In order to rinse the system of exhaled C02 and meet the inspiratory flow rate requirements of infants placed on B-CPAP, the flow rate of humidified gas should be set at:_________.

Answer: option A. It is much greater

Explanation:

To calculate the molar concentration of HNO3 in the solution, use the balanced chemical equation to determine the mole ratio and then calculate the moles of HNO3 used. Finally, divide the moles of HNO3 by the volume of the solution to get its molar concentration.

To calculate the molar concentration of HNO3, we first need to use the balanced chemical equation to determine the mole ratio between HNO3 and LiOH. The balanced equation is 2HNO3 + 2LiOH → 2LiNO3 + H2O. From the equation, we can see that 2 moles of HNO3 react with 2 moles of LiOH. Since we know the molar concentration of LiOH and the volume used, we can calculate the moles of LiOH used: (12.5 mL)(1.0x10^-4 M) = 0.00125 moles of LiOH. Since the mole ratio is 1:2, the moles of HNO3 used would be half of that, which is 0.000625 moles of HNO3. Now, we can use the volume of the HNO3 solution to calculate its molar concentration:

Molar concentration of HNO3 = (0.000625 moles) / (25.0 mL) = 0.025 M HNO3

Final answer:

The molar concentration of HNO3 is 5.00x10-5 M.

Explanation:

To calculate the molar concentration of HNO3, we first need to determine the number of moles of LiOH used in the titration. The molarity of the LiOH solution is given as 1.0x10-4 M, and the volume used is 12.5 mL (0.0125 L).

moles of LiOH = Molarity x Volume = (1.0x10-4 M) x (0.0125 L) = 1.25x10-6 mol

Since the balanced chemical equation shows a 1:1 ratio between LiOH and HNO3, the number of moles of HNO3 is also 1.25x10-6 mol.

To find the molar concentration of HNO3, we use the volume of the HNO3 solution, which is 25.0 mL (0.025 L).

Molarity of HNO3 = moles of HNO3 / Volume = (1.25x10-6 mol) / (0.025 L) = 5.00x10-5 M HNO3

Final answer:

To balance the equation C₂H₆ + O₂ → CO₂ + H₂O, we end up with a balanced chemical equation of 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O, where the coefficient of O₂ is 7.

Explanation:

The question concerns the balancing of a chemical equation involving ethane (C₂H₆) and oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O). Balancing chemical equations requires ensuring that the number of atoms of each element is the same on both sides of the equation. Starting with the unbalanced equation C₂H₆ + O₂ → CO₂ + H₂O, we first balance the carbons (C) and hydrogens (H) by adjusting the coefficients of the products. For example, the equation can be balanced by placing a coefficient of 3 before H₂O and 2 before CO₂, resulting in C₂H₆ + O₂ → 2CO₂ + 3H₂O. However, this leads to seven oxygen atoms on the product side. We need an even number of oxygen atoms to balance them with the O₂ reactant, so we use a fractional coefficient of 3.5 (which is 7/2) in front of O₂.

To remove the fractional coefficient, we can multiply all coefficients by 2, resulting in the balanced chemical equation: . Thus, the coefficient of O₂ when the equation is balanced is 7.

Answer:

0.3758moles

Explanation:

moles of kcl = mass of kcl/ molar mass of kcl = 28/74.5 = 0.3758moles

Answer:

We have 0.376 moles in 28.0 grams of KCl

Option C is correct.

Explanation:

Step 1: Data given

Mass KCl = 28.00 grams

Molar mass KCl = 74.55 g/mol

Step 2: Calculate moles KCl

Moles KCl = mass KCl / moalr mass KCl

Moles KCl = 28.0 grams / 74.55 g/mol

Moles KCl = 0.376 moles KCl

We have 0.376 moles in 28.0 grams of KCl

Option C is correct.

The larger atomic radius of potassium, as compared to sodium, is due to the increase in the principal quantum number, n, as you move down a group in the periodic table. This denotes an increase in electron shells, which increases the atomic radius. A less effective nuclear charge on more distant electrons also contributes to the larger size.

Compared to sodium, the atomic radius of a potassium atom is larger primarily because of the increase in the principal quantum number, n, which leads to larger radii. Potassium, like other elements in its group, has more electron shells than sodium, which causes its size to be larger.

As one moves down the groups in the periodic table, the number of electron shells increases. This means that the principal quantum number (n), which denotes the electron shell number (energy level), increases. As a result, the atomic radius also increases. In the case of potassium and sodium, Potassium is placed below Sodium on the periodic table, thus it has a higher principal quantum number, i.e., more electron shells, which increases the atomic radius.

This phenomenon can also be explained using the concept of effective nuclear charge. This refers to the pull exerted by the nucleus on the outermost electrons. As you move down a group, more shells of electrons are being added and hence the outermost electrons are not as strongly pulled by the nucleus, thus increasing the atomic radius.

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The initial temperature of the ethylene glycol is equal to 40.5°C.

The specific heat capacity is defined as the amount of heat required to raise the temperature of one unit of material by one degree Celsius. The specific heat capacity of the material depends upon the nature of the material.

The mathematical expression is used to calculate the specific heat is equal to:

[tex]Q = mC \triangle T[/tex]

Given, the mass of the sample of ethylene glycol, m = 31.4 g

The final temperature of the sample, T₂ = 32.5°C = 305.5 K

The specific heat capacity of ethylene glycol, C =  2.42 J/g·K

The heat lost from the sample, Q = - 607 J

The initial temperature of the sample:

- 607 = 31.4 × 2.42 × (305.5 - T₁)

305.5 - T₁ = - 7.988

T₁ = 313.49 K

T₁ = 40.5°C  

Therefore, the initial temperature of the ethylene glycol is 40.5°C.

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The initial temperature of ethylene glycol can be found by rearranging the equation q = mcΔT and dividing the heat lost by the product of the mass and specific heat capacity, considering the known final temperature and that the ethylene glycol is cooling down.

To calculate the initial temperature of ethylene glycol when it loses heat, we can use the equation that relates heat loss to temperature change, q = mcΔT, where q is the heat lost, m is the mass, c is the specific heat capacity of the substance, and ΔT is the change in temperature. Given that ethylene glycol loses 607 J of heat (q), has a mass of 31.4g (m), and a specific heat capacity of 2.42 J/g·°C (c), and the final temperature is 32.5°C, we can rearrange the equation to solve for the initial temperature, T_initial.

The heat lost is negative because the ethylene glycol is cooling down, so we have: -607 J = 31.4g × 2.42 J/g·K × (32.5°C - T_initial). Solving for T_initial we find that the initial temperature of the ethylene glycol is higher than the final temperature of 32.5°C by an amount resultant from dividing the heat lost by the product of the mass and the specific heat capacity.

Answer:

Covalent bond

Explanation:

Functional groups can be defined as a group of atoms or ions responsible for the properties particular to a certain group of organic compounds. What we are saying in essence is that it is the functional groups that decides the behavior of the organic compound in question.

Covalent linkages are the mechanism through which these functional groups are linked to the carbon skeleton of the compound to which they belong. Covalent bonding is a type of chemical bonding which in there is sharing of electrons between atoms

Covalent bond

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Answer:

1.49×10²² atoms of H are contained in the sample

Explanation:

TNT → C₇H₅N₃O₆

1 mol of this has 7 moles of C, 5 moles of H, 3 moles of N and 6 moles of O

Let's determine the mass of TNT.

Molar mass is = 227 g/mol

As 1 mol has (6.02×10²³) NA atoms, how many moles are 8.94×10²¹ atoms.

8.94×10²¹ atoms / NA = 0.0148 moles

So this would be the rule of three to determine the mass of TNT

3 moles of N are in 227 g of compound

0.0148 moles of N are contained in (0.0148 .227) / 3 = 1.12 g

Now we can work with the hydrogen.

227 grams of TNT contain 5 moles of H

1.12 grams of TNT would contain (1.12 .5) / 227 = 0.0247 moles

Finally let's convert this moles to atoms:

0.0247 mol . 6.02×10²³ atoms / 1 mol = 1.49×10²² atoms

The number of Hydrogen atoms in the TNT is 1.51×10²² atoms.

We'll begin by calculating the number of mole of nitrogen that contains 8.94×10²¹ atoms of nitrogen.

From Avogadro's hypothesis,

6.02×10²³ atoms = 1 mole of N

Therefore,

8.94×10²¹ atoms = 8.94×10²¹ / 6.02×10²³  

8.94×10²¹ atoms = 0.015 mole of N

Next, we shall determine the number of mole of TNT (C₇H₅N₃O₆) that contains 0.015 mole of N

3 moles of N are present in 1 mole of TNT (C₇H₅N₃O₆).

Therefore,

0.015 mole of N will be present in = 0.015 / 3 = 0.005 mole of TNT (C₇H₅N₃O₆).

Next, we shall determine the number of mole of H in 0.005 mole of TNT (C₇H₅N₃O₆).

1 mole of TNT (C₇H₅N₃O₆) contains 5 moles of H.

Therefore,

0.005 mole of TNT (C₇H₅N₃O₆) will contain = 0.005 × 5 = 0.025 mole of H

Finally, we shall determine the number of atoms in 0.025 mole of H.

From Avogadro's hypothesis,

1 mole of H = 6.02×10²³ atoms

Therefore,

0.025 mole of H = 0.025 × 6.02×10²³

0.025 mole of H = 1.51×10²² atoms.

Thus, the number of atoms of Hydrogen in the sample of the TNT is 1.51×10²² atoms.

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Explanation:

The mass of carbon and hydrogen is calculated from the mass of their oxides ([tex]CO_{2}[/tex] and [tex]H_{2}O[/tex]) as follows.

    Mass of C = [tex]3.00 g CO_{2} \times \frac{12.01 g C}{44.0 g CO_{2}}[/tex]

                      = 0.818 g C

    Mass of H = [tex]1.23 g CO_{2} \times \frac{2.016 g H}{18.02 g H_{2}O}[/tex]

                      = 0.137 g H

So, the mass of C + mass of H is as follows.

                        0.818 g + 0.137 g = 0.955 g

This mass is actually less than the mass of sample. And, the missing mass must be caused by O. Hence, the mass of O will be calculated as follows.

            Mass of O = 1.50 g - 0.955 g

                              = 0.545 g

Now, we convert masses to moles and find their moles and ratio as follows.

  Element     Mass/g      Moles    Ratio    [tex]\times 2[/tex]     Integers

       C             0.818       0.068       1        2              2

       H             0.137        0.068      1        2               2

       O             0.545       0.0340    0.5   1                1

Thus, we can conclude that simplest formula for the given compound is [tex]C_{2}H_{2}O[/tex].

Answer:

C2H4O (ut quest)

Answer : The true statements are:

(a) Emulsions are a type of colloid

(b) The particles of a colloid are larger than the particles of a solution

(d) Many colloids scatter light (tyndal effect)

Explanation :

Colloid : It is defined as the solution in which the one substance is insoluble in another solution that means the insoluble substance rotating in the solution.

The particles of a colloid are larger than the particles of a solution.

Colloid do not separate on standing.

Cannot be separated by filtration.

Scatter light (Tyndall effect).

For example :

Milk is considered as a colloid because various substances (fats, proteins etc..) are present in milk which are suspended in a solution.

Suspension : It is a heterogeneous mixture in which some of the particles are settle down in the mixture on standing or over time.

The particles in a suspension are far larger than those of a solution.

Emulsion : It is a mixture of two or more liquids that are normally immiscible.

Emulsions are a type of colloid.

A barium fluoride solution with a concentration of 1.32 grams per liter, equivalent to its solubility, would be considered saturated. This means that no more barium fluoride can be dissolved in the water at the same temperature.

The solubility of barium fluoride in water is 1.32 grams per liter. Therefore, if a barium fluoride solution had a concentration of 1.32 grams per liter, it would be referred to as being saturated. A saturated solution is a solution in which no more solute can be dissolved in the solvent at a given temperature. In this case, the maximum amount of barium fluoride that can be dissolved in water at the given temperature has been reached.

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Answer:

The answer to your question is m = 3.2

Explanation:

Molality is defined as the number of moles of a solute dissolved in a mass of solvent (kg).

Data

moles of solute = 43.6

mass of solvent = 13.5 kg

Formula

Molality = [tex]\frac{number of moles}{Kg of solvent}[/tex]

Substitution

Molality = [tex]\frac{43.6}{13.5}[/tex]

Simplification and result

Molality = 3.2

Answer : The change in enthalpy of the coffee is negative.

Explanation :

Endothermic reaction : It is defined as the chemical reaction in which the energy is absorbed from the surrounding.

In the endothermic reaction, the energy of reactant are less than the energy of product.

Exothermic reaction : It is defined as the chemical reaction in which the energy is released into the surrounding.

In the exothermic reaction, the energy of reactant are more than the energy of product.

Enthalpy change : It is the difference between the energy of product and the reactant. It is represented as [tex]\Delta H[/tex].

As per question, if hot coffee sits on your desktop and loses 50 kJ of energy during cooling, the change in enthalpy of the coffee is negative.

Hence, change in enthalpy of the coffee is negative.

The noble gases undergo very few chemical reactions. They are stable and unreactive due to their full valence electron shells.

The noble gases d. undergo very few chemical reactions. The noble gases, such as helium, neon, and argon, are characterized by their high stability and low reactivity. They have full valence electron shells, which makes them unreactive and unlikely to form compounds with other elements. Because of their stable electron configurations, noble gases do not readily lose or gain electrons, meaning they do not undergo many chemical reactions.

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Noble gases are elements that are chemically very stable and undergo very few chemical reactions due to their complete outer electron shell. They do not easily give or accept electrons, thereby making them generally nonreactive. However, under extreme conditions, some noble gases like xenon can form compounds.

The noble gases are elements in group 18 of the periodic table including helium, neon, argon, krypton, xenon, and radon. They're named 'noble' due to their unique characteristics of being largely inert, or unreactive. Their outer electron shell is completely filled, which makes them high in ionization energy and resistant to forming compounds under normal conditions. This means that they do not readily give or accept electrons - hence they are chemically very stable, or in other words, they undergo very few chemical reactions.

However, there are a few exceptions to the rule. For example, under high pressure and temperature conditions, some noble gases like xenon can be forced to create compounds such as xenon hexafluoride (XeF6).

Despite their general chemical inactivity, noble gases have various practical applications. They are used in neon signs, as inert atmospheres in certain industrial processes, and as coolants due to their low boiling and melting points compared to other substances of similar atomic or molecular masses.

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Answer:

Question 1

B) Volume decreases

Question 2

A) Higher pressure

Question 3

A and B

Explanation:

Question 1

B) Volume decreases

Charles' Law: The Temperature-Volume Charles Law. It states that at constant pressure, the volume of a given amount of gas is directly proportional to temperature in Kelvin. As the temperature goes down, the volume also goes down, and vice-versa.

ogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules." For a given mass of an ideal gas, the volume and amount (moles) of

Question 2

A) Higher pressure

Avogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules." For a given mass of an ideal gas, the amount (moles) and the volume of the gas are directly proportional at constant temperature and pressure

Increasing the number of moles increases the volume to maintain the same volume of the container with less moles the gas has to be compressed, increasing its pressure

Boyle's law is a gas law, that states that at constant temperature, the pressure and volume of a gas are inversely proportional. increasing the volume , decreases the pressure and vice versa.

Question 3

A and B

A) Doubling pressure while keeping temperature constant. B) Doubling absolute temperature while keeping pressure constant.

The volume of a given mass of gas decreases as pressure increases but increases as absolute temperature increases.

In the first question, a helium balloon initially at room temperature is dunked into a bucket of liquid nitrogen, the volume of the immediately decreases because volume and absolute temperature are directly proportional according to Charles law.

In the second question, two separate containers A and B have the same volume and temperature, but container A has more gaseous molecules than B, then container A will have higher pressure because the pressure of a gas is directly proportional to the number of molecules of a gas present.

In the third question, the change that would cause the volume of a gas to double, assuming moles were held constant is doubling absolute temperature while keeping pressure constant. This follows from Charles law where the volume of a given mass of gas is directly proportional to its volume at constant pressure.

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Answer:

The answer to your question is the original dose of Ba-142 was 1184 μg

Explanation:

Data

Total time = 1.25 hours

The Final amount of Ba = 9.25 μg

The Half-life of Ba = 10.6 minutes

Process

1.- Convert total time to minutes

                      1 h ----------------- 60 min

                       1.25 h ------------ x

                        x = 75 min

2.- Draw a table of this process

                   Final amount of Ba               Time

                           9.25                                 75 min

                           18.5                                  64.4 min

                           37                                     53.8 min

                           74                                     43.2 min

                          148                                     32.6 min

                          296                                    22  min

                          592                                    11.4 min

                         1184                                      0.8 min        

Answer:

0.09425

Explanation:

The reactant is in solid phase and therefore has zero partial pressure.

The products have the same mole ratio (1:1) and will have the same partial pressure = 1/2 × 0.614 atm = 0.307 atm

Kp =  (NH3)(H2S) = 0.09425

The equilibrium constant expression for the given reaction is K = [NH3][H2S]. Since NH4HS is a solid, it is not included in the equilibrium constant expression. If the concentration of H2S triples, the concentration of NH3 should decrease by a factor of 3 to maintain equilibrium.

The given reaction involving NH4HS(s)⇌NH3(g)+H2S(g) is an equilibrium reaction. At equilibrium, the total pressure of NH3 and H2S combined is 0.614 atm. In this equilibrium expression, NH4HS(s) does not appear since it is a solid. Therefore, the equilibrium constant, K, is given by the expression K = [NH3][H2S]. Since the concentrations of the products are inversely proportional, if the H2S concentration triples, the NH3 concentration must decrease by a factor of 3 to keep the system at equilibrium so that the product of the concentrations equals K.

Answer:

The answer to your question is 4.07 x 10²² atoms

Explanation:

Process

1.- Get the molecular weight of Cocaine hydrochloride

C = 12 x 17 = 204 g

H = 22 x 1 = 22 g

Cl = 36 x 1 = 36 g

N = 14 x 1 = 14 g

O = 16 x 4 = 64 g

Molecular mass = 340 g

2.- Calculate the moles of the mass given

                             340 g -------------------  1 mol

                               23 g -------------------  x

                              x = (23 x 1) / 340

                              x = 0.068 moles of Cocaine

3.- Calculate the atoms

                            1 mol -------------------- 6 .023 x 10 ²³ atoms

                   0.068 moles -------------  x

                              x = (0.068 x 6.023 x 10²³) / 1

                              x = 4.07 x 10²² atoms

To find the heat of combustion for cyclopropane in kcal/gram, divide the heat of reaction (499.8 kcal/mol) by the molar mass of cyclopropane (42.08 g/mol).

The heat of combustion for cyclopropane in kcal/gram can be calculated using its molar mass and the given heat of reaction. The molar mass of cyclopropane, C3H6, is 42.08 g/mol. With a heat of reaction of 499.8 kcal/mol, we can calculate the heat of combustion per gram by dividing the heat of reaction by the molar mass:

Heat of combustion (kcal/g) = Heat of reaction (kcal/mol) / Molar mass (g/mol)

Thus,

Heat of combustion (kcal/g) = 499.8 kcal/mol / 42.08 g/mol

Once you perform the division, you'll have the heat of combustion for cyclopropane in kcal/gram.

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Dalton described the relationship between atoms and elements in terms of the composition of elements and the combination of atoms to form compounds.

Dalton described the relationship between atoms and elements in the following way:

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Dalton described elements as being made up of a single, unique type of atom. Atoms of different elements differ in properties. Atoms combine in fixed, whole-number ratios to form compounds.

John Dalton, the English chemist, proposed a theory that is fundamental to understanding chemical elements and atoms. According to Dalton's atomic theory:

Hence, Dalton described the relationship between atoms and elements by stating that each element is composed of one kind of atom unique to that element, which can combine to form more complex structures.

The Rydberg formula can calculate the wavelengths of light emitted by hydrogen atoms. However, with some modifications, it can also be used to calculate the wavelengths of light emitted by atoms of other elements. Hence, the initial energy level is n = 3

Rydberg equation:

1/λ = f/c = R(1/m² - 1/n²)

where c is the speed of light, f is frequency, λ is the wavelength, R is Rydberg constant, and n and m are the quantum numbers of the energy levels.

The initial energy level has quantum number n.

The final energy level is a ground state with the quantum number m = 1.

2.924 x 1015/2.998 x 108 = 1.097 x 107 x (1/12 - 1/n²)

(1/12 - 1/n²) = 0.8891

1/n² = 0.1109

=> n²

= 9

=> n = 3

Thus, the initial energy level is n = 3

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The electron in the hydrogen atom was in the 9th energy level (n = 9) before emitting the photon.

When a hydrogen atom transitions from an excited state to the ground state, it emits a photon with a certain frequency. In this case, the frequency is given as ν = 3.084 x 10^15 s^-1. We can use the equation ν = E/h, where E is the energy of the photon and h is Planck's constant, to find the energy of the emitted photon. Once we have the energy, we can determine the energy level the electron was in before the photon was emitted.

Using the energy-frequency relationship and Planck's constant, we have:

E = hν = (6.626 x 10^-34 J s)(3.084 x 10^15 s^-1) ≈ 2.041 x 10^-18 J

From this energy, we need to find the corresponding energy level. The energy levels of hydrogen are given by the formula: E = -13.6 eV/n^2, where E is the energy of the level and n is the principal quantum number. Rearranging the formula, we have:

n^2 = -13.6 eV/E ≈ -13.6 eV/(2.041 x 10^-18 J) ≈ -6.662 x 10^17

Taking the square root of both sides, we find:

n ≈ -8.159 x 10^8

Since n must be a positive integer, we can conclude that the electron was in the 9th level (n = 9) before emitting the photon.

Answer:

108.25 ºC

Explanation:

The boiling point elevation for a given solute in water is given by the expression:

ΔTb = i Kbm

where ΔTb is the boiling point elevation

           i is the van´t Hoff factor

           Kb the boiling constant which for water is 0.512 ºC/molal

           m is the molality of the solution

The molality of the solution is the number of moles per kilogram of solvent. Here we run into a problem since we are not given the identity of the coolant, but a search in the literature tells you that the most typical are ethylene glycol and propylene glycol. The most common is ethylene glycol and this it the one we will using in this question.

Now the i factor in the equation above is 1 for non ionizable compounds such as ethylene glycol.

Our equation is then:

ΔTb =  Kbm

So lets calculate the molality and then ΔTb:

m = moles ethylene glycol / Kg solvent

Converting gal to L

4.90 g x 3.785 L/gal = 18.55 L

in a 50/50 blend by volume we have 9.27 L of ethylene glycol, and 9.27 L of water.

We need to convert this 9.27 l of ethylene glycol to grams assuming a solution density of 1 g/cm³

9.27 L x 1000 cm³ / L = 9273.25 cm³

mass ethylene glycol = 9273.25 cm³ x 1 g/cm³ = 9273.25 g

mol ethylene glycol = 9273.25 g/ M.W ethylene glycol

                                 = 9273.25 g / 62.07 g/mol =149.4mol

molality solution =  149.4 mol / 9.27 Kg H₂O = 16.12 m

( density of water 1 kg/L )

Finally we can calculate ΔTb:

ΔTb =  Kbm = 0.512  ºC/molal x 16.12 molal = 8.25 ºC

boiling point = 100 º C +8.25 ºC = 108.25 ºC

( You could try to solve for propylene glycol the other popular coolant which should give around 106.7 ºC )

Answer:

Option 3.

364.1 g of CO₂ are produced by the reaction

Explanation:

Let's verify the balance equation for this reaction:

Fe₃O₄  +   4 CO  →  3 Fe   +   4CO₂

Let's convert the mass of Fe₃O₄ to moles (mass / molar mass)

478.9 g / 231.55 g/mol  = 2.07 moles

Ratio is 1:4, so let's make a rule of three to determine the moles of CO₂ produced and then its mass.

1 mol of Fe₃O₄ is needed to produce 4 moles of CO₂

Then, 2.07 moles of Fe₃O₄ will produce (2.07  .4) /1 = 8.27 moles of CO₂.

Molar mass . moles = mass

44 g/mol . 8.27 mol = 364 g of CO₂

Answer:

a. 25.8 g of water

b. 1720.4 g of water

Explanation:

A percent by mass, means the grams of solute in 100 g.  of solution. So, 16.2 g of urea are contained in 100 g of solution.

Then, solute mass + solvent mass = solution mass

16.2 g of urea + solvent mass = 100 g

solvent mass = 100 g - 16.2 g → 83.8 g

Now we can make the rule of three:

16.2 g of urea use 83.8 g of water

then, 5 g of urea would use (5 . 83.8) / 16.2 = 25.8 g of water

b. 1.5 % by mass, means 1.5 g of solute in 100 g of solution.

So water mass, for this solution will be 100 g - 1.5 g = 98.5 g

Now, we apply the rule of three:

1.5 g of solute use 98.5 of water

26.2 g of solute will use (26.2  . 98.5)/1.5 = 1720.4 g

The 5.00 g of urea in the preparation of a 16.2 percent by mass solution need 25.8 gram of water and 26.2 g of magnesium Chloride in the preparation of a 1.5 percent by mass solution needs 1720.4 g of water.

Percent by mass solution, means the grams of solute in 100 g. of solution.

(A) So, 16.2 g of urea are contained in 100 g of solution.

So, 83.8 water added to make 16.2 g solution.

Thus, 5 g of urea need

[tex]\bold {\dfrac { (5\times 83.8)} { 16.2} = 25.8 g}[/tex]

(B). 1.5 % by mass, means 1.5 g of solute in 100 g of solution.  

So, 98.5 g of water added to prepare 1.5 % solution.      

26.2 g of Magnesium Chloride will use

[tex]\bold {\dfrac {26.2 \times 98.5}{1.5} = 1720.4 g}[/tex]

Therefore, the 5.00 g of urea in the preparation of a 16.2 percent by mass solution need 25.8 gram of water and 26.2 g of magnesium Chloride in the preparation of a 1.5 percent by mass solution needs 1720.4 g of water.

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Answer:

Option A, The Rutherford experiment proved the Thomson "plum-pudding" model of the atom to be essentially correct.

Explanation:

Thomson's plum pudding model:

Plum pudding model was proposed by J.J Thomson. In Thomson's model, atoms are proposed as sea of positively charge in which electrons are distributed through out.

Result of Rutherford experiment:

As per Rutherford's experiment:

Most of the space inside the atom is empty.

Positively charge of the atom are concentrated in the centre of the atom known as nucleus.

Electrons are present outside the nucleus and revolve around it.

As it is clear that, result of Rutherford experiment did not supported the Thomson model.

The Rutherford experiment did not prove the Thomson 'plum-pudding' model of the atom to be essentially correct. Instead, it disproved this model by showing that atoms have a small, dense, positively charged nucleus.

The statement 'The Rutherford experiment proved the Thomson 'plum-pudding' model of the atom to be essentially correct.' (option A) is not accurate. In fact, the Rutherford experiment disproved the Thomson 'plum-pudding' model of the atom. Rutherford's experiment revealed that atoms have a small, dense, positively charged nucleus, contradicting Thomson's model which proposed that positive charge was spread evenly throughout the atom. All other listed experiments (options B, C, and D) did indeed provide the results described.

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Answer:b

Explanation:

Answer: B. Compressing 2 liters of water into a 1 liter volume.

Explanation: Liquids have little compressibility as compared to gases. Their molecules are quite near each other compared to gases which far apart. When liquids are compressed up to their capacity it will begin to resist. But gases have higher compressibility rate than liquids.

Answer:

The answer to your question is 1.46 moles of KMnO₄

Explanation:

Data

number of moles = ?

mass = 230.8 g

molecular mass of KMnO₄ = 39 + 55 + (16 x 4) = 158 g

Process

1.- Use proportions and cross multiplication to answer this problem.

                          158 g of KMnO₄ ----------------  1 mol

                           230.8 g of KMnO₄ -----------    x

                           x = (230.8 x 1) / 158

2.- Simplifying

                           x = 230.8 / 158

3.- Result

                          x = 1.46 moles

Answer:

80 liters

Explanation:

At STP, 1 mole of ideal gas has a volume of 22.4 liters.

Therefore, since liters and moles are directly proportional, we can use stoichiometry directly.

40L CH₄ × (2L H₂ / 1L CH₄) = 80L H₂

Answer:

Never leave an open flame unattended

Explanation: if an open flame is left unattended it can cause a fire outbreak so we have to watch it at all times to prevent the fire outbreak

Answer:

False.

Explanation:

The given statement is false because for hot vacuum filtration, the filter paper should be wet rather than dry when pouring the hot solution into the Buchner funnel. This is because The possible explanation the filter paper needs to be wetted is not only to allow it to adhere to the funnel, but also to promote the solute to filter down readily across its pores of the paper without wetting it (this is true for organic and aqueous solvents).