Which of the following is an example of a mechanical wave?
O A. A light ray
B. A seismic wave
C. A radio wave
D. An X-ray

Answer: Fossil fuels

Explanation:

Fossil fuels such as petroleum, oil,  and natural gas, are non-renewable energy resources which are formed from the remains of  prehistoric ancient  plants and animals beneath layers of rock of the earth surface.

By analyzing and studying fossil fuels using Radiocarbon analyses by archaeologists, earth scientists  etc, Information about the history of the earth can be obtained from the decomposition of dead organisms present in fossil fuels.

Answer:

The enthalpy of reaction per mole of HBr for this reaction = ΔrH = -40.62 kJ/mole.

Explanation:

2HBr(g) + Cl2(g) → 2HCl(g) + Br2(g)

When 23.9 g HBr(g) reacts with sufficient Cl2(g), 12.0 kJ of heat is evolved, calculate the value of ΔrH for the chemical reaction.

Note that ΔrH is the enthalpy per mole for the reaction.

Molar mass of HBr (g) = 80.91 g/mol.

Hence, 1 mole of HBr = 80.91 g

23.9 g of HBr led to the reaction giving off 12.0 kJ of heat

80.91 g of HBr will lead to the evolution of (80.91 × 12/23.9) = 40.62 kJ heat is given off.

Hence, 40.62 kJ of heat is given off per 80.91 g of HBr.

This directly translates to that 40.62 kJ of heat is given off per 1 mole of HBr

Hence, the heat given off per mole of HBr for this reaction is 40.62 kJ/mole.

But since the reaction liberates heat, it means the reaction is exothermic and the enthalpy change for the reaction (ΔHrxn) is negative.

Hence, ΔrH = -40.62 kJ/mole.

Hope this Helps!!!

Answer:

The percentage yield is 50%

Answer:

For our assumed experiment; the expected  volume of Hcl acid needed to neutralize the bicarbonate is 0.13 mL

Explanation:

We are going attempt this question experimentally.

We know that benzoic acid originate from the relationship between  benzene and a carboxylic group. So basically , the functional group of a carboxylic acid (-COOH) joins with a benzene ring(C₆H₆) to form a simple aromatic carboxylic acid known as Benzoic acid. (C₇H₆O₂)

However, it is possible to isolate benzoic acid from  a bicarbonate solution in the presence of an acidified concentrated hydrochloric acid.

Let assume that ;

0.20 g of benzoic acid was reacted with 2 mL of a 20% solution of NaHCO₃, the amount of the  excess NaHCO₃ can be determined by subtracting the amount of benzoic acid from the amount of NaHCO₃.

Let first calculate the number of moles in 0.20 g of benzoic acid

we know that the standard  molar mass of benzoic acid is 122.12 g/mol

number of moles of benzoic acid = mass of benzoic acid/molar mass of benzoic acid =

number of moles of benzoic acid = 0.20/ 122.12

number of moles of benzoic acid = 0.0016 mol

number of moles of bicarbonate  solution = mass of bicarbonate solution/ molar mass of bicarbonate solution

number of moles of bicarbonate  solution =  0.2/84.00654 g/mol

number of moles of bicarbonate  solution =  0.00238 mol

∴

(0.00238 - 0.0016) mol

= 7.8 × 10⁻⁴ mol

Let assume that the concentrated HCl is 12  M

Also. HCl and NaHCO₃ react together at the ratio of 1:1; thus the  volume of Hcl acid needed to neutralize the bicarbonate is:

[tex]= ( 7.8 * 10^{-4} \ \ mol )* ( \dfrac{2\ L}{ 12 \ M})*( \dfrac{10^3 \mL}{1 \ L})[/tex]

= 0.13 mL

Thus; for our assumed experiment; the expected  volume of Hcl acid needed to neutralize the bicarbonate is 0.13 mL

Answer:

The correct answer is 66.35 kilograms.

Explanation:

Based on the data given in the question, the energy consumed by the body of a human being is 50%. Based on the given data, the energy consumed in a day is 8000 kJ, 50 percent is the energy transfer efficiency. Thus, the consumption of total energy is 4000 kJ, and for the transformation of ADP to ATP, the energy involved is 30.6 kJ per mole.  

Hence, the total ATP produced in the process is,  

ATP = 4000 kJ / 30.6 kJ/mol

= 130.7189 mol.  

Thus, with the energy transfer efficiency of 50 percent, the total moles of ATP produced is 130.7 mol.  

The mass of ATP can be calculated by using the formula,  

moles = mass/molecular mass

The molecular mass of ATP is 507.18 g per mol

Now by putting the values we get,  

mass of ATP = 130.7189 mol * 507.18 g/mol

= 66298.011 g or 66.298 kg

It is mentioned that human comprise 50 g of ATP or 0.05 kg of ATP. Therefore, the sum of the available ATP will be.  

= Total production of ATP + Total ATP available

= 66.298 kg + 0.05 kg

= 66.348 kg

Hence, the sum of the ATP that is turned over by a resting human in a day is 66.35 kg.  

Answer:

6.91 × 10¹⁴ s⁻¹

Explanation:

Step 1: Given data

Wavelength of the radiation absorbed by the cone (λ): 434 nm

Step 2: Convert the wavelength to meters

We will use the relationship 1 m = 10⁹ nm.

[tex]434nm \times \frac{1m}{10^{9}nm } =4.34 \times 10^{-7} m[/tex]

Step 3: Calculate the frequency (ν) of the radiation

We will use the following expression.

[tex]c = \lambda \times \nu[/tex]

where,

c is the speed of light (3.00 × 10⁸ m/s)

[tex]c = \lambda \times \nu\\\nu = \frac{c}{\lambda} = \frac{3.00 \times 10^{8}m/s }{4.34 \times 10^{-7}m }= 6.91 \times 10^{14} s^{-1}[/tex]

Answer:

[CH₂Cl₂] = 7.07x10⁻² M

[CH₄] = 0.319 M

[CCl₄] = 0.164 M  

Explanation:

The equilibrium reaction is the following:

2CH₂Cl₂(g) ⇄ CH₄(g) + CCl₄(g)  

The equilibrium constant of the above reaction is:

[tex] K = \frac{[CH_{4}][CCl_{4}]}{[CH_{2}Cl_{2}]^{2}} = \frac{0.173 M*0.173 M}{(5.35 \cdot 10^{-2} M)^{2}} = 10.5 [/tex]

When 0.155 mol of CH₄(g) is added to the flask we have the following concentration of CH₄:

[tex] C = \frac{\eta}{V} = \frac{0.155 mol}{1.00 L} = 0.155 M [/tex]

[tex]C_{CH_{4}} = 0.328 M[/tex]      

Now, the concentrations at the equilibrium are:

2CH₂Cl₂(g)   ⇄   CH₄(g)  +  CCl₄(g)

5.35x10⁻² - 2x   0.328 + x   0.173 + x    

[tex]K = \frac{[CH_{4}][CCl_{4}]}{[CH_{2}Cl_{2}]^{2}} = \frac{(0.328 + x)(0.173 + x)}{(5.35 \cdot 10^{-2} - 2x)^{2}}[/tex]

[tex]10.5*(5.35 \cdot 10^{-2} - 2x)^{2} - (0.328 + x)*(0.173 + x) = 0[/tex]

Solving the above equation for x:  

x₁ = 0.076 and x₂ = -0.0086

Hence, the concentration of the three gases once equilibrium has been reestablished is:

[CH₂Cl₂] = 5.35x10⁻² - 2(-0.0086) = 7.07x10⁻² M

[CH₄] = 0.328 + (-0.0086) = 0.319 M

[CCl₄] = 0.173 + (-0.0086) = 0.164 M  

We took x₂ value because the x₁ value gives a negative CH₂Cl₂ concentration.

I hope it helps you!

Answer:

Mass of Al2S3 remaining is 17.212 g

Explanation:

Equation of the reaction is given below:

Al2S3 + 6H2O -----> 2Al(OH)3 + 3H2S

From the balanced equation above

6 mole of H20 reacts with 1 mole of Al2S3

i.e. 6 * 18.02 g of H2O reacts with 1 * 150.71 g of Al2S3

= 108.12 g of H2O reacts with 150.71 g of Al2S3

Therefore 2.0 g of water will react with 2.0 * (150.71/108.12) g of Al2S3

= 2.788 g of Al2S3

Mass of Al2S3 remaining = 20.0 g - 2.788 g = 17.212 g

According to the properly balanced chemical equation, the amount (mass) of [tex]AL_2S_3[/tex] that remains after the chemical reaction is 17.22 grams.

Given the following data:

To calculate the amount (mass) of [tex]AL_2S_3[/tex] that remains after the chemical reaction:

First of all, we would write a properly balanced chemical equation for this chemical reaction.

                        [tex]Al_2S_3 + 6H_2O ---> 2Al(OH)_3 + 3H_2S[/tex]

By stoichiometry:

1 mole of [tex]AL_2S_3[/tex] reacts with 6 moles of [tex]H_2O[/tex]

Next, we would calculate the mass of each compound.

For [tex]AL_2S_3[/tex]:

[tex]Mass = Number\;of\;moles \times molar\;mass\\\\Mass = 1 \times 150.17[/tex]

Mass = 150.17 grams

For [tex]H_2O[/tex]:

[tex]Mass = Number\;of\;moles \times molar\;mass\\\\Mass = 6 \times 18.02[/tex]

Mass = 108.12 grams

108.12 grams of [tex]H_2O[/tex] = 150.17 grams of [tex]AL_2S_3[/tex]

2.00 grams of [tex]H_2O[/tex] = X grams of [tex]AL_2S_3[/tex]

Cross-multiplying, we have:

[tex]108.12 \times X = 150.17 \times 2\\\\108.12X = 300.34\\\\X = \frac{300.34}{108.12}[/tex]

X = 2.78 grams of [tex]AL_2S_3[/tex]

Remaining mass = [tex]20.00 - 2.78[/tex]

Remaining mass = 17.22 grams of [tex]AL_2S_3[/tex]

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Answer:

B. the molecules in liquid are loosely packed and scattered thus, they cannot be arranged

Answer:

Oxygen is the limiting reactant.

Explanation:

Based on the reaction:

C₁₂H₂₂O₁₁ + 12O₂ → 12CO₂ + 11H₂O

1 mole of sucrose reacts with 12 moles of oxygen to produce 12 moles of CO₂ and 11 moles of H₂O.

10.0g of sucrose (Molar mass: 342.3g /mol) are:

10.0g C₁₂H₂₂O₁₁ × (1mole / 342.3g) = 0.0292 moles of C₁₂H₂₂O₁₁

And moles of 10.0g of oxygen (Molar mass: 32g/mol) are:

10.0g O₂ × (1mole / 32g) = 0.3125 moles of O₂

For a complete reaction of 0.0292 moles of C₁₂H₂₂O₁₁ you need (knowing 12 moles of oxygen react per mole of sucrose):

0.0292 moles of C₁₂H₂₂O₁₁ × (12 moles O₂ / 1 mole C₁₂H₂₂O₁₁) = 0.3504 moles of O₂

As you have just 0.3125 moles of O₂, oxygen is the limiting reactant.

Answer:

The solubility is  [tex]S = 0.0014 \ g[/tex]

Explanation:

From the question we are told that

   The volume of the solution is  [tex]V = 600 mL[/tex]

    The initial temperature is  [tex]T_i = 37 ^oC[/tex]

     The final temperature is [tex]T_f = 21^oC[/tex]

      The additional precipitate is [tex]m = 0.084 \ kg = 84 \ g[/tex]

Yes because the solubility of the substance X is the amount of X needed to saturate a unit volume of the solvent  (for solubility of a solute to be calculated the solute must be able to saturate the solvent)

now we see that the substance X saturated the solvent because a precipitate was formed which the student threw away

   The solubility at  21 ° C is mathematically represented as

            [tex]S = \frac{m}{m_w * 100 g \ of water }[/tex]

Mass of water([tex]m_w[/tex]) in the solution is mathematically represented as

       [tex]m_w = V * \rho_w[/tex]

Where  [tex]\rho = 1 \frac{g}{mL}[/tex]

So  

      [tex]m_w =600 * 1[/tex]

       [tex]m_w =600g[/tex]

So  

    [tex]S = \frac{84}{600 * 100 g \ of water }[/tex]

    [tex]S = 0.0014 \ g[/tex]

Answer:

$11.81

Explanation:

27 lb cost $16

27/16=$1.69 per pound

$1.69*7=$11.81 for 7 lbs

Answer:

The correct answer is 5.447 × 10⁻⁵ vacancies per atom.

Explanation:

Based on the given question, the at 750 degree C the number of vacancies or Nv is 2.8 × 10²⁴ m⁻³. The density of the metal is 5.60 g/cm³ or 5.60 × 10⁶ g/m³. The atomic weight of the metal given is 65.6 gram per mole. In order to determine the fraction of vacancies, the formula to be used is,  

Fv = Nv/N------ (i)  

Here Nv is the number of vacancies and N is the number of atomic sites per unit volume. To find N, the formula to be used is,  

N = NA×P/A, here NA is the Avogadro's number, which is equivalent to 6.022 × 10²³ atoms per mol, P is the density and A is the atomic weight. Now putting the values we get,  

N = 6.022 × 10²³ atoms/mol × 5.60 × 10⁶ g/m³ / 65.6 g/mol

N = 5.14073 × 10²⁸ atoms/m³

Now putting the values of Nv and N in the equation (i) we get,  

Fv = 2.8 × 10²⁴ m⁻³ / 5.14073 × 10²⁸ atoms/m^3

Fv = 5.44669 × 10⁻⁵ vacancies per atom or 5.447 × 10⁻⁵ vacancies/atom.  

Answer:

ΔH of the reaction is -802.3kJ.

Explanation:

Using Hess's law, you can know ΔH of reaction by the sum of ΔH's of half-reactions.

Using the reactions:

(1) Cgraphite(s)+ 2H₂(g) → CH₄(g) ΔH₁ = −74.80kJ

(2) Cgraphite(s)+ O₂(g) → CO₂(g) ΔH₂ = −393.5k J

(3) H₂(g) + 1/2 O₂(g) → H₂O(g) ΔH₃ = −241.80kJ

The sum of (2) - (1) produce:

CH₄(g) + O₂(g) → CO₂(g) + 2H₂(g) ΔH' = -393.5kJ - (-74.80kJ) = -318.7kJ

And the sum of this reaction with 2×(3) produce:

CH₄(g) + 2 O₂(g) → CO₂(g) + 2H₂O(g) And ΔH = -318.7kJ + 2×(-241.80kJ) =

-802.3kJ

Answer:

Explanation:

The objective here is mainly drawing the diagrams of every stereoisomer for 1-bromo-2-chloro-1,2-difluorocyclopentane.

Stereoisomerism is the difference of the spatial arrangement of atoms in a molecule or a compound with the same molecular formula.

For 1-bromo-2-chloro-1,2-difluorocyclopentane.

We have the stereoisomers as follows:

(1R,2S)-1-bromo-2-chloro-1,2-difluorocyclopentane.

(1S,2R)-1-bromo-2-chloro-1,2-difluorocyclopentane.

(1S,1S)-1-bromo-2-chloro-1,2-difluorocyclopentane.

(1R,1R)-1-bromo-2-chloro-1,2-difluorocyclopentane.

Their diagrams are drawn and shown in the attached file below in the order with which they are listed above.

Answer: 125 g

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} B_2H_6=\frac{36.1g}{17}=1.30moles[/tex]

The balanced reaction is:

[tex]B_2H_6+3O_2\rightarrow 2HBO_2+2H_2O[/tex]

According to stoichiometry :

1 mole of [tex]B_2H_6[/tex] require = 3 moles of [tex]O_2[/tex]

Thus 1.30 moles of [tex]B_2H_6[/tex] will require=[tex]\frac{3}{1}\times 1.30=3.90moles[/tex]  of [tex]O_2[/tex]

Mass of [tex]O_2=moles\times {\text {Molar mass}}=3.90moles\times 32g/mol=125g[/tex]

Thus 125 g of [tex]O_2[/tex] will be needed to burn 36.1 g of [tex]B_2H_6[/tex]

Answer:

[tex]B_2O_3[/tex]

Explanation:

First, we have to find the reaction:

[tex]B_2O_3~+~C~+~Cl_2~->~BCl_3~+~CO[/tex]

The next step is to balance the reaction:

[tex]B_2O_3~+~3C~+~3Cl_2~->~2BCl_3~+~3CO[/tex]

Now, we have to calculate the molar mass for  each compound, so:

[tex]B_2O_3=~69.62~g/mol[/tex]

[tex]C=~12~g/mol[/tex]

[tex]Cl_2=~70.96~g/mol[/tex]

With these values, we can calculate the moles of each compound:

[tex]95.7~g~B_2O_3\frac{1~mol~B_2O_3}{69.62~g~B_2O_3}=1.37~mol~B_2O_3[/tex]

[tex]75.7~g~C\frac{1~mol~C}{112~g~C}=6.30~mol~C[/tex]

[tex]369~g~Cl_2\frac{1~mol~Cl_2}{70.96~g~C}=5.20~mol~Cl_2[/tex]

Now we can divide by the coefficient of each compound in the balanced equation:

[tex]\frac{1.37~mol~B_2O_3}{1}=~1.37[/tex]

[tex]\frac{6.30~mol~C}{3}=~2.1[/tex]

[tex]\frac{5.20~mol~Cl_2}{3}=~1.73[/tex]

The smallest values are for  [tex]B_2O_3[/tex], so this is our limiting reagent.

I hope it heps!

Answer:

  1140 mmHg

Explanation:

1 atmosphere is 760 mmHg, so 1.5 atmospheres is ...

  1.5×760 mmHg = 1140 mmHg

Answer: [tex]y = \frac{1}{4}x-3[/tex]

Explanation:

To find the inverse of an equation, follow these steps:

We are given the equation [tex]f(x) = 4x + 12[/tex] , so replace f(x) with x.

Then, replace x with y.

Your new equation:

[tex]x = 4y + 12[/tex]

Now, solve for y:

[tex]x = 4y + 12\\\\4y = x - 12\\\\y = \frac{1}{4}x-3[/tex]

This equation is the inverse of f(x), or g(x).

Answer:

Plasmas and solids are both made up of cation-anion pairs. Solids and plasmas are both made up of electrons and cations. Solids are made up of cation-anion pairs, but plasmas are not.

Answer:

Solids are made up of cation-anion pairs, but plasmas are not.

Explanation:

Answer:

That its small pointed. Pink(Himalayan salt)or white(normal salt)

Explanation:

Summa dees questions are so stupid, deys makin me salty.

Answer: (e) The pressure in the container increases but does not double.

Explanation:

To solve this, we need to first remember our gas law, Boyle's law states that the pressure and volume of a gas have an inverse relationship. That is, If volume increases, then pressure decreases and vice versa, when temperature is held constant. Therefore, increasing the volume in this case does not double the pressure owning to out gas law, but an increase in pressure would be noticed if temperature is constant

The pressure in the container increases but does not double.

At constant temperature and volume, the pressure of a given mass of gas is directly proportional to the number of moles of gas present.

Number of moles of He = 10 g/4 g/mol = 2.5 moles

Number of moles of Ne = 10 g/20 g/mol  = 0.5 moles

We can see that the number of moles only increases by 1/5 of its initial value therefore, the pressure in the container increases but does not double.

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Answer:

0.287 mole of PCl5.

Explanation:

We'll begin by calculating the number of mole in 51g of Cl2. This is illustrated below:

Molar mass of Cl2 = 2 x 35.5 = 71g/mol

Mass of Cl2 = 51g

Number of mole of Cl2 =..?

Mole = Mass /Molar Mass

Number of mole of Cl2 = 51/71 = 0.718 mole

Next, we shall write the balanced equation for the reaction. This is given below:

P4 + 10Cl2 → 4PCl5

Finally, we determine the number of mole of PCl5 produced from the reaction as follow:

From the balanced equation above,

10 moles of Cl2 reacted to produce 4 moles of PCl5.

Therefore, 0.718 mole of Cl2 will react to produce = (0.718 x 4)/10 = 0.287 mole of PCl5.

Therefore, 0.287 mole of PCl5 is produced from the reaction.

Answer: if im not wrong the relations are that the electrochemistry can detect the cancer and any other sickness

just like it does with chemical phenomena

=)

Answer:

FOR EVERY 10 DEGREE CELSIUS INCREASE IN TEMPERATURE, THE ACTIVATION ENERGY THAT SHOWS THIS IS 52.4 KJ/MOL

Explanation:

From Arrhenius equation, the relationship between the rate constant and the temperature is as shown below:

k = Ae^ -Ea/RT

At initial temperature T1, the initial rate constant is (k1)

At final temperature T2, the final rate constant is k2

For the reaction rate to be doubled, we must double the rate constant which shows that the ratio of k2 / k1 must be equal to 2.

That is, k2 / k1 = 2 (rate is doubled)

Equating this into the Arrhenius equation, we have:

k2 / k1 = Ae^ (-Ea / R ) (1/ T2 - 1/T1)

2 = e^ (-Ea / R) (1 / T2 = 1 / T1)

Taking the natural logarithm of both sides:

ln 2 = - (Ea / R) (1 / T2 - 1 / T1)

Making Ea the subject of the formula, we obtain:

Ea = - (ln 2 R / (1 / T2- 1 / T1))

Let T1 = 25 C = 25 + 273 K = 298 K

T2 = 35 C = 35 + 273 K = 308 K

R = 8.314

So,

Ea = - (ln 2 * 8.314 / ( 1/308 - 1 / 298))

Ea = - (0.693 * 8.314 / 0.00324 - 0.00335)

Ea = - 5.7616 / -0.00011

Ea = 52 378,18 J / mol

So therefore, the activation energy Ea is 52.4 kJ/mol.

Answer:

Explanation:

use gas law eqation

P1 * V1  / T1 = P2 * V2 /T2

600*V1/227 = 300*800/23

V1 = 300*800*227 / 23*600 = ............ can you solve this and get the answer?

Answer:

Amorphous solids .

Explanation:

They have no particular order or pattern.Each particle is in a particular spot, but the particles are in no organized pattern.

Complete Question

The diagram for this question is shown on the second uploaded image

Answer:

The organic product obtained is  shown  on the first uploaded image

Explanation:

The process that lead to this product formation is known as oxidative cleavage   which is a reaction that involves the cleavage of a carbon to carbon bond at the same time this carbon which formed the carbon bond are oxidized i.e oxygen is been added to them