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

Explanation:

Ecology -

It is the branch of biology , which studies the interactions between the organisms and their biophysical environment , including both abiotic and biotic components .

Environmental science -

It is an interdisciplinary field , of information science , physical , biological science , which helps to study the environment and the environmental problems .

Environmentalism -

It is the rights regarding the concern for the protection of the environment and the to improve the health of the environment .

The main focus on environment and nature-related aspects of green ideology and politics .

.Answer:

Explanation:

Environmentalism -

It is the rights regarding the concern for the protection of the environment and the to improve the health of the environment .

The main focus on environment and nature-related aspects of green ideology and politics.

Ecology -

It is the branch of biology , which studies the interactions between the organisms and their biophysical environment , including both abiotic and biotic components .

Environmental science -

It is an interdisciplinary field , of information science , physical , biological science , which helps to study the environment and the environmental problems .

The question involves calculating the kinetic energy and speed of photoelectrons ejected from metallic caesium due to incident light using the photoelectric effect. The photoelectron kinetic energy is found using the difference between the photon energy and the work function, and the speed is calculated from the kinetic energy.

The question asks to calculate the kinetic energy and speed of electrons ejected from metallic caesium when exposed to light of different wavelengths, applying the principles of the photoelectric effect. The work function given for caesium is 2.14 eV.

To find the kinetic energy of the ejected electrons, we use the equation:

where KEmax is the maximum kinetic energy of the photoelectrons, E is the energy of the incident photons, and φ is the work function. The energy of the photons (E) can be calculated using:

where h is Planck's constant (4.135667696 × 10-15 eV·s), c is the speed of light (≈ 3.00 × 108 m/s), and λ is the wavelength of the incident light.

After finding KEmax, we can find the speed of the electrons using the equation:

where v is the speed of the electron, m is the mass of the electron (9.10938356 × 10-31 kg), and KE is the kinetic energy in joules. Remember to convert electron volts to joules (1 eV = 1.602 × 10-19 J) to use SI units.

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For 750 nm light, no electrons are ejected. For 250 nm light, the ejected electrons have a kinetic energy of 4.52 ×  [tex]10^{-19[/tex] J and a speed of 996 km/s.

Let's solve this problem step by step for both parts (a) and (b) using the photoelectric effect equation:

hf = Φ + KE

Where:

KE = kinetic energy of ejected electrons

convert eV to Joules: 1 eV = 1.602 × [tex]10^{-19[/tex] J

So, Φ = 2.14 eV = 2.14 × (1.602 ×  [tex]10^{-19[/tex] ) J = 3.42828 × [tex]10^{-19[/tex] J

Also, use c = λf, where c is the speed of light (3 × [tex]10^{8[/tex] m/s)

1. Calculate frequency:

  f = [tex]\frac{C}{\lambda}[/tex] = (3 × [tex]10^{8[/tex] m/s) ÷ (750  × [tex]10^{-19[/tex]) = 4 × [tex]10^{14[/tex] Hz

2. Calculate energy of photon:

  E = hf = (6.626 × [tex]10^{-34[/tex])(4 ×[tex]10^{14[/tex]) = 2.6504 × [tex]10^{-19[/tex] J

3. Calculate kinetic energy:

  KE = E - Φ = 2.6504 ×  [tex]10^{-19[/tex] - 3.42828 ×  [tex]10^{-19[/tex] = -0.77788 ×  [tex]10^{-19[/tex] J

The negative KE means no electrons are ejected for this wavelength.

1. Calculate frequency:

   f = [tex]\frac{C}{\lambda}[/tex] = (3 × [tex]10^{8[/tex]) ÷ (250 × [tex]10^{-9[/tex]) = 1.2 × [tex]10^{15[/tex] Hz

2. Calculate energy of photon:

  E = hf = (6.626 × [tex]10^{-34[/tex])(1.2 × [tex]10^{15[/tex]) = 7.9512 × [tex]10^{-19[/tex] J

3. Calculate kinetic energy:

  KE = E - Φ = 7.9512 ×  [tex]10^{-19[/tex] - 3.42828 ×  [tex]10^{-19[/tex] = 4.52292 ×  [tex]10^{-19[/tex]

Therefore Kinetic energy is 4.52 ×  [tex]10^{-19[/tex] J.

4. Calculate speed:

Therefore the speed is approximately 996 km/s.

Answer:

The correct option is: A. saturation pressure of water increases

Explanation:

Relative humidity, at a given temperature, is defined as ratio of partial pressure of the water vapor preset in the air mixture and equilibrium vapor pressure above pure water.

Relative humidity of a system is dependent upon the pressure and temperature. When the amount of water vapor in a system is constant, the higher relative humidity is higher at low temperature and lower at high temperature. This is because at high temperatures, the air capacity and the saturation pressure increases.

Therefore, in the morning when the temperature is low in fali, the relative humidity is high. As the temperature increases, the relative humidity in fali decreases.

Answer:

[tex]T_f = 76.46°C[/tex]

Explanation:

Given data:

Mass of mixture = 454 kg

Initial temperature is 10°C

Heat added is Q = 121300 kJ

Heat capacity (Applesuace) at 32.8°C is 4.02kJ/kg K

From heat equation we have

[tex]Q = mCp(T_f -T_i)[/tex]

[tex]\frac{Q}{mCp} = (T_f -T_i)[/tex]

[tex]T_f = T_i + \frac{Q}{mCp}[/tex]

Putting all value to get required final temperature value

[tex]T_f = \frac{121300}{454\times 4.02} + 10[/tex]

[tex]T_f = 76.46°C[/tex]

Answer:

(a) No. Ground-state carbon has only 2 half-filled orbitals that could be used for bonding.

(b) No. The bond angles would be incorrect as the p-orbitals are all perpendicular to each other (90°).

Explanation:

See attachment for the ground-state and excited-state electron orbital diagrams of carbon.

A methane molecule has all four CH bonds the same length and at 109.5° from each other. Hybridization of the s and p orbitals to sp³ orbitals is necessary.

Answer:

The overall standard deviation, s = 6.46 %

Given:

Sampling variance, [tex]s_{b} = \pm 6.0% = 0.06[/tex]

Analytical variance, [tex]s_{a} = \pm 2.4% = 0.024[/tex]

Solution:

Variance additive is given by:

[tex]s^{2} = s_{a}^{2} + s_{b}^{2}[/tex]                        (1)

where

s = overall variance

Also, we know that:

Standard Deviation, [tex]\sigma = \sqrt{variance}[/tex]

Therefore the standard deviation of the sampling, analytical and overall sampling is given by taking the square root of eqn (1) on both the sides:

[tex]s = \sqrt{s_{a}^{2} + s_{b}^{2}}[/tex]

[tex]s = \sqrt{0.024^{2} + 0.06^{2}} = 0.0646[/tex]

s = 6.46 %

The overall standard deviation considering both the sampling method (±6.0%) and analytical method (±2.4%) is calculated by squaring each to get variances, summing these, and taking the square root of the sum, resulting in an overall standard deviation of ±6.464%.

To calculate the overall standard deviation when given two separate standard deviations from different sources (sampling and analytical method), you should first square each standard deviation to find the variances. Then, add the variances together, and finally, take the square root of the total to find the overall standard deviation.

The standard deviation of the sampling method is ±6.0%, and this value squared gives 36.0%. The analytical method's standard deviation is ±2.4%, which squared gives 5.76%. Adding these two variances together, we get 36.0% + 5.76% = 41.76%. The overall standard deviation is then the square root of 41.76%, which is approximately ±6.464%.

Therefore, the combined or overall standard deviation considering both the sampling and analytical methods is ±6.464%.

Answer:

The correct option is: c. petroleum jelly, d. Polyethylene glycol 4000/600 mixture                

Explanation:

Topical medications are used for the treatment of ailments and include ointments, gels, lotions creams etc. that can applied directly on the surface of the body i.e. skin.

An ointment base medication gets rapidly absorbed into the skin. Some of the examples of ointment bases include water-soluble bases: polyethylene glycol, hydrocarbon bases: petroleum jelly, paraffin wax.

Answer:

yo mama

Explanation:

Final answer:

Radon-219 decays into radon-218 by emitting an alpha particle, a process that reduces its atomic mass and number, leading to a new isotope.

Explanation:

Radon-219 decays to radon-218 by releasing an alpha particle (Helium nucleus). When radium-226 undergoes alpha decay, it forms radon-222 and an alpha particle. The process involves the nucleus of an atom releasing two protons and two neutrons, which together form an alpha particle. In the case of radon decay, the alpha particle is emitted, decreasing the atomic mass by 4 units and atomic number by 2, resulting in a new isotope. Therefore, the correct answer is b. a muon. The emission of an alpha particle is a common mode of decay for heavy, unstable nuclei such as radon-219.

Answer:

Weight of solution produced = 5135 kg

Amount of water removed = 4865 kg

Explanation:

For the balance of mass, the incoming mass of sugar must be equal to the outgoing mass. So, the incoming mass (mi) is 38% of 10000 kg

mi = 0.38x10000 = 3800 kg

The outgoing mass (mo) must be 3800 kg, and it is 74% of the total mass (mt)

mo = 0.74xmt

0.74xmt = 3800

mt = 3800/0.74

mt = 5135 kg

This is the mass of solution produced.

The amount of water removed (wr) is the amount of water incoming (wi) less the amount of water outgoing (wo). Both will be the total mass less the mass of sugar :

wi = 10000 - 3800 = 6200 kg

wo = 5135 - 3800 = 1335 kg

wr = wi - wo

wr = 6200 - 1335

wr = 4865 kg

From the data provided and the concept of balance of mass,, the mass of solution produced is 5135 kg and the water removed is  4865 kg

Evaporation is the process by which molecules of a liquid turn to gas.

To calculate the weight of solution produced and amount of water removed:

Using the concept of balance of mass, the incoming mass of sugar must be equal to the outgoing mass.

Incoming mass (Mi) = 38% of 10000 kg

Mi = 3800 kg

Therefore, the outgoing mass (Mo)= 3800 kg

Mo= 74% of the total mass, Mt

Mo = 0.74 x Mt

Mt = 3800/0.74

Mt = 5135 kg

Thus, the mass of solution produced is 5135 kg

The amount of water removed (Wr) = water incoming (Wi) - water outgoing (wo).

Wi = 10000 - 3800 = 6200 kg

Wo = 5135 - 3800 = 1335 kg

Wr = Wi - Wo

Wr = 6200 - 1335

Wr = 4865 kg

Therefore, the water removed is  4865 kg

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Answer:

Mass percentage of analyte = 0.10%

Explanation:

The mass of analyte in the solution is calculated as follows:

n = CV = (21.1 mgL⁻¹)(100 mL)(1L/1000mL) = 2.11 mg

The mass percentage of the analyte is calculated as follows:

(g analyte)/(g sample) x 100%

(2.11 mg)(1g/1000g) / (2.05g) x 100% = 0.10%

Answer:

Explanation:

Given parameters:

Molar mass of compound = 60.1g/mol

Mass of fuel used in the combustion process = 2.859g

Mass of carbon dioxide produced = 4.190g

Mass of oxygen produced= 3.428g

Unknown parameters;

Empirical and molecular formula of the compound

Solution

The empirical formula of a compound is its simplest formula. The molecular formula is the actual formula of the compound showing the proportions of the atoms.

We can derive the empirical formula of a compound from its molecular formula and vice versa.

Here, we have to work from empirical formula to molecular formula.

Solving

For Carbon, we can determine the mass from the amount of carbon dioxide produced:

       mass of carbon in compound = [tex]\frac{12}{44}[/tex] x 4.19g = 1.14g

For Hydrogen, we can determine the mass from the amount of water produced:      

   mass of hydrogen in compound = [tex]\frac{2}{18}[/tex] x 3.428g = 0.38g

To determine the mass of N in the compound:

 mass of compound = mass of C + mass of H + mass of N

   mass of N = mass of compound - (mass of C + mass of H)

    mass of N = 2.859g - (1.14g + 0.38g) = 1.34g

2. we now proceed to find the empirical formula using the process below:

Elements                C                          H                       N

mass of the

elements                1.14                      0.38                 1.34

Atomic mass

of elements             12                           1                        14

Number of

moles                     1.14/12                  0.38/1                1.34/14

                              0.095                    0.38                   0.095

Dividing by

the smallest        0.095/0.095       0.38/0.095          0.095/0.095

                                1                                4                           1

The empirical formula of the compound is CH₄N

To obtain the molecular formula, we need to find the number of times the empirical formula must have repeated itself in the original form.

 Molar mass of CH₄N = 12 + 4 + 14 = 30g/mol

  Ratio = [tex]\frac{molar mass of molecular formula}{molar mass of empirical formula}[/tex] = [tex]\frac{60.1}{30}[/tex] = 2

Molecular formula = 2(CH₄N) = C₂H₈N₂

Answer : The moles of [tex]O_2[/tex] left in the products are 0.16 moles.

Explanation :

First we have to calculate the moles of [tex]CH_4[/tex].

Using ideal gas equation:

[tex]PV=nRT[/tex]

where,

P = pressure of gas = 1 atm

V = volume of gas = 10 L

T = temperature of gas = [tex]27^oC=273+27=300K[/tex]

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

[tex](1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)[/tex]

[tex]n=0.406mole[/tex]

Now we have to calculate the moles of [tex]O_2[/tex].

The balanced chemical reaction will be:

[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]

From the balanced reaction we conclude that,

As, 1 mole of [tex]CH_4[/tex] react with 2 moles of [tex]O_2[/tex]

So, 0.406 mole of [tex]CH_4[/tex] react with [tex]2\times 0.406=0.812[/tex] moles of [tex]O_2[/tex]

Now we have to calculate the excess moles of [tex]O_2[/tex].

[tex]O_2[/tex] is 20 % excess. That means,

Excess moles of [tex]O_2[/tex] = [tex]\frac{(100 + 20)}{100}[/tex] × Required moles of [tex]O_2[/tex]

Excess moles of [tex]O_2[/tex] = 1.2 × Required moles of [tex]O_2[/tex]

Excess moles of [tex]O_2[/tex] = 1.2 × 0.812 = 0.97 mole

Now we have to calculate the moles of [tex]O_2[/tex] left in the products.

Moles of [tex]O_2[/tex] left in the products = Excess moles of [tex]O_2[/tex] - Required moles of [tex]O_2[/tex]

Moles of [tex]O_2[/tex] left in the products = 0.97 - 0.812 = 0.16 mole

Therefore, the moles of [tex]O_2[/tex] left in the products are 0.16 moles.

Answer:

c. electron-pair donor.

Explanation:

Lewis base -

A Lewis base is an electron rich species , which is available for donation .

Hence , Lewis base is a electron - pair donor .

The indication of a Lewis base , is it has a negative charge or lone pairs of electrons .

Hence , the species with a lone pair or negative charge act as a Lewis base .

for example , OH ⁻ is a Lewis base , due to its negative charge which is available for donation .

H₂O is also a Lewis base , due to its lone pairs of electrons , that are available for donation .

Answer : The mass of sodium sulfate needed is 5.7085 grams.

Explanation : Given,

Concentration of sodium ion = 0.148 mol/L

Volume of solution = 2.29 L

Molar mass of sodium sulfate = 142 g/mole

First we have to determine the moles of sodium ion.

[tex]\text{Concentration of sodium ion}=\frac{\text{Moles of sodium ion}}{\text{Volume of solution}}[/tex]

[tex]0.184mol/L=\frac{\text{Moles of sodium ion}}{2.29L}[/tex]

[tex]\text{Moles of sodium ion}=0.08035mole[/tex]

Now we have to calculate the moles of sodium sulfate.

The balanced chemical reaction will be,

[tex]Na_2SO_4\rightarrow 2Na^++SO_4^{2-}[/tex]

As, 2 moles of sodium ion produced from 1 moles of [tex]Na_2SO_4[/tex]

So, 0.08035 moles of sodium ion produced from [tex]\frac{0.08035}{2}=0.040175[/tex] moles of [tex]Na_2SO_4[/tex]

Now we have to calculate the mass of sodium sulfate.

[tex]\text{Mass of }Na_2SO_4=\text{Moles of }Na_2SO_4\times \text{Molar mass of }Na_2SO_4[/tex]

[tex]\text{Mass of }Na_2SO_4=0.040175mole\times 142g/mole=5.7085g[/tex]

Therefore, the mass of sodium sulfate needed is 5.7085 grams.

Answer: The frequency of the radiation is 33.9 THz

Explanation:

We are given:

Wave number of the radiation, [tex]\bar{\nu}=1130cm^{-1}[/tex]

Wave number is defined as the number of wavelengths per unit length.

Mathematically,

[tex]\bar{\nu}=\frac{1}{\lambda}[/tex]

where,

[tex]\bar{\nu}[/tex] = wave number = [tex]1130cm^{-1}[/tex]

[tex]\lambda[/tex] = wavelength of the radiation = ?

Putting values in above equation, we get:

[tex]1130cm^{-1}=\farc{1}{\lambda}\\\\\lambda=\frac{1}{1130cm^{-1}}=8.850\times 10^{-4}cm[/tex]

Converting this into meters, we use the conversion factor:

1 m = 100 cm

So, [tex]8.850\times 10^{-4}cm=8.850\times 10^{-4}\times 10^{-2}=8.850\times 10^{-6}m[/tex]

[tex]\nu=\frac{c}{\lambda}[/tex]

where,

c = the speed of light = [tex]3\times 10^8m/s[/tex]

[tex]\nu[/tex] = frequency of the radiation = ?

Putting values in above equation, we get:

[tex]\nu=\frac{3\times 10^8m/s}{8.850\times 10^{-4}m}[/tex]

[tex]\nu=0.339\times 10^{14}Hz[/tex]

Converting this into tera Hertz, we use the conversion factor:

[tex]1THz=1\times 10^{12}Hz[/tex]

So, [tex]0.339\times 10^{14}Hz\times \frac{1THz}{1\times 10^{12}Hz}=33.9THz[/tex]

Hence, the frequency of the radiation is 33.9 THz

Answer:

37.275g must react

Explanation:

Step 1: Balance the reaction

The reaction is as follow:

2 K + Cl2---------> 2KCl

This means that for 2 moles K we have 1 mole Cl2 ( and 2 moles KCl)

Step 2: Calculate moles

Moles of potassium:  moles = mass / Molar mass

Moles of potassium = 41g / 39.1g/mol  

Moles of potassium = 1.05 moles

for each 2 moles potassium we have 1 mole Cl2 ( and 2 moles KCl)

Moles of Chlorine : 1.05 / 2 = 0.525 moles ( and 1.05 moles of KCl)

Step 3: Calculate mass of chlorine

mass chlorine = moles x Molar mass  = 0.525 x 35.5 x 2

       = 37.275g Chlorine

Answer:

[tex]density=1.8x10^{-3}g/mL[/tex]

Explanation:

Hello,

Considering the ideal equation of state:

[tex]PV=nRT[/tex]

The moles are defined in terms of mass as follows:

[tex]n=\frac{m}{M}[/tex]

Whereas [tex]M[/tex] the gas' molar mass, thus:

[tex]PV=\frac{mRT}{M}[/tex]

Now, since the density is defined as the quotient between the mass and the volume, we get:

[tex]P=\frac{m}{V} \frac{RT}{M}[/tex]

Solving for [tex]m/V[/tex]:

[tex]density= m/V=\frac{PM}{RT}[/tex]

Thus, the result is given by:

[tex]density=\frac{(1atm)(44g/mol)}{[0.082atm*L/(mol*K)]*298.15K} \\density=1.8g/L=1.8x10^{-3}g/mL[/tex]

Best regards.

Answer:

When you prepared the solution, you will find that the molarity is 0.73M

Explanation:

First of all you should get by the periodic table, molar mass in the anhydrouds calcium.

CaCl2 · 0H20 = 110.98 g/m

Now we have to find out how many mols are 15 g.

So 15g / 110.98 g/m = 0.135 moles

This moles are in 185 ml of water. Molarity as you should know are moles of solute in 1 L of solution (either 1000 ml)

185 ml ______ 0.135 moles

1000 ml _____ x    x = (1000*0.135) /185 = 0.730M

Answer: Option (b) is the correct answer.

Explanation:

In material bonding, there occurs Vander waal foces between the molecules in which their is either an induced or permanent dipole moment that attract molecules towards each other.

And, due to these forces the molecules are held together.

On the other hand, in a ionic bond there will always be transfer of electrons from one atom to another. This is because on atom which loses its valence electrons acquires a positive charge and another atom which gains the electrons acquires a negative charge.

Hence, these opposite charges strongly gets attracted towards each other forming a strong bond.

Whereas in a covalent bond, there will be sharing of electrons between the combining atoms.

In a metallic bond, there occurs a sea of electrons which is uniformly distributed throughout the solid substance or material.

Thus, we can conclude that the statement, Van der Waals bonds are formed by Van der Waals forces in which molecules or atoms have either an induced or permanent dipole moment to attract each other, about material bonding is correct.

Final answer:

The correct statement is that Van der Waals bonds are formed by Van der Waals forces due to induced or permanent dipoles, unlike ionic or covalent bonds.

Explanation:

The correct statement about material bonding is option b: Van der Waals bonds are formed by Van der Waals forces in which molecules or atoms have either an induced or permanent dipole moment to attract each other. Unlike ionic and covalent bonds, Van der Waals bonds involve weaker, more temporary attractions. In contrast, ionic bonds occur when electrons are transferred from one atom to another, usually between metals and nonmetals, creating a positive and a negative ion that attract each other.

Finally, covalent bonds involve the sharing of electron pairs between atoms, and in metallic bonds, valence electrons are not donated to other atoms; instead, they form a 'sea of electrons' that surrounds the metal ions.

The heat released (ΔH) when one mole of manganese is burned to form Mn3O4 can be derived from a thermochemical equation and molar enthalpies. For specific values, consult a standard enthalpy of formation table.

The calculation for the heat released when one mole of manganese is burned to form Mn3O4 at standard state condition involves understanding of chemistry concepts such as molar enthalpies and thermochemical equations. Unfortunately, without specifying the thermochemical equation for the formation of Mn3O4 from manganese or the molar enthalpy of this specific reaction, an exact value cannot be given.

However, as a general guide, the heat released (also known as enthalpy change, ΔH) can be found from the formula ΔH = -q, where q represents the heat absorbed. A negative value indicates heat is being released. In thermochemical equations, the ΔH value is often given per mole of a substance involved in the reaction, so you would usually directly obtain the heat released when one mole of a substance is involved from the molar enthalpy.

For specific values, refer to a standard enthalpy of formation table, a resource often found in chemistry textbooks or scientific literature, to find the molar enthalpy for the formation of Mn3O4 from manganese.

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Answer:

3,78 mL of 12,0wt% H₂SO₄

Explanation:

The equilibrium in water is:

H₂O (l) ⇄ H⁺ (aq) + OH⁻ (aq)

The initial concentration of [H⁺] is 10⁻⁸ M and final desired concentration is [H⁺] = [tex]10^{-7,2}[/tex]

Thus, you need to add:

[H⁺] = [tex]10^{-7,2} - 10^{-8}[/tex] = 5,31x10⁻⁸ M

The total volume of the pool is:

7,00 m × 18,0 m ×1,50 m = 189 m³ ≡ 189000 L

Thus, moles of H⁺ you need to add are:

5,31x10⁻⁸ M × 189000 L = 1,00x10⁻² moles of H⁺

These moles comes from

H₂SO₄ → 2H⁺ +SO₄²⁻

Thus:

1,00x10⁻² moles of H⁺ × [tex]\frac{1 H_{2}SO_{4} moles}{2H^+ mole}[/tex] = 5,00x10⁻³ moles of H₂SO₄

These moles comes from:

5,00x10⁻³ moles of H₂SO₄ × [tex]\frac{98,1g}{1mol}[/tex] × [tex]\frac{100 gsolution}{12 g H_{2}SO_{4} }[/tex] × [tex]\frac{1mL}{1,080 g}[/tex] = 3,78 mL of 12,0wt% H₂SO₄

I hope it helps!

Answer:

A. 56 kJ

B. 75.8 kJ

C. 11 ˚C

Explanation:

A. The heat of vaporization, ∆Hvap = 40.7 kJ/mol, gives the amount of energy per mole of water required to vaporize water to steam. The molar mass of water is 18.02 g/mol.

Q = M·∆Hvap = (25 g)(mol/18.02g)(40.7 kJ/mol) = 56 kJ

B. Five steps are necessary in this process. First, the ice will be warmed to 0 °C, then melted to water. The water will be heated to 100 °C, then vaporized. Finally, the vapor will be heated from 100 °C to 105 °C.

We calculate the heat required to warm the ice from -4.0 °C to 0 °C:

Q₁ = mcΔt = (25 g)(2.06 J∙g⁻¹˚C⁻¹)(0 °C - (-4.0 °C)) = 206 J

Then we calculate the heat required to melt the ice to water:

Q₂ = M∙∆Hfus = (25 g)(mol/18.02 g)(6.02 kJ/mol) = 8.35 kJ

Then, we calculate the heat required to warm the water from 0 °C to 100 °C.

Q₃ = mcΔt = (25 g)(4.184 J∙g⁻¹˚C⁻)(100 °C - 0 °C) = 10460 J

Then we calculate the heat required to vaporize the water:

Q₄ = M∙∆Hvap = (25 g)(mol/18.02 g)(40.7 kJ/mol)  = 56.5 kJ

Finally, the vapor is heated from 100 °C to 105 °C.

Q₅ = mcΔt = (25 g)(2.03 J∙g⁻¹˚C⁻)(105 °C - 100 °C) = 254 J

The total heat required is the sum of Q₁ through Q₅

Qtotal = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Qtotal = (206 J)(1 kJ/1000J) + 8.35 kJ + (10460 J)(1 kJ/1000J) + 56.5 kJ + (254 J)(1 kJ/1000J)

Qtotal = 75.8 kJ

C. The heat required to melt the ice is provided by the water as it decreases in temperature.  

First, we calculate the energy required to melt ice to water

Q = M∙∆Hfus = (8.32 g)(mol/18.02 g)(6.02 kJ/mol) = 2.779 kJ

There are at least two ways to solve this problem. Here, we will calculate the heat lost when all the water is brought to a temperature of 0 °C:

Q = mc∆t = (55 g)(4.184 J∙g⁻¹˚C⁻¹)(25 °C - 0°C) = 5753 J

We see that the water has enough energy to melt all of the ice. The residual heat energy of the water after melting all the ice is:

5753 J - (2.779 kJ)(1000J/kJ) = 2974 J

Now the problem becomes that we have (8.32 g + 55 g) = 63.32 g of water at 0 °C that will be raised to some final temperature by the residual heat of 2974 J:

Q = mcΔt ⇒ Δt = Q/(mc)

Δt = (2974 J) / (63.32 g)(4.184 J∙g⁻¹˚C⁻¹) = 11 ˚C

T(final) - T(inital) = 11 ˚C

T(final) = 11 ˚C + T(inital) = 11 ˚C + 0 ˚C  = 11 ˚C

Thus, the final temperature will be 11 ˚C.

The heat required to vaporize 25g of water at 100˚C is 56.529 kJ. To convert 25g of ice at -4.0˚C to water vapor at 105˚C, approximately 83.8 kJ of heat is required. The final temperature of the water sample after the melting of an 8.32g ice cube is about 24.856˚C.

A. To find the heat required to vaporize 25 g of water at 100˚C, we need to use the given heat of vaporization (∆Hvap) which is 40.7 kJ/mol. However, we need the molar mass of water which is approximately 18.015 g/mol. The calculation is as follows: (25 g) / (18.015 g/mol) * (40.7 kJ/mol) = 56.529 kJ. Therefore, the heat needed is 56.529 kJ.

B. To convert 25 g of ice at -4.0˚C to water vapor at 105˚C we first calculate the heat needed to melt the ice to water at 0˚C, then to heat the water from 0˚C to 100 ˚C, and finally to vaporize the water. When all of these are added up, the total heat can be calculated as approximately 83.8 kJ.

C. To find the final temperature of the entire water sample after the ice is melted, we set the heat gained by the ice equal to the heat lost by the water. Solving gives a final temperature of about 24.856˚C.

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Answer:

This question begins with something, you should know: molar mass from water is aproximately 18 g/m, so if 18 grams of water are contained in 1 mole, the 40 grams occuped 2.22 moles. As you see, opcion a is the best!

Explanation:

Explanation:

Mass of fructose = 33.56 g

Mass of water =  18.88  g

Total mass of the solution =  Mass of fructose + Mass of water = M

M = 33.56 g + 18.88  g =52.44 g

Volume of the solution = V = 40.00 mL

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

a) Density of the solution:

[tex]\frac{M}{V}=\frac{52.44 g}{40.00 mL}=1.311 g/mL[/tex]

b) Molar mass of fructose = 180.16 g/mol

Moles of fructose = [tex]n_1=\frac{ 33.56 g}{180.16 g/mol}=0.1863 mol[/tex]

Molar mass of water = 18.02 g/mol

Moles of water= [tex]n_2=\frac{ 18.88 g}{18.02 g/mol}=1.0477 mol[/tex]

Mole fraction of fructose in this solution:[tex]\chi_1[/tex]

[tex]\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1863 mol}{0.1863 mol+1.0477 mol}[/tex]

[tex]\chi_1=0.1510[/tex]

Mole fraction of water = [tex]\chi_2=1-\chi_1=0.8490[/tex]

c) Average molar mass of of the solution:

=[tex]\chi_1\times 180.16 g/mol+\chi_2\times 18.02 g/mol[/tex]

[tex]=0.1510\times 180.16 g/mol+0.8490\times 18.02 g/mol=42.50 g/mol[/tex]

d) Mass of 1 mole of solution = 42.50 g/mol

Density of the solution = 1.311 g/mL

d) Specific molar volume of the solution:

[tex]\frac{\text{Average molar mass}}{\text{Density of the mass}}[/tex]

[tex]=\frac{42.50 g/mol}{1.311 g/mL}=32.42 mL/mol[/tex]

Answer:

The correct options are: 1. Ca²⁺ and 4. Mg²⁺      

Explanation:

Hard water is the water with high mineral content. Temporary hardness and permanent hardness are the two types of hardness of water.

Temporary hardness is due to the presence of dissolved bicarbonate minerals. These minerals present in the water, dissociate to give multivalent calcium cations (Ca²⁺) and magnesium cations (Mg²⁺).

Therefore, the presence of metal cations such as calcium cations (Ca²⁺) and magnesium cations (Mg²⁺) makes the water hard.

Answer:

the maximum volume that can be burned per minute is: 0,895 mL of ethanol.

Explanation:

The combustion of ethanol is:

C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O

With gas law:

PV/RT = n

Where P is pressure (1,00 atm)

V is volume (4,73 L of air per minute)

R is gas constant (0,082 atmL/molK)

T is temperatue(25°C≡298,15K)

And n are moles, replacing:

n = 0,193 moles of air per minute.

These moles of air contain:

0,193 moles air ×[tex]\frac{21 molesO_2}{100 molesAIR}[/tex] = 0,0406 moles O₂

Thus, the maximum volume that can be burned per minute is:

0,046 moles O₂[tex]\frac{1molC_{2}H_{5}OH}{3molesO_2} \frac{46,07 g}{1mol} \frac{1mL}{0,789g}[/tex] = 0,895 mL of ethanol per minute

I hope it helps!

Answer : The new density and new volume of carbon dioxide gas is 0.2281 g/L and [tex]7.2m^3[/tex] respectively.

Explanation :

First we have to calculate the new or final volume of carbon dioxide gas.

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas = 10 kPa

[tex]P_2[/tex] = final pressure of gas = 15 kPa

[tex]V_1[/tex] = initial volume of gas = [tex]10m^3[/tex]

[tex]V_2[/tex] = final volume of gas = ?

[tex]T_1[/tex] = initial temperature of gas = [tex]50^oC=273+50=323K[/tex]

[tex]T_2[/tex] = final temperature of gas = [tex]75^oC=273+75=348K[/tex]

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

[tex]\frac{10kPa\times 10m^3}{323K}=\frac{15kPa\times V_2}{348K}[/tex]

[tex]V_2=7.2m^3[/tex]

The new volume of carbon dioxide gas is [tex]7.2m^3[/tex]

Now we have to calculate the new density of carbon dioxide gas.

[tex]PV=nRT\\\\PV=\frac{m}{M}RT\\\\P=\frac{m}{V}\frac{RT}{M}\\\\P=\rho \frac{RT}{M}\\\\\rho=\frac{PM}{RT}[/tex]

Formula for new density will be:

[tex]\rho_2=\frac{P_2M}{RT_2}[/tex]

where,

[tex]P_2[/tex] = new pressure of gas = 15 kPa

[tex]T_2[/tex] = new temperature of gas = [tex]75^oC=273+75=348K[/tex]

M = molar mass of carbon dioxide gas = 44 g/mole

R = gas constant = 8.314 L.kPa/mol.K

[tex]\rho[/tex] = new density

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

[tex]\rho_2=\frac{(15kPa)\times (44g/mole)}{(8.314L.kPa/mol.K)\times (348K)}[/tex]

[tex]\rho_2=0.2281g/L[/tex]

The new density of carbon dioxide gas is 0.2281 g/L

Answer :

(a) The 60 feet of water is equal to 179344.2 Pa.

(b) The 220 psi is equal to [tex]31680lbf/ft^2[/tex]

(c) The 120 torr is equal to 15998.6 Pa.

(d) The 1.0 atm to inches of glycerin  is equal to 323.07 inches.

(e) The 1050 mm Hg is equal to [tex]376.95lbf/ft^2[/tex]

Explanation :

(a) The conversion used from feet to pascal is:

[tex]1\text{ feet}=2989.07Pa[/tex]

As we are given that 60 feet of water. Now we have to convert into Pa.

As, [tex]1\text{ feet}=2989.07Pa[/tex]

So, [tex]60\text{ feet}=\frac{60\text{ feet}}{1\text{ feet}}\times 2989.07Pa=179344.2Pa[/tex]

The 60 feet of water is equal to 179344.2 Pa.

(b) The conversion used from [tex]psi[/tex] to [tex]lbf/ft^2[/tex] is:

[tex]1\text{ psi}=144lbf/ft^2[/tex]

As we are given that 220 psi. Now we have to convert into [tex]lbf/ft^2[/tex].

As, [tex]1\text{ psi}=144lbf/ft^2[/tex]

So, [tex]220\text{ psi}=\frac{220\text{ psi}}{1\text{ psi}}\times 144lbf/ft^2=31680lbf/ft^2[/tex]

The 220 psi is equal to [tex]31680lbf/ft^2[/tex]

(c) The conversion used from torr to pascal is:

[tex]1\text{ torr}=133.322Pa[/tex]

As we are given that 120 torr. Now we have to convert into Pa.

As, [tex]1\text{ torr}=133.322Pa[/tex]

So, [tex]120\text{ torr}=\frac{120\text{ torr}}{1\text{ torr}}\times 133.322Pa=15998.6Pa[/tex]

The 120 torr is equal to 15998.6 Pa.

(d) 1.0 atm to inches of glycerin

Formula used : [tex]P=\rho gh[/tex]

where,

P = pressure of glycerin = 1.0 atm = 101325 Pa

g = acceleration due to gravity = [tex]9.8m/s^2[/tex]

[tex]\rho[/tex] = density of glycerin = [tex]1260kg/m^3[/tex]

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

[tex]101325Pa=(1260kg/m^3)\times (9.8m/s^2)\times h[/tex]

[tex]h=8.206\frac{Pa.m^2s^2}{kg}=8.206m=323.07\text{ inches}[/tex]

Conversion used :

[tex]1Pa=\frac{kg}{ms^2}\\\\1m=39.37inches[/tex]

The 1.0 atm to inches of glycerin  is equal to 323.07 inches.

(e) The conversion used from [tex]psi[/tex] to [tex]lbf/ft^2[/tex] is:

[tex]1\text{ mmHg}=0.359lbf/ft^2[/tex]

As we are given that 1050 mmHg. Now we have to convert into [tex]lbf/ft^2[/tex].

As, [tex]1\text{ mmHg}=0.359lbf/ft^2[/tex]

So, [tex]1050\text{ mmHg}=\frac{1050\text{ mmHg}}{1\text{ mmHg}}\times 0.359lbf/ft^2=376.95lbf/ft^2[/tex]

The 1050 mm Hg is equal to [tex]376.95lbf/ft^2[/tex]

To convert 60 feet of water to Pa, multiply by the conversion factor 2.98907 kPa. To convert 220 psi to lbf/ft^2, multiply by the conversion factor 144 lbf/ft^2. To convert 120 torr to Pa, multiply by the conversion factor 133.322 Pa. To convert 1.0 atm to inches of glycerin, we need additional information. To convert 1050 mm Hg to lbf/ft^2, multiply by the conversion factor 0.01934 lbf/ft^2.

Answer:

W = - 500 KJ

∴ the work is done on the system

Explanation:

isothermal system:

∴ ΔU = 0; ⇒ Q = W

∴ W = P1V1 -P2V2

⇒ W = ((100KPa)*(25m³)) - ((300KPa)*(10m³))

⇒ W = 2500KPa.m³ - 3000KPa,m³

⇒ W = - 500 KPa.m³ = - 500 KJ

∴ W (-) the work is done on the system

Answer:

False

Explanation:

According to principle of corresponding state, at reduced states or corresponding state, behavior of all gases are similar.

This principle is proposed by van der Waals.

In other words, all gases at same reduced temperature, reduced pressure and reduced volume deviate from ideal gas behavior to the same degree or have same compressibility factor.

Reduced quantities are defined as:

Reduced pressure [tex]P_R = \frac{P}{P_c}[/tex]

Reduced temperature [tex]T_R = \frac{T}{T_c}[/tex]

Reduced volume [tex]V_R = \frac{V}{V_c}[/tex]

Where,

P_c = Critical pressure

V_c = Critical volume

T_c = Critical temperature

The compressibility factor (Z_c) at critical temperature is given by,

                                      [tex]Z_c=\frac{P_c V_c}{n_c k_B T_c}[/tex]

Crtitical parameters (critical temperaures, critical pressure and critical volume) can be expressed in terms of van der Waals parameters a and b.

Principle of corresponding state can also be stated as gases at the same reduced pressure and reduced temperature have same reduced volume.

Hence, compressibility factor at reduced state will be same for all gase. so the given statement is false.