How many moles of iron is equivalent to 4.45 x 10^22 atoms of iron

Answers

Answer 1

Answer:

0.074 moles

Explanation:

For every mole (of any element), there are 6.022 x 10^23 atoms.

There are 4.45 x 10^22 atoms of iron.

To find the moles we divide the number of atoms by Avogadro's number

4.45 x 10^22 / 6.022 x 10^23 = 0.0738957

Don't forget sig figs

Answer 2

Answer:

[tex]\boxed {\boxed {\sf 0.0739 \ mol \ Fe}}[/tex]

Explanation:

We are asked to convert a number of iron atoms to moles of iron.  

We will use Avogadro's Number for this, which is 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. For this problem, the particles are atoms of iron. There are 6.022 ×10²³ atoms of iron in 1 mole of iron.  

We will also use dimensional analysis to solve this problem. To do this, we use ratios. Set up a ratio using the underlined information.

[tex]\frac {6.022 \times 10^{23} \ atoms \ Fe} {1 \ mol \ Fe}[/tex]

Since we are converting 4.45 × 10²² atoms of iron to moles, we multiply the ratio by that value.

[tex]4.45 \ \times 10^{22} \ atoms \ Fe *\frac {6.022 \times 10^{23} \ atoms \ Fe} {1 \ mol \ Fe}[/tex]

Flip the ratio. The value is the same, but it allows us to cancel the units of atoms of iron.

[tex]4.45 \ \times 10^{22} \ atoms \ Fe *\frac {1 \ mol \ Fe}{6.022 \times 10^{23} \ atoms \ Fe}[/tex]

[tex]4.45 \ \times 10^{22} *\frac {1 \ mol \ Fe}{6.022 \times 10^{23}}[/tex]

Condense into 1 fraction.

[tex]\frac {4.45 \ \times 10^{22} }{6.022 \times 10^{23}} \ mol \ Fe[/tex]

[tex]0.07389571571 \ mol \ Fe[/tex]

The original measurement of atoms ( 4.45 × 10²²) has 3 significaint figures, so our answer must have the same. For the number we calculated, that is the ten-thousandths place. The 9 to the right of this place (0.07389571571) tells us to round the 8 up to a 9.

[tex]0.0739 \ mol \ Fe[/tex]

There are approximately 0.0739 moles of iron in 4.45 × 10²² atoms of iron.  


Related Questions

A sample of Kr gas is observed to effuse through a pourous barrier in 8.15 minutes. Under the same conditions, the same number of moles of an unknown gas requires 4.53 minutes to effuse through the same barrier. The molar mass of the unknown gas is ____________ g/mol.

Answers

Answer:

25.88 g/mol

Explanation:

Graham's law is a famous law which states that the diffusion rate or the effusion rate of any gas varies inversely to the square root of the molecular weight the gas.

So from Graham's law, we have,

[tex]$\frac{\text{time}}{M^{1/2}}=\text{constant}$[/tex]

Using the sample of Kr gas having M = 83.8

[tex]$\frac{8.15}{(83.8)^{0.5}}= \frac{4.53}{M^{0.5}}$[/tex]

[tex]$M^{0.5}= 5.088$[/tex]

M = 25.88 g/mol

melting points of lipids are strongly influenced by the length and degree of unsaturation of hydrocarbon chain.justify this statement? ​

Answers

Answer

The properties of fatty acids and of lipids derived from them are markedly dependent on chain length and degree of saturation. Unsaturated fatty acids have lower melting points than saturated fatty acids of the same length. For example, the melting point of stearic acid is 69.6°C, whereas that of oleic acid (which contains one cis double bond) is 13.4°C. The melting points of polyunsaturated fatty acids of the C18 series are even lower. Chain length also affects the melting point, as illustrated by the fact that the melting temperature of palmitic acid (C16) is 6.5 degrees lower than that of stearic acid (C18). Thus, short chain length and unsaturation enhance the fluidity of fatty acids and of their derivatives.

In an ELISA, the compound 4-chloro-1-naphthol is used because:_______
a. it turns color in the presence of an enzyme that is bound to the secondary antibody
b. it helps the primary antibody bind to the protein
c. it helps the secondary antibody to bind to the protein
d. all of the choices

Answers

Answer:

a. It turns color in the presence of an enzyme that us bound to the secondary antibody.

Explanation:

The compound chloronapthenel is used in the reaction because it changes the color in the presence of an enzyme. It is strong organic compound which is used in biochemical processes.

Explain why
when suger
is heated it
does not boil but rather break down
into carbon and water?

Answers

Answer:

When simple sugars such as sucrose (or table sugar) are heated, they melt and break down into glucose and fructose, two other forms of sugar. Continuing to heat the sugar at high temperature causes these sugars to lose water and react with each other producing many different types of compounds.

Calculate the mass percent of each component in the following solution.
159 g NiCl2 in 500 g water
% Nicla
% water

Answers

Answer:

% NiCl2 = 24.13%

% water = 78.57%

Explanation:

Mass percentage = mass of solute/mass of solution × 100

According to this question, a solution contains 159 g of NiCl2 in 500 g of water. Hence, mass of the solution is calculated as follows:

Mass of solution = 159g + 500g

Mass of solution = 659g

Therefore;

A) % Mass of NiCl2 in solution = mass of NiCl2/mass of solution × 100

% Mass of NiCl2 in solution = 159/659 × 100

% Mass of NiCl2 in solution = 0.2413 × 100

= 24.13%

B) % Mass of water in solution = mass of water/mass of solution × 100

% Mass of water in solution = 500/659 × 100

% Mass of water in solution = 0.7587 × 100

% Mass of water in solution = 75.87%

write down the different uses of water that you know about​

Answers

Answer:

The various uses of water :

1. Water is used for daily purpose like cooking , bathing , cleaning and drinking.

2. Water is used as a universal solvent.

3. water maintains the temperature of our body.

4. Water helps in digestion in our body.

5 .water is used in factories and industries.

6. Water is used to grow plants , vegetables and crops.

Answer:

The various use of water are;

I) Cooking.

ii) Drinking

III) Bathing

iv) Generating hydro- electricity

v) Construction work etc

Aqueous hydrobromic acid HBr will react with solid sodium hydroxide NaOH to produce aqueous sodium bromide NaBr and liquid water H2O. Suppose 55.8 g of hydrobromic acid is mixed with 17. g of sodium hydroxide. Calculate the minimum mass of hydrobromic acid that could be left over by the chemical reaction. Round your answer to 2 significant digits.

Answers

Answer:

21.4g of HBr is the minimum mass that could be left over.

Explanation:

Based on the reaction:

HBr + NaOH → NaBr + H2O

1 mole of HBr reacts per mole of NaOH

To solve this question we need to find the moles of both reactants. If moles NaOH > moles HBr, the difference in moles represents the minimum moles of HBr that could be left over because this reaction is 1:1. Using the molar mass we can find the minimum mass of HBr that could be left over, as follows:

Moles NaOH -40.0g/mol-

17g * (1mol/40.0g) = 0.425 moles NaOH

Moles HBr -Molar mass: 80.91g/mol-

55.8g * (1mol/80.91g) = 0.690 moles HBr

The difference in moles is:

0.690 moles - 0.425 moles =

0.265 moles of HBr could be left over

The mass is:

0.265 moles * (80.91g/mol) =

21.4g of HBr is the minimum mass that could be left over.

Methyl pentanoate condensed structural formula

Answers

Answer:

C6H12O2 is the formula for Methyl pentanoate

In the Bohr model of the hydrogen atom, the electron occupies distinct energy states.

a. True
b. False

Answers

Answer:

a. True

Explanation:

The Bohr model shows the n-2 as the first transitions between the energy states and hydrogen atom which is shown by changes in the electron structure n = 1 1. In this transition period, the electron moves from the n level to the next level, and energy d transmitted.

How many grams of NaCl (MM = 58.44g/mol) are in 250mL of a 0.75 molar solution?

Answers

Answer:

[tex]\boxed {\boxed {\sf 11 \ grams \ NaCl}}[/tex]

Explanation:

We are asked to find how many grams of sodium chloride are in a solution.

1. Moles of Solute

Molarity is a measure of concentration in moles per liter.

[tex]molarity= \frac{ moles \ of \ solute}{liters \ of \ solution}[/tex]

We know the molarity is 0.75 molar. 1 molar is the same as 1 mole per liter, so the solution contains 0.75 moles of sodium chloride per liter.

There are 250 milliliters of solution but molarity uses liters for volume. We must convert milliliters to liters. Remember that 1 liter contains 1000 milliliters. Set up a ratio and use dimensional analysis to convert.

[tex]250 \ mL * \frac{1 \ L} {1000\ mL} = \frac{ 250}{1000} \ L = 0.250 \ L[/tex]

Now we know the molarity and the liters of solution, but the moles of solute are unknown.

molarity = 0.75 mol NaCl/L moles of solute =x liters of solution = 0.25 L

Substitute the values into the formula.

[tex]0.75 \ mol \ NaCl/L = \frac{x}{0.250 \ L}[/tex]

We are solving for the moles of solute, so we must isolate the variable x. It is being divided by 0.250 liters. The inverse of division is multiplication, so multiply both sides of the equation by 0.250 L.

[tex]0.250 \ L *0.75 \ mol \ NaCl/L = \frac{x}{0.250 \ L} * 0.250 \ L[/tex]

[tex]0.250 \ L *0.75 \ mol \ NaCl/L = x[/tex]

The units of liters cancel.

[tex]0.250 * 0.75 \ mol \ NaCl[/tex]

[tex]\bold {0.1875 \ mol \ NaCl}[/tex]

2. Grams of Solute

Now that we have calculated the moles of solute, we must convert this to grams. We use the molar mass or the mass of 1 mole of a substance. Sodium chloride's molar mass is given and it is 58.44 grams per mole. This means there are 58.44 grams of sodium chloride in 1 mole of sodium chloride.

Set up a ratio so we can convert using dimensional analysis.

[tex]\frac {58.44 \ g \ NaCl}{1 \ mol \ NaCl}[/tex]

Multiply by the number of moles we calculated.

[tex]0.1875 \ mol \ NaCl *\frac {58.44 \ g \ NaCl}{1 \ mol \ NaCl}[/tex]

The units of moles of sodium chloride cancel.

[tex]0.1875 *\frac {58.44 \ g \ NaCl}{1}[/tex]

[tex]\bold {10.9575 \ g \ NaCl}[/tex]

3. Round using Significant Figures

The original measurements of molarity and volume have 2 significant figures, so our answer must have the same. For the number we calculated, that is the ones place. The 9 in the tenths place tells us to round the 0 up to a 1.

[tex]11 \ g \ NaCl[/tex]

There are approximately 11 grams of sodium chloride in 250 mL of a 0.75 molar solution.

pplication)Using multiple models simultaneously: This FNT refers to a processinvolving three moles of a diatomic gas (which behaves as an ideal gas).The PV curve at right describes this process.Assume, as is typical near room temperature,vibrational modes are frozen out.a)Determine the energy transferred as work, the change in internal energy, and the energy transferred as heat in this process. b)Could youhave stillansweredthe questions in a) if the temperature was not provided on the plot

Answers

Complete Question

Questions Diagram is attached below

Answer:

*  [tex]W=1142.86Joule[/tex]

*  [tex]Q=997.7J[/tex]

*  [tex]H=2140.5J[/tex]

Explanation:

From the question we are told that:

Temperature [tex]T=337K[/tex]

Pressure [tex]P=(60-55)Pa*10^5[/tex]

Volume[tex]V=(1.6-1.4)m^3*10^{-3}[/tex]

Generally the equation for gas Constant is mathematically given by

[tex]\frac{P_2}{P_1}=\frac{V_1}{V_2}^n[/tex]

 [tex]\frac{55*10^5}{60*10^5}=\frac{1.4*10^{-3}}{1.6*10^{-3}}^n[/tex]

 [tex]n=0.65[/tex]

Therefore

Work-done

 [tex]W=\int{pdv}[/tex]

 [tex]W=\frac{55*10^5*1.6*10^{-3}*60*10^5*1.4*10^{-3}}{1-0.65}[/tex]

 [tex]W=1142.86Joule[/tex]

Generally the equation for internal energy is mathematically given by

 [tex]Q=mC_vdT\\\\Q=\frac{3*1*3.314*16}{1.4-1}[/tex]

 [tex]Q=997.7J[/tex]

Therefore

 [tex]H=Q+W[/tex]

 [tex]H=997.7J-11.42.9[/tex]

 [tex]H=2140.5J[/tex]

By using photons of specific wavelengths, chemists can dissociate gaseous HI to produce H atoms with certain speeds. When HI dissociates, the H atoms move away rapidly, whereas the heavier I atoms move more slowly. If a photon of 231 nm is used, what is the excess energy (in J) over that needed for dissociation

Answers

Answer:

The excess energy over that needed for dissociation is 3.712 × 10⁻¹⁹ J

Explanation:

Given the data in the question;

wavelength of proton λ = 231 nm = 231 × 10⁻⁹ m

we determine the energy of the proton;

E = hc / λ

where h is plank constant ( 6.626 × 10⁻³⁴ JS )

and c is the speed of light ( 3 × 10⁸ m/s )

we substitute

E = [ ( 6.626 × 10⁻³⁴ JS ) × ( 3 × 10⁸ m/s ) ] / [ 231 × 10⁻⁹ m ]

E = 8.61 × 10⁻¹⁹ J

we know that, bond energy for H-I is 295 kJ/mol

so, H = 295 × 10³ J/mol

Now, energy to dissociate HI will be;

⇒ H / N

where N is the Avogadro's number ( 6.023 × 10²³ mol⁻¹ )

energy to dissociate HI = ( 295 × 10³ J/mol ) / ( 6.023 × 10²³ mol⁻¹ )

= 4.898 × 10⁻¹⁹ J

Therefore, Excess energy over dissociation will be;

⇒ ( 8.61 × 10⁻¹⁹ J ) - ( 4.898 × 10⁻¹⁹ J )

= 3.712 × 10⁻¹⁹ J

The excess energy over that needed for dissociation is 3.712 × 10⁻¹⁹ J

Critique this statement: Electrons can exist in any position
outside of the nucleus.

Answers

Answer:

However, there has to be 2 electrons on the first shell, and 8 on the others.

Explanation:

Hope this helps :)

list two uses of H2SO4

Answers

Drink water

Consume food
It is used in manufacture of fertilizers, pigments, dyes, drugs, explosives, detergents, and inorganic salts and
acid

It is used in preparation of OLEUM

Bond length is the distance between the centers of two bonded atoms. On the potential energy curve, the bond length is the internuclear distance between the two atoms when the potential energy of the system reaches its lowest value. Given that the atomic radii of H and Br are 37.0 pm and 115 pm , respectively, predict the upper limit of the bond length of the HBr molecule. Express your answer to three significant figures and include the appropriate units. View Available Hint(s)for Part C

Answers

Answer:

The answer is "152 pm".

Explanation:

The bond length from the values inside the atomic radii is calculated according to the query. This would be the upper limit of a molecule's binding length.

The atomic radius of [tex]H= 37.0 \ pm[/tex]

The atomic radius of [tex]Br = 115.0 \ pm[/tex]

[tex]\text{Bond length = Atomic radius of H + Atomic radius of Br}[/tex]

                    [tex]= 37.0\ pm + 115.0 \ pm\\\\= 152\ pm[/tex]

Given the standard enthalpy changes for the following two reactions: (1) 2Fe(s) + O2(g)2FeO(s)...... ΔH° = -544.0 kJ (2) 2Zn(s) + O2(g)2ZnO(s)......ΔH° = -696.6 kJ what is the standard enthalpy change for the reaction:

Answers

Answer:

-76.3 kJ

Explanation:

Here is the complete question

Given the standard enthalpy changes for the following two reactions:

(1) 2Fe(s) + O₂(g) → 2FeO(s)......ΔH° = -544.0 kJ

(2) 2Zn(s) + O₂(g) → 2ZnO(s)......ΔH° = -696.6 kJ. What is the standard enthalpy change for the reaction:

(3) FeO(s) + Zn(s) → Fe(s) + ZnO(s)......ΔH° = ?

Solution

Since (1) 2Fe(s) + O₂(g) → 2FeO(s)......ΔH° = -544.0 kJ

reversing the reaction, we have

2FeO(s) → 2Fe(s) + O₂(g) ......ΔH° = +544.0 kJ (4)

Adding reactions (2) and (3), we have

2FeO(s) → 2Fe(s) + O₂(g) ......ΔH° = +544.0 kJ (4)

2Zn(s) + O₂(g) → 2ZnO(s)......ΔH° = -696.6 kJ  (2)

This gives

2FeO(s) + 2Zn(s) → 2Fe(s) + 2ZnO(s)......ΔH° =

The enthalpy change for this reaction is the sum of enthalpy changes for reaction (2) and (3) = ΔH° = +544.0 kJ + (-696.6 kJ)

= +544.0 kJ - 696.6 kJ)

= -152.6 kJ

Since the required reaction is (3) which is FeO(s) + Zn(s) → Fe(s) + ZnO(s)

we divide the enthalpy change for reaction (4) by 2 to obtain the enthalpy change for reaction (3).

So, ΔH° = -152.6 kJ/2 = -76.3 kJ

So, the standard enthalpy change for the reaction

FeO(s) + Zn(s) → Fe(s) + ZnO(s) is -76.3 kJ

A solution is prepared by dissolving 6.60 g of an nonelectrolyte in water to make 550 mL of solution. The osmotic pressure of the solution is 1.84 atm at 25 °C. The molecular weight of the nonelectrolyte is ________ g/mol.

Answers

Answer:

160 g/mol

Explanation:

Step 1: Calculate the molarity of the solution

We will use the following expression.

π = M × R × T

where,

π: osmotic pressure of a nonelectrolyteM: molarityR: ideal gas constantT: absolute temperature (25 °C = 298 K)

M = π / R × T

M = 1.84 atm / (0.0821 atm.L/mol.K) × 298 K = 0.0752 mol/L

Step 2: Calculate the moles of solute in 550 mL (0.550 L)

0.550 L × 0.0752 mol/L = 0.0413 mol

Step 3: Calculate the molecular weight of the nonelectrolyte

0.0413 moles weigh 6.60 g.

6.60 g/0.0413 mol = 160 g/mol

The Bohr effect:_____.
a. explains through the Bohr model of the atom why Fe2+ will bind O2 in heme but Fe3+ will not.
b. contributes to binding of O2 by hemoglobin in lungs and release of O2 from hemoglobin in tissues.
c. applies to both myoglobin and hemoglobin.
d. relates [H+] to [CO2].

Answers

Answer:

b. contributes to binding of O2 by hemoglobin in lungs and release of O2 from hemoglobin in tissues.

Explanation:

The Bohr effect is a phenomenon described by Christian Bohr. Is an affinity that binds oxygen and hemoglobin and is inversely related to the concentration of carbon dioxide. As CO2 reacts with water and an increase in CO2 results in a decrease in blood ph.

FILL IN THE BLANK:
The rate of a reaction is measured by how fast a (Product Or Reactant)
is used up or how fast a
(Reactant Or Product) is formed?

Answers

Answer:

the rate of a reaction is measured by how fast a REACTANT is used up or how fast a PRODUCT is formed

For an atoms electrons, how many energy sublevels are present in the principal energy level n = 4?
A. 4
B. 9
C. 10
D. 16
E. 32

Answers

Answer:

by the own's formula energy sublevels are 2 the power of n or principal quantum number this means 2 the power of 4 equal to 32

The solubility of lithium fluoride, LiF, is 1.6 g/L, or 6.2 x 10â2 M.
a. Write the balanced solubility equilibrium equation for LiF.
b. Determine the molar concentration of the lithium ion and the fluoride ion.
c. Write the Ksp expression for the reaction.
d. Calculate Ksp for lithium fluoride.

Answers

Answer:

a. LiF(s) ⇄ Li⁺(aq) + F⁻(aq)

b. [Li⁺] = [F⁻] = 6.2 x 10⁻² M

c. Ksp = [Li⁺] [F⁻]

d. Ksp = 3.8 × 10⁻³

Explanation:

The solubility (S) of lithium fluoride, LiF, is 1.6 g/L, or 6.2 x 10⁻² M.

a. The balanced solubility equilibrium equation for LiF is:

LiF(s) ⇄ Li⁺(aq) + F⁻(aq)

b. We will make an ICE chart.

        LiF(s) ⇄ Li⁺(aq) + F⁻(aq)

I                       0           0

C                     +S         +S

E                       S           S

Then, [Li⁺] = [F⁻] = S = 6.2 x 10⁻² M

c. The solubility product constant, Ksp​, is the equilibrium constant for a solid substance dissolving in an aqueous solution.

Ksp = [Li⁺] [F⁻]

d.

Ksp = [Li⁺] [F⁻] = (6.2 x 10⁻²)² = 3.8 × 10⁻³

Phosphorus-32 is radioactive and has a half life of 14.3 days. Calculate the activity of a 3.5mg sample of phosphorus-32. Give your answer in becquerels and in curies. Round your answer to 2 significant digits.

Answers

Answer:

The activity of P-32 is 3.7x10¹³ becquerels = 1.0x10³ curies.

Explanation:

The activity of P-32 can be calculated with the following equation:

[tex] A = \lambda N [/tex]   (1)

Where:

N: is the number of atoms of P-32

λ: is the decay constant

We can find the number of atoms of P-32 as follows:

[tex] N = \frac{N_{A}*m}{M} [/tex]  (2)

Where:

[tex]N_{A}[/tex]: is the Avogadro's number = 6.022x10²³ atoms/mol

m: is the mass of P-32 = 3.5x10⁻³ g

M: is the molar mass of the radionuclide (P-32) = 32 g/mol    

Now, the decay constant is given by:

[tex] \lambda = \frac{ln(2)}{t_{1/2}} [/tex]   (3)

Where:

[tex]{t_{1/2}} [/tex]: is the half-life of P-32 = 14.3 days

Finally, we can find the activity of P-32 by entering equations (2) and (3) into (1):

[tex] A = \lambda N = \frac{ln(2)}{t_{1/2}}*\frac{N_{A}*m}{M} = \frac{ln(2)}{14.3 d*\frac{24 h}{1 d}*\frac{3600 s}{1 h}}*\frac{6.022 \cdot 10^{23} mol^{-1}*3.5 \cdot 10^{-3} g}{32 g/mol} = 3.7 \cdot 10^{13} dis/s [/tex]      

Since a becquerel (Bq) is defined as a disintegration (dis) per second, the activity in Bq is:

[tex] A = 3.7 \cdot 10^{13} Bq [/tex]

And, since a Curie (Ci) is 3.7x10¹⁰ Bq, the activity in Ci is:

[tex] A = 3.7 \cdot 10^{13} Bq*\frac{1 Ci}{3.7 \cdot 10^{10} Bq} = 1.0 \cdot 10^{3} Ci [/tex]

Therefore, the activity of P-32 is 3.7x10¹³ becquerels = 1.0x10³ curies.  

               

I hope it helps you!

A) Draw a conclusion about the relationship between enzyme activity and reaction temperature.
B) Provide an explanation for why we see a flattening of the curve for the 75 degree reaction after 3 minutes.

Answers

Answer:

Direct relationship.

Explanation:

There is direct relationship between enzyme activity and temperature of reaction. Direct relationship means if one factor is increases the other factor is also increase and vice versa. In chemical reactions, the rate of an enzyme action increases as the temperature of the chemical reaction also increases. we see a flattening of the curve for the 75 degree reaction after 3 minutes because the enzyme action is not working at that temperature or in other words, this temperature is not suitable for the enzyme activity.

An analytical chemist is titrating of a solution of benzoic acid with a solution of . The of benzoic acid is . Calculate the pH of the acid solution after the chemist has added of the solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of solution added.

Answers

The question is incomplete. The complete question is :

An analytical chemist is titrating 148.9 mL of a 1.100 M solution of benzoic acid [tex]$HC_6H_5CO_2$[/tex]  with a 0.3600 M solution of KOH. The [tex]pK_a[/tex]  of benzoic acid is 4.20. Calculate the pH of the acid solution after the chemist has added 232.0 mL of the KOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added.

Solution :

Number of moles of [tex]$C_6H_5OCOOH$[/tex] [tex]$=148.9 \ mL \times \frac{L}{1000\ mL} \times \frac{1.100 \ mol}{L}$[/tex]

                                                         = 0.16379 mol

Number of moles of NaOH added [tex]$=232.0 \ mL \times \frac{L}{1000\ mL} \times \frac{0.3600 \ mol}{L}$[/tex]

                                                         = 0.08352 mol

ICE table :

                     [tex]C_6H_5OCOOH \ \ \ + \ \ \ OH^- \ \ \ \rightarrow \ \ C_6H_5OCOO^- \ \ \ \ + \ \ H_2O[/tex]

I (mol)              0.16379                 0.08352              0

C (mol)           -0.08352              -0.08352            +0.08352

E (mol)            0.08027                    0                    0.08352

Total volume = (148.9 + 232) mL

                      = 380.9 mL

                     = 0.3809 L

Concentration of [tex]$C_6H_5OCOOH, [C_6H_5OCOOH]$[/tex] [tex]$=\frac{0.08027 \ mol}{0.3809 \ L}$[/tex]

                                                                               = 0.211 M

Concentration of [tex]$C_6H_5OCOO^- , [C_6H_5OCOO^-] =\frac{0.08352 \ mol}{0.3809 \ L}[/tex]

                                                                              = 0.219 M

[tex]pK_a[/tex] of [tex]C_6H_5OCOOH = 4.20[/tex]

According to Henderson equation,

[tex]$pH = pK_a + \log \frac{[C_6H_5OCOO^-]}{[C_6H_5OCOOH]}[/tex]

     [tex]$=4.20 + \log \frac{0.219}{0.211}$[/tex]

     = 4.22

Therefore, the pH of the acid solution is 4.22


(a) (i) What is the name of apparatus used to measure conductivity of water​

Answers

Answer:

An electrical conductivity meter (EC meter) measures the electrical conductivity in a solution. It has multiple applications in research and engineering, with common usage in hydroponics, aquaculture, aquaponics, and freshwater systems to monitor the amount of nutrients, salts or impurities in the water.

0. When measuring tert-butyl alcohol for this experiment, a student first weighs an empty graduated cylinder, then pours 15 mL of the alcohol into the graduated cylinder and weighs the cylinder again. He records the amount of alcohol used as the difference in these two masses. What is wrong with this method

Answers

Answer:

Both have solutions in the  graduated cylinder.

Explanation:

Recording the amount of alcohol used as the difference between two masses is the wrong method used for measuring tert-butyl alcohol for the experiment.  For measuring tert-butyl alcohol for this experiment, the student has to measure the two masses when both the graduated cylinders has solution of tert-butyl alcohol not when one of it is empty (having no tert-butyl alcohol ).

The wrong aspect is that the liquid didn't need to be weighed. Instead the volume should have been recorded with the aid of the graduated cylinder.

What is a Graduated cylinder?

This is a cylinder with marked readings and is used to measure the volume of liquids in the laboratory.

The graduated cylinder will accurately measure the amount of  alcohol used due to it being volatile and the mass fluctuating during the measurement.

Read  more about Graduated cylinder here https://brainly.com/question/24869562

HBr can be added to an alkene in the presence of peroxides (ROOR). What function does the peroxide serve in this reaction

Answers

Answer:

Radical chain initiator

Explanation:

The peroxide here serves as a radical chain initiator. In the field of chemistry the radical initiatives are those substances that are used in industrial processes like polymer synthesis. These initiatives have weak bonds generally and they're mostly used to create free radicals. These radicals are atoms that have odd numbers of electrons. Peroxide is an example of such.

The mole fraction of NaCl in an
aqueous solution is 0.132. How
many moles of NaCl are present in
1 mole of this solution?
Molar Mass
NaCl: 58.44 g/mol
H2O: 18.016 g/mol

Answers

Answer:

Moles of water are 0.868

Explanation:

Determine the total pressure of a mixture that contains 5.25 g of He and 3.25 g of N2 in a 7.75-L flask at a temperature of 27ºC.

Answers

Answer:

4.54 atm

Explanation:

Step 1: Calculate the total number of gaseous moles

We will calculate the moles of each gas using its molar mass.

He: 5.25 g × 1 mol/4.00 g = 1.31 mol

N₂: 3.25 g × 1 mol/28.01 g = 0.116 mol

The total number of moles is:

n = 1.31 mol + 0.116 mol = 1.43 mol

Step 2: Convert 27 °C to Kelvin

We will use the following expression.

K = °C + 273.15 = 27 + 273.15 = 300 K

Step 3: Calculate the total pressure of the mixture

We will use the ideal gas equation.

P × V = n × R × T

P = n × R × T / V

P = 1.43 mol × (0.0821 atm.L/mol.K) × 300 K / 7.75 L = 4.54 atm

sublimation is a change from the solid phase to the____phase

Answers

Answer:

solid to gaseous or gaseous to solid

Explanation:

Sublimation is the transition of a substance directly from the solid to the gas state, without passing through the liquid state. sublimation is most often used to describe the process of snow and ice changing into water vapor in the air without first melting into water.

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