Balancing PbO2 + ClO → PbO + ClO3

Answer:

When a body covers equal distances in a straight line, equal intervals of time are called uniform motion

When a body covers unequal distance in equal intervals of time in a straight line is called non-uniform

Answer:

Uniform motion- is motion in which the

velocity does not change.

Non-uniform motion-occurs when the velocity of an object is not

constant : the object speeds up or slows down during its motion, or changes direction.

Hope it will help you !!

Answer:

They make a rice Krispe gramcracker bar baby

Explanation:

Answer: When they meet they instantly fall in love, (lol) and have a baby.... They name the baby S'more because it has features from the graham cracker and the rice krispie.

Stay safe and have a Merry Christmas!!!!!!!! :D  

Maybe a S'more rice krispie.....

Explanation:

1. Measure your waste

It is easier to manage and track your improvement on something that is measured. Start by recording the types and quantities of waste materials generated within your business.

This could just be a quick visual assessment and estimation of amount and types of waste per week. As you change how you manage your waste, you can compare and see what difference your changes make to the amount of waste disposed to landfill.

2. Reduce

To reduce the amount of waste going to landfill, consider:

can waste be avoided or reduced by the way you procure goods and services or by changing your business operations?

does another local business have a use for the waste materials produced by your business?

what materials can be targeted for recycling?

3. Recycle

Find out what options are available locally for business recycling.

4. Collection services

It is a good idea to find out how levy fees will be managed by your waste service provider, and find out more about your waste collection services—how it impacts your ability to recycle and what options are available to you.

5. Separating waste

It’s important to ensure that waste is correctly separated for recycling. Putting the wrong waste in contaminates the recycling. Check out the free-to-download recycling signage to help businesses communicate correct recycling and waste separation behaviours.

Answer:

2.088464 grams.

Explanation:

The density of neon is 0.9002 g/L. Multiplying this value by the number of liters gives how many grams of neon it contains.

0.9002(2.32) = 2.088464

Answer:

-water is collected from nearby water holes and put in a water tower after it's clean

Answer:

Your water comes from rivers and streams.

Explanation:

The water is collected from rivers and stream and gets delivered from the storage tank to the pipes of our house

To explain the changes that the balloon underwent using the ideal gas law, we need to consider the variables of the ideal gas law: pressure (P), volume (V), temperature (T), and the number of moles of gas (n).

Here is a table to identify the variable that was changed and the primary effect it had:

| Change | Variable | Effect |

|--------|----------|--------|

| Change 1 | Volume (V) | Increase |

| Change 2 | Temperature (T) | Increase |

| Change 3 | Pressure (P) | Decrease |

| Change 4 | Number of moles (n) | Increase |

1. Change 1: Increase in Volume (V)

  - Effect: The balloon expanded.

  - Explanation: Increasing the volume of the balloon allows more space for the gas molecules, causing the balloon to expand.

2. Change 2: Increase in Temperature (T)

  - Effect: The balloon expanded further.

  - Explanation: Increasing the temperature of the gas increases the kinetic energy of the gas molecules, causing them to move faster and exert greater pressure on the balloon, leading to further expansion.

3. Change 3: Decrease in Pressure (P)

  - Effect: The balloon deflated.

  - Explanation: Decreasing the pressure on the balloon reduces the force exerted by the gas molecules on the balloon, causing it to deflate.

4. Change 4: Increase in Number of moles (n)

  - Effect: The balloon expanded again.

  - Explanation: Increasing the number of moles of gas increases the amount of gas inside the balloon, leading to further expansion.

Overall, these changes demonstrate the relationship between the variables of the ideal gas law and the corresponding effects on the balloon's volume, pressure, and temperature.

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Answer:

1) 3Fe + 2O(2, this is the subscript) - Fe(3)O(4)

2) 2Sr + O(2) - 2SrO

3) Sn + 2NaOH - Na(2)SnO(2) + H(2)

4) 2K + Br(2) -2KBr

5) 2C(8)H(18) + 25O(2) - 16CO(2) + 18H(2)O

The CEC for this soil sample is 675.2 cmol(+)/kg.

The % Base Saturation for this soil sample is approximately 136.62%.

The % Acid Saturation for this soil sample is approximately 60.55%.

To    calculate the CEC, % Base Saturation (B.S.), and % Acid Saturation for the given soil sample:

a. Calculation of CEC (Cation Exchange Capacity):

CEC is the sum of exchangeable cations in the soil. From the given results, we have:

CEC = Ca2+ + Mg2+ + K+ + Al3+

CEC = (100 mg/L + 45 mg/L + 85.5 mg/L + 94.2 mg/L) / (25 g / 1000)

CEC = 168.7 mg / (25 g / 1000)

CEC = 675.2 cmol(+)/kg

b. Calculation of % Base Saturation (B.S.):

% B.S. represents the percentage of CEC occupied by base cations. In this case, we consider Ca2+, Mg2+, and K+ as base cations. The formula to calculate % B.S. is:

% B.S. = (Ca2+ + Mg2+ + K+) / CEC * 100

% B.S. = (100 mg/L + 45 mg/L + 85.5 mg/L) / (168.7 cmol(+)/kg) * 100

% B.S. = 230.5 mg / (168.7 cmol(+)/kg) * 100

% B.S. = 136.62%

c. Calculation of % Acid Saturation:

% Acid Saturation represents the percentage of CEC occupied by acid cations, in this case, H+ and Al3+. The formula to calculate % Acid Saturation is:

% Acid Saturation = (H+ + Al3+) / CEC * 100

% Acid Saturation = (8.0 mg/L + 94.2 mg/L) / (168.7 cmol(+)/kg) * 100

% Acid Saturation = 102.2 mg / (168.7 cmol(+)/kg) * 100

% Acid Saturation = 60.55%

Please note that the given values were in milligrams per liter (mg/L), and the CEC and % Saturation values were calculated assuming a conversion from mg/L to cmol(+)/kg using the mass of the soil sample (25 g).

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Answer:

In both natural and built systems, stability and change are an important focus of study for both scientists and engineers. Stability refers to a system that is unchanging. ... A dynamic equilibrium exists when chemical reactions or physical movements occur at rates that balance out, creating no net change in a system.

Answer:

In both natural and built systems, stability and change are an important focus of study for both scientists and engineers. Stability refers to a system that is unchanging. ... A dynamic equilibrium exists when chemical reactions or physical movements occur at rates that balance out, creating no net change in a system.

Explanation:

The molecular weight of cerium(IV) nitrate hexahydrate is 446.24 g/mol. Therefore, one mole of cerium(IV) nitrate hexahydrate contains one mole of cerium(IV) ions, which will combine with four moles of nitrate ions to form one mole of cerium(IV) nitrate hexahydrate.

The formula for the concentration of ions in a solution is C = n/V where C is the concentration of ions, n is the number of moles of ions, and V is the volume of the solution in liters. The first step in solving this problem is to calculate the number of moles of cerium(IV) nitrate hexahydrate in 150.0 mL of a 0.565 M solution. This can be done using the following formula:n = M x V n = 0.565 mol/L x 0.150 L= 0.08475 mol of cerium(IV) nitrate hexahydrate This amount contains four times as many moles of nitrate ions as cerium(IV) ions.

Therefore, the number of moles of nitrate ions is: nitrate ions = 4 x 0.08475 militate ions = 0.339 molThe volume of the solution is 150.0 mL, which is equal to 0.150 L. Using the formula given above, we can calculate the concentration of nitrate ions :C = n/V= 0.339 mol/0.150 LC = 2.26 M Therefore, the concentration of nitrate ions in the solution is 2.26 M.

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Answer:

b. false

Eugenol is typically extracted from cloves using methods such as steam distillation or solvent extraction, but not with CO2 extraction. CO2 extraction is a technique commonly used to extract essential oils from various plant materials, but it is not the preferred method for isolating eugenol from cloves.

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The total potential of the given system after 120 mL of titrant is added cannot be determined using the given information.

The given reaction is:Fe²⁺ + Ce⁴⁺ → Ce³⁺ + Fe³⁺

The balanced equation is:

2 Fe²⁺ + Ce⁴⁺ + 2 H₂O → 2 Fe³⁺ + Ce³⁺ + 4 H⁺

The number of moles of Fe²⁺ present in 150 mL of 0.025 M Fe²⁺ in 1 M HClO₄ solution is:

0.025 mol L⁻¹ × 150 × 10⁻³ L = 0.00375 mol

The number of moles of Ce⁴⁺ present in 120 mL of 0.15 M Ce⁴⁺ in 1 M HClO₄ solution is:

0.15 mol L⁻¹ × 120 × 10⁻³ L = 0.018 mol

As per the balanced equation, the stoichiometry of Fe²⁺ and Ce⁴⁺ is 2:1 i.e. 2 moles of Fe²⁺ will react with 1 mole of Ce⁴⁺. Therefore, the number of moles of Fe²⁺ that reacts with 0.018 moles of Ce⁴⁺ is:0.018 mol Ce⁴⁺ × 2/1 = 0.036 mol Fe²⁺The number of moles of Fe²⁺ remaining unreacted is:

0.00375 mol - 0.036 mol = -0.03225 mol

The negative sign indicates that Fe²⁺ is present in excess. Thus, after 120 mL of titrant is added, the potential of the given system will be governed by the concentration of Fe²⁺ i.e. the potential will depend upon the half-cell reaction occurring at the anode. Hence, the total potential of the given system after 120 mL of titrant is added cannot be determined using the given information.

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A 1200 mL IV drip is labelled to contain 12% of a laxative. The number of mL of the IV drip that should be administered if a patient is to receive 21 mL of the laxative is 175 mL.

To find the number of mL of the IV drip that should be administered to provide 21 mL of the laxative, we can set up a proportion using the concentration of the laxative.

Let x represent the number of mL of the IV drip to be administered.

The concentration of the laxative is given as 12% or 0.12, meaning that in 100 mL of the IV drip, there are 12 mL of the laxative.

Using the proportion:

(12 mL of laxative) / (100 mL of IV drip) = (21 mL of laxative) / (x mL of IV drip)

Cross-multiplying, we get:

12 mL * x mL of IV drip = 100 mL * 21 mL of laxative

12x = 210 IV drip0

Dividing both sides by 12:

x = 2100 / 12

x = 175

Therefore, 175 mL of the IV drip should be administered to provide 21 mL of the laxative.

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The coefficient for OH- (aq) in the balanced equation in basic aqueous solution is 12.

To balance this equation in basic aqueous solution, we first balance the atoms that are not hydrogen or oxygen. We start by balancing the Fe atoms on both sides, which requires multiplying Fe2+ on the reactant side by 3 to get 3Fe2+. Next, we balance the Mn atoms on both sides, which requires multiplying MnO4- on the reactant side by 2 to get 2MnO4-.
The balanced equation in basic solution is:
2MnO4- + 6Fe2+ + 8OH- → 2Mn2+ + 6Fe3+ + 4H2O
To find the coefficient for OH- (aq), we look at the number of OH- ions on both sides of the equation. On the reactant side, there are 8 OH- ions. On the product side, there are 4 H2O molecules, each of which contains 2 H+ ions and 1 OH- ion, so there are a total of 8 H+ ions and 4 OH- ions.
To balance the OH- ions, we add 4 OH- ions to the reactant side to get a total of 12 OH- ions, and the balanced equation in basic solution is:
2MnO4- + 6Fe2+ + 12OH- → 2Mn2+ + 6Fe3+ + 4H2O
Therefore, the coefficient for OH- (aq) in the balanced equation in basic aqueous solution is 12.

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Answer:

Powerade has the most dissolved ions, followed by Gatorade and lastly vitamin water

Explanation:

How conductive a certain solution is, is based on the kind of ions that are present or dissolved in the solution. Hence, we have it that, the higher the concentration of the ions that are present in a solution, the better it will conduct.

There is thus, a direct relationship between what is dissolved or the kind of present ions and how conductive the solution is.

A solution with more dissolved ions is a better conductor of electricity and thus has a conductivity value that is higher

Thus, we can conclude that Powerade has more dissolved ions, Gatorade has less and Vitamin water has the least

Answer: This is because the atomic radius increases with increasing electronic energy. This reduces the attraction for the valence electrons of other atoms, reducing reactivity.

To determine the vapor pressure of a solution at 25°C containing 76.6 g of glucose, we need to know additional information such as the solvent used and the concentration of the glucose solution. The vapor pressure of a solution is influenced by the solvent properties and the concentration of the solute.

Assuming we have a glucose solution in water, we can use the concept of molality to calculate the vapor pressure. Molality is the number of moles of solute per kilogram of solvent.

Here are the steps to calculate the vapor pressure of the glucose solution:

1. Calculate the number of moles of glucose:

To do this, divide the mass of glucose by its molar mass. The molar mass of glucose (C6H12O6) is approximately 180.16 g/mol.

Number of moles of glucose = mass of glucose / molar mass of glucose

Number of moles of glucose = 76.6 g / 180.16 g/mol

2. Determine the molality of the solution:

Molality (m) is the number of moles of solute per kilogram of solvent. In this case, the solvent is water.

Molality = moles of solute / mass of solvent (in kg)

Since we don't have the mass of the water, we cannot calculate the molality or the vapor pressure without that information.

To determine the vapor pressure of the solution, we need to know the concentration of the solution or have the mass of the solvent (water) available.

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The estimation of the acidic and the basic character of a solution is called and defined by the pH. The \(\rm [H^{+}]\) concentration of the solution will be \(\rm 1.77\times 10^{-4}.\)

pH is a measurement of the free hydrogen and hydroxyl ions in the solution of the substance.

It can be calculated as:

\(\rm pH = - \rm log[H^{+}]\)

Given,

pH = 3.75

Substituting values in the equation:

\(\begin{aligned}\rm 3.75 &= -\rm log[H^{+}]\\\\\rm [H^{+}] &= 10^{-3.75}\\\\&= 1.7782 \times 10^{-4}\end{aligned}\)

Therefore,  \(\rm 1.77\times 10^{-4}\) is the concentration of the hydrogen ions.

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Answer:

1.78 x 10^nM

n= -4

Explanation:

the answer after that is A

Answer:

Molecule is the general term used to describe any atoms that are connected by chemical bonds. Every combination of atoms is a molecule. A compound is a molecule made of atoms from different elements. All compounds are molecules, but not all molecules are compounds.

Explanation:

Answer:

Atoms are the thing that make up molecules and compounds. Molecules contain two or more atoms and are held together by covalent bonds, whereas compounds are held together by ionic bonds.

The atoms within a compound also must be different from eachother, whereas a molecule can consist of only one element.

Given;At t = 0 seconds, the concentration of S2O8^-2 is 0.010 MThe half-life for the decomposition of S2O8^-2 is 75 seconds. The concentration of S2O8^-2 remaining at 1600 s is___3.635 x 10^-7 ___ M.

To solve this, we need to use the first-order reaction equation, which is:k = 0.693/t1/2Wherek is the rate constantt1/2 is the half-life of the reaction from the equation, the rate constant (k) can be calculated by;k = 0.693/t1/2 = 0.693/75 = 0.00924s^-1Let's find the concentration of S2O8^-2 after 1600 s using the first-order rate law equation. The equation is;ln([S2O8^-2]t/[S2O8^-2]0) = -ktWhere[S2O8^-2]t is the concentration at the remaining time[S2O8^-2]0 is the initial concentration oft = the remaining timek = the rate constantln = natural logSubstituting the given values in the equation;ln([S2O8^-2]t/0.010) = -0.00924 x 1600ln([S2O8^-2]t) = -14.784[S2O8^-2]t = e^-14.784[S2O8^-2]t = 3.635 x 10^-7 MConsequently, the concentration of S2O8^-2 remaining at 1600 s is 3.635 x 10^-7 M.

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Orbits do the proton complete:

A proton exits the cyclotron 1. 0 ms after starting its spiral trajectory in the center of the cyclotron and 1.8 x \(10^{4}\) orbits does the proton complete during this 1. 0 ms

Calculation:

(a) The kinetic energy of the proton is given by, K.E. = \(\frac{1}{2} m v^{2}\)

v = \(\sqrt{\frac{2(K E)}{m}}\)

v = \(\sqrt{\frac{2(6.5 \mathrm{MeV})\left(1.6 \times 10^{-19} \mathrm{~J} / \mathrm{eV}\right)}{1.672 \times 10^{-27} \mathrm{~kg}}}\)= 3.5 \(\times 10^{7} \mathrm{~m} / \mathrm{s}\)

The radius of the orbit is given by,

r = \(\frac{m v}{B q}=\frac{1.672 \times 10^{-27} \mathrm{~kg}\left(3.5 \times 10^{7} \mathrm{~m} / \mathrm{s}\right)}{(1.2 \mathrm{~T})\left(1.6 \times 10^{-19} \mathrm{C}\right)}\) = 0.30 m

Hence, the diameter of the largest orbit just before the proton leaves the orbit is 0.60 m.

(b) The time required to complete one revolution is given by, T=\(\frac{2 \pi m}{B q}=\frac{2 \pi\left(1.672 \times 10^{-27} \mathrm{~kg}\right)}{(1.2 \mathrm{~T})\left(1.6 \times 10^{-19} \mathrm{C}\right)}\) =5.48 x \(10^{-8}\) s

The number of revolutions made by the proton is given by,

Tn = t

Therefore, n = t/T = \(\frac{1 \mathrm{~ms}}{5.48 \times 10^{-8} \mathrm{~s}}\) = 1.8 x \(10^{4}\)

Note: your question is incomplete, but most probably your full question was, A medical cyclotron used in the production of medical isotopes

accelerates protons to 6.5 MeV. The magnetic field in the

cyclotron is 1.2 T.

a. What is the diameter of the largest orbit, just before the protons

exit the cyclotron?

b. A proton exits the cyclotron 1.0 ms after starting its spiral

trajectory in the center of the cyclotron. How many orbits

does the proton complete during this 1.0 ms?

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Approximately 172.222 mL of a 0.250 M NaOH solution would be required to neutralize 155 mL of a 0.225 M H3PO4 solution.

To determine the volume of a 0.250 M NaOH solution required to neutralize 155 mL of a 0.225 M H3PO4 solution, we need to use the concept of stoichiometry and the balanced chemical equation between NaOH and H3PO4.

The balanced equation between NaOH and H3PO4 is:

3NaOH + H3PO4 -> Na3PO4 + 3H2O

From the balanced equation, we can see that it takes three moles of NaOH to neutralize one mole of H3PO4. This means that the mole ratio between NaOH and H3PO4 is 3:1.

To calculate the volume of NaOH solution needed, we need to use the molarity and volume of the H3PO4 solution and set up a proportion based on the stoichiometry:

0.225 moles H3PO4 / 155 mL H3PO4 = 0.250 moles NaOH / x mL NaOH

Cross-multiplying and solving for x:

x = (155 mL H3PO4 * 0.250 moles NaOH) / 0.225 moles H3PO4

x = 172.222 mL NaOH

Using the stoichiometry and the balanced equation between NaOH and H3PO4, we find that approximately 172.222 mL of a 0.250 M NaOH solution is needed to neutralize 155 mL of a 0.225 M H3PO4 solution. The calculation involves setting up a proportion based on the mole ratio between NaOH and H3PO4.

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Answer:

The answer is Observation.

For you to observe the things around you then you need to use all your sense organs....

Answer: making an observation.

The periodic table is an arrangement of the elements in order of increasing reactivity. The elements that are more reactive are found at the left hand side of the table among the metals. The reactive non metals are found at the right hand side of the table.

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\(CrO_{4} ^{2-} (aq) + 3H_{2} O (l)\)→ \(Cr(OH)_{3} (s) + 4OH^{-} (aq)\) is the chemical equation for the reaction between chromate ion and water.

The decent synthetic condition for the response between chromate particle (\(CrO_{4} ^{2-}\)) and water to deliver chromium (III) hydroxide (\(Cr(OH)_{3}\)) is as per the following:

\(CrO_{4} ^{2-} (aq) + 3H_{2} O (l)\) → \(Cr(OH)_{3} (s) + 4OH^{-} (aq)\)

In this situation, the chromate particle (\(CrO_{4} ^{2-}\)) responds with water (\(H_{2} O\)) to create chromium (III) hydroxide (\(Cr(OH)_{3}\)) and hydroxide particles (Goodness ). This response is an illustration of a precipitation response, where an insoluble strong is framed when two watery arrangements are combined as one.

The chromium (III) hydroxide (\(Cr(OH)_{3}\)) framed in this response is a green strong, which can be utilized as a color or in the assembling of different synthetic compounds. The hydroxide particles (Goodness ) created in the response can likewise respond with different particles to frame different mixtures or take part in corrosive base responses.

Generally, the reasonable compound condition for the response between chromate particle and water to deliver chromium (III) hydroxide is \(CrO_{4} ^{2-} (aq) + 3H_{2} O (l)\)→ \(Cr(OH)_{3} (s) + 4OH^{-} (aq)\).

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The molar mass of the compound is 128.4 g/mol.

The empirical formula is an empirical formula that represents the lowest whole-number ratio of the atoms present in a compound. The empirical formula for the molecular compound is calculated using the percentage composition of the elements present in the compound. The steps used to find the empirical formula are as follows:

Find the mass of each element present in the compound.Convert each mass to moles.Divide each mole value by the smallest number of moles.Round to the nearest whole number and write the subscripts.The molar mass is the mass of a substance that contains 6.02 × 10²³ atoms or molecules. To calculate the molar mass of a compound, add the masses of all the atoms present in the compound.

C=0.707g,12.01 g/mol=0.0588 molCnH=0.2372 =1.01g/m=0.235 mol H

nH=4nC

The empirical formula of the compound is CH4. The molar mass of the compound can be calculated using the empirical formula.

M=12.01 g/mol+4(1.01 g/mol)=16.05 g/mol

The molar mass of the compound is 8 times the empirical mass, so the actual molar mass is;

M=8(16.05{g/mol})=128.4g/mol. The molar mass of the compound is 128.4 g/mol.

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The standard reduction potential for the half-reaction of Be^2+ + 2e^- -> Be is given as E^0 = 3.83 V.

For the half-cell Hg^2+ | Hg, the standard reduction potential is not provided in the given information. To calculate the electric potential for the voltaic cell, we need the reduction potential for the Hg^2+ | Hg half-cell.

A voltaic cell, also known as a galvanic cell or an electrochemical cell, is an electrochemical device that generates electrical energy through a spontaneous chemical reaction. It consists of two half-cells connected by an external circuit and a salt bridge or porous barrier that allows the flow of ions between the two half-cells.

Each half-cell consists of an electrode immersed in an electrolyte solution. The electrode is typically made of a metal or a conductive material, and the electrolyte is a solution containing ions that can participate in the redox (reduction-oxidation) reaction.

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