CHEMISTRY COLLEGE

Please ignore my answers because they are wrong.

Answers

Answer 1

Answer:

Answer:

The wavelength of the radio wave is 3.003 m.

The energy of the radio wave is $6.6194* 10^(-26) J$ .

Explanation:

Frequency of the radio waves, ν = 99.9 MHz = $99.9* 10^6 Hz$

Wavelength and frequency are related to each other by realtion:

$\lambda =(c)/(\nu )$

$\lambda$ = Wavelength of the wave

c = speed of the light

ν = Frequency of the wave

$\lambda =(3* 10^8 m/s)/(99.9* 10^6 s^(-1))=3.003 m$

The wavelength of the radio wave is 3.003 m.

The energy of the electromagnetic wave is given by Planck's equation:

$E=h* \nu$

h = Planck's constant = $6.626* 10^(-34) Js$

The energy of the radio wave with 99.9 MHz frequency will be:

$E=6.626* 10^(-34) Js* 99.9* 10^6 s^(-1)$

$E = 6.6194* 10^(-26) J$

The energy of the radio wave is $6.6194* 10^(-26) J$ .

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Part D 2ClO2(g)+2I−(aq)→2ClO−2(aq)+I2(s) Drag the appropriate labels to their respective targets. ResetHelp e−→e Superscript- rightarrow ←e−leftarrow e Superscript- Cathode Cathode Anode Anode II Superscript- ClO−2C l O Subscript 2 Superscript- Request Answer Part E Indicate the half-reaction occurring at Anode. Express your answer as a chemical equation. Identify all of the phases in your answer. nothing

Answers

The half reaction occurring at anode is:

$2I^-(aq)---- > I_2(s)+2e^-$

Half reaction for the cell:

The substance having highest positive potential will always get reduced and will undergo reduction reaction.

Balanced chemical equation:

$2ClO_2(g)+2I^-(aq)----- > 2ClO^(2-)(aq)+I_2(s)$

The half reaction follows:

Oxidation half reaction: $2I^-(aq)---- > I_2(s)+2e^-$ , Reduction potential is 0.53V

Reduction half reaction: $ClO_2(g)+e^----- > ClO_2^-$ ( × 2 ), Oxidation potential is +0.954 V

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

Hence, the half reaction occurring at anode is :

$2I^-(aq)---- > I_2(s)+2e^-$

Find more information about Reduction potential here:

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Answer: The half reaction occurring at anode is $2I^-(aq.)\rightarrow I_2(s)+2e^-$

Explanation:

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

For the given chemical equation:

$2ClO_2(g)+2I^-(aq)\rightarrow 2ClO^(-2)(aq.)+I_2(s)$

The half reaction follows:

Oxidation half reaction: $2I^-(aq.)\rightarrow I_2(s)+2e^-;E^o_(I_2/I^-)=0.53V$

Reduction half reaction: $ClO_2(g)+e^-\rightarrow ClO_2^-(aq.);E^o_(ClO_2/ClO_2^-)=+0.954V$ ( × 2 )

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

Hence, the half reaction occurring at anode is $2I^-(aq.)\rightarrow I_2(s)+2e^-$

Why is predicting our future oil supply controversial and involves some uncertainty?I just need some explaining.

Answers

Natural resources such as oil that are as yet undiscovered are very difficult to predict or estimate; this creates huge uncertainty and large errors in even the most rigorous scientific efforts to predict future supply. Changes in oil drilling and discovery technology can and will completely change the eventual results in the search for undiscovered resources. The controversy lies in the desire to eliminate oil from the world's energy mix in order to prevent a perceived but uncertain risk of global warming and the desire to encourage the use of alternative energy in order to reduce that risk, even though so far no alternative energy source other than natural gas, coal, and nuclear energy has been able to compete with oil economically.

The Problem with predicting things like our natural supply of oil can be picked out to a LOT of different variables. For starters, we find a lot of oil in a myriad of different places. The frequency in which we find this oil could theoretically be linked to a computer and predicted like Stocks. But then again, since the rate changes almost as much as the finding sites, You can easily be slapped aside by prehistoric biology and geology. The other problem with predicting our supply is often controversial; "How much do we use" well, you have to look at individual countries or the entire planet. And that alone is a lot of work. Imagine knocking on your neighbor's door and asking exactly how much hot water he uses.... Every day.... For a year! Pretty creepy right? Not to mention intrusive. But the internet has some of these things, So lets say you managed to find the frequency of oil findings globally, and the rate of use for these areas. Well now you have even more problems. Because there are many different people looking for oil. And when they find it. They sell it. Or they sue eachother over who found it first. Because underground supplies are huge at times. Often companies will be drilling the same one. To get an exact count from one company you would need to track all of the oil possessed by them and the buyers, which is paperwork. Which is Highly variable depending on who you are counting from. Without Tainting your next variable. "Current processed supply." ie. The stuff already out of the ground. If you have got all of that counted and punched into the smartest computer you can find. Then you still have some data to collect. Because the numbers are always changing. And everyone uses a different amount every day. This alone can stop any predictions cold for obvious. Reasons. So in conclusion, There literally is a large amount of Static variables and a few constant variables to consider when predicting future supply. This is why simply internet searching these things often gives numbers that are highly different from eachother. Being both controversial and uncertain.

How much heat is required to change the temperature of two cups of water (500 g) from room temperature (25◦C) to boiling? Specific heat of water is c=4.184 J/(g oC) a 78.5 kJ b 15.7 kJ c 157 kJ d 1.57 kJ

Answers

The heat that is required to change the temperature of two cups of water (500 g) from room temperature (25◦C) to boiling

C) 157 kJ

Heat

Heat required= Mass of water x specific heat capacity of water x change in temperature of water required

Q=m* c* delta T

M = 500g

C = 4.184 g°C

Delta T = 100 - 25(room temp) = 75°C

Heat = 500 x 4.184 x 75

Heat = 156900 J

Heat = 156.9 KJ

Heat ~ 157.0 KJ (3.D.P)

Thus, the correct answer is C.

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Answer:

Explanation:

Heat required= Mass of water x specific heat capacity of water x change in temperature of water required

$q \: = m * c * delta \: t$

M = 500g

C = 4.184 g°C

Delta T = 100 - 25(room temp) = 75°C

Heat needed= 500 x 4.184 x 75

= 156900 J

= 156.9 KJ

~ 157.0 KJ (3.D.P)

To measure the solubility product of lead (II) iodide (PbI2) at 25°C, you constructed a galvanic cell that is similar to what you used in the lab. The cell contains a 0.5 M solution of a lead (II) nitrate in one compartment that connects by a salt bridge to a 1.0 M solution of potassium iodide saturated with PbI2 in the other compartment. Then you inserted two lead electrodes into each half-cell compartment and closed the circuit with wires. What is the expected voltage generated by this concentration cell? Ksp for PbI2 is 1.4 x 10-8. Show all calculations for a credit.

Answers

Answer:

0.2320V

Explanation:

Voltage can be defined as the amount of potential energy available (work to be done) per unit charge, to move charges through a conductor.

Voltage can be generated by means other than rubbing certain types of materials against each other.

Please look at attached file for solution to the problem.

Final answer:

The expected voltage generated by this concentration cell is approximately 0.113 V.

Explanation:

To calculate the voltage generated by the concentration cell, we can use the Nernst equation. The Nernst equation relates the concentration of the ions in the two compartments to the voltage of the cell. The equation is:

E = E° - (RT/nF) ln(Q)

Where:

E is the voltage of the cell
E° is the standard cell potential
R is the gas constant (8.314 J/mol·K)
T is the temperature in Kelvin (25 + 273 = 298 K)
n is the number of moles of electrons transferred (2 in this case)
F is Faraday's constant (96,485 C/mol)
ln(Q) is the natural logarithm of the reaction quotient

The reaction quotient (Q) can be calculated using the concentrations of the lead (II) and iodide ions in each compartment.

Since this is a concentration cell, the standard cell potential (E°) for this system is 0 V. Therefore, the equation simplifies to:

E = - (RT/nF) ln(Q)

Now we can calculate the voltage:

Calculate Q:

The solubility product constant (Ksp) for PbI2 is 1.4 x 10-8. Because PbI2 is in a saturated solution, the concentration of Pb2+ ions and I- ions are both equal to the solubility of PbI2. We can substitute these values into the equation to calculate Q:

Q = [Pb²+] x [I-]

Q = (1.4 x 10-8) x (1.4 x 10-8) = 1.96 x 10-16

Calculate E:

Now we can calculate the voltage using the given values:

For the Nernst equation, we need to convert the temperature to Kelvin:

T = 25°C + 273 = 298 K

Substitute the values into the equation:

E = - (8.314 J/mol·K x 298 K / 2 x 96,485 C/mol) ln(1.96 x 10-16)

E ≈ 0.113 V

Therefore, the expected voltage generated by this concentration cell is approximately 0.113 V.

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Consider the combination reaction of samarium metal and oxygen gas. If you start with 33.7 moles of samarium metal, how many moles of oxygen gas would be required to react completely with all of the samarium metal? For this reaction, samarium has a +3 oxidation state within the samarium/oxygen compound.

Answers

Answer:

25.275 moles of oxygen gas will be required to completely react with all the samarium metal.

Explanation:

$4Sm+3O_2\rightarrow 2Sm_2O_3$

Number of moles samarium metal = 33.7 moles

According to reaction, 4 moles of samarium reacts with 3 moles of oxygen gas.

Then 33.7 moles of samarium will react with:

$(3)/(4)* 33.7 mol=25.275 mol$ of oxygen gas.

25.275 moles of oxygen gas will be required to completely react with all the samarium metal.

Answer:

Moles of oxygen gas required to react completely with 33.7 moles of samarium metal is $\fbox{25.3 \text{ mol}}$ .

Explanation:

A chemical equation in which the number of atoms of each element is the same on the reactant and product side is called a balanced chemical equation.

The balanced chemical equation can be used to determine the stoichiometric ratio between the reactant and the product. The stoichiometric ratio thus enables us to calculate:

1. Amount of one reactant required to react completely with the other reactant.

2. Amount of the product that can be produced from the given amount of the reactant.

Step 1: Write the chemical equation for the reaction between samarium metal and oxygen gas.

The chemical formula for oxygen gas is $\text{O}_(2)$ .

Samarium has +3 oxidation state within the samarium/oxygen compound. So, the chemical formula of the samarium oxygen compound is $\text{Sm}_(2)\text{O}_(3)$ .

The chemical equation is as follows:

$\fbox{\text{Sm}+\text{O}_(2) \rightarrow \text{Sm}_(2)\text{O}_(3)\n\end{minipage}}$

Step 2: Balance the chemical equation for the reaction between oxygen and samarium metal.

The number of oxygen atoms in the product side is 3 and in the reactant side is 2. Put coefficient 2 in front of $\text{Sm}_(2)\text{O}_(3)$ and 3 in front of $\text{O}_(2)$ to balance the oxygen atoms.

$\fbox{\text{Sm}+3\text{O}_(2) \rightarrow 2 \text{Sm}_(2)\text{O}_(3)\n\end{minipage}}$

The number of samarium atoms in the product side is 4 and in the reactant side is 1. Put coefficient 4 in front of Sm in the reactant side.

$\fbox{\n4\text{Sm}+3\text{O}_(2) \rightarrow 2 \text{Sm}_(2)\text{O}_(3)\n\end{minipage}}$

Step 3: Determine the stoichiometric ratio between samarium and oxygen from the above balanced chemical equation.

According to the balanced chemical equation, we can see that the stoichiometric ratio between samarium and oxygen is 4:3.

Step 4: Use unitary method and calculate the moles of oxygen required to completely react with the given moles of samarium metal as follows:

$\text{moles of O}_(2) = \left( \text{moles of Sm} \right)\left( \frac{3 \text{ mol of O}_(2)}{4 \text{ mol of Sm}} \right)$ ...... (1)

Step 5: Substitute 33.7 mol for moles of Sm in equation (1).

$\text{Moles of O}_(2) = \left( \text{33.7 mol} \right) \left( \frac{3 \text{ mol of O}_(2)}{4 \text{ mol of Sm}} \right)\n\text{Moles of O}_(2)= 25.275 \text{ mol}\n\text{Moles of O}_(2)= 25.3 \text{ mol}$

Note:

Do not forgot to balance the reaction. The reaction must be balanced in order to calculate the amount (mol) of oxygen required to completely react with the given amount of samarium.

Learn more:

1. Balanced chemical equation brainly.com/question/1405182

2. Learn more about how to calculate moles of the base in given volume brainly.com/question/4283309

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Some basic concept of chemistry

Keywords: samarium, oxygen gas, samarium/oxygen compound, 33.7 moles, 25.3 mol, balanced equation, stoichiometric ratio, coefficient, balance, moles, completely react.

What is the empirical formula of a compound composed of 36.7 g 36.7 g of potassium ( K K ) and 7.51 g 7.51 g of oxygen ( O O )? Insert subscripts as needed.

Answers

Answer:

K₂O

Explanation:

Given parameters:

Mass of K = 36.7g

Mass of O = 7.51g

Unknown:

Empirical formula of the compound

Solution:

The empirical formula of a compound is it's simplest ratio by which the elements in the compound combines. It differs from the molecular formula that shows the actual atomic ratios.

To find the empirical formula, follow this process;

Elements K O

Mass 36.7 7.51

Molar

mass 39 16

Number of

moles 36.7/39 7.51/16

0.94 0.47

Divide by

the smallest 0.94/0.47 0.47/0.47

2 1

Empirical formula is K₂O

Final answer:

The empirical formula of the compound composed of 36.7 g of potassium and 7.51 g of oxygen is K2O.

Explanation:

To determine the empirical formula of a compound, we need to find the ratio of the elements present. In this case, we have 36.7 g of potassium and 7.51 g of oxygen. To find the ratio, we need to convert these masses to moles by dividing them by the molar masses of potassium and oxygen. The molar mass of potassium is 39.10 g/mol and the molar mass of oxygen is 16.00 g/mol. Dividing the masses by the molar masses gives us 0.939 mol potassium and 0.469 mol oxygen. The ratio between these two elements is approximately 2:1, so the empirical formula of the compound is K2O.

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