 600 K in a 1 L container. When the system comes to equilibrium it contains 0.17 mol of H2, 0.17 mol of CO, 0.74 mol of H2O, and some graphite.

The equilibrium constant [tex]K_c= \dfrac{[CO][H_2]}{[H_2O]}[/tex]

The equilibrium constant  [tex]K_c= \dfrac{(0.17 )(0.17)}{0.74}[/tex]

The equilibrium constant [tex]K_c= 0.03905[/tex]

Some O2 is added to the system and a spark is applied so that the H2 reacts completely with the O2.

The equation for the reaction is :

[tex]H_2 + \dfrac{1}{2}O_2 \to H_2O \\ \\ 0.17 \ \ \ \ \ \ \ \ \ \to0.17[/tex]

Total mole of water now = 0.74+0.17

Total mole of water now = 0.91 moles

Again:

[tex]K_c= \dfrac{[CO][H_2]}{[H_2O]}[/tex]

[tex]0.03905 = \dfrac{[0.17+x][x]}{[0.91 -x]}[/tex]

0.03905(0.91 -x) = (0.17 +x)(x)

0.0355355 - 0.03905x = 0.17x + x²

0.0355355 +0.13095 x -x²

x² - 0.13095 x - 0.0355355 = 0

By using quadratic formula

x = 0.265  or   x = -0.134

Going by the value with the positive integer; x = 0.265 moles

Total moles of CO in the flask when the system returns to equilibrium is :

= 0.17 + x

= 0.17 + 0.265

= 0.435 moles

=0.44 moles (to two significant figures)

Answer:

c = 250.58 J/kg/[tex]^{0}C[/tex]

Explanation:

The specific heat of a substance is the required quantity of heat to increase or decrease the temperature of its unit mas by 1 kelvin.

Q = mcΔθ

where: Q is the quantity of heat absorbed or released, m is the mass of the substance, c is its specific heat and Δθ is the change in temperature of the substance.

Given that; m = 1.0 kg, Q = 1303 J and Δθ = 5.2 [tex]^{0}C[/tex], then;

c = Q ÷ (mΔθ)

  = 1303 ÷ (1.0 × 5.2)

  = 1303 ÷ 5.2

  = 250.58 J/kg/[tex]^{0}C[/tex]

The specific heat of the object is 250.58 J/kg/[tex]^{0}C[/tex].

Answer:

0.25

Explanation:

Answer:

1.42 × 10⁻⁹ M

Explanation:

Step 1: Given data

Step 2: Write the reaction for the solution of BaSO₄

BaSO₄(s) ⇄ Ba²⁺(aq) + SO₄²⁻(aq)

Step 3: Calculate the concentration of SO₄²⁻

We will use the following expression.

Ksp = 1.08 × 10⁻¹⁰ = [Ba²⁺] × [SO₄²⁻]

[SO₄²⁻] = 1.08 × 10⁻¹⁰ / [Ba²⁺] = 1.08 × 10⁻¹⁰ / 0.0758 = 1.42 × 10⁻⁹ M

Answer:

CH3CH2C≡CH

Explanation:

The particular reaction under study is known as the alkykation of acetylide ions. An acetylide ion can be alkykated using a suitable alkyl halide. The overall scheme of the reaction is;

CH≡C^- + RX -----> RC≡CH + X^-

This reaction is most effective when primary alkyl halides are used. It involves SN2 substitution of a halide in the alkyl halide by an acetylide ion. Secondary, tertiary or even bulky primary substrates are known to yield alkenes and alkynes owing to elimination by E2 mechanism.

A decomposition reaction occurs when one reactant breaks down into two or more products. This can be represented by the general equation: AB → A + B. Examples of decomposition reactions include the breakdown of hydrogen peroxide to water and oxygen, and the breakdown of water to hydrogen and oxygen.

Answer:

The mass of the element is 141.03701 amu

Explanation:

The catch here is that it notes a " newly found element. " Otherwise you could just refer to the average atomic mass of the element in the periodic table, and receive your solution in a much faster way.

The first isotope has an atomic mass of 139.905 amu, and a respective percent abundance of 37.25%. The second isotope has an atomic mass of 141.709 amu, and the remaining percent abundance, 100% - 37.25% = 62.75% ( given ). We can calculate the mass of the unknown element by associating each percentage with the mass of their respective isotope, over 100%.

Mass = ( ( 139.905 amu )( 37.25% ) + ( 141.709 amu )( 62.75% ) )/ 100,

Mass = ( ( 5211.46125 ) + ( 8892.23975 ) ) / 100,

Mass = ( 14103.701 ) / 100 = 141.03701 amu