The process where bacteria takes nitrogen out of the air and makes it usable for other organisms is called

The work done on a piston going from 0.130 liters to 0.600 liters under an external pressure of 8.00 atm is 380.882 Joules. This is calculated using the formula for work done under constant pressure, which is W = PΔV.

The subject of this question is regarding the computation of work done on a piston due to change in volume under constant external pressure, and this is a concept in Physics. To compute the work done when a gas expands or compresses, we can use the formula W = PΔV, where W is work done, P is the external pressure, and ΔV is the change in volume.

Here, the external pressure (P) is 8.00 ATM. But to obtain the work done in Joules, we first need to convert this pressure from ATM to Pa (Pascals): 1 ATM = 101325 Pa, thus 8.00 ATM = 8.00 * 101325 = 810600 Pa.

The change in volume (ΔV) is the final volume minus the initial volume, which is 0.600 liters - 0.130 liters = 0.470 liters. But again to match units, we should convert this volume from liters to cubic meters: 1 liter = 0.001 cubic meters, so 0.470 liters = 0.470 * 0.001 cubic meters = 0.00047 m^3.

Therefore, substituting the values into the formula, we get: Work W = PΔV = 810600 Pa * 0.00047 m^3 = 380.882 Joules.

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In order to calculate how much work has been done, it is crucial to convert the provided values to SI units before using the formula for work done by a gas at constant pressure. In this particular case, the amount of work done is approximately 380.9 joules.

The work done by a gas when it expands or contracts at constant pressure can be calculated using the formula W = PΔV, where W is the work done, P is the pressure, and ΔV is the change in volume. However, the pressure is given in atm and the volume in liters, we need to convert these to SI units. The conversion factors are 1 atm = 101.3 kPa = 101,300 Pa and 1 liter = 1 x 10-3 m3. Using these conversion factors, the pressure is 8.00 atm x 101,300 Pa/atm = 810,400 Pa and the change in volume is 0.600 liters - 0.130 liters = 0.470 liters = 0.470 x 10-3 m3. Substituting these values into the formula gives W = (810,400 Pa)(0.470 x 10-3 m3) = 380.9 J.

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Answer: The increasing wavelength of colors:

Red > Green > Blue

Explanation:

Wavelength: This is the property of wave which includes the distance between two consecutive crests or trough. This is denoted by the Greek letter Lambda and it is found by dividing the velocity of the wave with its frequency.

Wavelength of colours are

Violet: 400 - 420 nm

Indigo: 420 - 440 nm

Blue: 440 - 490 nm

Green: 490 - 570 nm

Yellow: 570 - 585 nm

Orange: 585 - 620 nm

Red: 620 - 780 nm

In visible light, the wavelength increases from violet to red. For the colors mentioned, in order of increasing wavelength, it is blue < green < red and blue < yellow < red for the additional colors provided.

The question asks to arrange the colors of visible light (green, red, and blue) in order of increasing wavelength and then provides a similar task with the colors yellow, blue, and red. For visible light, wavelengths increase from violet through to red. Hence, using the mnemonic ROY G BIV (Red, Orange, Yellow, Green, Blue, Indigo, Violet), we can deduce that blue has a shorter wavelength than green, which in turn has a shorter wavelength than red.

Part A: For the colors yellow, blue, and red, in order of increasing wavelength, it would be: blue < yellow < red.

Part B: Since the frequency of light waves is inversely proportional to their wavelength, the order according to frequency, from lowest to highest, would inversely mirror the wavelengths: red < yellow < blue.

The enthalpy of formation, ΔHf, for CsCl, can be determined using the Born-Haber cycle by summing up the energy changes in various steps including the sublimation of Cs, ionization of Cs, dissociation of Cl2, and the formation of CsCl. The lattice energy of CsCl is an exothermic process and is equal to the negative of the enthalpy of formation.

The enthalpy of formation, ΔHf, for CsCl(s) can be determined using the Born-Haber cycle. The cycle involves several steps including the sublimation of Cs(s), ionization of Cs(g), dissociation of Cl2(g), and the formation of CsCl(s). The lattice energy of CsCl is an exothermic process and is equal to the negative of the enthalpy of formation. By summing up the energy changes in all the steps, we can calculate the enthalpy of formation for CsCl.

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Final answer:

To find the enthalpy of formation for CsCl, the enthalpy changes from sublimation, ionization, bond dissociation, electron affinity, and lattice energy are combined resulting in an enthalpy of formation of -390.5 kJ/mol.

Explanation:

To determine the enthalpy of formation (ΔHf) of cesium chloride (CsCl), we must use the Born-Haber cycle which involves several energy changes related to the formation of ionic compounds. These are the steps to calculate ΔHf for CsCl(s):

The enthalpy of sublimation of Cs (ΔHsub)

The first ionization energy of Cs (IE1)

The bond energy of Cl2 (BE)

The electron affinity of Cl (ΔHEA)

The lattice energy of CsCl (ΔHlattice)

To calculate the enthalpy of formation, first, we need to break the Cl2 bond energy into two Cl atoms which takes 1/2 of the bond energy (1/2 x BE) for one mole of Cl atoms. The total enthalpy change of formation is:

ΔHf = ΔHsub + IE1 + (1/2 x BE) - ΔHEA + ΔHlattice

Substituting the given values:

ΔHf = 76.5 + 376 + (1/2 x 243) - (-349) - 657
=76.5 + 376 +121.5 + 349 - 657

= -657 + 266.5 kJ/mol

= -390.5 kJ/mol

Answer:

2.12 grams of primary-standard-grade sodium carbonate are necessary to prepare 800.0 mL of this solution.

Explanation:

Molarity of sodium ions = [tex][Na^+][/tex] = 0.0500 M

Moles of sodium ions = n

Volume of the solution = V = 800.0 mL = 0.800 L

[tex]Molarity=\frac{n}{V(L)}[/tex]

[tex][Na^+]=\frac{n}{V}[/tex]

[tex]n=[Na^+]\times V=0.0500 M\times 0.800 L=0.04 mol[/tex]

[tex]Na_2SO_4(aq)\rightarrow 2Na^+(aq)+CO_3^{2-}(aq)[/tex]

1 mole sodium carbonates gives 2 moles of sodium ion and 1 mole of carbonate ions.

Then 0.04 moles of sodium ions will be obtained from:

[tex]\frac{1}{2}\times 0.04 mol=0.02 mol[/tex] of sodium m carbonation.

Mass of 0.02 moles of sodium carbonate = 0.02 mol × 106 g/mol= 2.12 g

2.12 grams of primary-standard-grade sodium carbonate are necessary to prepare 800.0 mL of this solution.

Answer:

The number of moles of hydrochloric acid are 0.861

Explanation:

In first solution, the [HCl] is 11.7 M, which it means that 11.7 moles are present in 1 liter.

So we took 25 mL and we have to know how many moles, do we have now.

1000 mL ____ 11.7 moles

25 mL _____ (25 . 11.7)/1000 = 0.2925 moles

This are the moles, we add to the solution where the [HCl] is 3.25 M

In 1000 mL __ we have __ 3.25moles

175 mL ____ we have __ (175 . 3.25)/1000 = 0.56875 moles

Total moles: 0.2925 + 0.56875 = 0.861 moles

The number of moles of hydrochloric acid from 175.00 mL of 3.25 M hydrochloric acid solution is 0.56875 mol.

To determine the number of moles of hydrochloric acid (HCl) in the 175.00 mL of 3.25 M hydrochloric acid solution, we can use the formula for molarity (M), which is:

[tex]\[ M = \frac{\text{moles of solute (mol)}}{\text{volume of solution (L)}} \][/tex]

 Rearranging the formula to solve for the moles of solute, we get:

[tex]\[ \text{moles of solute (mol)} = M \times \text{volume of solution (L)} \][/tex]

 Given that the molarity (M) of the hydrochloric acid solution is 3.25 M and the volume is 175.00 mL, we first need to convert the volume from milliliters to liters because molarity is defined in terms of moles per liter.

[tex]\[ \text{Volume in liters (L)} = \frac{175.00 \text{ mL}}{1000 \text{ mL/L}} = 0.175 \text{ L} \][/tex]

Now, we can calculate the moles of HCl:

[tex]\[ \text{moles of HCl} = 3.25 \text{ M} \times 0.175 \text{ L} \][/tex]

[tex]\[ \text{moles of HCl} = 0.56875 \text{ mol} \][/tex]

 Therefore, the number of moles of hydrochloric acid from 175.00 mL of 3.25 M hydrochloric acid solution is 0.56875 mol.

Answer:

The value of gold in the oceans is 4.07x10¹⁴ $

Explanation:

A problem with relation between the units.

This are the data, 3800 m (depth of the ocean)

The total surface 3.63×10⁸ km² (total surface)

5.8×10⁻⁹g/L (gold concentration)

$1595 (value of each troy oz)

31.1 g (mass of each troy oz in grams)

19.3 g/cm³ (density of gold)

As we have a depth (a kind of height) and the total surface we can know the volume that the ocean occupies. This height is in m, the surface in km².

We should convert eveything in dm to work with concentration.

3800 m to cm = 3800 . 10 = 3800 → 3.8 x10⁴ dm

3.63×10⁸ km² to dm² = 3.63×10⁸ . 1x10⁸ = 3.63×10¹⁶ dm²

(1 km² = 1x10⁸dm)

3.8 x10⁴ dm  . 3.63×10¹⁶ dm² = 1.37 x10²¹ dm³

This is the volume that the ocean occupies. By using concentration, we can know the mass of gold in all the ocean.

1L = 1 dm³

1L _____ 5.8×10⁻⁹g

1.37 x10²¹ L ____ 7.9 x10¹² g

So 1 troy oz pays $1595 and 1 troy oz is 31.1 grams, so 31.1 grams pay $1595.

The final rule of three will be

31.1 g __ pay ___ $1595

7.9 x10¹² g ___ pay (8 x10¹¹ g . $1595) / 31.1 g = 4.07x10¹⁴ $

This detailed answer explains how to calculate the value of gold in Earth's oceans based on surface area, depth, and gold concentration. The value of gold in the oceans is found to be around $8.1 trillion.

Earth's oceans have a total surface area of 3.63×[tex]10^8 km^2[/tex] and an average depth of 3800 m. The oceans have an average concentration of dissolved gold of 5.8×[tex]10^{-9}[/tex] g/L. Given the price of gold as $1595 per troy oz, we can calculate the value of gold in the oceans.

To calculate the value of gold in the oceans, we first determine the total mass of gold present in oceans, which is approximately 1.4 × [tex]10^{14}[/tex] g. By using the density of gold as 19.3 [tex]g/cm^{3}[/tex] and the conversion factor of 1 troy oz = 31.1 g, we can find the total value of gold in the oceans.

The value of gold in the oceans is found to be around $8.1 trillion based on the given data points and calculations.

Answer:

55.7%

Explanation:

The reaction that takes place is:

With the volume and concentration of NO₃⁻ solution, we can calculate the moles of Sn²⁺ that reacted:

Now we convert moles of Sn to mass, using its atomic weight:

Finally we calculate the percent by mass of Sn:

Answer : The volume of [tex]H_2[/tex] gas formed at STP is 7.86 liters.

Explanation :

The balanced chemical reaction will be:

[tex]2NaOH(s)+2Al(s)+6H_2O(l)\rightarrow 2AnAl(OH)_4(s)+3H_2(g)[/tex]

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

[tex]\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}[/tex]

Molar mass of Al = 27 g/mole

[tex]\text{Moles of }Al=\frac{6.32g}{27g/mole}=0.234mole[/tex]

Now we have to calculate the moles of [tex]H_2[/tex] gas.

From the reaction we conclude that,

As, 2 mole of [tex]Al[/tex] react to give 3 mole of [tex]H_2[/tex]

So, 0.234 moles of [tex]Al[/tex] react to give [tex]\frac{0.234}{2}\times 3=0.351[/tex] moles of [tex]H_2[/tex]

Now we have to calculate the volume of [tex]H_2[/tex] gas formed at STP.

As, 1 mole of [tex]H_2[/tex] gas contains 22.4 L volume of [tex]H_2[/tex]  gas

So, 0.351 mole of [tex]H_2[/tex] gas contains [tex]0.351\times 22.4=7.86L[/tex] volume of [tex]H_2[/tex] gas

Therefore, the volume of [tex]H_2[/tex] gas formed at STP is 7.86 liters.

The vapor pressure of the solution can be calculated using Raoult's law. First, we need to calculate the mole fraction of glycerol and water in the solution. Then, we can use the mole fraction to calculate the vapor pressure of the solution.

In order to find the vapor pressure of the solution, we can use Raoult's law. Raoult's law states that the vapor pressure of a solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent.

First, we need to calculate the mole fraction of glycerol in the solution. To do this, we need to convert the mass percent of glycerol to moles. Since the molar mass of glycerol is 92.1 g/mol and the solution contains 34.4% glycerol by mass, we can calculate the moles of glycerol:

Moles of glycerol = (34.4 g / 92.1 g/mol)

Next, we need to calculate the mole fraction of water in the solution. Since the solution is 34.4% glycerol, the mass percent of water is 100% - 34.4% = 65.6%. Using the molar mass of water (18.0 g/mol), we can calculate the moles of water:

Moles of water = (65.6 g / 18.0 g/mol)

Now we can calculate the mole fraction of glycerol and water:

Mole fraction of glycerol = (moles of glycerol) / ((moles of glycerol) + (moles of water))

Mole fraction of water = (moles of water) / ((moles of glycerol) + (moles of water))

Finally, we can calculate the vapor pressure of the solution using Raoult's law:

Vapor pressure of solution = (mole fraction of water) * (vapor pressure of water)

Answer:

B

Explanation:

Firstly, we will need to calculate the number of moles. To do this, we make use of the ideal gas equation

PV = nRT

n = PV/RT

The parameters have the following values according to the question:

P = 780mmHg, we convert this to pascal.

760mHG = 101325pa

780mmHg = xpa

x = (780 * 101325)/760 = 103,991 Pa

V= 400ml = 0.4L

T = 135C = 135 + 273.15 = 408.15K

n = ?

R = 8314.463LPa/K.mol

Substituting these values into the equation yields the following:

n = (103991 * 0.4)/(8314.463 * 408.15)

= 0.012 moles

Now we know 1 mole contains 6.02 * 10^23 molecules, hence, 0.012moles will contain = 0.012 * 6.02 * 10^23 = 7.38 * 10^21 molecules

Answer: The number of nitrogen molecules in the container are [tex]7.38\times 10^{21}[/tex]

Explanation:

To calculate the moles of gas, we use the equation given by ideal gas which follows:

[tex]PV=nRT[/tex]

where,

P = pressure of the gas = 780 mmHg  

V = Volume of the gas = 400.0 mL = 0.4 L     (Conversion factor:  1 L = 1000 mL)

T = Temperature of the gas = [tex]135^oC=[135+273]K=408K[/tex]

R = Gas constant = [tex]62.364\text{ L.mmHg }mol^{-1}K^{-1}[/tex]

n = number of moles of nitrogen gas = ?

Putting values in above equation, we get:

[tex]780mmHg\times 0.4L=n\times 62.364\text{ L. mmHg }mol^{-1}K^{-1}\times 408K\\\\n=\frac{780\times 0.4}{62.364\times 408}=0.01226mol[/tex]

According to mole concept:

1 mole of a compound contains [tex]6.022\times 10^{23}[/tex] number of molecules

So, 0.01226 moles of nitrogen gas will contain = [tex](0.012\times 6.022\times 10^{23})=7.38\times 10^{21}[/tex] number of molecules

Hence, the number of nitrogen molecules in the container are [tex]7.38\times 10^{21}[/tex]

Answer:

2Fe + 3Br2 = 2FeBr3

Explanation:

The density of a metal that crystallizes in a face-centered cubic unit cell can be calculated by determining the number of atoms per unit cell, calculating the volume of a unit cell, and finally calculating the mass of the atoms in the unit cell using the given molar mass. The density is then found by dividing the mass by the volume.

To answer this question, we need to know a couple of key pieces of information. First, a face-centered cubic unit cell consists of four atoms: one-eighth of an atom at each of the eight corners and half of an atom on each of the six faces (1/8*8+1/2*6=4 atoms).

Secondly, we need to find the volume of the atom that is engulfed in the cubic unit cell. Given that the radius of the metal is 234 picometers (or 234*10^-12 meters), the volume of the unit cell can be found by applying the formula for the volume of a cube (side^3) where the side equals 2*sqrt(2)*r.

We can then calculate the number of moles of atoms in the unit cell, convert them to grams using the molar mass, and finally calculate the density by dividing the calculated mass by the calculated volume.

We first convert the volume of the unit cell from cubic meters to cubic centimeters (1m^3 = 10^6 cm^3).

Then, we calculate the mass of the atoms in the unit cell using the molar mass (195.08 g/mol).

Finally, we calculate the density. Density = Mass/Volume.

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To calculate the density of a metal in a face-centered cubic structure, you need to determine the volume of the unit cell and the molar mass of the metal.

The density of a metal can be calculated using the formula:

Density = (Molar Mass ×Avogadro's Number) / (Volume of Unit Cell)

First, let's calculate the volume of the face-centered cubic (FCC) unit cell. The FCC unit cell consists of 4 atoms at the corners and 8 atoms at the face centers. The radius of the atom can be used to calculate the edge length of the unit cell using the formula:

Edge Length = 4×Radius / √2

Once we have the edge length, we can calculate the volume of the unit cell using the formula:

Volume of Unit Cell = Edge Length³

Finally, we can plug the values into the density formula and calculate the density of the metal.

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Answer:

(a)The molar mass of the gene fragment is 18220.071g/mol = 18.22 kg/mol

(b)The freezing point for the aqueous solution is [tex]-3.413\times10^{-5}[/tex] C

Explanation:

The osmotic pressure (π) is given by the following equation:

[tex]\pi =cRT[/tex]

[tex]c[/tex]= Concentration of solution

R = universal gas constant = 62.364 [tex]\frac{L\times torr}{mol\times K}[/tex]

T = temperature

Weight of solute = w = 10.0 mg

Let the molecular weight of the solute be m g/mol.

Concentration = [tex]c=\frac{n}{V}\\ n=\frac{w}{m}\\ n=\frac{10\times10^{-3}}{m}\\c=\frac{10\times10^{-3}}{m\times30\times10^{-3}}M[/tex]

[tex]m=\frac{RT}{3\pi}[/tex]

m = 18220.071g/mol

Therefore, the molar mass of the gene fragment is 18220.071g/mol = 18.22 kg/mol

[tex]\Delta T_{f}=K_{f}m[/tex]

m is the molality of the solution.

m = [tex]1.835\times10^{-5}[/tex] mol/kg

[tex]\Delta T_{f}=1.86\times m[/tex]

[tex]\Delta T_{f}[/tex] = [tex]3.413\times10^{-5}[/tex] C

The freezing point for the aqueous solution is [tex]-3.413\times10^{-5}[/tex] C

Answer:

pH = 7.45

Explanation:

This is a buffer solution and we can solve it by using the Henderson-Hasselbalch equation:

pH = pKa + log ((A⁻)/(HA))

Here we will first have to calculate the  A⁻ formed  in the 1. 0 L solution which is formed by the reaction of  HClO with the strong base NaOH and add  it to the original mol of NaClO

mol NaClO = mole NaCLO originally present in the 1L of M solution + 0.030 mol produced in the reaction of HCLO with NaOH

0.350 mol + .030 mol = 0.380 mol

New concentrations :

HClO = 0. 350 mol-0.030 mol  = 0.320 M (have to sustract the 0.030 mol reacted with NaOH)

NaClO = 0.380 mol/ 1 L = 0.380 M

Now we have all the values required and we can plug them into the equation

pH = -log (2.9 x 10^-8) + log (0.380/.320) = 7.45

The pH of 1.00 L of the solution is 7.45.

This is defined as the power of hydrogen and it measures how acidic or basic a substance is.

Using Henderson-Hasselbalch equation:

pH = pKa + log ((A⁻)/(HA))

mol of NaClO = mole NaCLO initially present in the 1L of M solution + 0.030 mol produced in the reaction of HClO with NaOH

0.350 mol + .030 mol = 0.380 mol

We can then calculate the new concentrations below:

HClO = 0. 350 mol-0.030 mol  = 0.320 M

NaClO = 0.380 mol/ 1 L = 0.380 M

Substitute the values into the equation

pH = -log (2.9 x 10⁻⁸) + log (0.380/.320)

     = 7.45

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Answer:

c. anode, cathode.

Explanation:

In a voltaic cell, electrons flow from the anode to the cathode.

In the anode takes place the oxidation, in which the reducing agent loses electrons. Those electrons flow to the cathode where reduction takes place, that is, the oxidizing agent gains electrons. The salt bridge has the function of maintaining the electroneutrality.

Answer:

Electrons will move across the salt bridge from the anode to the cathode.

Explanation:

Educere/ Founder's Education Answer

The thermosphere is a layer of Earth's atmosphere located 80-700km above sea level. It supports long-distance communication by reflecting radio waves back to Earth and is also where auroras form due to the collision of charged particles and gas molecules.

The thermosphere is the fourth layer of Earth's atmosphere, located above the mesosphere and extend from about 80 to 700 kilometers above sea level. This layer holds a unique property due to the presence of electrically charged particles, or ions which allows it to support long-distance communication by reflecting radio waves back to Earth, bypassing the need for satellites. This is also the layer in which auroras form. Auroras occur when these charged particles from the sun collide with gas molecules in the Earth's atmosphere, releasing energy that manifests as visible, colorful light.

The Earth's atmosphere is divided into five main layers: The troposphere, the stratosphere, the mesosphere, the thermosphere and the exosphere. The Troposphere being the one closest to the Earth's surface and is where weather is generally observed. This layer expands to a height of roughly 12 km above the sea level and makes up around 80% of the atmosphere's mass. The Thermosphere is not the layer of atmosphere closest to the Earth’s surface where weather typically occurs, this property represents the troposphere.

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The third option is correct.

3. The thermosphere is the layer where auroras form when electrically charged particles from the sun collide with gas molecules releasing energy visible as light of different colors.

Answer:

a. If the reaction mixture initially contains only OF2(g), then at equilibrium, the reaction mixture will consist of essentially only O2(g) and F2(g).

Explanation:

The answer is a) because the value for Kp is really close to zero (having x10⁻¹⁵), this means that at equilibrium O₂ and F₂ will be significantly more present than OF₂.

Final answer:

For the equilibrium O2(g) + 2F2(g) ↔ 2OF2(g) with Kp = 2.3×10^-15, the reaction favors the reactants, making the correct answer that a mixture initially containing only OF2(g) will consist of essentially only O2(g) and F2(g) at equilibrium.

Explanation:

The question considers the equilibrium O2(g) + 2F2(g) ↔ 2OF2(g); Kp = 2.3×10-15 and asks which statement is true. Given the extremely low value of Kp, this indicates a strong preference for reactants at equilibrium. Therefore, the correct answer is: If the reaction mixture initially contains only OF2(g), then at equilibrium, the reaction mixture will consist of essentially only O2(g) and F2(g). This is because a very small Kp value means the equilibrium lies heavily on the side of the reactants, making statement (a) true.

Answer:can change as new researches and experiments are done

Explanation:

Explanation:

It is known that relation between partial pressure, mole fraction and pressure is as follows.

      Partial pressure of gas = mole fraction of gas × Pressure of gas

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

   Partial pressure of gas = mole fraction of gas × Pressure of gas

                                          = [tex]0.21 \times 1.13 atm[/tex]

                                           = 0.237 atm

According to Henry's law,

       Concentration of oxygen = Henry's law constant × partial pressure of oxygen

             = [tex]1.3 \times 10^{-3} M/atm \times 0.2373 atm[/tex]

             = [tex]3.08 \times 10^{-4}[/tex] M

Therefore, calculate moles of oxygen in 5.00 L present as follows.

   Moles of oxygen in 5.00 L = volume × concentration

                                                 = [tex]5.00 \times 3.0849 \times 10^{-4}[/tex]

                                                 = [tex]1.542 \times 10^{-3}[/tex] mol

Now, we will calculate the mass of oxygen as follows.

        Mass of oxygen = moles × molar mass of oxygen

                                    = [tex]1.542 \times 10^{-3} mol \times 32 g/mol[/tex] mol    

                                    = 0.0494 g

or,                                 = 49.4 mg           (As 1 g = 1000 mg)

thus, we can conclude that the mass of given oxygen (in mg) is 49.4 mg.

The mass of oxygen dissolved in water has been 49.4 mg.

The partial pressure of oxygen in the air:

Partial pressure = Mole fraction [tex]\times[/tex] Pressure of gas

Partial pressure = 0.21 [tex]\times[/tex] 1.13 atm

Partial pressure of oxygen = 0.237 atm.

The concentration of oxygen can be given by Henry's law.

Concentration of Oxygen = Henry's constant ([tex]\rm k_H[/tex]) [tex]\times[/tex] Partial pressure

Concentration = 1.3 [tex]\rm \times\;10^-^3[/tex] M/atm [tex]\times[/tex] 0.237 atm

Concentration of oxygen = 3.08 [tex]\rm \times\;10^-^4[/tex] M

Moles can be given by:

Moles = Molarity [tex]\times[/tex] Volume

Moles of oxygen =  3.08 [tex]\rm \times\;10^-^4[/tex] M [tex]\times[/tex] 5

Moles of oxygen = 1.542  [tex]\rm \times\;10^-^3[/tex] mol

Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]

Weight of oxygen = Moles [tex]\times[/tex] Molecular weight

Weight of oxygen =  1.542  [tex]\rm \times\;10^-^3[/tex] mol [tex]\times[/tex] 32 g/mol

Weight of oxygen = 0.0494 grams

Weight of oxygen = 49.4 mg.

The mass of oxygen dissolved in water has been 49.4 mg.

For more information about the mass of gas, refer to the link:

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Answer:

Covalent network solids typically have high melting points and high boiling points

Explanation:

Covalent solids are the solids which have covalent bonds as intermolecular or interatomic interactions.

As covalent bonds are very strong, these solids are generally characterized by high melting points and high boiling points.

The examples of such solids are: Diamond and Graphite.

The other classes of solids are

i) ionic

ii) molecular

iii) metallic.

Answer:

The grams of MgCl₂ that are needed are 63.1 g

Explanation:

First of all, try to think the equation and ballance it. This is it:

3MgCl₂  +  2Na₃PO₄  →  Mg₃(PO₄)₂  +  6NaCl

Let's convert our f.u in mol

1 mol =  6.02x10²³ formula units (Avogadro's number)

So f.u / Avogadro = mol

1.33x10²³ / 6.02x10²³ = 0.221 moles.

So 1 mol of phosphate sodium comes from 3 mol of magnesium chloride.

How many mol of magnesium chloride are necessary, for 0.221 mol of phosphate.?

0.221 moles .3 = 0.663 moles.

Molar mass of MgCl₂ is 95.2 g/m

0.663 moles . 95.2 g/m = 63.1 g

The balanced chemical equation between magnesium chloride and sodium phosphate is 3 MgCl2 + 2 Na3PO4 -> Mg3(PO4)2 + 6 NaCl. To determine the grams of magnesium chloride needed to produce a specific number of formula units of magnesium phosphate, stoichiometry calculations can be used.

The balanced chemical equation between magnesium chloride (MgCl2) and sodium phosphate (Na3PO4) is:

3 MgCl2 + 2 Na3PO4 → Mg3(PO4)2 + 6 NaCl

To determine the grams of magnesium chloride needed to produce 1.33 x 10^23 formula units of magnesium phosphate (Mg3(PO4)2), we need to use stoichiometry.

To determine the spontaneity of the reaction between N2 and O2 to form 2NO and the equilibrium constant K, one would use the Gibbs Free Energy (ΔG) formula and the equilibrium constant (K) formula. While exact values can't be computed without data for the standard enthalpy and entropy changes, it can be inferred that this reaction is typically not spontaneous at room temperature (298 K), but becomes more spontaneous as temperature increases toward 2000 K.

The given chemical reaction is between nitrogen and oxygen to form nitrogen monoxide: N2(g) + O2(g) = 2 NO(g). To answer your question, we'll need to use the Gibbs Free Energy (ΔG) formula and the equilibrium constant (K) formula which are linked by the equation ΔG = -RTlnK. These formulas are what we use to determine the spontaneity of a reaction and the equilibrium state at a given temperature.

Without the specific values of the standard enthalpy (ΔH) and entropy (ΔS) changes for this reaction, it's impossible to calculate ΔG and K exactly. However, at room temperature (298 K), the reaction is typically not spontaneous because the formation of NO requires high-energy conditions - usually above 2000 K. This is inferred by the presence of NO only in extreme conditions such as lightning strikes or in high-temperature combustion processes.

At 2000 K, although I can't give an exact numerical estimate without the ΔH and ΔS values, we can say that the reaction becomes more spontaneous as temperature increases since high energy and high temperature favor the formation of NO. It's always important to remember that whether a chemical reaction is spontaneous or not depends not only on the change in enthalpy (ΔH), but also on the change in entropy (ΔS) and the absolute temperature (T).

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The reaction N2(g) + O2(g) = 2NO(g) is not spontaneous at 298 K, with a very small equilibrium constant (K). However, at 2000 K, the reaction becomes more spontaneous as the free energy change (ΔG) is negative.

The reaction N2(g) + O2(g) = 2NO(g) has a standard free energy change (ΔG°) of 173.4 kJ/mol at 298 K. To find the equilibrium constant (K) at this temperature, we use the equation ΔG° = -RT ln(K), where R is the gas constant and T is the temperature in Kelvin. Rearranging the equation, we get K = e-ΔG°/(RT).

Substituting the values into the equation and solving, we find K to be 6.95 x 10-31, which is a very small value. Since K is less than 1, the reaction is not spontaneous at 298 K.

To estimate ΔG° at 2000 K, we can use the equation ΔG = ΔH - TΔS, where ΔH is the enthalpy change and ΔS is the entropy change. Assuming ΔH and ΔS do not change significantly with temperature, we can use the same values as at 298 K.

Substituting the values into the equation, we find ΔG to be -545.4 kJ/mol at 2000 K. Since ΔG is negative, the reaction becomes more spontaneous as temperature increases.

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Answer:

coefficient 3 is needed t with water molecule to balance the equation.

Explanation:

Chemical equation:

Fe₂O₃ + H₂SO₄ →  Fe₂(SO₄)₃ + H₂O

Balanced chemical equation:

Fe₂O₃ + 3H₂SO₄ →  Fe₂(SO₄)₃ + 3H₂O

The equation is balanced. There are two iron three sulfate six hydrogen and three oxygen atoms are present on both side of equation.

The given equation completely follow the law of conservation of mass.

According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.

Coefficient with reactant and product:

Fe₂O₃             1

H₂SO₄             3

Fe₂(SO₄)₃        1

H₂O                 3

So coefficient 3 is needed t with water molecule to balance the equation.

The jar contains A Mixture

Explanation:

A mixture consists of different atoms that are not chemically bonded. It is of two types of the heterogeneous mixture and homogeneous mixture. The chemical substances can be differentiated visually in the heterogeneous mixture.

In a mixture, the proportion of atoms can be slightly changed without changing  or modifying any substance. This is because the individual substances in mixture keep their properties when mixed together. Further, mixtures have variable compositions and substances. these mixtures can be are separated using physical methods.  

The jar contains A Mixture

Explanation:

A mixture consists of different atoms that are not chemically bonded. It is of two types of the heterogeneous mixture and homogeneous mixture. The chemical substances can be differentiated visually in the heterogeneous mixture.

In a mixture, the proportion of atoms can be slightly changed without changing  or modifying any substance. This is because the individual substances in mixture keep their properties when mixed together. Further, mixtures have variable compositions and substances. these mixtures can be are separated using physical methods.

Answer:

The change in energy for the combustion of a mole of naphthalene is 79 kJ

Explanation:

An excersise where you have to apply the Heat Capacity formula

C =  Q . ΔT

where C is Heat Capacity and Q is Heat

ΔT means (T° final - T° initial)

5.11 kJ/°C = Q (32.33°C - 24.25°C)

5.11 kJ/°C = Q (32.33°C - 24.25°C)

5.11 kJ/°C = Q . 8.08°C

5.11 kJ / 8.08°C = Q

0.632 kJ = Q

This heat is released by 1.025 grams of naphtalene.

Molar mass Naphtalene : 128.17 g/m

1.025 g / 128.17 g/m = 0.008 moles

This are the moles, so we have to divide heat/moles to get the change in energy for one mole.

0.632 kJ/0.008 mol = 79 kJ/m

Answer:

A) Work (positive)

B) Heat (negative) and work (positive)

C) Work (negative) and heat (negative)

Explanation:

A) This energy change is work given that the first ball is using this energy to move the other one. The system (fist ball) is doing the work so it has + sign.

B) The book loses part of its energy by friction (heat) and part by applying a force to the floor (work). The heat is taken from the system so it has - sign and the work is done by the system and it has + sign.  

C) The daughter is receiving the work done by the father. This work done to the system has - sign. Also, you can say that the system (daughter and swing) loses energy by friction (heat) and because of that it slows down.

Note: the sign of work and heat is defined by convention.

Answer:

[tex]K_{sp}[/tex] of [tex]Zn(CH_{3}COO)_{2}[/tex] is 0.0513

Explanation:

Solubility equilibrium of [tex]Zn(CH_{3}COO)_{2}[/tex]:

[tex]Zn(CH_{3}COO)_{2}\rightleftharpoons Zn^{2+}+2CH_{3}COO^{-}[/tex]

Solubility product of [tex]Zn(CH_{3}COO)_{2}[/tex] ([tex]K_{sp}[/tex]) is written as-            [tex]K_{sp}=[Zn^{2+}][CH_{3}COO^{-}]^{2}[/tex]

Where [tex][Zn^{2+}][/tex] and [tex][CH_{3}COO^{-}][/tex] represents equilibrium concentration (in molarity) of [tex]Zn^{2+}[/tex] and [tex]CH_{3}COO^{-}[/tex] respectively.

Molar mass of [tex]Zn(CH_{3}COO)_{2}[/tex] = 183.48 g/mol

So, solubility of [tex]Zn(CH_{3}COO)_{2}[/tex] = [tex]\frac{43.0}{183.48}M[/tex] = 0.234M

1 mol of [tex]Zn(CH_{3}COO)_{2}[/tex] gives 1 mol of [tex]Zn^{2+}[/tex] and 2 moles of [tex]CH_{3}COO^{-}[/tex] upon dissociation.

so,   [tex][Zn^{2+}][/tex] = 0.234 M and [tex][CH_{3}COO^{-}][/tex] = [tex](2\times 0.234)M=0.468M[/tex]

so, [tex]K_{sp}=(0.234)\times (0.468)^{2}=0.0513[/tex]          

Based on the data provided, the Ksp of zinc acetate is 0.051 M^{2}.

The solubility product, Ksp, of zinc acetate is derubed from the equation for the dissolution of Zinc acetate given below:

The solubility product, is given below:

Molar concentration of the zinc acetate = mass concentration/molar mass

Molar concentration of Zinc acetate = 43.0/183.48

Molar concentration of Zinc acetate = 0.234 M

From the equation of the reaction:

1 mole of Zn(CH_{3}COO)_{2} produces 1 mole Zn^{2+} and 2 CH_{3}COO^{-}

Hence;

Ksp = 0.234 × 0.468^{2}

Ksp = 0.051 M^{2}

Therefore, the Ksp of zinc acetate is 0.051 M^{2}.

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Answer:

Lean Flammability limit ,or lean limit.

Explanation:

The point is know as Lean Flammability limit ,or lean limit. The lean limit is usually expressed in volume percent. It can be defined as the lower range of concentration of over which a flammable mixture of gas and vapor can be fired at a constant temperature and pressure.

Here in this case also As more nitrogen (or any other inert gas) is added to a flame, the oxidation reaction stops as the concentration has dropped below the  Lean Flammability limit.

Answer:

-1.71 J/K

Explanation:

To solve this problem we use the formula

ΔS = n*ΔH/T

Where n is mol, ΔH is enthalpy and T is temperature.

ΔH and T are already given by the problem, so now we calculate n:

Molar Mass C₂H₅OH = 46 g/mol

2.71 g C₂H₅OH ÷ 46g/mol = 0.0589 mol

Now we calculate ΔS:

ΔS = 0.0589 mol * −4600 J/mol / 158.7 K

ΔS = -1.71 J/K