How many liters of solution will be produced when diluting 1.65 L of a 0.432 M KI solution to 0.212 M KI?

Answers

Answer 1

All questions where they ask you about concentration or volume of a solution prepared with another solution can be solved using the following equation:

[tex]C_1V_1=C_2V_2[/tex]

Where C1 is the concentration of the first solution, V1 is the volume of the first solution, C2 is the concentration of the second solution and V2 is the volume of the second solution. You only need to be careful to use the rigth units.

This particular question ask for the Volume of the second solution, V2. Then we need to solve for V2 in the previous equation:

[tex]V_2=\frac{C_1V_1}{C_2}=\frac{0.432\text{ M}\times1.65L}{0.212M}=3.36L[/tex]


Related Questions

Calculate the number of mol corresponding to 18.8 g Na2SO4.

Answers

Answer: 0.132 moles of Na2SO4

Explanation:

Multiply the grams of Na2SO4 by the 1/molar mass of Na2SO4, which is the sum of all the elements 2(Na)+S+4(O)= 142.04.

Explain how you would calculate the total change in bond energy for the reaction H2+Cl2->2HCI. How would you know if the reaction was endothermic or exothermic?

Answers

To calculate the total change in binding energy we must apply the following equation:

[tex]\Delta H=\Delta H_{f(\text{Reagents)}}-=\Delta H_{f(\text{Products)}}[/tex]

This equation tells us that the energy change will be equal to the sum of the potential energy of the product bonds minus the sum of the potential energy of the reagent bonds.

Now, let's calculate each term separately.

[tex]\begin{gathered} \Delta H_{f(\text{Reagents)}}=n_{H2}\times\text{Energy bond H-H + }n_{Cl2}\text{Energy bond Cl-Cl} \\ \Delta H_{f(\text{Reagents)}}=1molH_2\times432\frac{kJ}{mol}+1molCl_2\times239\frac{kJ}{mol} \\ \Delta H_{f(\text{Reagents)}}=671kJ \end{gathered}[/tex][tex]\begin{gathered} \Delta H_{f(\text{Products)}}=n_{\text{HCl}}\times EnergyBond\text{ H-Cl} \\ \Delta H_{f(\text{Products)}}=2molHCl\times427\frac{kJ}{mol} \\ \Delta H_{f(\text{Products)}}=854kJ \end{gathered}[/tex]

So, the change in bond energy will be:

[tex]\begin{gathered} \Delta H=\Delta H_{f(\text{Reagents)}}-=\Delta H_{f(\text{Products)}} \\ \Delta H=671kJ-854kJ=-183kJ \end{gathered}[/tex]

We have a negative value in the result, when this happens it means that the reaction is exothermic, that is to say, that it releases heat and the energy of the products is greater than that of the reagents.

When we have a positive value the reaction will be endothermic, this means that it needs energy.

In aqueous solution, 5.2 mg of iron(III) chloride reacts with excess ammonium hydroxide.A. Write and kalance the equation, including phasesB. Is this a limiting reactant problem? Why or why not?C. Calculate the moles of each product that are formedD. Calculate the grams of each product that are formedI already answered part a and b. I need help on part c and d.

Answers

With a balanced reaction, we can determine the moles of products. The balanced equation will be:

FeCl3 + 3NH4OH → Fe(OH)3 + 3NH4Cl

They says that NH4OH is in excess, so, the limiting reactant will be FeCl3 and we will do all the calculations with this reactant.

We have to calculate the moles of FeCl3, we will use the molar mass:

[tex]\begin{gathered} molFeCl_3=5.2mg\times\frac{1g}{1000mg}\times\frac{1molFeCl_3}{MolarMass,gFeCl_3} \\ molFeCl_3=5.2mg\times\frac{1g}{1,000mg}\times\frac{1molFeCl_3}{162.2gFeCl_3}=3.2\times10^{-5}molFeCl_3 \end{gathered}[/tex]

Now, to calculate the moles of the products we must take into account the product/reactive ratios, for this we are guided by the coefficients that accompany the molecules.

Ratio Fe(OH)3 to FeCl3 = 1/1

Ratio NH4Cl to FeCl3 = 3/1

Moles of each product

Moles of Fe(OH)3

[tex]molFe(OH)_3=3.2\times10^{-5}molFeCl_3\times\frac{1molFe(OH)_3}{1molFeCl_3}=3.2\times10^{-5}molFe(OH)_3[/tex]

Moles of NH4Cl

[tex]molNH_4Cl=3.2\times10^{-5}molFeCl_3\times\frac{3molNH_4Cl}{1molFeCl_3}=9.6\times10^{-5}molNH_4Cl[/tex]

The grams of each product we will find by multiplying the moles by the molar mass. So we have.

g of Fe(OH)3

[tex]\begin{gathered} gFe(OH)_3=3.2\times10^{-5}molFe(OH)_3\times106.87g/molFe(OH)_3 \\ gFe(OH)_3=3.4\times10^{-3}g=3.4mg \end{gathered}[/tex]

g of NH4Cl

[tex]\begin{gathered} gNH_4Cl=molNH_4Cl\times MolarMassNH_4Cl \\ gNH_4Cl=9.6\times10^{-5}molNH_4Cl\times53.491g/molNH_4Cl=5.1\times10^{-3}g=5.1mg \end{gathered}[/tex]

Find the volume in liters of the 0.505 molar NaOH solution needed to react with 40 milliliters of the 0.505 molar H2SO4 solution.

A//: 80 mililiters

Please, it's urgent, I need it ASAP

Answers

The volume of NaOH required is 80ml.

Sulfuric acid is a dibasic acid and NaOH is monoacidic base.

M1V1 = M2V2

where, M1 = initial concentration,

V1 = initial volume,

M2 = concentration after mixing

V2 = total final volume.

This formula is used for calculating the final volume or molarity after mixing two solutions.

Given the question,

M1 = 0.505M

V1 =?

M2 = 0.505M

V2 = 40 ml

M1V1 = 2×M2V2

0.505 × V2 = 2 × 0.505 × 40

V2 = 2×0.505×40

            0.505

V2 = 80ml

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2Al + 3Cl₂ 2AlCl₃molar mass of AlCl3 => 133 g/molIf 25.0 moles of chlorine is used, how many grams of aluminum chloride can be produced?

Answers

Answer

Explanation

Given:

2Al + 3Cl₂ 2AlCl₃

Molar mass of AlCl3 => 133 g/mol

Moles of Cl₂ = 25.0 mol

What to find:

The grams of aluminum chloride that can be produced.

Step-by-step solution:

Step 1: Determine the mole of AlClproduced.

From the given chemical equation for the reaction,

3 mol Cl₂ produce 2 mol AlCl

So 25.0 mol Cl₂ will produce

[tex]\frac{25.0mol\text{ }Cl_2\times2mol\text{ }AlCl₃}{3mol\text{ }Cl_2}=16.67mol\text{ }AlCl₃[/tex]

16.67 moles of AlCl₃ produced.

Step 2: Calculated the grams of AlClproduced.

Using the mole formula

[tex]undefined[/tex]

How many grams of Br are in 395 g CaBr2 ?

Answers

In order to answer this question we will need to use the molar mass of Calcium bromide (CaBr2), which is 199.89 g/mol, now we can find how much of this mass is only Bromide, so what is the percentage of Bromide that makes up this whole compound, to do so we will also use the molar mass of Bromide (Br2) which is 159.8 g/mol

199.89 g/mol = 100% of the compound

159.8 g/mol = x %

x = roughly 80% (is like 79.9%)

So, regardless of the mass of CaBr2, bromide needs to be 80% of its mass, now solving our question

395 g = 100%

x grams = 80%

x = 316 grams is the mass of Bromide in the compound

Phthalic acid ( H2C8H4O4) is a diprotic acid with a1=1.12×10−3 and a2=3.90×10−6. Determine the pH of a 0.294M phthalic acid ( H2C8H4O4 ) solution.

Answers

pH of the acid can be calculated from the first and second ionization constants of the acid. The pH of 0.294 M of Phthalic acid is 0.53.

What is pH?

pH of a solution is the measure of its H+ ion concentration.It measures the acidity or basicity of the solution. Mathematically it is the negative logarithm of hydrogen ion concentration.

pH = -log [H+]

Give that the first ionization constant of the acid is 1.12 × 10 ⁻³ and second ionization constant is 3.90 × 10⁻⁶. The diprotic acid can be represented as H2A and after its first ionization it produce HA- and H+ On the second ionization the base A- and H+ ion is produced.

The second ionization constant a₂  is written as follows:

a₂ =  [A-] [H+] / [HA - [A]] = 3.90 × 10⁻⁶

From this simply we can get [A] = 3.90 × 10⁻⁶

Thus [H+] = H₂A - [A]

The concentration of the diprotic acid H₂A is given 0.294 M.

Thus [H+] = 0.294 - 3.90 × 10⁻⁶  

                = 0.2933 M

Now, the pH is calculated as follows:

pH = -log (0.2933)

    = 0.53.

Therefore, the pH of 0.294 M of Phthalic acid is 0.53.

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You have 0.834 moles of Potassium. The NIH recommends 2,600mg of potassium as a daily intake recommendation. Convert the moles to mg.

Answers

Answer:

0.834 mole = 13293.32 mg

2,600 mg = 0.16 mole

Explanation:

1 mole of K = 39.098 g

=> 0.34 mole has (0.34)(39.098) = 13.29332g

13.29332 g x 10^3 = 13293.32 mg

so 0.834 mole of K = 13293.32 mg

mole of 2,600mg = (0.834)(2,600)/(13293.32) = 0.16311952168 or 0.16 mole

Which is an example of plasmas in nature?

Answers

Answer: nbhhvyyvub

Explanation:

Scientists are looking for ways to help increase the amount of calcium carbonate in oceans so that coral reefs can be healthy ecosystems again. Why is the amount of calcium carbonate in the ocean decreasing? (1 point)
Responses

Increased amounts of carbon dioxide cause reactions to happen in the water that decrease the pH and the amount of calcium carbonate.


Decreased amounts of carbon dioxide cause reactions to happen in the water that decrease the pH and the amount of calcium carbonate.


Increased amounts of carbon dioxide cause reactions to happen in the water that increase the pH and decrease the amount of calcium carbonate.

Decreased amounts of carbon dioxide cause reactions to happen in the water that increase the pH and decrease the amount of calcium carbonate.

Answers

Scientists are looking for ways to help increase the amount of calcium carbonate in oceans so that coral reefs can be healthy ecosystems again mount of calcium carbonate in the ocean decreasing because increased amounts of carbon dioxide cause reactions to happen in the water that decrease the pH and the amount of calcium carbonate

Calcium carbonate is a compound CaCO₃ found in nature as calcite and aragonite and in plant ashes, bones and shells and used especially in making lime and portland cement and as a gastric antacid and as ocean acidification increases available carbonate ion bond with excess hydrogen resulting in fewer carbonate ion available for calcifying organisms to build and maintain their shells, skeletons, and other calcium carbonate structures

Ocean acidification describe the lowering of sweater pH and carbonate saturation that result from increasing atmospheric CO₂ concentration and that's why increased amounts of carbon dioxide cause reactions to happen in the water that decrease the pH and the amount of calcium carbonate.

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Answer: Its C I did my research and that's what makes sense to me. because increased amount of carbon dioxide decreases the PH and the amount of calcium carbonate.

What is the oxidation number or Br in NaBrO?

Answers

The oxidation number of Br is +1

can atoms of different elements have the same atomic number?

Answers

No, Atoms of different element can not have same atomic number because only same type of atoms combine to form element.  Atoms belonging to different element can have different atomic number.

What is element?

Element generally consist of atoms or we can  atoms combine to form element. Atoms of an element is always same, means all the properties of all atoms of one type of element is same. Two or more than two atoms with different physical or chemical properties can not combine together to form an element.

So we can say that atomic number of all atoms that is constituting an element is same. It can never be different.

Thus Atoms of different element can not have same atomic number.

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Which of the following changes are chemical changes?

Answers

Answer:

flammability, toxicity etc..

Explanation:

are chemical changes

How does temperature of land far from water compare to that of land near water?

The temperature of land far from water is hotter than land near water.
The temperature of land far from water is less stable compared to land near water.
The temperature is similar for land both near and far from water.
The temperature is either too hot or too cold near water.

Answers

The temperature of land far from water is hotter than land near water.

Why the temperature of land far away from water different from land near water?

The lower heat capacity of land often allows them to cool the nearby water temperature so it takes less energy to change the temperature of land compared to water bodies. This means that land heats and cools more quickly as compared to water. This difference affects the climate of different areas on Earth. Large bodies of water like oceans, seas and large lakes can affect the climate of the nearby regions such as coastal regions. Water heats and cools more slowly than land regions. The coastal regions will stay cooler in summer season and warmer in winter season, which creates moderate climate on the coastal regions.

So we can conclude that the temperature of land far from water is different from land near water because of cooling effect of water bodies.

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Give the names of following compounds (in aqueous solution):
HBr
7 HBrO
2 HCN
3 H₂CO3
4 HC₂H302
5 H₂SO4
6 H3ASO3
8 HNO2
9 HC104
10 H₂C2O4
11 H3PO4
12 H₂CrO4_

Answers

Here are all names of compounds in aqeous solution.

What is aqeous solution?

An aqeous solution is one in which the solvent is liquid water .

In this solution water act as a solvent.

Sol-

HBr- Hydrogen bromide

1-HBrO- hydrobromic acid

2-HCL-hydrochloric acid

3-H2O3- hydrogen peroxide

4- HC2H3O2- glacial acetic acid

5-H2SO4-sulfuric acid

6- H3ASO3- Arsenous acid

7- HNO2- Nitrous acid

8- HC104- Perchloric acid

9-H2C2O4- oxalic acid

10- H3pO4- phosporic acid

11-H2CrO4- chromic acid

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How should the electrode of a pH meter be preserved? Explain your answer.

Answers

We can care for the pH meter electrode by;

1) Making sure that the electrode remains moist

2) The electrode should be stored in a 4M solution of KCl

3) Electrodes should not come in contact with deionized water.

What is pH?

The term pH has to do with the negative logarithm of the hydrogen ion concentration. Now we know that the pH of a solution tells us the amount of the hydrogen ions or the hydroxyl ions that is present in the solution. The pH mete is the instrument that we could use to be able to measure the pH of the solution.

The pH scale runs between 0 - 14. The points on the scale that have been labeled from 0 - 6 tells us that the solution is an acidic solution and contains more hydrogen ions.  If the solution has a pH of 7, then it is neutral and contains equal concentration of hydrogen and hydroxyl ions. A solution of pH 8 - 14 is basic and contains more hydroxyl than hydrogen ions.

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If carbon has undergone neutron capture what happened?

Answers

Answer:

It would generate a electro lithium nucleous

Gaseous ethane (CH3CH3) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). if 2.46 g of carbon dioxide is produced from the reaction of 2.71 g of ethane and 16.7 g of oxygen gas, calculate the percent yield of carbon dioxide. Round to 3 sig figs

Answers

Taking into account definition of percent yield, the percent yield of carbon dioxide is 30.90%

Reaction stoichiometry

In first place, the balanced reaction is:

2 CH₃CH₃ + 7 O₂ → 4 CO₂ + 6 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

CH₃CH₃: 2 molesO₂: 7 molesCO₂: 4 molesH₂O: 6 moles

The molar mass of the compounds is:

CH₃CH₃: 30 g/moleO₂: 32 g/moleCO₂: 44 g/moleH₂O: 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

CH₃CH₃: 2 moles ×30 g/mole= 60 gramsO₂: 7 moles ×32 g/mole= 224 gramsCO₂: 4 moles ×44 g/mole= 176 gramsH₂O: 6 moles ×18 g/mole= 108 grams

Limiting reagent

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

Limiting reagent in this case

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 224 grams of O₂ reacts with 60 grams of CH₃CH₃, 16.7 grams of O₂ reacts with how much mass of CH₃CH₃?

mass of CH₃CH₃= (16.7 grams of O₂× 60 grams of CH₃CH₃)÷ 224 grams of O₂

mass of CH₃CH₃= 4.47 grams

But 4.47 grams of CH₃CH₃ are not available, 2.71 grams are available. Since you have less mass than you need to react with 16.7 grams of O₂, CH₃CH₃ will be the limiting reagent.

Percent yield

The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.

The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:

percent yield= (actual yield÷ theoretical yield)×100%

where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

Theoretical yield of CO₂

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 60 grams of CH₃CH₃ form 176 grams of CO₂, 2.71 grams of CH₃CH₃ form how much mass of CO₂?

mass of CO₂= (2.71 grams of CH₃CH₃× 176 grams of CO₂)÷ 60 grams of CH₃CH₃

mass of CO₂= 7.96 grams

Then, the theoretical yield of CO₂ is 7.96 grams.

Percent yield for the reaction in this case

In this case, you know:

actual yield= 2.46 gramstheorical yield= 7.96 grams

Replacing in the definition of percent yields:

percent yield= (2.46 grams÷ 7.96 grams)×100%

Solving:

percent yield= 30.90%

Finally, the percent yield for the reaction is 30.90%.

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Convert 0.05090Kg/mol to dg/mmol

Answers

Answer: .509

Explanation: Im pretty sure

Uranium-238 is an unstable nuclide that emits an alpha particle, three neutrons, and
gamma particles. A = 231 and Z = 90. What daughter nuclide is formed from this
reaction?
OA) Ra
232U2a + 2x + 3 n + Y
-
B) Pu
C) Th
D) Fr

Answers

Uranium-238 is an unstable nuclide that emits an alpha particle, three neutrons, and

gamma particles and produces a daughter nuclide Thorium.

The atomic mass of alpha particle is 4g and atomic number 2. Since an alpha particle emits, so there is reduction in the atomic mass 4 and atomic number 2.

The atomic number of neutron is 0 and the atomic mass of neutron is 1. When three neutrons emit the atomic mass reduces to 3 while atomic number remains same.

Gamma particles have no mass and charge. So, atomic number and atomic mass remains same.

Total reduction in atomic mass : 238 - 4 - 3 = 231.

Total reduction in atomic number = 92- 2 = 90

Therefore, the daughter nuclide is Thorium.

Thus, we concluded that the Uranium-238 is an unstable nuclide that emits an alpha particle, three neutrons, and gamma particles and produces a daughter nuclide Thorium.

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What is the volume, in mL, of a solution that is 0.421 M K2SO3 and was prepared by dissolving 16.7 g of solid potassium sulfite in water?

Answers

The volume is 251mL. The final answer is rounded to the nearest whole number.

PLEASE HELP!
In a monomolecular reaction A->B , at t =250C, the initial concentration decrease at 25% in t =52 min. Calculate:

a) the constant rate;

b) the time after the initial concentration decrease with 75%;

c) the initial reaction rate, if the initial concentration of the reactant is 2.5 mol/L·s

Answers

1) The calculate rate constant is  9.22 * 10^-5 s-1

2) In the period of  is 15033 s 75% is used up

3) From the calculation, there is an initial 9.22 * 10^-5 M.

How can we find reaction rate?

In chemistry, the rate of reaction would give the idea that the reaction is proceeding quickly or slowly. If a reaction has a large rate of reaction then it tends to move on to completion.

From the question, it  is clear that there is 25% in t =52 min.

Initial concentration [A]o =  [A]o

Final concentration =  [A]o - 0.25  [A]o = 0.75 [A]o

Time taken = 52 min or 3120 s

From the formula that can be applied to a first order reaction;

ln[A] = -kt + ln[A]0

k = -(ln[A] -  ln[A]0)/t

k = - (ln0.75 [A]o/A]0)/3120

k = 9.22 * 10^-5 s-1

b)

Then we have  to find  the time after the initial concentration decrease with 75%

[A] =  [A]o - 0.75  [A]o = 0.25 [A]o

ln[A] = -kt + ln[A]0

t = -(ln[A] -  ln[A]0)/k

t = - (ln0.25 [A]o/A]0)/9.22 * 10^-5

t = 15033 s

c)  Given that in this case, the initial concentration of the reactant is 2.5 mol/L·s

k = - (ln0.25 (2.5)/ln(2.5))/15033

k = 9.22 * 10^-5 M

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At equilibrium, a reaction vessel contains 4.50 atm of Br₂ and 1.10 atm of NBr₃. According to the reaction: 2 NBr₃ (g) ⇌ N₂ (g) + 3 Br₂ (g) Kp = 4.8
Determine the equilibrium partial pressure of N₂.

Answers

Equilibrium constant is the only concept that is to be used here to calculate partial pressure of nitrogen. The partial pressure of Nitrogen comes out to be  1.17atm

What is equilibrium constant?

Equilibrium constant is a rate constant at equilibrium shows the values of reaction with respect to the atmospheric pressure and concentration

[tex]K_{c}[/tex]  is equilibrium constant with respect to concentration.

[tex]K_{p}[/tex] is equilibrium constant with respect to atmospheric pressure.

Mathematically,

[tex]K_{p}=\frac{[N_{2}][Br_{2} ] }{[NBr_{3}] }[/tex]

Substituting values

(4.8×1.10atm)÷ 4.50atm=[tex]{[N_{2}][/tex]

1.17atm=[tex]{[N_{2}][/tex]

Thus the equilibrium partial pressure of N₂ is 1.17atm.

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13. A gas of unknown molecular mass was allowed to effuse through a small opening under
constant pressure conditions. It required 72 s for the gas to effuse. Under identical
experimental conditions, it required 28 s for O₂ gas to effuse. Determine the molar mass
of the unknown gas.

Answers

The molar mass of unknown gas is 211.59 g/mol.

What is effusion of gases?Effusion occurs when a gas pass through an opening that is smaller than the mean free path of the particles, which is the average distance traveled between collisionsGraham's law is an empirical relationship which states that ratio of the rates of diffusion or effusion of two gases is the square root of the inverse ratio of their molar masses.Higher the molar mass of a gas, slower the effusion

Since both the gases are present in identical experimental condition, using Grahams law of effusion:

      [tex]\frac{time req for unknown gas}{time of oxygen}[/tex] = [tex]\sqrt{\frac{Molar mass of unknown gas}{molar mass of Oxygen} }[/tex]

  Given:

Time of unknown gas = 72 sec

Time of Oxygen = 28 sec

we know the molar mass of oxygen = 32 g/mol

Now substituting:

    [tex]\frac{72 sec}{28 sec} = \sqrt{\frac{M.M of unknown}{32g/mol} }[/tex]

M.M of unknown = (72 / 28)² × 32 g/mol

                            = 211.59 g/mol

So the molar mass of unknown gas is 211.59 g/mol

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on a solid mixture, X. The inferences made are recorded in Table 2. Complete Table 2 by filling in the observations based on the inferences made. TABLE 2: TESTS ON MIXTURE X Test Observation Inferences a) Distilled water was added to a portion of X and the resulting mixture stirred and filtered. (The residue was set aside for use later.) The filtrate was divided into 3 equal portions and tests (b) to (d) done on separate portions. Cl- ions are present. b) Dilute nitric acid followed by a few drops of silver nitrate solution was added. Ammonium hydroxide solution was added to the resulting mixture. (3 marks)

Answers

Answer:

a) Upon adding silver nitrate, a white precipitate is observed

b) Upon adding ammonium hydroxide, the white precipitate dissolves to give a clear, colorless solution

Explanation:

Here, we want to state the observations when testing for chloride ions

From what we have:

a) When silver nitrate is added, a white precipitate is formed

This is as a result of the following chemical reaction:

[tex]Ag\placeholder{⬚}_{(aq)}^+\text{ + Cl}_{(aq)}^-\text{ }\rightarrow\text{ AgCl}_{(s)}[/tex]

The AgCl is the white precipitate formed

b) Upon the addition of the ammonium hydroxide solution, a colorless and clear solution is observed showing that the white precipitate has dissolved

Acetone has a density of 0.7857 g/cm^3. What is the volume in mL of 5.52 g of acetone?

Answers

Let's see that the formula to find the volume using density and mass is:

[tex]V=\frac{m}{d}\begin{cases}V=\text{volume} \\ m=\text{mass} \\ d=\text{density}\end{cases},[/tex]

In this case, the problem is asking about the volume in mL but remember that mL is the same that cm^3, so using the formula we're going to obtain:

[tex]V=\frac{5.52\text{ g}}{0.7857\text{ }\frac{g}{mL}}=7.02\text{ mL.}[/tex]

The volume of 5.52 g of acetone is 7.02 mL.

If 335g water at gains 102.3 J of heat, how much does the temperature of the water change? The specific heat of water is 4.184 J/g*C

Answers

Temperature is a measure of how hot a substance or radiation is expressed numerically.

There are three different types of temperature scales:

those that depend only on macroscopic properties and thermodynamic principles, like Kelvin's original definition;

those that depend on practical empirical properties of particles rather than theoretical principles;

and those that are defined by the average translational kinetic energy per freely moving microscopic particle, like an atom, molecule, or electron, in a body, like the SI scale.

How much heat is gained or lost by a sample can be calculated using the equation q = mcΔT, where m is the mass of the sample, c is the specific heat, and T is the temperature change (q).

Therefore,

q = m*c*ΔT

102.3 = 335 * 4.184 * ΔT

ΔT = 0.073 °c

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In the following equation, how many grams of O2 are needed to react with 24.0 g of NH3?

4 NH3(g)+302(g) → 2N2(g)+6H₂O(1)

a) 45.1
b) 22.6
c) 33.9
d) 45.0
e) 135.5

Answers

33.6 g grams of O2 are needed to react with 24.0 g of NH3.

The molecular mass of NH3 = 14 + 3 = 17g

Given mass of NH3 = 24g

Firstly, we will calculate the number of moles.

Moles is defined as the ratio of given mass of substance to the molecular mass of substance.

Moles = given mass/ molecular mass

Number of moles of NH3 = 24/17

= 1.4 mole

Chemical reaction

4 NH3(g)+302(g) → 2N2(g)+6H₂O(1)

4 moles of NH3 require 3 moles of O₂ to react.

1 moles of NH3 require 3/4 moles of O₂ to react.

1.4 moles of NH3 require 1.05 moles of O₂ to react.

Now we calculate the grams of O₂.

1.05 × 32 = 33.6 g.

Thus, we concluded that the 33.6 g grams of O2 are needed to react with 24.0 g of NH3.

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Following the Experiment 3 procedure, you combine 50.0 mL H2O, 50.0 mL of 2.2 M HCl(aq), and 4.039 g NaOH(s) (molar mass of NaOH = 40.00 g/mol) in a Styrofoam calorimeter, recording a maximum solution temperature change of 22.9 oC and a final solution mass of 104.153 g (csolution = 4.184 J/g.oC). Calculate the molar change in enthalpy of reaction

Answers

The molar change in enthalpy of reaction is 99.6 kJ/mol.

the formula for the specific heat capacity expressed as :

Q = mcΔT

where,

m , mass = 104.153 g

c, specific heat = 4.184 J/g °C

dt, change in temperature = 22.9  °C

Q = 104.153 × 4.184 × 22.9

Q = 9969.7 J = 9.96 kJ

now, the molar change in enthalpy is give by:

Q = ΔH / n

n is no. of moles

morality of HCl = 2.2 M

v = 50 mL = 0.05 L

n = 0.05 × 2.2

   =  0.11 moles

no. of moles of NaOH = mass / molar mass

                                    = 4.039 / 40

                                    = 0.100

NaOH is limiting reactant.

using the formula we get:

Q = ΔH / n

ΔH = Q / n

ΔH  =  9.96 kJ/ 0.10

ΔH, change in enthalpy  = 99.6 kJ/mol

Thus, combine 50.0 mL H₂O, 50.0 mL of 2.2 M HCl(aq), and 4.039 g NaOH(s) (molar mass of NaOH = 40.00 g/mol) in a Styrofoam calorimeter, recording a maximum solution temperature change of 22.9 °C and a final solution mass of 104.153 g (c solution = 4.184 J/g°C).  the molar change in enthalpy of reaction is 99.6 kJ/mol.

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What are the strengths, weaknesses, and implications of risk/benefit analysis as a method for deciding whether a technology should be employed?

Answers

A risk-benefit analysis compares the risks and benefits of a situation and determines whether the advantages outweigh the disadvantages. 

What is Risk-benefit analysis in technology?

Strengths

Risk-benefit analysis calculates the amount of time will be worth it to the production of technology and whether the technology will have a healthy impact on the industry or not.

Weaknesses

Risk-benefit analysis cannon determine product implementation and the outcomes of real life experiences of individual customers. It has some drawbacks like benefits of customer is take under consideration but pollution in nature is not calculated.

Implication

Risk-benefit analysis is implied in almost all technical industry as it the decision maker of any developing team to work on a particular project or not. Some examples are automobile industry and smartphone industry.

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