Determine whether the bond between each of the following pairs of atoms is purely covalent, polar covalent, or ionic.
- I and Cl
-C and O
- Ca and F
- P and P

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

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

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:

The molar mass of the compounds is:

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

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.

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.

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.

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.

In this case, you know:

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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The half-life of N₂O₅ at 70°C then it is 50s or 0.8min

Half life is the the interval of time required for one half of the atomic nuclei of a radioactive sample to decay

The first-order rate constant for the decomposition of N₂O₅ and reaction is

2N₂O₅(g) → 4NO₂(g)+O₂(g)

The half life is the time at which the concentration of  N₂O₅ is the half initial concentration that is [ N₂O₅ ] = 1/2 [ N₂O₅ ]₀

Using this same equation then

ln ([N₂O₅]) = -2kt + ln ([N₂O₅]₀)

We replace [ N₂O₅ ] by 1/2[ N₂O₅ ]₀ and solve for t

ln (1/2 [N₂O₅]₀) = -2kt + ln ([N₂O₅]₀) applying logarithmic property

ln 1/2 + ln([N₂O₅]₀) = -2kt + ln ([N₂O₅]₀)  solving for t and applying logarithmic property (ln 1/2 = ln 1 - ln 2 = 0 - ln 2 = - ln 2)

t = ln 2/2k

t = 50s or 0.8min

The half-life of N₂O₅ at 70°C then it is 50s or 0.8min

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The mass of the piece of porcelain is 27.3 oz.

1st) It is necessary to calculate the mass of the porcelain with the density formula:

2nd) Now it is necessary to convert the gram unit to oz. Here we need to use the relation that 1 g is equal to 0.035 oz:

So, the mass of the piece of porcelain is 27.3 oz.

In order to separate the mixture of toluic acid, p-bromo-aniline, and anthracene, the difference in their physical properties such as solubility in organic or inorganic solvents is utilized.

The flow chart for the separation method is found in the attachment.

Mixtures are substances that consist of two or more substances that are physically combined together.

The physical properties of the components of the mixture are employed in the separation of the constituents of the mixture

In separating the given mixture, it is first dissolved in dichloromethane, and then dilute NaOH is added to separate the toluic acid in the aqueous phase. The toluic acid is then precipitated by adding HCl.

Dilute HCl is then added to the organic phase to separate p-bromo-aniline into an aqueous phase and then precipitated by adding 10% NaOH.

Anthracene is left in the organic phase.

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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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Answer:

flammability, toxicity etc..

Explanation:

are chemical changes

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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Answer: .509

Explanation: Im pretty sure

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

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 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

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.

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

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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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:

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

Moles of NH4Cl

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

g of Fe(OH)3

g of NH4Cl

• We are given that Lidocaine is available as 0.5%(w/v) , 0.5%(w/v) represents the amount of Lidocaine in grams per 100mL. so, 0.5% is the same as 0.5g of Lidocaine in 100ml.

• Since our solution has a volume of 6.0mL:

mass of Lidocaine = 6mL * (0.5g /100mL)

= 0.03g of Lidocaine in 6.0mLsolution.

This means that, in 6.0ml of this injection solution, we have 0.03g of Lidocaine .

Answer:

It would generate a electro lithium nucleous

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

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:

The volume of 5.52 g of acetone is 7.02 mL.

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.

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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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.

Atomic mass of zinc is 65. Simply dip one strip into the water for 3 seconds, remove it and shake once to remove excess water, wait 20 seconds for the color to develop, and compare it to the closest color match to determine your zinc concentration.

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

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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Answer:

b.Carbon has a total of four bonded pairs of electrons around it.

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.

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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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 AlCl₃ produced.

From the given chemical equation for the reaction,

3 mol Cl₂ produce 2 mol AlCl₃

So 25.0 mol Cl₂ will produce

16.67 moles of AlCl₃ produced.

Step 2: Calculated the grams of AlCl₃ produced.

Using the mole formula

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

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.

So, the change in bond energy will be:

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.

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.

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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Answer:

Explanation:

Here, we want to know the volume of chlorine gas necessary to complete the reaction

We can start by writing a balanced equation of reaction as follows:

From the balanced equation, we can see that 1 mole of chlorine gas requires 2 moles of sodium metal

Now, let us get the number of moles of sodium metal that actually reacted

We have that as the mass of sodium metal divided by the atomic mass

The atomic mass of sodium metal is 23 amu

Thus, we have the number of moles as:

Since the mole ratio is 2 to 1, half of this number of moles is needed by the chlorine gas

Mathematically, 1 mole of a gas occupies a volume of 22.4 L at STP

the volume occupied by the calculated number of moles will be: