how many half-lives did a radioactive sample complete if 2.5 g of a sample remains from 20 g of initial material?

Answer: .00829 atm

Explanation:

1 atm is = 101325 Pa

Take 840 Pa and divide by 101325

840 Pa = .00829 atm

Answer:

4.2 gigagrams

Explanation:

kilo = 1,000

giga = 1,000,000,000

milli = .001

Physical chemistry a  major branch of chemistry would be most concerned with studying a series of chemical reactions in order to measure the amount of heat being released in each one.

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The expression relating the concentration of a gas in \(\mu g/m^-^3 (X)\) with the concentration expressed in ppm (Y) is \(Y = (X/MW) * 10^6 / V\)

To derive an expression relating the concentration of a gas in \(\mu g/m^3 (X)\) with the concentration expressed in ppm (Y), we can use the ideal gas law. The ideal gas law states that \(PV = nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. Rearranging the equation, we have \(n = PV/RT\)

To convert the concentration from \(\mu g/m^-^3 (X)\) to moles (n), we divide X by the molecular weight (MW) of the gas. Thus, \(n = X/MW\)

Combining the two equations, we have \(X/MW = PV/RT\)

Since the concentration expressed in ppm (Y) is the same as the number of moles per million parts of air, we can write \(Y = n * 10^6 / V\)

Substituting \(n = X/MW\), we get \(Y = (X/MW) * 10^6 / V\)

Therefore, the expression relating the concentration of a gas in \(\mu g/m^3 (X)\) with the concentration expressed in ppm (Y) is:

\(Y = (X/MW) * 10^6 / V\)

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Complete question is:

Using the ideal Gas Law, derive an expression relating the concentration of a gas \(\mu_g^-^3 (X)\) with the concentration expressed in ppm (Y). (10 pts) Hint:- Ideal gas law \(PV= nRt -R = 0.0821 atm mol^-^1\)  −MW=X= molecular weight

Answer:

Hydrogen

Explanation:

An example of this can be In this example, hydrogen is the limiting reagent and oxygen is the excess reagent. The amount of product formed is limited by the amount of hydrogen.

Answer:

The answer is 32. 29

Explanation:

Answer:

Two design criteria for the design of the ammonia making process are;

1) Sustainable hydrogen production by renewable energy

2)  Sustainable use of the ammonia-water thermal absorption cycle to retrieve the ammonia present in the waster water

Explanation:

Ammonia is produced from natural gas in the following process;

H₂ + RSH → RH + H₂S

H₂S + ZnO → ZnS + H₂O (Waste water)

CH₄ + H₂O → CO + 3H₂

CO + H₂O → CO₂ + H₂

Ammonia is then produced by reacting the produced hydrogen with nitrogen in the presence of a catalyst in a process known as the Haber-Bosch process as follows;

3H₂ + N₂ → 2NH₃

The

The design criteria for the design of the ammonia making process therefore as follows;

1) Sustainable use of renewable energy to produce hydrogen gas from the electrolysis of water

2)  Use of the ammonia-water thermal absorption cycle to retrieve the ammonia present in the waster water.

The branch of science which deals with chemicals and their bond is called chemistry.

The two ways by which we can design ammonia making process is as follows:-

Two design criteria for the design of the ammonia making process are:-

The usage of ammonia is important in these ways:-

Hence, these are the design for the ammonia-making process.

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When the amount of reactant A is doubled, the rate of reaction is also doubled. When the amount of reactant B is halved, the rate of reaction reduce 1/4 times. When reactant C is tripled, the rate become 9 times.

a) Reactant A undergoes a first order reaction. The equation for the rate of the reaction is

     R = k[A]¹

When the concentration is doubled,

   R' = k[2A]¹ = 2k[A] = 2R

So the rate of reaction is doubled.

b) The reactant B follows second order kinetics.

  Rate , R = k[B]²

When [B] is halved it becomes [B/2]

R' = k[\(\frac{B}{2}\)]² = \(\frac{1}{4}\) k[B]² = \(\frac{R}{4}\)

So the rate is decreased 1/4 times.

c) The reactant C follows second order kinetics.

   So,  R = k[C]²

When the concentration is tripled,

  R' = k [3C]² = 9k [C]² = 9R

So the rate increases nine folds.

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Answer:

13.50 moles of AgNO₃ are needed to produce 6.75 moles of Cu(NO₃)₂

Explanation:

We state the reactants:

AgNO₃

Cu

Then, the products are:

Cu(NO₃)₂

Ag

The reaction is:  2AgNO₃ + Cu  → 2Ag + Cu(NO₃)₂

Ratio is 1:2.

The copper is in excess, so the limiting reactant is the silver nitrate:

1 mol of Cu(NO₃)₂ is produced by 2 moles of AgNO₃

Then, 6.75 moles of copper (II) nitrate would be produced by, the double of moles

(6 75  . 2) /1  = 13.50 moles

The enthalpy of reaction for the decomposition of calcium carbonate is 177.4  KJ/mole .

We know that the enthalpy of reaction can be obtained from the enthalpy of formation of the reactants and that of the products using the formula;

ΔHrxn = ΔHf (products) - ΔHf (reactants)

The heat of formation of the species are as follows;

ΔHf (CaCO3(s)) = −1206.9 KJ/mol

ΔHf (CaO(s)) =  - 636 KJ/mole

ΔHf (CO2(g)) = -393.5  KJ/mole

ΔHrxn = [( -393.5  KJ/mole ) + ( - 636 KJ/mole)] - [ −1206.9 KJ/mol]

ΔHrxn = 177.4  KJ/mole

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Answer:

First one is 656.3 and second is endothermic.

Explanation:

Edge Thermochemical Equations assignment

Decolorization is an important step in the Gram staining process that removes the crystal violet stain from the Gram-negative bacteria, leaving them colorless. If the decolorization step is prolonged, it can result in the removal of the crystal violet stain from the Gram-positive bacteria as well, leading to a false Gram-negative classification.

During the final step of the Gram staining procedure, Gram-positive bacteria retain the crystal violet stain and appear purple, while Gram-negative bacteria do not retain the crystal violet stain and appear red/pink after counterstaining with safranin. Therefore, if the decolorization step is prolonged, the Gram-positive bacteria may appear red/pink in the final step, leading to inaccurate results and misinterpretation of the bacterial morphology and classification.

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Answer:0

Explanation: zero because it is the most stable form of oxygen in its standard state

The "block flow" of the basic steps of converting natural gas steam to syngas is described below: 1. Pre-treatment, 2. Reforming, 3. Cooling, 4. Cleaning.

The "block flow" of the basic steps of converting natural gas steam to syngas is described below:

The conversion of natural gas steam to syngas is divided into four stages, each of which has a different purpose. Here are the steps in detail.

1. Pre-treatment: The gas is first pre-treated to eliminate all impurities such as sulfur. The pre-treatment process helps to improve the quality of the syngas produced.

2. Reforming: The reforming process occurs at high temperatures (above 800°C) and in the presence of a catalyst. The goal is to convert natural gas into syngas, which contains carbon monoxide, hydrogen, and other components.

3. Cooling: Once the syngas has been created, it is necessary to cool it to a lower temperature. The gas will be condensed and cooled by a heat exchanger.

4. Cleaning: To remove any remaining impurities from the syngas, the final step is purification. The syngas is purified in a scrubber or filter tower using a process known as pressure swing adsorption (PSA). After this stage, the syngas is ready to be used or sent to the downstream processing plant.

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Based on the recipe for preparing the food, the limiting ingredient is the 2 cans of tomatoes.

The recipe for the food is given below:

Hence, 2 cans of diced tomatoes require 1 cup of green beans, 1/2 cup of pasta, 1 can of beans, and 4 of cups broth.

Hence, after, the 2 cans of diced tomatoes are used up, no more food can be prepared.

A limiting reagent or limiting ingredient is used up when cooking, hence, the 2 cans of tomatoes are the limiting ingredient.

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E°cell for each reaction, and rank the strengths of the oxidizing and reducing agents:

Complete the given reaction

\($$\mathrm{N}_{2} \mathrm{O}(g)+2 \mathrm{H}^{+}(a q)+2 \mathrm{Au}(s) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{Au}^{+}(a q)$$\)

Write the oxidation reaction

\($$\mathrm{Au}(s) \longrightarrow \mathrm{Au}^{+}(a q)+e^{-}$$\)

\($$E_{a c d}=-1.69 \mathrm{~V}$$\)

Write the reduction reaction

\($$3 \mathrm{~N}_{2} \mathrm{O}(g)+2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \quad E_{r e d}=+1.77 \mathrm{~V}$$\)

Calculate the standard cell voltage

\($$E_{c ell}=E_{r e d}+E_{o x d}$$\)

Substitute the value

\($$\begin{aligned}&E_{c e l l}=1.77-1.69 \\&E_{c e l l}=0.08 \mathrm{~V}\end{aligned}$$\)

Hence the standard cell potential is \($0.08 \mathrm{~V}$\)

Oxidizing agent in this reaction is

\($$\mathrm{N}_{2}(g), \mathrm{N}_{2} \mathrm{O}(g)$$\)

Reducing agent in this reaction is

\($$\mathrm{Au}^{+}(a q), \mathrm{Au}(s)$$\)

The potential difference between the two electrodes of an electrochemical cell, which develops when electrons are sent via the external circuit of a cell that has not reached equilibrium, is known as the cell potential.

A chemical process known as an oxidation-reduction (redox) reaction includes the exchange of electrons between two substances. Any chemical reaction in which the oxidation number of a molecule, atom, or ion changes due to the gain or loss of an electron is referred to as an oxidation-reduction reaction.

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Answer:

True

Explanation:

Answer:

4. The amount of caffeine in the drinks

5. The heart rate of the participants of the experiments

6. Water

7. i) The volume of drink taken should be constant

ii) The frequency of taking their drink is constant

iii) The time of drinking by the brothers is constant

Explanation:

In the question, Jeffery intends to find the caffeine drink that will result in the heart rate increasing most

The variables (varieties) of drinks tested by Jeffery = Pepsi, espresso, and water

Drink variable arranged by the order of increasing Caffeine content are presented as follows;

Caffeine content of water < Caffeine content of Pepsi < Caffeine content of espresso

The triplet with the double shot of espresso = The triplet with the highest heart rate

4. The independent variable is the variable which is suspected to be the cause of the specified observation

Therefore, in the question, the independent variable are the drinks with different amount of caffeine

5. The dependent variable is the effect or the outcome of the independent variable

The dependent variable in the question is the heart rate of the subjects in the study

6. The control group is the independent variable or input that is expected to give the minimum effect or output compared to other independent variables in the study such that the control group does not contain the suspected cause of the observation or effect under investigation

The control group (or variable) in the question is water which does not contain caffeine

7. Three constants that all three brothers should be doing are;

i) The three brothers should be taking a constant or the same quantity of their preferred drink

ii) The frequency at which they take their drinks should be constant

iii) The time at which the brothers take the drink should be the same

The following statement that incorrectly describes mechanisms of CO2 transport is "7-10% of CO2 is dissolved directly into the plasma as bicarbonate ions in plasma."

The transport of carbon dioxide in the blood is done in three ways which are:

Dissolved CO2 Carbaminohaemoglobin Bicarbonate ions in the plasma

Dissolved CO2: This is the smallest fraction of carbon dioxide carried by the blood. Dissolved CO2 combines with water in the blood to form carbonic acid (H2CO3). This reaction is facilitated by an enzyme known as carbonic anhydrase. Carbonic acid dissociates to form H+ and bicarbonate ions (HCO3-).

Carbaminohaemoglobin: Just like oxygen, carbon dioxide can also bind with hemoglobin. Carbon dioxide binds with hemoglobin to form carbaminohaemoglobin. Carbaminohaemoglobin accounts for about 20% of the carbon dioxide transported by the blood.

Bicarbonate ions in the plasma: This is the largest fraction of carbon dioxide transported by the blood. Carbon dioxide diffuses from the tissues into the blood where it combines with water to form carbonic acid. Carbonic acid dissociates to form H+ and bicarbonate ions (HCO3-). The bicarbonate ions diffuse into the plasma where they are transported to the lungs. In the lungs, the bicarbonate ions diffuse back into the red blood cells. The bicarbonate ions are converted back to carbon dioxide, which is then exhaled.

The statement that incorrectly describes mechanisms of CO2 transport is "7-10% of CO2 is dissolved directly into the plasma as bicarbonate ions in plasma." This is incorrect because about 70% of CO2 is transported as bicarbonate ions, not 7-10%. Therefore, the statement is false.

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The mistake in the question's descriptions concerning CO2 transport involves the part of hemoglobin that carbaminohemoglobin binds to (it's the globin part, not the heme part), and the proportion of CO2 carried as bicarbonate ions in the plasma (it's about 70%, not 7-10%).

The mechanisms described here for the transport of CO2 in the body are partially incorrect. In the human body, around 20% of CO2 is indeed transported as carbaminohemoglobin, but it binds to the globin part, not the heme part, of the hemoglobin protein. Additionally, the primary way that CO2 is transported is as bicarbonate ions, but these bicarbonate ions constitute around 70%, not 7-10%, of the CO2 carried in the plasma. Therefore, the incorrect descriptions mentioned in the question are that carbaminohemoglobin binds to the heme part of hemoglobin and that only 7-10% of CO2 is carried as bicarbonate ions in the plasma.

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Answer: A. H 0  μ1 = μ2 = μ3

Ha  μ1 ≠ 2μ ≠ μ3

2. Test Statistics is 95%

3. Critical F-Value is 3.76.

4. P-Value is 2.32.

5. Conclusion  Reject the null hypothesis.

6. Type of vehicle does effect the amount of green house gas emissions.

The correct order of the steps of a hypothesis test is given below.

1. Determine the null and alternative hypothesis.

2. Select a sample and compute the critical value F-test for the sample mean.

3. Determine the probability at which you will conclude that the sample outcome is very unlikely.

4. Make a decision about the unknown population.

All steps are performed in the given sequence to test a hypothesis.

The null hypothesis is rejected or accepted on the basis of level of significance. When the p-value is greater than level of significance we fail to reject the null hypothesis and null hypothesis is then accepted. It is not necessary that all null hypothesis will be rejected at 95% level of significance. To determine the criteria for accepting or rejecting a null hypothesis we should also consider p-value.

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Explanation:

The color of the solution changed when the person blew into the tube because blowing into the tube introduced carbon dioxide, which reacted with water to form carbonic acid.The indicator used in this experiment is Bromothymol blue.

Bromothymol blue is a pH indicator. It is a blue solution that turns yellow when the pH drops to 6 or less. When the pH rises to 7.6 or above, it turns green. Bromothymol blue has a pKa of 6.3. The buffer system used in this lab is the bicarbonate buffer system. The bicarbonate buffer system maintains the pH of the blood in humans. The system works by shifting the equilibrium between carbon dioxide, bicarbonate ions, and carbonic acid.

When carbon dioxide is produced in the body, it diffuses into the blood and reacts with water to form carbonic acid. The carbonic acid then dissociates to form bicarbonate ions and hydrogen ions. The bicarbonate ions act as a buffer by reacting with any excess hydrogen ions to form carbonic acid, which can then dissociate to form more bicarbonate ions.

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The sentence makes it clear that none of these are true.

Iodine shortage is treated or prevented using sodium iodide. Iodine is required by the body in proper development and growth. Potassium iodide may be required for people who need additional iodine or cannot acquire adequate copper from their normal diet. Your thyroid gland needs iodine to operate correctly.

A complement used in complete parenteral feeding is sodium iodide. A water-soluble ionic bond having a crystal lattice is sodium iodide. Iodine can be obtained from sodium iodide, which can be given as a supplement for complete parenteral nutrition but is most frequently utilized in

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Answer:

Mixing chemicals can be dangerous because you are handling chemicals that are often hazardous.*

Always wear the correct personal protective clothing and equipment when mixing chemicals.

Follow these steps to mix chemicals safely:

Choose a suitable mixing area. The site should be away from plants, people, animals and waterways. It needs to have good lighting and must be well ventilated. There should be first aid equipment and materials to clean up any spills that may occur. Make sure you have lots of clean, fresh water available. The water that you are using in your mixture should be as clean as possible.

Check the product label for the chemical to find out how to mix a solution, including the proportions or volumes needed. If you are unsure, contact the manufacturer or retailer for instructions. I hope it helps. Be safe.

Answer:

Chlorine (Cl)

D. A mountain range with folded layers of rock.

Intense compression can cause the rock layers to fold, creating a mountain range. This type of feature forms suddenly in the geological timescale, as a result of tectonic activity, and is known as a fold mountain.

The intense pressure causes the rock layers to buckle and deform, resulting in folds, faults, and other features. The Appalachian Mountains and the Rocky Mountains are examples of fold mountains in the United States.

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Answer:

F=Air

Explanation:

Because I took the test hope it helps

The process in which entropy decreases is the water freezing process. When water freezes, it changes from a liquid state to a solid state, resulting in a decrease in entropy. This is because the particles in a solid are more ordered and have less freedom of movement compared to those in a liquid. In the liquid state, the particles move more randomly and have more freedom, leading to higher entropy.

During freezing, the water molecules arrange themselves in a regular pattern, forming a solid crystal lattice. This orderly arrangement reduces the randomness of the molecules, resulting in a decrease in entropy.

In contrast, the other processes listed—ice melting, water evaporating, salt dissolving in water, and dry ice subliming—all involve an increase in entropy. These processes involve the molecules becoming more disordered and having more freedom of movement, leading to higher entropy.

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130 ppm will contain  0.12985167 grams of chlorine present in per liter of pool water.

This is an abbreviation for "parts per million" and it also can be expressed as milligrams per liter (mg/L).

1001.142303 ppm is equal to 1 g/L

So, 130 ppm will contain  \(\frac{130}{1001.142303}\) grams of chlorine present in per liter of pool water which is equal to 0.12985167 gram/L.

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Answer:

rock star

Explanation:

The balanced equation for the reaction between hydrogen and oxygen to form water is:

2H2 + O2 -> 2H2O

To react 4 moles of hydrogen with 3 moles of oxygen, we will also need 1.5 moles of oxygen.

Since we have enough hydrogen and not enough oxygen to react completely, we can calculate the theoretical yield of water produced.

4 moles of H2 will react to produce 2 moles of H2O and 1.5 moles of O2 will react to produce 0.75 moles of H2O. Therefore, the total theoretical amount of water produced is 2 + 0.75 = 2.75 moles of water.

percent yield = (actual yield / theoretical yield) x 100

percent yield = (3 / 2.75) x 100 = 109.09%.

So the percent yield is 109.09%. This means that 109.09% of the theoretical yield was actually produced, and the reaction was more efficient than expected.

It's worth noting that percent yield can't be more than 100% because it implies that more than the theoretical amount of product was produced, which is not possible. In this case, the percent yield is not a realistic value, therefore, the actual yield and the theoretical yield should be rechecked.