A city in Laguna generates 0.96 kg per capita per day of Municipal Solid Waste (MSW). Makati City in Metro Manila generates 1.9 kg per capita per day of MSW. If both cities have population of 20000 inhabitants, determine the following
i. amount of MSW generated in each city
ii. number of trucks needed to collect the MSW twice weekly, if each truck has a
capacity of 4.4 tons and operates 5 days per week. Two loads per day at 75%
capacity is the travel quota per day
iii. -the number and volume of MSW in tons that enter the landfill

Answer:

4.076g of HCl is neutralized by a dose of milke of magnesia containing 3.26g of Mg(OH)2

Explanation:

Step 1: The balanced equation

Mg(OH)₂(aq) + 2 HCl(aq) ⇔ 2 H₂O(l) + MgCl₂(aq)

This means that for 1 mole of Magnesium hydroxide consumed, there i s consumed 2 moles of HCl and there is produced 1 mole of MgCl2 ( and 2 moles of H2O)

Step 2: Calculating moles of Mg(OH)2

moles of Mg(OH)2 = mass of Mg(OH)2 / Molar mass of Mg(OH)2

moles Mg(OH)2 = 3.26g / 58.32g/mole = 0.0559 mole Mg(OH)2

Step 3: Calculating moles of HCl

Since there is for 1 mole Mg(OH)2 consumed, there is consumed 2 moles of HCl

There is for each 0.0559 moles of Mg(OH)2, 2*0.0559 = 0.1118 moles of HCl consumed.

Step 4: Calculating mass of HCl

mass of HCl = moles of HCl x Molar mass of HCl

mass of HCl = 0.1118 moles * 36.46 g/mole

mass of HCl = 4.076 g

4.076g of HCl is neutralized by a dose of milke of magnesia containing 3.26g of Mg(OH)2

To find the mass of HCl neutralized by Mg(OH)2, calculate the number of moles in 3.26 g of Mg(OH)2 and multiply by the molar mass of HCl, taking account that each mole of Mg(OH)2 neutralizes two moles of HCl.

The subject of this question is stoichiometry, which is a part of Chemistry. The equation provided showcases the reaction between magnesium hydroxide (Mg(OH)2, the active ingredient in milk of magnesia) and stomach acid (HCl). To find the mass of HCl that is neutralized by 3.26 g of Mg(OH)2, we first need to determine the molar mass of Mg(OH)2, which is 58.3197 g/mol. Then, calculate the number of moles in 3.26 g of Mg(OH)2. Given the balanced formula, 1 mole of Mg(OH)2 neutralizes 2 moles of HCl. Therefore, the moles of HCl that can be neutralized is twice the moles of Mg(OH)2. The molar mass of HCl is 36.461 g/mol. Multiply the moles of HCl by the molar mass to acquire the mass of the HCl that can be neutralized.

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Answer:

[α] = -77.5° / [tex]\frac{\textup{dm-g}}{\textup{mL}}[/tex]

Explanation:

Given;

Mass of optically pure substance in the solution = 10 g

Volume of water = 500 mL

Length of the polarimeter, l = 20 cm = 20 × 0.1 dm = 2 dm

measured rotation = - 3.10°

Now,

The specific rotation ( [α] ) is given as:

[α] = [tex]\frac{\alpha}{c\times l}[/tex]

here,

α is the measured rotation = -3.10°

c is the concentration

or

c = [tex]\frac{\textup{Mass of optically pure substance in the solution}}{\textup{Volume of water}}[/tex]

or

c =  [tex]\frac{10}{500}[/tex]

or

c = 0.02 g/mL

on substituting the values, we get

[α] = [tex]\frac{-3.10^o}{0.02\times2}[/tex]

or

[α] = -77.5° / [tex]\frac{\textup{dm-g}}{\textup{mL}}[/tex]

Explanation:

The electronic configuration of magnesium with Z = 12 is : 2, 8, 12

The electronic configuration of bromine with Z = 35 is : 2, 8, 18, 7

The Lewis structure is drawn in such a way that the octet of each atom is complete.  

Thus, magnesium losses two electrons to bromine and 2 atoms of bromine accepts the electron.

Thus, the valence electrons are shown by dots in Lewis structure. The reaction is shown in image below.

Lewis symbols represent electron transfer between Mg and Br atoms in the formation of an ionic compound.

Lewis symbols, also known as Lewis dot diagrams or electron dot diagrams, depict the valence electrons of atoms using dots around the symbol of the element. In the formation of an ionic compound between Mg and Br atoms, magnesium (Mg) donates two electrons to each bromine (Br) atom, resulting in the transfer of electrons and the creation of Mg2+ and Br- ions.

Explanation:

It is known that the Van der Waals equation is a description of real gases, as in this equation there are all those interactions which we previously ignore in the ideal gas law.

In Vander Waals equation, the repulsion and collision, between molecules of gases are being considered. They are no longer ignored and they also are not considered a "point" particle.

According to the ideal gas law, PV = nRT

or,                [tex]P(\frac{V}{n})[/tex] = RT

and, let [tex]\frac{V}{n}[/tex] = v; which is called molar volume

Hence,                  P × v = RT

As, the van der Waals equation corrects pressure and volume  as follows.

                  [tex](P+ \frac{a}{v^{2}}) \times (v - b)[/tex] = RT

where,   R = idel gas law; recommended to use the units of a and b; typically bar/atm and dm/L

              T = absolute temperature, in K

              v = molar volume, v = [tex]\frac{\text{Volume of gas}}{\text{moles of gas}}[/tex]

              P = pressure of gas

Now, substitute the data in Vander Waal,s equation as follows.

For argon,    [tex](P+ \frac{a}{v^{2}}) \times (v - b)[/tex] = RT

            [tex](P+ \frac{1.355}{(0.45)^{2}}) \times (0.45- 0.0320)[/tex] = 0.08314 [tex]bar dm^{3}/molK \times (200)K[/tex]

                    (P+ 6.691) = [tex]0.08314 \times \frac{200}{(0.45- 0.0320)}[/tex]

                            P = (39.7799 - 6.691) bar

                            P = 33.0889 bar

or,                       P = 33.09 bar (approx)

Thus, we can conclude that the pressure exerted by Ar in the given situation is 33.09 bar.

Explanation:

The given data is as follows.

        Mass of gold = 645 lb,       Volume = 0.5348 ft3

         Density of water = 62.4 lbs/ft3

It is known that specific gravity is defined as density of substance divided by the density of standard fluid.

Mathematically,       Specific gravity = [tex]\frac{\text{density of gold}}{\text{density of water}}[/tex]

               Specific gravity = [tex]\frac{\text{density of gold}}{62.4 (lbs.ft^{-3})}[/tex]

Now, calculate the density of gold then from density we will calculate specific gravity as follows

Since,             Density = [tex]\frac{mass}{volume}[/tex]

                        Density = [tex]\frac{645 lbs}{0.5348 ft^{3}}[/tex]

                                    = 1206.06 [tex]lbs/ft^{3}[/tex]

As,        Specific gravity = [tex]\frac{\text{density of gold}}{\text{density of water (standard fluid)}}[/tex]

                                      = [tex]\frac{1206.06 (lbs/ft^{3})}{62.4 (lbs/ft^{3})}[/tex]

                                      = 19.32

Therefore, the value of specific gravity is 19.32.

Specific gravity has no units as it is density divided by density. Hence, all the units get canceled out.

Explanation:

STP means standard temperature and pressure where values are as follows.

                      T = 273.15 K,         P = 1 atm

According to ideal gas equation, PV = nRT. Since, it is given that mass is 48.3 g and we have to find the volume as follows.

                  n = [tex]\frac{mass}{\text{molar mass}}[/tex]

So,                       PV = nRT

               V = [tex]\frac{mass}{\text{molar mass}} \times \frac{RT}{P}[/tex]                  

                   = [tex]\frac{48.3 g}{28 g/mol} \times \frac{0.0821 L atm/mol K \times 273.15 K}{1 atm}[/tex]             (molar mass of CO = 28 g/mol)

                   = 38.68 L                

Thus, we can conclude that volume of  48.3 g of carbon monoxide at STP is 38.68 L.

Answer:

A liquid with a sharp contact angle (e.g., water on glass) will form a concave meniscus, and the liquid pressure under the meniscus will be smaller than the atmospheric pressure

Explanation:

The phenomenon of capillarity is produced by the action of the surface tension of the fluids and is observed when a small diameter tube is immersed within the fluid. If we pay attention to the result, we can see that, depending on the fluid, two different things can happen, that the liquid rises through the tube and that the level inside the tube is greater than that of the liquid or that the opposite happens.

The case in which the liquid rises above the tube occurs when the liquid "wets". This occurs when the adhesion forces with the walls exceed those of cohesion between the fluid molecules. In this case, the concave side is out of the fluid.

The case where the level of the liquid inside the tube is lower than the level of the liquid occurs when the liquid does not get wet. We remember that the liquid does not get wet when the cohesion forces are greater than those of adhesion. This phenomenon is called capillary depression and the concave angle is for the liquid side and is said to be convex.

Answer: The molarity of sulfuric acid solution is 2.3 M

Explanation:

The relationship between specific gravity and density of a substance is given as:

[tex]\text{Specific gravity}=\frac{\text{Density of a substance}}{\text{Density of water}}[/tex]

Specific gravity of sulfuric acid solution = 1.13

Density of water = 1.00 g/mL

Putting values in above equation we get:

[tex]1.13=\frac{\text{Density of sulfuric acid solution}}{1.00g/mL}\\\\\text{Density of sulfuric acid solution}=(1.13\times 1.00g/mL)=1.13g/mL[/tex]

We are given:

20% (m/m) sulfuric acid solution. This means that 20 g of sulfuric acid is present in 100 g of solution

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

Density of solution = 1.13 g/mL

Mass of Solution = 100 g

Putting values in above equation, we get:

[tex]1.13g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{100g}{1.13g/mL}=88.5mL[/tex]

[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]

We are given:

Mass of solute (sulfuric acid) = 20 g

Molar mass of sulfuric acid = 98 g/mol

Volume of solution = 88.5 mL

Putting values in above equation, we get:

[tex]\text{Molarity of solution}=\frac{20g\times 1000}{98g/mol\times 88.5mL}\\\\\text{Molarity of solution}=2.3M[/tex]

Hence, the molarity of sulfuric acid solution is 2.3 M

Answer:

The pump work is 3451 kJ/s

Explanation:

Pump work (W) is calculated as

[tex] W = (h_f - h_i) \times \dot{m}[/tex]

where

[tex] h_f [/tex] is the enthalpy of water at its final state

[tex] h_i [/tex] is the enthalpy of water at its initial state

[tex]\dot{m} = 10 kg/s [/tex] is water mass flow

For liquids, properties are evaluated as saturated liquid. From the figure attached, it can be seen that

[tex] h_f = 762.81 kJ/kg [/tex]

[tex] h_i = 417.46 kJ/kg [/tex]

Replacing

[tex] W = (762.81 kJ/kg - 417.46 kJ/kg) \times 10 kg/s[/tex]

[tex] W = 3451 kJ/s [/tex]

Answer:

Ka = [tex]2.32 \times 10^{-4}[/tex]

pH = 2.12

Explanation:

Calculation of Ka:

% Dissociation = 3% = 0.03

Concentration of solution = 0.25 M

HA dissociated as:

       [tex]HA \rightarrow H^+ + A^{-}[/tex]

      C(1 - 0.03)    C×0.03   C×0.03

HA] after dissociation = 0.25×0.97 = 0.2425 M

[tex][H^+]= 0.25\times0.03 = 0.0075 M[/tex]

[tex][A^{-}]= 0.25 \times 0.03 = 0.0075 M[/tex]

[tex]Ka= \frac{[H^+][A^{-}]}{[HA]}[/tex]

[tex]Ka = \frac{(0.0075)^2}{0.2475} =2.32 \times 10^{-4}[/tex]

Calculation of pH of the solution

[tex]pH = -log [H^+][/tex]

[tex]H^+[\tex] = 0.0075 M[/tex]

[tex]pH = -log 0.0075 = 2.12[/tex]

The Ka of the weak acid is calculated using the concentrations of the dissociated and non-dissociated parts of the weak acid and the pH is derived from the concentration of H+ ions.

The subject in concern relates to the dissociation of a weak acid and the calculation of the Acid dissociation constant (Ka) and the pH of the solution.

We are given that 3% of the weak acid, HA, is dissociated. Meaning, 3% of 0.25 M HA dissociates into H+ and A-. This will be 0.03 * 0.25 M = 0.0075 M.

The equilibrium for the reaction is HA <--> H+ + A-. Due to water autoionization, the concentration of water is approximately taken as constant in the denominator of Ka calculation. Hence, Ka for HA can be approximated as [H+][A-] / [HA]. We know that [H+] = [A-] = x, and [HA] = 0.25 - x. We approximate x as 0.0075 since only 3% of HA dissociates. Hence Ka=~(0.0075)^2/(0.25-0.0075).

The pH of the solution is -log[H+] = -log(0.0075).

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The significant figures are always:

Different from zero except there are only zeros before the point.

You can round them to the previous significant.

In scientific notation, you have one figure point two more figures.

Examples:

You have 4.21

All different from zero and only two decimals.

Those are all significant figures.

if you have 000.231555

You will shorten this to two significant figures.

Before the point, you only have zeros, so you need to keep only one of them to say its less than one.

After the point, you have a lot of figures, but you need to round this to two.

Because you have a one before the three, you'll keep the three. If you have a five or bigger number, you round it.

In this case, you'll have 0.23 with two significant figures.

Answer:

[tex]\frac{N_C}{N} = 1.42\times 10^{-4}[/tex]

Explanation:

given data:

temperature  = 552 degree celcius  = 825 Kelvin

energy for vacancy formation is given as 0.63 eV/atom

fraction of atom can be obtained from following formula

[tex]N_C = N e^{\frac{-Q_V}{kT}}[/tex]

Where, K IS BOLTZMAN CONSTANT [tex]= 8.62\times 10^{-5}eV/K[/tex]

[tex]\frac{N_C}{N} =e^{\frac{-0.63}{8.62\times 10^{-5} \times 825}[/tex]

[tex]\frac{N_C}{N} = e^{-8.8858}[/tex]

[tex]\frac{N_C}{N} = 1.42\times 10^{-4}[/tex]

The fraction of atom sites that are vacant for silver at 552°C is approximately [tex]\( 7.347 \times 10^{-5} \).[/tex]

The fraction of atom sites that are vacant for silver at a given temperature can be calculated using the formula derived from the statistics of a canonical ensemble:

[tex]\[ f = \exp\left(-\frac{\Delta G_{\text{form}}}{k_B T}\right) \][/tex]

Now, we can plug in the values:

[tex]\[ f = \exp\left(-\frac{0.63 \text{ eV/atom}}{(8.617 \times 10^{-5} \text{ eV/K}) \times (825.15 \text{ K})}\right) \] \[ f = \exp\left(-\frac{0.63}{8.617 \times 10^{-5} \times 825.15}\right) \] \[ f = \exp\left(-\frac{0.63}{0.0658}\right) \] \[ f = \exp\left(-9.574\right) \] \[ f \approx \exp\left(-9.574\right) \] \[ f \approx 7.347 \times 10^{-5} \][/tex]

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory:

[tex]HA+H_2O\rightarrow A^-+H_3O^+[/tex]

Suppose acid Ha is getting dissociated in its solution and after dissociation it donates its proton to water molecule and forms conjugate base. Where as water (acting as a base) accepts protons and forms conjugate acid.

HA = Acid

[tex]H_2O[/tex] = Base

[tex]A^-[/tex] = Conjugate base

[tex]H_3O^+[/tex] = Conjugate acid

For example:

[tex]H_2SO_4+2H_2O\rightarrow SO_4^{2-}+2H_3O^+[/tex]

Sulfuric acid dissociating in its solution to form conjugate base and conjugate acid.

Sulfuric acid  = Acid

[tex]H_2O[/tex] = Base

[tex]SO_^{2-}[/tex] = Conjugate base

[tex]H_3O^+[/tex] = Conjugate acid

Answer:

5.45 g is 0.0221 moles of magnesium sulfate heptahydrate.

Explanation:

The molecular formula of magnesium sulfate heptahydrate is Mg SO₄·7H₂O.

The molar mass of this compound is calculated adding the molar mass of each element:

Mg: 24. 3 g

S: 32. 1 g

O: 16 g

H: 1 g

Then the mass of a mole of Mg SO₄·7H₂O is:

24.3 g + 32.1 g +4(16 g) + 7(2(1) + 16) = 246.4 g.

if 246.4 g is 1 mole of Mg SO₄·7H₂O, then 5.45 g will be:

5.45 g *(1 mol / 246.4 g) = 0.0221 mol.

Answer: The number of moles of disulfur decafluoride is [tex]6.62\times 10^{-20}[/tex]

Explanation:

We are given:

Number of disulfur decafluoride molecules = [tex]3.99\times 10^4[/tex]

According to mole concept:

[tex]6.022\times 10^{23}[/tex] number of molecules are contained in 1 mole of a compound.

So, [tex]3.99\times 10^4[/tex] number of molecules will be contained in = [tex]\frac{1mol}{6.022\times 10^{23}}\times 3.99\times 10^{4}=6.62\times 10^{-20}mol[/tex] of disulfur decafluoride.

Hence, the number of moles of disulfur decafluoride is [tex]6.62\times 10^{-20}[/tex]

Answer:

The molar mass of the unknown solute is 106,9 g/m

Explanation:

Cryoscopic descent formula to solve this

ΔT = Kf . m

Be careful because units in Kfp are K/m, so let's get the ΔT degrees °C in K

3,91°C  = 3,91 K

It's a difference, in the end it does not matter

For example you can have 5° C as the final temperature and as initial, 1,09 °C -- ΔT is 5 - 1.09 = 3.91

What happens in Kelvin?

5°C + 273 = 278 K

1,09° C + 273 = 274,09 K

ΔT = 278 K - 274,09 K = 3,91 K

3,91 K = 20,1 K/m * m

3,91 K / 20,1 m/K = m

0,194 = m (molality)

Molality means moles from solute in 1 kg of solvent.

1kg = 1000 g

1000 g ________  0,194 moles

50 g _________ x

x = (50 g * 13,77 moles) / 1000 g = 9,72 *10-3 moles

Moles = mass / molar mass

Molar mass = mass / moles

Molar mass = 1,04 g / 9,72 *10-3 moles

Molar mass = 106,9 g/m

Answer: Lithium is getting oxidized and is a reducing agent. Hydrogen is getting reduced and is oxidizing agent.

Explanation:

Oxidation reaction is defined as the reaction in which an atom looses its electrons. Here, oxidation state of the atom increases.

[tex]X\rightarrow X^{n+}+ne^-[/tex]

Reduction reaction is defined as the reaction in which an atom gains electrons. Here, the oxidation state of the atom decreases.

[tex]X^{n+}+ne^-\rightarrow X[/tex]

Oxidizing agents are defined as the agents which oxidize other substance and itself gets reduced. These agents undergoes reduction reactions.

Reducing agents are defined as the agents which reduces the other substance and itself gets oxidized. These agents undergoes reduction reactions.

For the given chemical reaction:

[tex]2Li(s)+2H_2O(l)\rightarrow 2LiOH(aq.)+H_2(g)[/tex]

The half reactions for the above reaction are:

Oxidation half reaction: [tex]2Li(s)\rightarrow 2Li^{+}(aq.)+2e^-[/tex]

Reduction half reaction: [tex]2H^(aq.)+2e^-\rightarrow H_2(g)[/tex]

From the above reactions, lithium is loosing its electrons. Thus, it is getting oxidized and is considered as a reducing agent.

Hydrogen is gaining electrons and thus is getting reduced and is considered as an oxidizing agent.

Answer:

The energy of the photon generated in the transition is 3.14*10⁻¹⁹ J

Explanation:

There are two equation that we need to use in order to solve this problem:

The first one is Planck's equation, which describes the relationship between energy and frequency:

E = h*v    eq. 1)

Where E is energy, h is Planck's constant (6.626 * 10⁻³⁴ J*s) and v is the radiation frequency.

In order to know the frequency, we use the second equation, which is the wave equation:

c = λ*v    eq. 2)

Where c is the speed of light in vacuum (aprx 3 * 10⁸ m/s), and λ is the wavelength. If we solve that equation for v we're left with

v=c/λ    eq. 3)

We replace v in eq. 1):

E= c*h/λ

Lastly we put the data we know and solve the equation, keeping in mind the correct use of units (converting 652 nm into m):

[tex]E=\frac{3*10^{8}ms^{-1} *6.626*10^{-34}Js}{633*10^{-9}m }=3.14*10^{-19}J[/tex]

Final answer:

The energy of the photon generated in the transition using the given formula. Calculating gives approximately 3.04 x 10^-19 joules.

Explanation:

The energy of the photon generated in the transition can be calculated using the formula:

E = hc / λ

Where E is the energy of the photon, h is the Planck's constant, c is the speed of light, and λ is the wavelength of the light. Substituting the given values:

E = (6.63 x 10^-34 J s x 3 x 10^8 m/s) / (652 x 10^-9 m)

Calculating this gives the energy of the photon as approximately 3.04 x 10^-19 joules.

Answer: The number of atoms of carbon present in given number of moles are [tex]2.350\times 10^{20}[/tex]

Explanation:

We are given:

Number of moles of carbon = [tex]3.900\times 10^{-4}mol[/tex]

According to mole concept:

1 mole of an element contains [tex]6.022\times 10^{23}[/tex] number of atoms.

So, [tex]3.900\times 10^{-4}mol[/tex] of carbon will contain = [tex]3.900\times 10^{-4}\times 6.022\times 10^{23}=2.350\times 10^{20}[/tex] number of atoms.

Hence, the number of atoms of carbon present in given number of moles are [tex]2.350\times 10^{20}[/tex]

Answer:

a. T and P remain the same (T=298 K and P=1 bar)

b. 11.23J/K

Explanation:

a. Since the mixing process of an idea gas doesn't present a change in the enthalpy, we could state that no change in neither temperature and pressure are given.

b. It is not necessary to know enthalpy data, the following formula is enough to compute the entropy change:

Δ[tex]S_{mix}=-n_{N_2}R ln(x_{N_2})-n_{O_2}R ln(x_{O_2})[/tex]

Thus, the molar fractions are equal to 0.5, and the result yields:

Δ[tex]S_{mix}=-(1mol)[(8.314J/(mol*K)]ln(0.5)-(1mol)[(8.314J/(mol*K)]ln(0.5)[/tex]

Δ[tex]S_{mix}=11.23J/K[/tex]

Best regards

Answer:

3 "digits" are required to code for the 20 different amino acids.

This means that in order to code for one amino acid, you require a group of 3 nucleotides, which is called a 'codon'

Explanation:

If each nucleotide determined one amino acid, we could only code for four different amino acids, since DNA contains only four kinds of nucleotides.

If an amino acid were to be coded by a group of two nucleotides, the total number of diniclueotides we could get would be 4^2 = 16. This means that we could only code for 16 amino acids, which is an inferior amount than the number of amino acids required for protein synthesis (20).

If an amino acid were to be coded by a group of three nucleotides, the total number of trinucleotides we could get would be 4^3 = 64. This means that the  total amount of triplets we could get is 64, which is more than enough to be able to code for the 20 different amino acids.

The given data suggests that the decay of Chlorine oxide (ClO) is a first-order reaction. The rate constant can be calculated using the first-order rate law equation, which in this case gives a value of approximately 70000 s⁻¹.

The reaction order is determined by the relationship between the rate of reaction and the concentration of the reactants. By observing the given data, it appears that the decay of Chlorine oxide (ClO) is halving approximately. This suggests that it could be a first-order reaction, where the rate of the reaction is directly proportional to the concentration of one reactant.

To calculate the rate constant of the reaction, we can use the first-order rate law equation: k = -1/[t]*ln([A]t/[A]0), where 'k' is the rate constant, '[t]' is the elapsed time, '[A]t' is the concentration at time 't' and '[A]0' is the initial concentration.

Using the initial and final concentrations given (at 4.26 × 10−3 s and 6.74 × 10−3 s), the equation for the rate constant becomes: k = -1/(6.74 × 10⁻³ - 4.26 × 10⁻³)*ln((4.01 × 10⁻⁶)/(7.73 × 10⁻⁶)). Calculating gives a rate constant value of approximately 70000 s-1. Remember, these values may vary depending on specific calculation approaches.

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The decomposition of ClO is determined to be first-order with a rate constant of approximately 0.263 s⁻¹. We used the method of initial rates and plotting ln[ClO] versus time to ascertain the reaction order and calculate the rate constant.

To determine the reaction order and rate constant for the decomposition of chlorine oxide (ClO) according to the equation 2ClO(g) → Cl₂(g) + O₂(g), we need to analyze the given concentration data over time.

The reaction is **first-order** with respect to ClO, and the rate constant, **k**, is approximately 0.263 s⁻¹.

Answer:

Hello my friend! The amount of 31g of KNO3 will precipitate!

Explanation:

At 40 ∘C the solution has 45 g of KNO3 per 100 g of water and it can contain up to 63 g of KNO3 per 100 g of water. At 0 ∘C the solubility is ~ 14 gKNO3per 100 g of water, so 31 gKNO3 per 100 g of water will precipitate

The cooling of a solution with 45g of KNO3 in 100g of water from 40 degrees Celsius to 0 degrees Celsius results in precipitation of excess KNO3 because of decreased solubility.

This question is related to the solubility of Potassium Nitrate (KNO3) in water at different temperatures. At 40 degrees Celcius, the solution already has 45g of KNO3 per 100g of water. The solubility of KNO3 at this temperature is approximately 60g per 100g of water. However, upon cooling to 0 degrees Celcius, the solubility of KNO3 drops to around 13g per 100g of water. As a result, the excess KNO3 which is around 32g per 100g of water, will precipitate out of the solution.

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Answer: The mass of water produced in the reaction is 47.25 grams.

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]      .....(1)

Given mass of ammonia = 29.7 g

Molar mass of ammonia = 17 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of ammonia}=\frac{29.7g}{17g/mol}=1.75mol[/tex]

The given chemical reaction follows:

[tex]4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)[/tex]

By stoichiometry of the reaction:

4 moles of ammonia produces 6 moles of water.

So, 1.75 moles of ammonia will produce = [tex]\frac{6}{4}\times 1.75=2.625mol[/tex] of water.

Now, calculating the mass of water by using equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 2.625 moles

Putting values in equation 1, we get:

[tex]2.625mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=47.25g[/tex]

Hence, the mass of water produced in the reaction is 47.25 grams.

The complete reaction of 29.7 grams of NH3 would produce 47.099 grams of water by converting the given mass of NH3 to moles, using the balanced chemical equation to find the ratio of NH3 to H2O, and then converting the moles of H2O to grams.

To find out how many grams of water are produced in the complete reaction of 29.7 grams of ammonia (NH3), we need to use stoichiometry, which involves several steps:

Therefore, the complete reaction of 29.7 grams of NH3 would produce 47.099 grams of water (H2O).

Answer:

a) mass flow = 56940 Kg/h

b) mass flow = 202.5 mol/s

Explanation:

∴ δ C6H6 = 876 Kg/m³,,,,,wwwcarlroth.com

⇒ 65m³/h * 876 Kg/m³ = 56940 Kg/h

⇒ 56940 Kg/h * ( 1000 g/Kg ) * ( mol/ 78.11 g) * ( h/3600s )= 202.5 mol/s

The mass flow rate of benzene is calculated using its density of 0.876 kg/m³, resulting in 56.94 kg/h. The molar flow rate is determined using the molar mass of benzene, 78.11 g/mol, which yields 20.34 mol/s.

Mass Flow and Molar Flow Calculations

To calculate the mass flow fed in kg/h for benzene, the density of benzene is required. Benzene has a density of approximately 0.876 kg/m³ at room temperature. Using this density, the mass flow can be calculated as follows:

Mass flow (kg/h) = Volumetric flow (m³/h) × Density (kg/m³)

Mass flow = 65 m³/h ×  0.876 kg/m³ = 56.94 kg/h

To calculate the molar flow in mol/s, we use the molar mass of benzene which is approximately 78.11 g/mol:

Molar flow (mol/s) = Mass flow (kg/h) × (1000 g/kg) / (Molar mass (g/mol) × (3600 s/h))

Molar flow = (56.94 kg/h × 1000 g/kg) / (78.11 g/mol ×  3600 s/h) = 20.34 mol/s

Note: The molecular formula of benzene is C₆H₆, not CH as mentioned in the information provided. Therefore, the combustion analysis would yield different amounts of CO₂ and H₂O than suggested by the empirical formula CH. It's important to use the correct molecular formula for accurate calculations.

Answer:

The correct answer is: PCl₃ and PCl₅

Explanation:

Law of multiple proportions, also known as the Dalton's Law, states that the when the two different chemical elements combine in order to form two or more than two chemical compounds, then the ratio of mass of the chemical element that combines with the fixed mass of other chemical element is the ratio of small whole numbers.

Example: PCl₃ and PCl₅

Answer:

Head loss in turbulent flow is varying as square of velocity.

Explanation:

As we know that head loss in turbulent flow given as

[tex]h_F=\dfrac{FLV^2}{2gD}[/tex]

Where

F is the friction factor.

L is the length of pipe

V is the flow velocity

D is the diameter of pipe.

So from above equation we can say that

[tex]h_F\alpha V^2[/tex]

It means that head loss in turbulent flow is varying as square of velocity.

We know that loss in flow are of two types

1.Major loss :Due to surface property of pipe

2.Minor loss :Due to change in momentum of fluid.

Answer:

The correct answer is head varies directly with square of velocity of flow

Explanation:

The head loss in a pipe as given by Darcy Weisebach equation is

[tex]h_L=\frac{flv^2}{2gD}[/tex]

where

'f' is friction factor whose value depends on the nature of flow (Laminar/turbulent)

'L' is the length of the section in which the head loss is calculated

'v' is the velocity of the flow

'D' is the diameter of the duct

Thus we can see that the head loss varies with square of velocity of the fluid.

Answer:

100.223°C is the boiling point of an aqueous solution.

Explanation:

Osmotic pressure of the solution = π = 10.50 atm

Temperature of the solution =T= 25 °C = 298 .15 K

Concentration of the solution = c

van'y Hoff factor = i = 1 (non electrolyte)

[tex]\pi =icRT[/tex]

[tex]c=\frac{\pi }{RT}=\frac{10.50 atm}{0.0821 atm L/mol K\times 298.15 K}[/tex]

c = 0.429 mol/L = 0.429 mol/kg = m

(density of solution is the same as  pure water)

m = molality of the solution

Elevation in boiling point = [tex]\Delta T_b[/tex]

[tex]\Delta T_b=iK_b\times m[/tex]

[tex]\Delta T_b=T_b-T[/tex]

T = Boiling point of the pure solvent

[tex]T_b[/tex] = boiling point of the solution

[tex]K_b[/tex] = Molal elevation constant

We have :

[tex]K_b=0.52^oC/m[/tex] (given)

m = 0.429 mol/kg

T = 100° C (water)

[tex]\Delta T_b=1\times 0.52^oC/m\times 0.429 mol/kg[/tex]

[tex]\Delta T_b=0.223^oC[/tex]

[tex]\Delta T_b=T_b-T[/tex]

[tex]T_b=\Delta T_b+T=0.223^oC+100^oC=100.223^oC[/tex]

100.223°C is the boiling point of an aqueous solution.

Answer:

[tex]T=-272.9^{o}C[/tex]

Explanation:

We have the ideal gasses equation [tex]PV=nRT[/tex] and the expression for the specific volume [tex]v=\frac{V}{m}[/tex], that is the inverse of the density, and for definition the number of moles is equal to the mass over the molar mass, that is [tex]n=\frac{m}{M}[/tex]

And we can relate the three equations as follows:

[tex]PV=nRT[/tex]

Replacing the expression for n, we have:

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

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

Replacing the expression for v, we have:

[tex]Pv=\frac{RT}{M}[/tex]

Now resolving for T, we have:

[tex]T=\frac{PvM}{R}[/tex]

Now, we should convert all the quantities to the same units:

-Convert 500kPa to atm

[tex]500kPa*\frac{0.00986923}{1kPa}=4.93atm[/tex]

-Convert 0.2[tex]\frac{m^{3}}{kg}[/tex] to [tex]\frac{L}{kg}[/tex]

[tex]0.2\frac{m^{3} }{kg}*\frac{1L}{1m^{3}}=0.2\frac{L}{kg}[/tex]

- Convert the molar mass M of the water from [tex]\frac{g}{mol}[/tex] to [tex]\frac{kg}{mol}[/tex]

[tex]18\frac{g}{mol}=\frac{1kg}{1000g}=0.018\frac{kg}{mol}[/tex]

Finally we can replace the values:

[tex]T=\frac{(4.93atm)(0.2\frac{L}{kg})(0.018\frac{kg}{mol})}{0.082\frac{atm.L}{mol.K}}[/tex]

[tex]T=0.216K[/tex]

[tex]T=0.216K-273.15\\T=-272.9^{o}C[/tex]