Nault 25000L 250 our = 25000L 15040 8. How many grams of CaCl2 are needed to make 150.0 mL of a 0.500 M CF solution? (Note: CaCl2 is a soluble salt. The molar mass of CaCl2 is 110.98 g/mol.)

Answers

Answer 1

Answer:

You need 8,324 g of CaCl₂ yo make this solution

Explanation:

Molarity is a way to express concentration in a solution, in units of moles of solute per liter of solution.

To know the grams of CaCl₂ it is necessary to know, first, the moles of this substance with the desired volume and concentration , thus:

0,1500 L × [tex]\frac{0,500 mol}{L}[/tex] = 0,075 CaCl₂ moles

Now, with the molar mass of CaCl₂ you will obtain the necessary grams, thus:

0,075 CaCl₂ moles  × [tex]\frac{110,98 g}{mol}[/tex] = 8,324 g of CaCl₂

So, you need 8,324 g of CaCl₂ to make 150,0 mL of a 0,500M solution

I hope it helps!


Related Questions

0.450 L of 0.0500 M HCl is titrated to the equivalence point with 21.45 mL of a NaOH solution. What is the concentration (in M) of the NaOH solution that was added?

Answers

Answer:

1.0489 M is the concentration (in M) of the NaOH solution that was added

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HCl[/tex]

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.

We are given:

[tex]n_1=1\\M_1=0.0500 M\\V_1=0.450 L\\n_2=1\\M_2=?\\V_2=21.45 mL=0.02145 L[/tex]

Putting values in above equation, we get:

[tex]1\times 0.0500 M\times 0.450 L=1\times M_2\times 0.02145 L\\\\M_2=1.0489 M[/tex]

1.0489 M is the concentration (in M) of the NaOH solution that was added

write a reaction to describe the behavior of the following substances in water. include all phases.
CH3OH (l)
HC2H3O2 (l)
C12H22O11 (s)

Answers

CH3OH (l)

Methanol.

This an alcohol, in ambient temperature is liquid, and in contact with water, they dissolve.

HC2H3O2 (l)

Acetic Acid.

This is a liquid acid, common in vinegar, giving them it's taste and sour flavor.

With water they do not repel, they can create a solution.

C12H22O11 (s)

Sucralose.

This is what we know as a common sugar.

It is a solid that dissolves in water.

.

400. mg of an unknown protein are dissolved in enough solvent to make 5.00 mL of solution. The osmotic pressure of this solution is measured to be 0.0861 atm at 25.0 °C. Calculate the molar mass of the protein. Round your answer to 3 significant digits. mol x 6 ?

Answers

Final answer:

To determine the molar mass of an unknown protein, first calculate the molar concentration using the osmotic pressure formula. Next, calculate the moles of solute from this molarity. Finally, determine the molar mass by dividing the mass of the protein by the moles of solute.

Explanation:

First, we need to calculate the molarity (M) of the solution using the osmotic pressure formula II = MRT. Here, 'II' represents the osmotic pressure in atm, which is 0.0861 atm, 'R' is the gas constant (0.08206 L atm/mol K), and 'T' is the temperature in Kelvin. Convert the given temperature 25.0 °C to Kelvin by adding 273.15, which results in 298.15 K.

Substituting the known values into the formula, we get M = II / RT. Now, compute the molar concentration (M) from the osmotic pressure.

Next, the molar mass of the protein can be determined. Divide the mass of the solute by the number of moles in that mass. In this case, the mass of the solute is 400mg, which equals 0.4 g (as 1 g = 1000 mg). By using the calculated molarity, you can get the moles of solute, from which the molar mass can be determined by dividing the mass of solute by moles of solute.

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To determine the molar mass of the protein, use the osmotic pressure equation. After solving for molarity and then moles, the molar mass is found to be approximately 22,700 g/mol. This process involves converting units and applying the ideal gas constant.

To find the molar mass of the protein, we use the formula for osmotic pressure (π):

π = MRT

π (osmotic pressure) = 0.0861 atm

M (molarity) = n/V (number of moles/volume)

R (gas constant) = 0.0821 L·atm/(K·mol)

T (temperature) = 25 + 273 = 298 K

First, convert the mass of the protein to grams:

400 mg = 0.400 g

The volume of the solution is 5.00 mL, which is 0.00500 L.

We can rearrange the formula to find the molarity (M):

M = π / (RT)

M = 0.0861 atm / (0.0821 L·atm/(K·mol) × 298 K)

M = 0.00352 mol/L

Now, the number of moles of the protein (n) can be related to its molarity and the volume of the solution:

n = M × V

n = 0.00352 mol/L × 0.00500 L = 1.76 x 10⁻⁵ mol

Finally, the molar mass (M) is given by the mass of the protein divided by the number of moles:

Molar mass = mass / n

Molar mass = 0.400 g / 1.76 x 10⁻⁵ mol ≈ 22727 g/mol

Rounded to three significant digits, the molar mass of the protein is 22,700 g/mol.

An imaginary element (X) on Mars is composed of three isotopes, 10.68% of isotope X-95 with a mass of 95.0 amu, 16.90% of isotope X-96 with a mass of 96.0 amu, and 72.42% of isotope X-97 with a mass of 97.0 amu. Calculate the atomic mass (in amu) of the element. Type in your answer with 3 significant figures.

Answers

Answer: The average atomic mass of element X is 96.6 amu.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex]    .....(1)

For isotope 1 (X-95):

Mass of isotope 1 = 95.0 amu

Percentage abundance of isotope 1 = 10.68 %

Fractional abundance of isotope 1 = 0.1068

For isotope 2 (X-96):

Mass of isotope 2 = 96.0 amu

Percentage abundance of isotope 2 = 16.90 %

Fractional abundance of isotope 2 = 0.1690

For isotope 3 (X-97):

Mass of isotope 3 = 97.0 amu

Percentage abundance of isotope 3 = 72.42 %

Fractional abundance of isotope 3 = 0.7242

Putting values in equation 1, we get:

[tex]\text{Average atomic mass of X}=[(95\times 0.1068)+(96\times 0.1690)+(97\times 0.7242)][/tex]

[tex]\text{Average atomic mass of X}=96.6amu[/tex]

Hence, the average atomic mass of element X is 96.6 amu.

Final answer:

To calculate the atomic mass of the element with three isotopes, you multiply the percentage abundance by the mass of each isotope and then add the values.

Explanation:

To calculate the atomic mass of the element, we need to multiply the percentage abundance of each isotope by its respective mass and then sum up the products.

For isotope X-95: (10.68/100) x 95.0 amu = 10.1376 amu

For isotope X-96: (16.90/100) x 96.0 amu = 16.224 amu

For isotope X-97: (72.42/100) x 97.0 amu = 70.2774 amu

Adding the values: 10.1376 amu + 16.224 amu + 70.2774 amu = 96.639 amu

Therefore, the atomic mass of the element is 96.639 amu (rounded to 3 significant figures).

Calculate the number of moles of Californium represented by 5.92 x 1024 atoms of Californium. Enter your answer in the provided box. mol Cf

Answers

Answer:

9.834 moles Cf.

Explanation:

The number of moles of a substance is an easy way to represents its amount. Avogadro has determined that the total amount in 1 mol is equal to 6.02x10²³(Avgadros' number), so 1 mol has 6.02x10²³ atoms, molecules, ions, or what we are measuring. So:

1 mol of Cf -------------------- 6.02x10²³ atoms

x -------------------- 5.92x10²⁴

By a simple direct three rule:

6.02x10²³x = 5.92x10²⁴

x = 5.92x10²⁴/6.02x10²³

x = 9.834 moles Cf

A chemist measures the amount of oxygen gas produced during an experiment. She finds that 4.87 g of oxygen gas is produced. Calculate the number of moles of oxygen gas produced. Be sure your answer has the correct number of significant digits.

Answers

Answer:

0.152 moles

Explanation:

Given that:

The mass of the oxygen gas produced = 4.87 g

Also, The molar mass of oxygen gas, [tex]O_2[/tex] = [tex]2\times 16\ g/mol[/tex] = 32 g/mol

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus,

[tex]Moles= \frac{4.87\ g}{32\ g/mol}[/tex]

[tex]Moles=0.152\ mol[/tex]

Both given values and the answer is in 3 significant digits.

Sulfur hexafluoride gas is collected at 14.0°C in an evacuated flask with a measured volume of 5.0L. When all the gas has been collected, the pressure in the flask is measured to be 0.090atm . Calculate the mass and number of moles of sulfur hexafluoride gas that were collected. Be sure your answer has the correct number of significant digits.

Answers

Answer: The mass and number of moles of sulfur hexafluoride gas is 2.8 g and 0.019 moles respectively.

Explanation:

To calculate the moles of sulfur hexafluoride gas, we use the equation given by ideal gas:

PV = nRT

where,

P = Pressure of sulfur hexafluoride = 0.090 atm

V = Volume of the sulfur hexafluoride gas = 5.0 L

n = number of moles of gas = ?

R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]

T = Temperature of sulfur hexafluoride gas = [tex]14^oC=[273+14]K=287K[/tex]

Putting values in above equation, we get:

[tex]0.090atm\times 5.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 287K\\\\n=0.019mol[/tex]

To calculate the mass of sulfur hexafluoride, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Molar mass of sulfur hexafluoride = 146.0 g/mol

Moles of sulfur hexafluoride = 0.0191 moles

Putting values in equation 1, we get:

[tex]0.019mol=\frac{\text{Mass of sulfur hexafluoride}}{146.0g/mol}\\\\\text{Mass of sulfur hexafluoride}=2.8g[/tex]

In case of multiplication and division, the number of significant digits is taken from the value which has least precise significant digits. Here, the least precise number of significant digits are 2.

Hence, the mass and number of moles of sulfur hexafluoride gas is 2.8 g and 0.019 moles respectively.

Final answer:

To calculate the mass and number of moles of sulfur hexafluoride gas, we use the ideal gas law equation PV = nRT. By substituting the given values and solving for n, we find that 0.1903 moles of sulfur hexafluoride were collected. The mass of the gas is determined by multiplying the moles by the molar mass of sulfur hexafluoride, resulting in approximately 27.8 grams.

Explanation:

To calculate the mass and number of moles of sulfur hexafluoride gas, we can use the ideal gas law equation: PV = nRT. Rearranging the equation to solve for moles (n), we have n = PV/RT. We are given the pressure (0.090 atm), volume (5.0 L), temperature (14.0°C which needs to be converted to Kelvin), and the ideal gas constant (R = 0.0821 L·atm/(mol·K)). After converting the temperature to Kelvin (287 K), we substitute the values and solve for n. n = (0.090 atm * 5.0 L) / (0.0821 L·atm/(mol·K) * 287 K). This gives us n = 0.1903 moles of sulfur hexafluoride gas.

To calculate the mass, we use the formula mass = moles * molar mass. The molar mass of sulfur hexafluoride (SF6) is  SF6 = (32.06 g/mol * 6) + (19.00 g/mol * 1). This gives us a total molar mass of 146.06 g/mol. We can now calculate the mass by multiplying the moles by the molar mass: mass = 0.1903 moles * 146.06 g/mol. The mass of sulfur hexafluoride gas that were collected is approximately 27.8 grams.

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How many moles are in 250.0 mL of a 0.15 M nitric acid solution?

Answers

Explanation:

It is given that molarity is 0.15 M and volume is 250.0 ml. As 1 ml equals 0.001 l.

Hence, 250.0 ml = 0.25 l.

Also, molarity is equal to the number of moles present in liter of solution. Therefore, calculate number of moles as follows.

                   Molarity = [tex]\frac{\text{no. of moles}}{volume}[/tex]

                      0.15 M = [tex]\frac{\text{no. of moles}}{0.25 l}[/tex]

             No. of moles = 0.0375 mol

Thus, we can conclude that moles present in given solution are 0.0375.

Final answer:

To find the number of moles in a 250.0 mL solution of 0.15 M nitric acid, you convert the volume to liters and multiply by the molarity, resulting in 0.0375 moles of HNO3.

Explanation:

To calculate the number of moles in a 0.15 M nitric acid (HNO3) solution with a volume of 250.0 mL, you would first convert the volume from milliliters (mL) to liters (L) because concentration (Molarity, M) is typically expressed in moles per liter. Therefore, 250.0 mL is equivalent to 0.250 L (since 1 L = 1000 mL).

Using the formula for molarity:

Molarity (M) = moles of solute / liters of solution

We can rearrange the formula to solve for the number of moles:

moles of solute = Molarity (M) × liters of solution

Thus, the number of moles of nitric acid in the solution is:

0.15 moles/L × 0.250 L = 0.0375 moles of HNO3

What is the mass, in pounds, of 389 mL of a gas that has a density of 1.29 g/L?

Answers

Answer: Mass of gas is 0.001 pounds.

Explanation:

Density is defined as the mass contained per unit volume.

[tex]Density=\frac{mass}{Volume}[/tex]

Given : Mass of gas = ?

Density of gas = [tex]1.29g/L[/tex]

Volume of gas = 389 ml = 0.389 L     (1L=1000ml)

Putting in the values we get:

[tex]1.29g/L=\frac{mass}{0.389L}[/tex]

[tex]mass=0.5grams[/tex]  

[tex]mass=0.5\times 0.002lb=0.001lb[/tex]       (1g =0.002 lb)

Thus the mass of gas is 0.001 pounds.

Final answer:

To find the mass in pounds of 389 mL of a gas with a density of 1.29 g/L, you convert the density to g/mL, calculate the mass in grams, and then convert that mass to pounds, resulting in approximately 0.001106 pounds.

Explanation:

To calculate the mass of the gas in pounds, we first need to convert the density from grams per liter (g/L) to grams per milliliter (g/mL) since the volume of gas given is in milliliters (mL). With the given density of 1.29 g/L, we can use dimensional analysis to calculate the mass of 389 mL of the gas:

Convert density to g/mL: 1.29 g/L = 0.00129 g/mL.

Calculate the mass: mass (in grams) = density (in g/mL)  imes volume (in mL) = 0.00129 g/mL  imes 389 mL = 0.50181 g.

Convert the mass to pounds using the conversion factor 1 lb = 453.59 g: mass (in pounds) = mass (in grams) / conversion factor = 0.50181 g / 453.59 g/lb

Carrying out the division, the mass is approximately 0.001106 pounds.

Gold is a very soft metal that can be hammered into extremely thin sheets known as gold leaf. If a 1.78 g piece of gold is hammered into a sheet whose area is 48.4 ft2, what is the average thickness of the sheet? (The density of gold is 19.32 g cm-3.) 9.917*10^-5 cm

Answers

Answer:

[tex]2.0489\times 10^{-6} cm[/tex] is the average thickness of the sheet.

Explanation:

Mass of gold ,m= 1.78 g

Volume of the gold = V

Density of the gold = D = [tex]19.32g/cm^3[/tex]

[tex]D=\frac{m}{V}=19.32g/cm^3=\frac{1.78 g}{V}[/tex]

[tex]V = 0.09213 cm^3[/tex]

Area of the hammered gold sheet,A = [tex]48.4 ft^2=44,965.052 cm^2[/tex]

Thickness of the hammered gold = h

([tex]1 ft^2=929.03 cm^2[/tex])

Volume = Area × thickness

[tex]V= A\times h[/tex]

[tex]0.09213 cm^3=44,965.052 cm^2\times h[/tex]

[tex]h=2.0489\times 10^{-6} cm[/tex]

[tex]2.0489\times 10^{-6} cm[/tex] is the average thickness of the sheet.

Calculate the frequency of the light emitted by ahydrogen atom
during a transition of its electron from the n = 4 tothe n = 1
principal energy level.

Answers

Final answer:

The frequency of light emitted by a hydrogen atom during a transition of its electron from the n = 4 to the n = 1 principal energy level is approximately -2.06 x 10^14 Hz.

Explanation:

The frequency of light emitted by a hydrogen atom during a transition of its electron can be calculated using the formula:

f = R(1/n1^2 - 1/n2^2)

where f is the frequency, R is a constant (R = 3.29 x 10^15 Hz), and n1 and n2 are the initial and final principal energy levels, respectively.

In this case, the electron transitions from n = 4 to n = 1. Plugging the values into the formula, we get:

f = 3.29 x 10^15 (1/4^2 - 1/1^2)

f = 3.29 x 10^15 (1/16 - 1/1)

f = 3.29 x 10^15 (0.9375 - 1)

f = 3.29 x 10^15 (-0.0625)

f = -2.06 x 10^14 Hz

The frequency of the light emitted by the hydrogen atom during this transition is approximately -2.06 x 10^14 Hz.

Limiting Reactant For the general reaction stoichiometry of 2A + 3B → D+E, which reactant is the limiting reactant (A or B), if the initial number of moles is 3 moles of A and 4 moles of B?

Answers

Answer:

B

Explanation:

Limiting reagent is the one which is present in small molar amount. It got exhausted at the end of the reaction and the formation of the products is governed by it.

The given reaction:

2A + 3B → D+E

2 moles of A react with 3 moles of B

1 mole of A react with 3 / 2 moles of B

Given moles of A = 3 moles

Moles of B = 4 moles

So,

3 moles of A will react with 1.5 * 3 moles of B

Moles of B = 4.5

Since, only we have 4 moles of B. So, B is the limiting reagent.

Spermine (structure shown) is a compound isolated from sperm. Which of the following statements correctly describes an aqueous solution of spermine?
NH2(CH2)3NH(CH2)4NH(CH2)3NH2

Question 2 options:

A)

The hydroxide ion concentration in the solution would be greater than that in pure water.
B)

The hydronium ion concentration in the solution would be greater than that in pure water.
C)

The solution would be acidic.
D)

The pH of the solution would be equal to that of pure water.

Answers

Answer:

A) The hydroxide ion concentration in the solution would be greater than that in pure water.

Explanation:

Spermine has four amino groups. In water, these amino groups will be protonate -In NH₂⁺ or NH₃⁺- decreasing the H⁺ concentration and, by water equilibrium, increasing OH⁻ concentration.

Thus, in an aqueous solution of spermine the hydroxide ion concentration in the solution would be greater than that in pure water.

I hope it helps!

Answer:

The correct answer is option A)  The hydroxide ion concentration in the solution would be greater than that in pure water.

Explanation:

Spermine is basic in nature as it contains di-amine group (NH2), which is basic in nature that is, it easily gives off electrons.

Thus being basic spermine solution will have more concentration of Hydroxide ion [OH-].

Ethylene enters a reversible, isothermal, steady-flow compressor at 1 bar and 280 K, and exits at 40 bar. Find the required compressor work (kJ/kmol) using the ideal gas equation of state.

Answers

Answer:

Required compressor work W=8560.44  KJ/Kmol

Explanation:

Given that

Initial pressure = 1 bar

[tex]P_1=1\ bar[/tex]

Final pressure = 40 bar

[tex]P_2=40\ bar[/tex]

Process is isothermal and T=280 K

We know that ,work done in isothermal process given as

[tex]W=P_1V_1\ln \dfrac{P_1}{P_2}[/tex]

given taht gas is ideal so

P V =m R T

[tex]W=mRT\ln \dfrac{P_1}{P_2}[/tex]

R for ethylene

R=0.296 KJ/kg.K

Now by putting the values

[tex]W=mRT\ln \dfrac{P_1}{P_2}[/tex]

[tex]W=1\times 0.296\times 280 \ln \dfrac{1}{40}[/tex]

W= -305.73 KJ/kg

Negative sign indicates that work done on the system.

Required compressor work W=305.73 KJ/kg

Molar mass of ethylene M= 28 Kg/Kmol

So W= 305.73 x 28  KJ/Kmol

W=8560.44  KJ/Kmol

Required compressor work W=8560.44  KJ/Kmol

What are the chemical formulas for the following compounds? magnesium carbonate dinitrogen monoxide sulfuric acid sodium acetate copper (II) hydroxide on the metal cation in each of the following compounds?

Answers

Explanation:

A chemical formula is defined as the symbolic representation of atoms present in a compound or molecule which also depicts the ratio in which the elements are combined to each other.

Chemical formula's for the given compounds are as follows.

Magnesium carbonate - [tex]MgCO_{3}[/tex]Dinitrogen monoxide - [tex]N_{2}O[/tex]Sulfuric acid - [tex]H_{2}SO_{4}[/tex]Sodium acetate - [tex]CH_{3}COONa[/tex]Copper (II) hydroxide - [tex]Cu(OH)_{2}[/tex]

Convert 3.99 g to kilograms. 3.99 g =

Answers

Answer: The given mass in kilograms is 0.00399 kg

Explanation:

Gram and Kilograms are the units which are used to express the mass of a substance. These units are inter changeable.

We are given:

Mass of a substance = 3.99 g

To convert the given mass into kilograms, we use the conversion factor:

1 kg = 1000 g

Converting the given value, we get:

[tex]\Rightarrow 3.99g\times (\frac{1kg}{1000g}=0.00399kg[/tex]

Hence, the given mass in kilograms is 0.00399 kg

5 moles of an ideal gas (Cp = 3R, Cv 2R ) are heated from 25°C to 300°C. Calculate (a) The change in internal energy of the gas (b) The change in enthalpy of the gas.

Answers

Answer :

(a) The change in internal energy of the gas is 22.86 kJ.

(b) The change in enthalpy of the gas is 34.29 kJ.

Explanation :

(a) The formula used for change in internal energy of the gas is:

[tex]\Delta U=nC_v\Delta T\\\\\Delta U=nC_v(T_2-T_1)[/tex]

where,

[tex]\Delta U[/tex] = change in internal energy = ?

n = number of moles of gas = 5 moles

[tex]C_v[/tex] = heat capacity at constant volume = 2R

R = gas constant = 8.314 J/mole.K

[tex]T_1[/tex] = initial temperature = [tex]25^oC=273+25=298K[/tex]

[tex]T_2[/tex] = final temperature = [tex]300^oC=273+300=573K[/tex]

Now put all the given values in the above formula, we get:

[tex]\Delta U=nC_v(T_2-T_1)[/tex]

[tex]\Delta U=(5moles)\times (2R)\times (573-298)[/tex]

[tex]\Delta U=(5moles)\times 2(8.314J/mole.K)\times (573-298)[/tex]

[tex]\Delta U=22863.5J=22.86kJ[/tex]

The change in internal energy of the gas is 22.86 kJ.

(b) The formula used for change in enthalpy of the gas is:

[tex]\Delta H=nC_p\Delta T\\\\\Delta H=nC_p(T_2-T_1)[/tex]

where,

[tex]\Delta H[/tex] = change in enthalpy = ?

n = number of moles of gas = 5 moles

[tex]C_p[/tex] = heat capacity at constant pressure = 3R

R = gas constant = 8.314 J/mole.K

[tex]T_1[/tex] = initial temperature = [tex]25^oC=273+25=298K[/tex]

[tex]T_2[/tex] = final temperature = [tex]300^oC=273+300=573K[/tex]

Now put all the given values in the above formula, we get:

[tex]\Delta H=nC_p(T_2-T_1)[/tex]

[tex]\Delta H=(5moles)\times (3R)\times (573-298)[/tex]

[tex]\Delta H=(5moles)\times 3(8.314J/mole.K)\times (573-298)[/tex]

[tex]\Delta H=34295.25J=34.29kJ[/tex]

The change in enthalpy of the gas is 34.29 kJ.

Write 0.00057824 in Engineering Notation with 3 significant figures.

Answers

Answer:

[tex]578.2\times 10^{-6}[/tex]

Explanation:

Scientific notation is the way of writing numbers which are either large or small. The number is written in the scientific notation when the number is between 1 and 10 and then multiplied by the power of 10. Engineering notation is the same version of the scientific notation but the number can be between 1 and 1000 and in this exponent of the ten is divisible by three.

For example, [tex]1000^2[/tex] is to be written as [tex]10^6[/tex] in engineering notation.

The given number:

0.00057824 can be written as [tex]578.24\times 10^{-6}[/tex]

Answer upto 4 significant digits = [tex]578.2\times 10^{-6}[/tex]

Calculate the mass, in grams, of Avogadro's number of fluorine atoms. Enter your answer in the provided box. I g/mol

Answers

Answer:

Every substance, whether an element or a compound that contains Avogadro's number, is said to form a mole of it. Therefore for the fluorine atoms that form the Avogadro number we say that we have one mole of fluorine atoms and this weighs 19 g, so 19 g/mol

Explanation:

You can search the F in the Periodic Table where you have the molar mass, 19 g/mol but if you don't have a Periodic Table you can calculate like this:

1 atom of F weighs 19 Da (unified atomic mass unit) and 1 Da is 1,661 x10*-24 g, so try to apply a rule of three, twice.

1 Da ........... 1,661x10*-24 g

19 Da .......... x

x= (19 Da . 1,661x10*-24 g ) / 1 Da = 3,1559x10*-23 g

1 atom F ..............  3,1559x10*-23 g

6,022x10*23 atoms F ............. x

X= (6,022x10*23 atoms F . 3,1559x10*-23 g) / 1 atom F = 19.0048 g

When will a precipitate form? a. When the concentration of the cation is greater than the concentration of the anion. b. When the concentration of the anion is greater than the concentration of the cation. c. When the reaction quotient is greater than the solubility-product constant. d. When the solubility-product constant is greater than the reaction quotient. e. None of these.

Answers

Answer:

c. when the raction quotient is greater than the Ksp

Explanation:

AxBy ↔ xA+ + yB-

∴ Ksp = [ A+ ]∧x  *  [ B- ]∧y.....solubility product constant

∴ I.P = xA+ * yB-............ionic product

⇒ the relationship between Ksp and I.P, will give us the precipitation conditions.

∴ I.P > Ksp ⇒ precipitation

∴ I.P < Ksp ⇒ no precipitation (disolution)

∴ I.P = Ksp ⇒ equilibrium

⇒ the ionic product is directly related to the reaction quotient Q, where Q = [A+] * [B-] / [AxBy] = PI / [AB], then the higher this product the higher the reaction quotient will be and therefore will be greater than the value of the constant and will have precipitated, the correct answer is the c

A major component of gasoline is octane . When liquid octane is burned in air it reacts with oxygen gas to produce carbon dioxide gas and water vapor. Calculate the moles of carbon dioxide produced by the reaction of of oxygen. Be sure your answer has a unit symbol, if necessary, and round it to significant digits.

Answers

Final answer:

The balanced equation for the combustion of octane is 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O. From this equation, we can determine the moles of carbon dioxide produced from a given amount of octane.

Explanation:

The balanced equation for the combustion of octane is:

2 C8H18 + 25 O2 → 16 CO2 + 18 H2O

From the equation, we can see that for every 2 moles of octane burned, 16 moles of carbon dioxide are produced. Therefore, if we know the number of moles of octane burned, we can calculate the moles of carbon dioxide produced.

Look at the equation below. What is the acid on the left-hand side of the equation? H30 СНЗСОО" CH3COOH + H20 O a. CH3COOH ОБ.Н20 O c. None of the above.

Answers

Answer:

c. None of the above

Explanation:

The equation is:

H₃O⁺ + CH₃COO⁻ ⇒ CH₃COOH + H₂O

An acid is a species that donates a proton. Thus, the acid is the species on the reactants' side that has one proton more than it's conjugate base species on the products's side.

The available options are:

a. CH₃COOH

b. H₂O

c. None of the above.

Both CH₃COOH and H₂O are on the right-hand side of the equation, so the answer must be c. The answer to the problem is H₃O⁺, since it has lost a proton and become H₂O.

Heat is transferred by conduction through a wall of thickness 0.87 m. What is the rate of heat transfer if the walls thermal conductivity is 13 W/m-K, the surface area is 5 m2, and the surface temperatures are 14 °C and 93 °C. Report your answer in Watts.

Answers

Explanation:

The given data is as follows.

         Thickness (dx) = 0.87 m,       thermal conductivity (k) = 13 W/m-K

        Surface area (A) = 5 [tex]m^{2}[/tex],       [tex]T_{1} = 14^{o}C[/tex]

        [tex]T_{2} = 93^{o}C[/tex]

According to Fourier's law,

                    Q = [tex]-kA \frac{dT}{dx}[/tex]

Hence, putting the given values into the above formula as follows.

                      Q = [tex]-kA \frac{dT}{dx}[/tex]

                          = [tex]-13 W/m-k \times 5 m^{2} \times \frac{(14 - 93)}{0.87 m}[/tex]

                          = 5902.298 W

Therefore, we can conclude that the rate of heat transfer is 5902.298 W.

Enter your answer in the provided box. The balanced equation for the combustion of ethanol (ethyl alcohol) is: C2H5OH() + 3O2(g) → 2CO2(g) + 3H2O(g) How many g of CO2 will be produced by the combustion of 4 mol of ethanol? g CO2

Answers

Answer: The mass of carbon dioxide produced is 352 grams

Explanation:

We are given:

Moles of ethanol = 4 mol

For the given chemical equation:

[tex]C_2H_5OH(g)+3O_2(g)\rightarrow 2CO_2(g)+3H_2O(g)[/tex]

By Stoichiometry of the reaction:

1 mole of ethanol produces 2 moles of carbon dioxide

So, 4 moles of ethanol will produce = [tex]\frac{2}{1}\times 4=8mol[/tex] of carbon dioxide

To calculate the mass of carbon dioxide, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Moles of carbon dioxide = 8 moles

Molar mass of carbon dioxide = 44 g/mol

Putting values in above equation:

[tex]8mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(8mol\times 44g/mol)=352g[/tex]

Hence, the mass of carbon dioxide produced is 352 grams

Final answer:

The combustion of 4 moles of ethanol produces 352 g of CO₂, based on the stoichiometry of the balanced chemical equation.

Explanation:

The question involves a stoichiometry calculation based on the balanced chemical equation for the combustion of ethanol, C₂H₅OH (l) + 3O₂ (g) → 2CO₂ (g) + 3H₂O (g). Given that 4 moles of ethanol are combusted, we need to determine the amount of CO2 produced.

From the equation, it is clear that 1 mole of ethanol produces 2 moles of CO₂. Therefore, 4 moles of ethanol will produce 4 * 2 = 8 moles of CO₂.

To calculate the weight of these CO₂ moles, we use the molar mass of CO₂, which is approximately 44 g/mol. Hence, the weight of CO₂ produced is 8 moles * 44 g/mol = 352 g.

For the metathesis reaction, K2CO3 + FeCl3, Write the balanced molecular equation. What evidence of reaction occurs in the mixture? Write the balanced ionic equation. Write the balanced net ionic equation for the reaction that occurs.

Answers

Final answer:

The metathesis reaction between potassium carbonate and iron (III) chloride results in the formation of a precipitate of Iron (III) Carbonate along with Potassium chloride. The chemical, full ionic, and net ionic equations for this reaction have been described in detail.

Explanation:

Firstly, the metathesis (or double replacement) reaction between potassium carbonate (K2CO3) and iron (III) chloride (FeCl3) can be represented as follows:

K2CO3(aq) + FeCl3(aq) -> Fe2(CO3)3(s) + 2 KCl(aq)

The evidence of the reaction is typically observed as the formation of a precipitate, in this case, Iron (III) Carbonate (Fe2(CO3)3).

The full ionic equation can be given as:

2 K+(aq) + CO32-(aq) + Fe3+(aq) + 3 Cl-(aq) -> Fe2(CO3)3(s) + 2 K+(aq) + 2 Cl-(aq)

From this, the net ionic equation becomes:

Fe3+(aq) + CO32-(aq) -> Fe2(CO3)3(s)

This reaction is characterized by a double exchange of ions between the reactants, resulting in the formation of a precipitate.

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The balanced molecular equation for the metathesis reaction between potassium carbonate (K₂CO₃) and iron(III) chloride (FeCl₃) is:

[tex]\[ \text{K}_2\text{CO}_3(aq) + 2\text{FeCl}_3(aq) \rightarrow 2\text{KCl}(aq) + \text{Fe}_2(\text{CO}_3)_3(s) \][/tex]

Evidence of reaction occurs in the mixture as the formation of a solid precipitate, which is iron(III) carbonate (Fe₂(CO₃)₃). The balanced ionic equation, taking into account that K₂CO₃ and FeCl₃ are strong electrolytes and dissociate completely in aqueous solution, is:

        [tex]\[ 2\text{K}^+ + \text{CO}_3^{2-} + 2\text{Fe}^{3+} + 6\text{Cl}^- \rightarrow 2\text{K}^+ + 2\text{Cl}^- + \text{Fe}_2(\text{CO}_3)_3(s) \][/tex]

To write the net ionic equation, we eliminate the spectator ions which are the ions that appear on both sides of the equation with the same charge:  [tex]\[ \text{CO}_3^{2-} + 2\text{Fe}^{3+} \rightarrow \text{Fe}_2(\text{CO}_3)_3(s) \][/tex]This is the balanced net ionic equation for the reaction that occurs. It shows that carbonate ions react with iron(III) ions to form the solid precipitate of iron(III) carbonate, while the potassium and chloride ions remain in solution as spectator ions and do not participate in the reaction.

A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% and 15%. A mixture containing 9.0 mole% methane in air flowing at a rate of 700 kg/h is to be diluted with pure air to reduce the methane concentration to the lower flammability limit. Calculate the required flow rate of air in mol/h and the percent by mass of oxygen in the product gas. (Note: Air may be taken to consist of 21 mole% O and 79% N2 and to have an average molecular weight of 29.0.) answer:

(a)n = 20,200 mol air/h; 0.225 kg O2/kg

Answers

Answer:

Explanation:

The air 9% mole% methane have an average molecular weight of:

9%×16,04g/mol + 91%×29g/mol = 27,8g/mol

And a flow of 700000g/h÷27,8g/mol = 25180 mol/h

In the reactor where methane solution and air are mixed:

In = Out

Air balance:

91% air×25180 mol/h + 100% air×X = 95%air×(X+25180)

Where X is the flow rate of air in mol/h = 20144 mol air/h

The air in the product gas is

95%×(20144 + 25180) mol/h = 43058 mol air× 21%O₂ = 9042 mol O₂ ×32g/mol = 289 kg O₂

43058 mol air×29g/mol 1249 kg air

Percent of oxygen is: [tex]\frac{289kg}{1249 kg}[/tex] =0,231 kg O₂/ kg air

I hope it helps!

Answer:

A) Mass flow rate of air = [tex]22.982[/tex] kmol/hr

B) percentage by mass of oxygen in the product gas = [tex]$22.52 \%$[/tex]

Explanation:

We are given that the mixture containing 9.0 mole% methane in air flowing.

Thus, we have [tex]0.09[/tex] mole of methane(CH4) and the remaining will be the air which is [tex]$(100 \%-9 \%)=91 \%$=0.91[/tex]

We are given the average molecular weight of air = [tex]29[/tex] g/mol Thus; Average molar mass of air and methane mixture is;

M-air- [tex]$=(0.09 \times 16)+(0.91 \times 29)$[/tex]

[tex]=$27.83 \mathrm{~g} / \mathrm{mol}$[/tex]

Mass fraction of oxygen in the product gas= [tex]=43.016 \mathrm{kmol} / \mathrm{h} \times(0.21 \mathrm{molO} 2 / 1 \mathrm{~mol}$ air $) \times(32 \mathrm{~kg}$ oxygen/1kmol oxygen $) \times(1 / 1283.596 \mathrm{~kg} / \mathrm{h})=0.2252$[/tex]

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Which of these is an example of a physical change?

a)Silver tarnishing
b)metal denting
c)iron rusting
d)gasoline burning

Answers

Answer:

b metal denting

Explanation:

Answer:

Choice b) Metal denting.

Explanation:

What's the difference between chemical and physical changes? In a chemical change, new substances are created. However, the types of substances stay the same in a physical change.

a)

When silver tarnishes, it reacts with a substance that contains sulfur (e.g., [tex]\rm SO_3[/tex]) to produce silver sulfide, which is a substance different from the other two.

Before the change:

Silver;A substance that contains sulfur.

After the change:

Silver sulfide.

A new substance is created in this change. As a result, this change is chemical.

b)

When metal is dented, atoms in the metal slide past each other. The shape of the metal might change, however the substance will still be the same the metal.

Before the change:

This particular metal.

After the change:

This particular metal.

No substance is created. As a result, this change is physical.

c)

When iron rusts in the air, it reacts with oxygen to produce oxides of iron.

Before the change:

Iron;Oxygen.

After the change:

Oxides of iron.

Oxides of iron are created. As a result, this change is chemical.

d)

When gasoline burns in the air, it reacts with oxygen to produce water and carbon dioxide (or carbon monoxide, or both.)

Before the change:

Gasoline;Oxygen.

After the change:

Water;Carbon dioxide, carbon monoxide, or both.

Water and carbon dioxide/monoxide are created. As a result, this change is chemical.

Question 5 Predict the reactants of this chemical reaction. That is, fill in the left side of the chemical equation. Be sure the equation you submit is balanced. (You can edit both sides of the equation to balance it, if you need to.) Note: you are writing the molecular, and not the net ionic equation. → + Ca Cl O 4 2 aq H 2 O l Clears your work. Undoes your last action. Provides information about entering answers.

Answers

Explanation:

A chemical reaction equation that contains equal number of atoms on both reactant and product side is known as a balanced chemical equation.

For example, when the products are given as [tex]Ca(ClO_{4})_{2}(aq) + H_{2}O(l)[/tex]

So, reactants for this equation will be as follows.

        [tex]HClO_{4}(aq) + Ca(OH)_{2}(aq) \rightarrow Ca(ClO_{4})_{2}(aq) + H_{2}O(l)[/tex]

Number of atoms present on reactant side are as follows.

H = 3

Cl = 1

O = 6

Ca = 1

Number of atoms present on product side are as follows.

H = 2

Cl = 2

O = 9

Ca = 1

Therefore, to balance this equation we multiply [tex]HClO_{4}[/tex] by 2 on reactant side and multiply [tex]H_{2}O[/tex] by 2 on product side.

Hence, the balanced chemical reaction equation will be as follows.

           [tex]2HClO_{4}(aq) + Ca(OH)_{2}(aq) \rightarrow Ca(ClO_{4})_{2}(aq) + 2H_{2}O(l)[/tex]

Final answer:

The reactants for the given chemical reaction are likely to be calcium carbonate (CaCO₃) and hydrochloric acid (2HCl). These react to form the given products: calcium chloride (CaCl₂) and water (2H₂O), with the balanced equation being: CaCO₃ + 2HCl --> CaCl₂ + H₂O + CO₂.

Explanation:

Predicting the reactants of the chemical reaction requires understanding of chemistry concepts. First, let's rearrange the given product side which is CaCl₂(aq) + 2H₂O(l). From this, we can find that the reactants are likely CaCO₃ + 2HCl, as calcium carbonate generally reacts with hydrochloric acid to form calcium chloride and water.

In terms of steps, the HCl will dissolve the limestone (CaCO₃), and the resulting calcium ions (Ca₂+) will combine with the chloride ions (Cl-) to form calcium chloride (CaCl₂). Meanwhile, the carbonate ions (CO₃₂-) will combine with the hydrogen ions (H+) from the HCl to form water (H₂O) and carbon dioxide (CO₂).

The balanced molecular equation would therefore be: CaCO₃ + 2HCl --> CaCl₂+ H₂O + CO₂.

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Question 6 1.75 pts The following reaction 2H2S(g)=2H2(g)+S2(g), Kc=1.625x10-7 at 800°C is carried out at the same temperature with the following initial concentrations: [H,S]=0.162M, [H2]=0.184 M, and [S2]=0.00 M. Find the equilibrium concentration of S2. Express the molarity to three significant figures. Answer in units of nM.

Answers

Answer:

[S₂] = 1.27×10⁻⁷ M

Explanation:

2 H₂S(g) ⇄ 2 H₂(g) + S₂(g), Kc=1,625x10⁻⁷

The equation of this reaction is:

1,625x10⁻⁷ = [tex]\frac{[H_2]^2[S_2]}{[H_{2}S]^2}[/tex]

The equilibrium concentrations are:

[H₂S] = 0,162 - 2x

[H₂] = 0,184 + 2x

[S₂] = x

Replacing:

1,625x10⁻⁷ = [tex]\frac{[0,184+2x]^2[x]}{[0,162-2x]^2}[/tex]

Solving:

4x³ + 0,736x² + 0,033856x - 4,3x10⁻⁹

x = 1.27×10⁻⁷

Thus, concentration of S₂ is:

[S₂] = 1.27×10⁻⁷ M

Perform unit conversions and determine Re for the case when a fluid with density of 92.8 lbm/ft3 and viscosity of 4.1 cP (centipoise) flows with a velocity of 237 ft/min in a pipe with 28 inch diameter.

Re = (density*velocity*diameter)/viscosity

Answers

Answer:

Re=309926.13

Explanation:

density=92.8lbm/ft3*(0.45kg/1lbm)*(1ft3/0.028m3)=1491.43kg/m3

viscosity=4.1cP*((1*10-3kg/m*s)/1cP)=0.0041kg/m*s

velocity=237ft/min*(1min/60s)*(0.3048m/1ft)=1.2m/s

diameter=28inch*(0.0254m/1inch)=0.71m

Re=(density*velocity*diameter)/viscosity=(1491.43kg/m3*1.2m/s*0.71m)/0.0041kg/m*s

Re=309926.13

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