Question 7 When of benzamide are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezing point of pure . On the other hand, when of iron(III) chloride are dissolved in the same mass of , the freezing point of the solution is lower than the freezing point of pure . Calculate the van't Hoff factor for iron(III) chloride in . Be sure your answer has a unit symbol, if necessary, and round your answer to significant digits.

Answers

Answer 1

Answer:

See explaination

Explanation:

moles of benzamide = mass / Molar mass of it = 70.4g / ( 121.14g/mol) = 0.58 mol

Molality = moles of solute ( benzamide) / ( solvent mass in kg) = 0.58 mol / ( 0.85kg) = 0.6837

we have formula dT = i x Kf x m , where dT = change in freezing point = 2.7C , i = vantoff factor = 1 for non dissociable solutes , Kf = freezing oint constant of solvent , m = 0.6837

hence 2.7C = 1 x Kf x 0.6837m

Kf = 3.949 C/m

we use this Kf value for calculating i for NH4Cl , where moles of NH4Cl = ( 70.4g/53.491g/mol) =1.316 mol

molality = ( 1.316mol) / ( 0.85kg) = 1.5484 , dT = 9.9

hence 9.9 = i x 3.949C/m x 1.5484 m

i = 1.62


Related Questions

g A mixture of gases contains 6.46 g of N2O, 2.74 g of CO, and 5.40 g of O2. If the total pressure of the mixture is 4.33 atm, what is the partial pressure of each component? a) P(N2O) = 0.635 atm, P(CO) = 0.424 atm, and P(O2) = 3.27 atm. b) P(N2O) = 2.31 atm, P(CO) = 0.622 atm, and P(O2) = 1.40 atm. c) P(N2O) = 1.54 atm, P(CO) = 1.02 atm, and P(O2) = 1.77 atm. d) P(N2O) = 0.999 atm, P(CO) = 0.371 atm, and P(O2) = 2.96 atm. e) P(N2O) = 1.28 atm, P(CO) = 1.93 atm, and P(O2) = 1.12 atm.

Answers

Answer:

The correct answer is c) P(N2O) = 1.54 atm, P(CO) = 1.02 atm, and P(O2) = 1.77 atm

Explanation:

In order to calculate the partial pressures of the mixture components, we have to first calculate the number of moles:

For N₂O:

Molecular weight (MW): (14 g/mol x 2) + 16 g/mol= 44 g/mol

Number of moles of N₂O (n) = mass/Mw = 6.46g/44 g/mol= 0.1468 mol

For CO:

Molecular weight (MW): 12 g/mol + 16 g/mol= 28 g/mol

Number of moles of CO (n) = mass/Mw = 2.74 g/28 g/mol= 0.0978 mol

For O₂:

Molecular weight (MW): 16 g/mol x 2= 32 g/mol

Number of moles of O₂ (n) = mass/Mw = 5.40 g/32 g/mol= 0.1687 mol

Once calculated the number of moles of each component, we can calculate the total number of moles (nt):

nt = 0.1468 mol + 0.0978 mol + 0.1687 mol = 0.4133 moles

The partial pressure of a gas in a mixture can be calculated from the molar fraction of the gas (X) and the total pressure of the mixture (Pt=4.33 atm):

P(N₂O) = X(N₂O) x Pt

           = (moles N₂O/nt) x Pt

           = 0.1468 moles/0.4133 moles x 4.33 atm

           = 1.538 atm

P(CO) = X(CO) x Pt

           = (moles CO/nt) x Pt

           = 0.0978 moles/0.4133 moles x 4.33 atm

           = 1.0246 atm

P(O₂) = X(O₂) x Pt

           = (moles O₂)/nt x Pt

           = 0.1687 moles/0.4133 moles x 4.33 atm

           = 1.767 atm

A tank contains helium gas at 191 mm HG nitrogen gas at 0.261 ATM and neon at 522 torr what is the total pressure in MM Hg

Answers

Answer:

911.36mmhg

Explanation:

1 torr is almost equivalent to 1mmhg.but to convert from atm to mmhg,multiply by 760

Final answer:

The total pressure in the tank, after converting all pressures to the same unit (mm Hg), is the sum of the partial pressures of helium, nitrogen, and neon, resulting in a value of 911.36 mm Hg.

Explanation:

To calculate the total pressure in a mixture of gases, you can use Dalton's Law of partial pressures which states that the total pressure exerted by a mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases.

Since the pressure values are given in different units, we first need to convert them all to millimeters of mercury (mm Hg). The pressure of helium gas is already given in mm Hg. The nitrogen gas pressure needs to be converted from atmospheres (ATM) to mm Hg by multiplying by 760 mm Hg/ATM, and the neon gas pressure in torr simply needs to be directly converted to mm Hg because 1 torr equals 1 mm Hg.

Now we can sum these values to find the total pressure in the tank.

The conversion and summation are as follows:

Helium: 191 mm Hg

Nitrogen: 0.261 ATM x 760 mm Hg/ATM = 198.36 mm Hg

Neon: 522 torr = 522 mm Hg

Adding these together, the total pressure in the tank is:

191 mm Hg + 198.36 mm Hg + 522 mm Hg = 911.36 mm Hg

Therefore, the total pressure in the tank in mm Hg is 911.36 mm Hg.



c. What predictions can you make about ΔH, ΔS, and ΔG for the general reaction A(s) + 2B(g) 3C(g) + D(g) + heat? What conclusions can you make about the spontaneity of the reaction? Explain your reasoning. (4 points)


d. The reaction 2NO2(g) N2(g) + 2O2(g) is spontaneous at all temperatures, yet it hardly happens under normal conditions. How can this be? (2 points)

Answers

ANSWER:

C.

1) Entropy (∆S) is spontaneous

2) Enthalpy (∆H) is not spontaneous

3) Gibbs free energy (∆G) is spontaneous

Therefore the reaction will be spontaneous at high temperature.

D.

Because Nitrogen is favourable to be produced under high temperature, and oxygen under low temperature. Which favours the product side of the equation. But when at a room temperature, which means the temperature is neither low nor high, the product side won't be favoured, and the reaction will not be spontaneous.

EXPLANATION:

C.

1) Entropy is the measure of disorderliness in a system, which increase more in gaseous substances, because the molecules of gases are not stable

Because almost all the reacting substances are in their gaseous state, the entropy of the reaction will be high. Therefore entropy will be positive, which makes the entropy of the system spontaneous.

2) Enthalpy is the measure of heat change in the system. Since their is an intake of heat in the system, therefore the reaction is endothermic and ∆H will be positive. Enthalpy of a system can only be spontaneous in an Exothermic reaction, where ∆H is negative. Therefore Enthalpy is not spontaneous.

3) Gibbs free energy is equal to the change in enthalpy minus the product of temperature and change in entropy. Since entropy and enthalpy are positive, the Gibbs free energy will be negative, which shows that the reaction can be spontaneous if some conditions are met. ∆G will be spontaneous because it is negative.

Therefore in summary, the reaction will favour the product side more, if the temperature of the system is increased, which will make the reaction to become more spontaneous.

D.

2NO2(g) --> N2(g) + 2O2(g)

This reaction is not spontaneous under atmospheric pressure and room temperature (normal conditions) because, Nitrogen can only be produced very fast at a high temperature, while oxygen production is preferable in a low temperature. For the reaction to favour the product side, the temperature should be increased or decreased. Because an increase or a decrease in the temperature will favour either nitrogen or oxygen, which are the product side of equation. This means that the reaction should not be spontaneous if you wish to achieve an equal proportion of the products.

What information about water is needed to calculate the enthalpy change for converting 1 mol H2O(g) at 100 °C to H2O(l) at 80 °C? (a) Heat of fusion, (b) heat of vaporization, (c) heat of vaporization and specific heat of H2O(g), (d) heat of vaporization and specific heat of H2O(l), (e) heat of fusion and specific heat of H2O(l).

Answers

Final answer:

The information needed includes the heat of vaporization and specific heat of H₂O(l).

Explanation:

The information needed to calculate the enthalpy change for converting 1 mol H₂O(g) at 100 °C to H₂O(l) at 80 °C is the (d) heat of vaporization and specific heat of H₂O(l).

The heat of vaporization is required because it accounts for the energy needed to convert water from a liquid to vapor state. The specific heat of H₂O(l) is needed to determine the amount of heat required to change the temperature of liquid water from 100 °C to 80 °C.

2. Carbon monoxide is a poisonous carbon oxygen compound (as opposed to carbon
dioxide). Provide the complete MO diagram including the atomic orbitals for both
carbon and oxygen as well as the MOs for CO assuming the bonding is the same as for
C2 in the reference. Also indicate the bond order for CO from the MO diagram. (14
points)

Answers

Answer:

Carbon monoxide has a bond order of 3

Explanation:

A molecular orbital diagram ( MO diagram ), is a qualitative descriptive tool tha explains chemical bonding in molecules in terms of the molecular orbital theory in general and particularly adopts the linear combination of atomic orbitals (LCAO) method in its approach. A fundamental principle of these theories is that as atoms bond to each other to form molecules, a certain number of atomic orbitals combine to form an equal number of molecular orbitals , although the electrons involved may be redistributed among the orbitals. This tool is very well suited for simple diatomic molecules such as hydrogen molecule , oxygen molecule, and carbon monoxide but becomes more complex when discussing even comparatively simple polyatomic molecules, such as methane. MO diagrams usually explains why some molecules exist and others do not exist. It is used to predict bond strength of molecules, as well as the electronic transitions that can take place.

The MO diagram for carbon monoxide is shown in the image attached. It can clearly be seen that carbon monoxide has a bond order of three. The combination of carbon and oxygen atoms to form molecular orbitals of appropriate energy is also seen in the image.

Given a balanced chemical equation below: 3Cu(s) + 2H3PO4 --- > Cu3(PO4)2 + 3H2 How many moles of copper are needed to react with 5 moles of phosphoric acid?

Answers

Answer:

7.5 moles

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

3Cu + 2H3PO4 —> Cu3(PO4)2 + 3H2

From the balanced equation above,

3 moles of Cu reacted with 2 moles of H3PO4.

Therefore, Xmol of Cu will react with 5 moles of H3PO4 i.e

Xmol of Cu = (3 x 5)/2

Xmol of Cu = 7.5 moles

Therefore, 7.5 moles of Cu are needed to react with 5 moles of H3PO4.

Which pair of elements can be used to determine the age of a fossil that is over one billion years old?

Answers

Answer:

Potassium and Argon (⁴⁰K-⁴⁰Ar)

Explanation:

Potassium-Argon dating is a method used in geology and archeology to date rocks or volcanic ash, generally the oldest.

It is based on the principle of radioactive decay, it is a process that has a half-life of 1248 million years, during which time the gas is concentrated in the rock crystals. Taking advantage of this known rhythm and half-life, the method lends itself to dating samples ranging from 100.000 years to several billion years.

Answer:

Its D to sum it up.

Explanation:

After the end of a normal inspiration, the volume of air in the lungs is about 2.8 L. Normally quiet inspiration is driven by a pressure difference of about 2 mm Hg. The air in the lungs is at 37C and after normal expiration it is at atmospheric pressure. Quiet inspiration is driven by the expansion of the chest cavity by contraction of the diaphragm, which expands the air in the lungs. How much is the air expanded to produce an decrease of 2 mmHg in pressure

Answers

Answer:

The Volume of the lungs that would produce 2 mmHg pressure decrease is

         [tex]V_2 = 2.81 \ L[/tex]

Explanation:

From the question we are told that

     The volume of air in the lungs is  [tex]V = 2.8 \ L[/tex]

     The pressure difference for quit normal inspiration is [tex]P = 2 \ mmHg[/tex]

      The temperature of air in the lungs [tex]T = 37^oC[/tex]

      The pressure  after normal  expiration is at  [tex]T = 760 \ mmHg[/tex]

     

From ideal gas law we have that

         [tex]PV= nRT[/tex]

Now since  nRT is constant we have that

        [tex]P_1 V_1 = P_2 V_2[/tex]

As the pressure decreased by 2 mmHg the volume becomes

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

        [tex]V_2 = \frac{2.8 * 760}{758}[/tex]

        [tex]V_2 = 2.81 \ L[/tex]

       

     

Final answer:

The change in the lung volume during the process of quiet inspiration can be calculated using Boyle's Law by determining the change in pressure and relating it to an equivalent change in volume. Once pressure values are converted to the same units, the Boyle's Law equation (P1V1 = P2V2) is used to solve for the final volume (V2) that represents the expanded lung volume.

Explanation:

The changes in volume and pressure in the lungs during inspiration (breath in) can be described using Boyle's Law, which states that the pressure and volume of a gas have an inverse relationship, when temperature is held constant. In this scenario, a decrease in pressure by 2 mmHg drives an expansion of the lungs, and we are asked to determine this corresponding change in volume.

Firstly, the pressure change needs to be converted into standard pressure units - the SI unit is Pascal. 1 mm Hg equals 133.322 Pa, so a 2 mm Hg difference equals 266.644 Pascal (Pa).

Using Boyle's Law equation (P1V1=P2V2), where P1=the initial pressure, V1=the initial volume, P2=the final pressure (P1 - 2mm of Hg) and V2=the final volume that we are trying to find, we can solve for V2. P1 is atmospheric pressure, which is approximately 101,325 Pa, and V1 is the initial volume of air in lungs, which is 2.8 Litres.

Solving the equation gives the final volume, V2, after the pressure decrease and resulting lung expansion.

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Determine the number of bonding electrons and the number of nonbonding electrons in the structure of SI2. Enter the number of bonding electrons followed by the number of nonbonding electrons, separated by a comma, in the dot structure of this molecule.

Answers

Final answer:

The molecule SI2 does not exist but if the student meant SiI2, it would have 8 bonding electrons and 14 nonbonding electrons.

Explanation:

The molecule SI2 does not exist. It may have been a typo. If the student meant Si2, this molecule also doesn't exist because silicon, a member of group 14 on the periodic table, needs 4 electrons to achieve a full outer shell, and thus generally forms 4 bonds. If we are talking about SiI2, or silicon diiodide, it would have 8 bonding electrons and 14 nonbonding electrons. This is because all four of the bonding electrons from silicon and four from iodine atoms make up the bonding electrons while the remaining 7 electrons from each iodine atom make up the nonbonding electrons.

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Which of the following is used to represent each element?
formula
equation
symbol
O
coefficient

Answers

pretty sure it would be formula
the corret one would be formula

What volume would 32.0 g of NO2 gas occupy at 3.12 atm and 18.0 Celsius

Answers

Answer:

Volume of 32.0 g [tex]NO_{2}[/tex] at 3.12 atm and 18.0 [tex]^{0}\textrm{C}[/tex] is 5.33 L.

Explanation:

Molar mass of [tex]NO_{2}[/tex] = 46.0055 g/mol

So, 32.0 g of [tex]NO_{2}[/tex] = [tex]\frac{32.0}{46.0055}[/tex] mol of [tex]NO_{2}[/tex] = 0.696 mol of [tex]NO_{2}[/tex]

Let's assume [tex]NO_{2}[/tex] behaves ideally.

Then [tex]P_{NO_{2}}V_{NO_{2}}=n_{NO_{2}}RT[/tex]    , where P, V, n, R and T represents pressure, volume, number of moles, gas constant and temperature (in kelvin scale) respectively.

Here [tex]P_{NO_{2}}[/tex] = 3.12 atm, [tex]n_{NO_{2}}[/tex] = 0.696 mol, R = 0.0821 L.atm/(mol.K) and T = (273+18.0)K = 291 K

So,       [tex]V_{NO_{2}}=\frac{n_{NO_{2}}RT}{P_{NO_{2}}}[/tex]

        or, [tex]V_{NO_{2}}=\frac{(0.696mol)\times (0.0821\frac{L.atm}{mol.K})\times (291K)}{3.12atm}[/tex]

        or, [tex]V_{NO_{2}}=5.33L[/tex]

So, volume of 32.0 g [tex]NO_{2}[/tex] at 3.12 atm and 18.0 [tex]^{0}\textrm{C}[/tex] is 5.33 L.

The Volume of  NO₂ is 542.8L

What is Ideal Gas Law ?

Ideal Gas Law  relates the macroscopic property of gas , It states that the product of pressure and volume of 1 mole of a gas is equal to the product of absolute temperature and universal gas constant.

For n mole of gas,

PV = nRT

In the question it is given that mass of NO₂ is 32 g

P = 3.12 atm

T = 18°Celsius = 291 K

Volume = ?

n = m/M = 32/46 = 0.7

3.12 * V = 0.7*0.0821*291

V = 5.36 L

Therefore the Volume of  NO₂ is 5.36L

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A student adds solid KCl to water in a flask. The flask is sealed with a stopper and thoroughly shaken until no more solid KCl dissolves. Some solid KCl is still visible in the flask. The solution in the flask is A) saturated and is at equilibrium with the solid KCl B) saturated and is not at equilibrium with the solid KCl C) unsaturated and is at equilibrium with the solid KCl D) unsaturated and is not at equilibrium with the solid KCl

Answers

Answer:

Option (A) saturated and is at equilibrium with the solid KCl

Explanation:

A saturated solution is a solution which can not dissolve more solute in the solution.

From the question given above, we can see that the solution is saturated as it can not further dissolve any more KCl as some KCl is still visible in the flask.

Equilibrium is attained in a chemical reaction when there is no observable change in the reaction system with time. Now, observing the question given we can see that there is no change in flask as some KCl is still visible even after thorough shaking. This simply implies that the solution is in equilibrium with the KCl solid as no further dissolution occurs.

What happens if you break a magnet in half?


a) Each half will be a new magnet, with both a north and south pole.


b) One half will have a north pole only and one half will have a south pole only.


c) Neither half will have a pole.


d) Neither half will be able to attract or repel

Answers

Answer:

neither half will have a pole

Competition occurs when two or more organisms within an ecosystem seek the same resource. Which of the following is an example of a resource that organisms might compete for?
habitat
water
sunlight
food

Answers

Answer:

the answer is all of them

Explanation:

it is all of them because organisms in and ecosystem compete for anything and every thing that they need. Hope this helps!

19.24 The mechanism for acetal hydrolysis has been heavily investigated. In one study, which explored rates as well as stereochemical aspects,[5] compound 1 was treated with aqueous acid to afford compound 2. Draw the structure of 2, clearly showing the configuration of any chiral center(s). c19s124

Answers

Answer:

Explanation:

check the attached file below for the diagram and better expalnation.

4.
What volume of 0.120 M HNO3(aq) is needed to
completely neutralize 150.0 milliliters of 0.100 M
NaOH(aq)?
A. 62.5 mL
B. 125 ml
C.
180. mL
D. 360. mL

Answers

Answer:

B) 125 mL

Explanation:

M1V1=M2V2

(0.120M)(x)=(150.0 mL)(0.100M)

x= 125 mL

*Text me at 561-400-5105 for private tutoring if interested: I can do homework, labs, and other assignments :)

Consider an electrolytic cell with a platinum anode and a silver cathode in a 1.0 M AgNO3(aq) solution.


a) (3 pts) What species can be reduced in this solution? Which species is preferentially reduced? Write the reduction half- reaction. (Note that oxyanions like nitrate are not commonly reduced in aqueous electrolysis due to kinetic reasons.)


b) (2 pts) Which species is oxidized during the electrolysis? Write the oxidation half-reaction. Note that acid will form as a byproduct of the oxidation.


c) (2 pts) Write the overall electrolysis reaction in net ionic and molecular forms.


d) (2 pts) Determine the external electric potential needed for the electrolysis under standard conditions.


e) (5 pts) If the electrolysis is carried out for 1.00 hour using 1.00 A current, how many grams of metal will be deposited at the cathode and how many liters of gas will form at the anode at 1.00 atm pressure and 25°C?

Answers

Answer:

Check the attached image below

Explanation:

The step by step solution to the question above can be seen in the attached image below.

A 475 ml sample of a gas was collected at room temperature of 23.5 °C and a pressure of

756 mm Hg. Calculate the volume of the gas if the conditions were altered to 10.0 °C and a

pressure of 722 mm Hg.

Answers

Answer:

The volume when the conditions were altered is 0.5109 L or 510.9 mL

Explanation:

Using the general gas equation,

P1 V1 / T1 = P2 V2 / T2

where;

P1 = 756 mmHg

V1 = 475 ml = 0.475 L

T1 = 23.5°C = 23.5 + 273K = 275.5 K

P2 = 722 mm Hg

T2 = 10°C = 10 + 273 K = 283 K

V2 = ?

Rearranging to make V2 the subject of the formula, we obtain:

V2 = P1 V1 T2 / P2 T1

V2 = 756 * 0.475 * 283 / 722 * 275.5

V2 = 101, 625.3 / 198911

V2 = 0.5109 L or  510.9 mL

Calculate the rate of emission of SO2 in g/s that results in a centerline (y = 0) concentration at ground level of 1.786 x 10-3 g/m3 one kilometer from the stack. The time of measurements was 1 P.M. on a clear summer afternoon. The wind speed was 2.0 m/s measured at a height of 10 m. The effective stack height is 96 m. No inversion is present.

Answers

Answer:

790 g/s

Explanation:

The first thing to do is to make sure to calculate the standard deviation of the y-axis and the z-axis and the values for both axis are; 215 m and 450 m for y-axis and z-axis respectively.

The next thing to do now is to find the rate of emission of SO2 in units of g/s using the relation by using the formula below;

Emission rate of pollutant, E = (x × π × by × bz × g) ÷ [ e^(-1/2) × (y/by)^2] × [e^(-1/2)× (h/bz)^2].

Where g = wind direction, y and h are the distance in metres.

Therefore, slotting in the values into the Emission rate of pollutant equation above, we have;

Emission rate of pollutant,E =[ 1.412 × 10^-3] × π × 215 × 450 × 1.8] ÷ [e^(-1/2)(0/215)^2] × e^ (-1/2)(94/450)^2.

Emission rate of pollutant,E = 790 g/s.

14 H+ + 6 Fe2+ + Cr2O72- . 2 Cr3+ + 6 Fe3+ + 7 H2O This reaction is used in the

titration of an iron solution. What is the concentration of the iron solution if it takes 45.20 mL

of 0.1000 M Cr2O72- solution to titrate 50.00 mL of an acidified iron solution?

a.

0.5424 M

b.

0.1000 M

c.

1.085 M

d.

0.4520 M

e.

0.2712 M

Answers

Answer:

Option A 0.5424 M

Explanation:

To do this, let's write the reaction again and see the mole ratio:

14H⁺ + 6Fe²⁺ + Cr₂O₇⁻ ---------> 2Cr³⁺ + 6Fe³⁺ + 7H₂O

According to this reaction, we have a mole ratio between the dicromate and iron of 1:6. In other words, 6 moles of dicromate reacts with only 1 mole of iron, therefore we can use the following expression:

nFe = 6 * nCr₂O₇

This is a REDOX reaction, where Iron is being oxidized while cromium is being reduced. Now, we have the concentrations and volume, let's express the moles in function of concentration and volume:

M₁V₁ = 6M₂V₂

Replacing the given data above and solving for concentration of iron (M₁)

M₁ = 6 * 45.20 * 0.1 / 50

M₁ = 0.5424 M

This is the concentration of the iron solution

What is the pH of a buffer solution that is 0.211 M in hypochlorous acid (HClO) and 0.111 M in sodium hypochlorite? The Ka of hypochlorous acid is 3.8 ⋅ 10-8. What is the pH of a buffer solution that is 0.211 M in hypochlorous acid (HClO) and 0.111 M in sodium hypochlorite? The Ka of hypochlorous acid is 3.8 x 10-8.


a. 7.7

b. 0

c. 7.14

d. 14.28

e. 9.05

f. 6.86

Answers

Answer:

c. 7.14

Explanation:

The buffer solution is formed by a weak acid ( hypochlorous acid, HClO) and its conjugate base (hypochlorite ClO⁻, coming from sodium hypochlorite NaClO). We can calculate the pH using the Henderson-Hasselbach equation.

pH = pKa + log [base]/[acid]

pH = -log 3.8 × 10⁻⁸ + log 0.111/0.211

pH = 7.14

A calorimeter weighing 123.7g has a quantity of 20C water added to it. The combined mass of the calorimeter + cold water is 198.3g.
61g of water is heated to 60C and is poured into the calorimeter. The temperature of the mixed cold and hot water and calorimeter is 38.5C. What is the calorimeter constant?

answer choices :
a ) 148.19 b) 91.29 c) 107.75 d) 161.58

Answers

Answer:

c) 107.75

Explanation:

Hot water lost = 61 g * 60C * (4.184 J g¯1 °C¯1) = 3,660

Cold water = 74.6g * 20C  * (4.184 J g¯1 °C¯1) = 1,492

The difference is 3,660 - 1,492 = 2,168

Calorimeter Constant = Heat released by burning / Change in temperature

Calorimeter Constant = 2,168 / 40C * (1.987 J g¯1 °C¯1)

Calorimeter Constant = 107.75

Some people must eat a low-sodium diet with no more than 2,000 mg of sodium per day. By eating 1 cracker, 1 pretzel, and 1 cookie, a person would ingest 149 mg of sodium. If a person ate 8 pretzels and 8 cookies, he or she would ingest 936 mg of sodium. By eating 6 crackers and 7 pretzels, the person would take in 535 mg of sodium. Find the amount of sodium in each.

Answers

Answer:

The amount of sodium is 32 mg per cracker, 49 mg per pretzel and 68 mg per cookie.

Explanation:

Let's assume amount of sodium is x mg per cracker, y mg per pretzel and z mg per cookie.

So, the following three equations can be written as per given information:

x+y+z = 149 ........(1)

8y+8z = 936 ........(2)

6x+7y = 535 .........(3)

From equation- (2), y+z = [tex]\frac{936}{8}[/tex] = 117

By substituting the value of (y+z) in equation- (1) we get,

                          x = 149-(y+z) = 149-117 = 32

By substituting the value of x into equation- (3) we get,

                           y = [tex]\frac{535-(6\times 32)}{7}[/tex] = 49

By substituting the value of y  into equation- (2) we get,

                           z = (117-49) = 68

So, the amount of sodium is 32 mg per cracker, 49 mg per pretzel and 68 mg per cookie.

Final answer:

To find the amount of sodium in each item, set up a system of equations and solve to get x = 9, y = 55, and z = 85.

Explanation:

To find the amount of sodium in each item, we can set up a system of equations.

Let x represent the amount of sodium in 1 cracker, y represent the amount of sodium in 1 pretzel, and z represent the amount of sodium in 1 cookie.

From the given information, we can create the following equations:

x + y + z = 1498y + 8z = 9366x + 7y = 535

Solving this system of equations, we can find the values of x, y, and z. The amount of sodium in 1 cracker is 9 mg, 1 pretzel is 55 mg, and 1 cookie is 85 mg.

At constant pressure, 50.0 mL of 0.100 M KOH and 45.0 mL of 0.100 M HNO3 are mixed in a styrofoam cup.The two solutions are initially at 25.00 °C and the final temperature is 25.65 °C. Calculate the ΔHneutralization (kJ/mol) in term of moles of HNO3 since it’s the limiting reagent. Assume a specific heat capacity of 4.184 J/(°C•g) and that the density of the solution is the same as water (0.997 g/mL).

Answers

Answer:

≈ -57.2 kJ/mol

Explanation:

Total volume of the solution = (50.0 + 45.0) mL = 95.0 mL.

Density of the solution = 0.997 g/mL.

Mass of the solution = (volume of the solution)*(density of the solution)

= (95.0 mL)*(0.997 g/mL)

= 94.715 g

Heat gained by the solution, qsoln = (mass of the solution)*(specific heat capacity)*(change in temperature)

= (94.715 g)*(4.184 J/ºC.g)*(25.65 – 25.00)ºC

= 257.5869 JAccording to the principle of thermochemistry,

qsoln + qrxn = 0

where qrxn denotes the heat change during the neutralization reaction.

Therefore,

(257.5869 J) + qrxn = 0

======> qrxn = -257.5869 J

HNO3 is the limiting reactant.

Moles of limiting reactant = (volume of the limiting reactant in L)*(molarity of limiting reactant)

= (45.0 mL)*(0.100 M)

= (45.0 mL)*(1 L)/(1000 mL)*(0.100 M)

= 0.0045 mole.

ΔHneutralization = qrxn/(moles of HNO3)

= (-257.5869 J)/(0.0045 mol)

= -57241.5 J/mol

= (-57241.5 J/mol)*(1 kJ)/(1000 J)

= -57.2415 kJ/mol

≈ -57.2 kJ/mol

A sample of helium gas occupies a volume of 152.0 mL at a pressure of 717.0 mm Hg and a temperature of 315.0 K. What will the volume be at a pressure of 463.0 mm Hg and a temperature of 777.0 K? The combined gas law equation is given below. P1V1 T1 = P2V2 T2 In this equation, P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 are the final pressure, volume, and temperature, respectively.

Answers

Answer:

0.581 L  or  581 mL

Explanation:

As stated in the question, the combined gas law is (P1*V1/T1) = (P2*V2/T2)

Write down the amounts you are given.

V1 = 0.152 L (I was taught to always convert milliliters to liters)

P1 = 717 mmHg

T1 = 315 K

V2 = ?

P2 = 463 mmHg

T2 = 777 K

The variable that is being solved for is final volume. Fill in the combined gas law equation with the corresponding amounts and solve for V2.

(717 mmHg*0.152 L) / (315 K) = (463 mmHg*V2) / (777 K)

0.346 = (463*V2) / (777)

0.346*777 = (463*V2) / (777)*777

268.842 = 463*V2

268.842/463 = (463*V2)/463

V2 = 0.581

Pressure and volume are indirectly proportional. This checks out because the volume increased while pressure decreased. Volume and temperature are directly proportional. This checks out because both volume and temperature increased. This is a good way to check your answers. You can also solve each side of the combined gas law equation to see if they are both the same.

How do neutrons stars generate the energy they need to survive

Answers

Ordinary stars maintain their spherical shape because the heaving gravity of their gigantic mass tries to pull their gas toward a central point, but is balanced by the energy from nuclear fusion in their cores, which exerts an outward pressure

The neutron stars are the smallest but heaviest density supergiant star. The surfaces of the neutron stars undergo thermonuclear fusion to produce energy to survive.

What is thermonuclear fusion?

Thermonuclear fusion or just simply nuclear fusion is a nuclear reaction that involves the combining (fusion) of the atomic nuclei together to produce different nuclei in the presence of high temperatures.

Unlike nuclear fission, it needs high temperature and energy to produce atomic nuclei. It is a process generally seen in the universe including the Sun, stars, galaxies, etc.

Neutron stars also use their energy to undergo thermonuclear fusion and produce heavy nuclei from lighter nuclei. The hydrogen atom combines fuses to form helium atoms resulting in high energy production.

Therefore, neutron stars generate energy by thermonuclear fusion.

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Butane (C4H10) burns completely with 110% of theoretical air entering at 74°F, 1 atm, 50% relative humidity. The dry air component can be modeled as 21% O2 and 79% N2 on a molar basis. The combustion products leave at 1 atm. For complete combustion of butane(C4H10) with theoretical amount of air, what is the number of moles of oxygen (O2) per mole of fuel?

Answers

To achieve complete combustion of one mole of butane (C₄H₁₀), 6.5 moles of oxygen (O₂) are necessary, as indicated by the balanced chemical equation for combustion.

To determine the number of moles of oxygen (O₂) required for the complete combustion of one mole of butane (C₄H₁₀), we must first write the balanced chemical equation for the combustion reaction:

C₄H₁₀ + O₂ → CO₂ + H₂O

After balancing the equation, we have:

2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

This balanced equation tells us that for every 2 moles of butane, we need 13 moles of oxygen for complete combustion, which means that for every mole of butane, we need 6.5 moles of oxygen.

9) 20 g of calcium chloride per 100 g of water at 90°C will be...
10 points
Сас,
KNO,
K₂ Cigar
Solubility (g of salt in 100 g H,0)
Pb(NO3)2
KCR
20
A
KCIO,
0
10
Ce (80)
20 30 40 50 60 70 80 90 100
Temperature (°C)
O
saturated
O
unsaturated
O
supersaturated

Answers

According to solubility, 20 g of calcium chloride per 100 g of water at 90°C will be saturated solution.

What is solubility?

Solubility is defined as the ability of a substance which is basically solute to form a solution with another substance. There is an extent to which a substance is soluble in a particular solvent. This is generally measured as the concentration of a solute present in a saturated solution.

The solubility mainly depends on the composition of solute and solvent ,its pH and presence of other dissolved substance in it . It is also dependent on temperature and pressure which is maintained through the process.Concept of solubility is not valid for chemical reactions which are irreversible. The dependency of solubility on various factors is due to interactions between the particles, molecule or ions.

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Consider the titration of a 20.0 mL sample of 0.500 M HCN (Ka =6.17x10-10) with 0.250 M KOH. a. (6pt) What is the initial pH? b. (7pt) What is the pH at 6.00 mL base added? c. (8pt) What is the pH when 40.00 mL of KOH is added to reach the equivalence point? d. (8pt) What is the pH if 42.00 mL of KOH is titrated with the solution?

Answers

Answer:

a. pH = 4.75

b. pH = 9.20

c. pH = 8.42

d. pH = 13.53

Explanation:

This is a titration between a strong base, the KOH and a weak acid, HCN.

The initial pH is the pH, when you did not add the base yet, so it is the pH of the HCN

          HCN + H2O ⇄  H₃O⁺  +  CN⁻

Initial    0.5                      -             -

Eq.      0.5-x                    x             x

Ka =  x² / (0.5-x) = 6.17ₓ10⁻¹⁰

Ka is really small, so we can say that 0.5-x = 0.5. Then,

x² = 6.17ₓ10⁻¹⁰ . 0.5

x = √(6.17ₓ10⁻¹⁰ . 0.5) = 1.75×10⁻⁵ → [H₃O⁺]

pH = - log [H₃O⁺]  →  - log 1.75×10⁻⁵ = 4.75

b. First of all, we determine the moles of base, we are adding.

0.250 mol/L . 0.006 L = 0.0015 moles

In conclussion we have 0.0015 moles of OH⁻

Now, we determine the moles of our acid.

0.500 mol/L . 0.020L = 0.01 moles

The  0.0015 moles of OH⁻ will be neutralized with the acid, so:

      HCN     +    OH⁻         →     H₂O   +    CN⁻

       0.01         0.0015                          0.0085

The hydroxides are neutralized with the proton from the weak acid, so we have 0.0085 moles of cyanide and 0.0085 moles of HCN. (0.01-0.0015)

Our new volume is 20 mL and 6mL that we added, so, 26mL

This is a buffer with the weak acid, and its conjugate base.

Our concentrations are 0.0085 moles / 0.026 L = 0.327 M

We apply Henderson-Hasselbach

pH = pKa + log (base/acid) → pH = 9.20 + log (0.327/0.327)

pH = pKa

c. When we add 40 mL, our volume is 20mL +40mL  = 60 mL

These are the moles, we add:

0.040 L . 0.250 mol/L = 0.01 moles of KOH (moles of OH⁻)

 HCN     +    OH⁻         →     H₂O   +    CN⁻

  0.01          0.01                                 0.01

All the hydroxides have neutralized all the moles from the HCN, so we only have in solution, cyanhide. This is the equivalence point.

0.01 moles / 0.060 L = 0.16 M → [CN⁻]

pH at this point will be

       CN⁻  +  H₂O ⇄  HCN + OH⁻             Kb = 1.62ₓ10⁻⁵ (Kw/Ka)

In.   0.16                        -          -

Eq. 0.16-x                     x          x

Kb = x² / (0.16-x)

We can also assume that 0.16-x = 0.16. Then:

[OH⁻] = √(Kb . 0.16) → √(1.62ₓ10⁻⁵ .  0.16) = 2.59×10⁻⁶

- log [OH⁻] = pOH → - log 2.59×10⁻⁶ = 5.58

pH = 14 - pOH  → 14 - 5.58 = 8.42

This is a basic pH, because the titration is between a weak acid and a strong base.

d. When we add 42 mL of base, our volume is 20mL + 42 mL = 62 mL

We add 0.5 mol/L . 0.062L = 0.031 moles

These are the moles of OH⁻ , so as we have neutralized all the acid with 40 mL, with 42 mL of base, we only have base in solution.

0.031 moles - 0.01 moles = 0.021 moles of OH⁻

[OH⁻] = 0.021 moles / 0.062L = 0.34M

- log [OH⁻]  = pOH → - log 0.34 = 0.47

pH = 14-pH → 14 - 0.47 = 13.53

Final answer:

This answer details the titration of HCN with KOH, explaining ways to calculate the pH values at different stages of the titration: the initial state, partway through, at the equivalence point when all acid has been neutralized, and after the equivalence point.

Explanation:

We are considering the titration of a solution of HCN with KOH. First, we calculate the initial pH using the formula pH=-log10[H+], where the H+ concentration can be determined from the Ka expression for HCN.

For question b, when 6.00 mL base KOH is added, we need to determine which ion is in excess, and then use its concentration to calculate the pH. In this case, there are still excess HCN ions leading to a buffer solution. The Henderson-Hasselbalch equation can be used to determine the pH.

For question c, at the equivalence point, 40.00 mL of base KOH has been added. All HCN has been titrated to form CN-, so the pH is determined by the OH- ions hydrolyzed from CN-. We use the Kb of CN- (which can be calculated using Kw/Ka) and the total volume to calculate [OH-], and then pH = 14 - pOH.

Finally, for question d, after 42.00 mL of KOH, OH- is in excess. We can calculate the [OH-] and then find the pH using pH = 14 - pOH.

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Name the organic compound CH4

Answers

Answer:

Methane

Explanation:

Methane is a potent greenhouse gas with the formula CH₄. Hope this helps!

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