Consider the gas reaction: H2 (g)2(g) 2HI (g) The equilibrium constant at 731 K is 50.3. Equal amounts of all three gases (0.100 M) are introduced in a container, calculate the concentration of each gas after the system reaches equilibrium. Express your results with the right number of significant figures.

Answer:

The mechanisms for the elimination reactions between NaOH in ethanol and the halogenoalkanes are demonstrated in the figure attached.

Explanation:

(i) 1-bromobutane will suffer elimination to for an alkene. The mechanism will be E2, which means that the attack and the elimination will occur simultaneously. This is the preferred mechanism because the bromine is in a primary carbon.

(ii) 2-bromo-2-methylpentane will suffer elimination to for an alkene. The mechanism will be E1, which means that the attack and the elimination will occur in two different steps. The bromine will be eliminated in the first step with the formation of a carbocation and in a second step the double bond will be formed after the anionic attack. This is the preferred mechanism because the bromine is in a terciary carbon which is able to stabilize the carbocation formed.

The rate law for the reaction BF3(g)+NH3(g)→F3BNH3(g) based on the input data is rate = k[BF3][NH3], where k is the rate constant, [BF3] and [NH3] denote the molar concentrations of BF3 and NH3, indicating a first order relationship for both reactants.

In order to determine the rate law for the reaction BF3(g)+NH3(g)→F3BNH3(g), we have to look at how the initial rate changes with respect to the change in concentration of the reactants. Using the given data, we can compare experiments where only one reactant's concentration is changed while the concentration of the other reactant remains constant. Upon analysis, we see that when the concentration of BF3 doubles, the rate also doubles, suggesting a first order relationship. Similarly, when the concentration of NH3 doubles, the rate doubles, indicating a first order relationship for NH3 as well. Hence, given that both BF3 and NH3 are first order, the rate law for the reaction should be: rate = k[BF3][NH3].

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The rate law for the reaction BF3(g) + NH3(g) → F3BNH3(g) is rate = k[BF3][NH3].

The rate law for the reaction BF3(g) + NH3(g) → F3BNH3(g) can be determined by analyzing the effect of changing concentrations on the initial rate of the reaction. By comparing the rates of reaction for different experiments, we can observe how changing the concentrations of the reactants affects the rate. In this case, it is clear that the rate of reaction is directly proportional to the concentration of BF3 and NH3, so the rate law for the reaction is rate = k[BF3][NH3].

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Answer : The time taken for the decomposition of drug will be 11.6 years.

Explanation :

Half-life = 5.00 years

First we have to calculate the rate constant, we use the formula :

[tex]k=\frac{0.693}{5.00}[/tex]

[tex]k=0.139\text{ years}^{-1}[/tex]

Now we have to calculate the time taken.

Expression for rate law for first order kinetics is given by:

[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]

where,

k = rate constant

t = time taken = ?

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process = 100 - 80 = 20 g

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

[tex]t=\frac{2.303}{0.139}\log\frac{100}{20}[/tex]

[tex]t=11.6\text{ years}[/tex]

Therefore, the time taken for the decomposition of drug will be 11.6 years.

Answer:

Volume of cube = [tex]L^3[/tex] = [tex]1 cm^3[/tex]

Density = [tex] \frac{m}{V} = 3.48 \frac{g}{cm^3}[/tex]

Answer:

Chemical formula for the black iron oxide sample is FeO

Explanation:

Given:

Mass ratio of black iron oxide sample, Fe/O = 3.491

To determine:

The chemical formula

Calculation:

The mass ratio is Fe:O = 3.491 : 1

Mass of Fe = 3.491 g

Mass of O = 1.000 g

Atomic mass of Fe =55.85 g/mol

Atomic mass of O = 16.00 g/mol

[tex]Moles\ Fe = \frac{3.491g}{55.85g/mol} =0.625\\\\Moles\ O = \frac{1.000g}{16.00g/mol} =0.0625\\[/tex]

Therefore, the molar ratio of Fe:O = 1:1

Hence, chemical formula is FeO

Answer:

[tex]\frac{\dot Q}{A} = 151.33 W/m^2[/tex]

[tex]T_c = 25.153°C[/tex]

Explanation:

Given data:

one side wall temperature 200°C

other side wall temperature 10°C

h = 100 W/m^2 °C

k = 2.6W/m °C

wall thickness L = 30 cm

we know that heat flux is given as

[tex]\frac{\dot Q}{A} = \frac{ T_A - T\infty}{\frac{L}{K} + \frac{1}{h}}[/tex]

[tex]\frac{\dot Q}{A} = \frac{ 20- 10} {\frac{0.30}{2.6} + \frac{1}{100}}[/tex]

[tex]\frac{\dot Q}{A} = 151.33 W/m^2[/tex]

[tex]1515.33 W/m^2 = \frac{ T_A - T_c}{\frac{L}{K}}[/tex]

solving for temperature for cold surface is given as

[tex]T_c = -1515.33 \times \frac{0.3}{2.6} + 200[/tex]

[tex]T_c = 25.153°C[/tex]

The mass of 11 nanomoles of the substance is 616 nanograms.

Given that one mole of a substance has a mass of 56.0 g, we can calculate the mass of 11 nanomoles of the substance as follows:

1 mole = 56.0 g

1 nanomole = 56.0 g / 1,000,000,000

=[tex]5.60 * 10^{-8} g[/tex]

Now, to find the mass of 11 nanomoles:

Mass = [tex]11 nanomoles * 5.60 * 10^{-8}g/nanomole[/tex]

Calculating this gives us:

Mass =[tex]6.16 * 10^{-7} g[/tex]

To express the answer in nanograms, we need to convert grams to nanograms:

1 g = 1,000,000,000 ng

So, [tex]6.16 * 10^{-7} g = 6.16 * 10^2 ng = 616 ng[/tex]

Therefore, the mass of 11 nanomoles of the substance is 616 nanograms.

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To find the mass of 11 nanomoles of a substance, multiply the number of nanomoles by the molar mass of the substance. In this case, the mass is 616.0 ng.

To calculate the mass of 11 nanomoles of a substance, we need to know the molar mass of the substance. If one mole of the substance has a mass of 56.0 g, then the molar mass is 56.0 g/mol. To find the mass of 11 nanomoles, we can use the conversion factor:

11 nmol * 56.0 g/mol = 616.0 ng

Therefore, the mass of 11 nanomoles of the substance is 616.0 nanograms (ng).

Answer:

Carbonic acid/bicarbonate; bicarbonate  

Explanation:

H₂CO₃ + H₂O ⇌ H₃O⁺ + HCO₃⁻; pKₐ₁ =   6.35

HCO₃⁻ + H₂O ⇌ H₃O⁺ + CO₃²⁻;  pKₐ₂ = 10.33

[tex]\text{pH}& = &\text{pK}_{\text{a}} + \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\7.4 & = &\text{pK}_{\text{a}} + \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}[/tex]

The best buffer is one for which pKₐ ≈ pH.  

6.35 is closer to 7.4, so the carbonic acid/bicarbonate form of the buffer predominates

The pH of the blood is higher (more basic) than the pKₐ of carbonic acid, so its basic form (bicarbonate, HCO₃⁻) predominates.

Final answer:

The bicarbonate form of carbonic acid predominates in the buffer system at pH 7.4.

Explanation:

The Henderson-Hasselbalch equation is used to determine the pH of a buffer solution. In this case, we are trying to determine which form of the carbonic acid buffer predominates at pH 7.4, which is the homeostasis pH of the blood.

The Henderson-Hasselbalch equation is:

pH = pKa + log ([base] / [acid])

Given that the pKa of carbonic acid is 6.35, we can plug in the values to calculate:

pH = 6.35 + log (0.024 / 0.0012)

Simplifying the equation, we get:

pH = 6.35 + log (20)

pH = 6.35 + 1.3010

pH = 7.65

So, at pH 7.4, the bicarbonate (HCO3-) form of carbonic acid predominates in the buffer system.

Answer:

Kc = 168.0749

Explanation:

initial mol:   0.822               0          0

equil. mol: 2(0.822 - x)         x           x

∴ [ HI ]eq = 0.055 mol/L = 2(0.822 - x) / (1.11 L )

⇒ 1.644 - 2x = 0.055 * 1.11

⇒ 1.644 = 2x + 0.06105

⇒ 2x = 1.583

⇒ x = 0.7915 mol equilibrium

⇒ [ H2 ] eq = 0.7915mol / 1.11L = 0.7130 M = [ I2 ] eq

⇒ Kc = ([ H2 ] * [ I2 ]) / [ HI ]²

⇒ Kc = ( 0.7130² ) / ( 0.055² )

⇒ Kc = 168.0749

The equilibrium constant (Kc) for the reaction where 2HI decomposes into H2 and I2 at a given temperature can be calculated using the ICE method. After setting up the Kc expression and solving for x, the concentration changes of HI, H2, and I2 can be used to find Kc, which is approximately 0.1176.

To calculate the equilibrium constant (Kc) for the decomposition of hydrogen iodide (HI) into hydrogen gas (H₂) and iodine gas (I₂), we must first write the Kc expression for the reaction:

2HI (g) ⇌ H₂ (g) + I₂ (g)

The Kc expression is Kc = [H₂][I₂] / [HI]^2.

Next, we use an ICE (Initial, Change, Equilibrium) table to keep track of the concentrations throughout the reaction:

To calculate x, solve the equation 0.7409 - 2x = 0.055, yielding x = 0.34295 M. Now we know at equilibrium:

Inserting these values into the Kc expression we get:Kc = (0.34295)(0.34295) / (0.055)^2

Simplifying gives us:Kc = 0.1176 to 4 decimal places.

Final answer:

To prepare a 0.05 M magnesium sulfate heptahydrate solution with a volume of 200 mL, you will need 2.46 grams of magnesium sulfate heptahydrate.

Explanation:

To calculate the amount of magnesium sulfate heptahydrate required to prepare 200 mL of 0.05 M solution, we need to use the formula:

moles = molarity x volume

First, we convert the volume from milliliters to liters:

200 mL x (1 L/1000 mL) = 0.2 L

Next, we substitute the given values into the formula:

moles = 0.05 mol/L x 0.2 L = 0.01 mol

Finally, we calculate the mass of magnesium sulfate heptahydrate using its molar mass:

moles = mass (g) / molar mass (g/mol)

0.01 mol = mass (g) / 246.48 g/mol

mass (g) = 0.01 mol x 246.48 g/mol = 2.46 g

Answer:

11.7 mL

Explanation:

Density of a substance is given by the mass of the substance divided by the volume of the substance .

Hence , d = m / V

V = volume

m = mass ,  

d = density ,

From the question ,

The density of the metal = 10.5 g/cm³

The mass of the metal = 5.25 g

Hence , the volume can be calculated from the above formula , i.e. ,

d = m / V  

V = m / d

V = 5.25 g / 10.5 g/cm³

V = 0.5 cm³

Since , The unit 1 mL = 1 cm³

V = 0.5 mL

The vessel has 11.2 mL of water ,

The new volume becomes ,

11.2 mL +  0.5 mL = 11.7 mL

Answer:

To prepare 20,0 mL of the liquid mixture you should mix 15,3 mL of CHCl₃ with 4,7 mL of CHBr₃

Explanation:

Here you have two variables: The volume of both CHCl₃ (X) and CHBr₃ (Y). To find these two variables you must have, at least, two equations.

You know total volume is 20,0 mL. Thus:

X + Y = 20,0 mL (1)

The other equation  is:

[tex]\frac{X}{20,0mL\\}[/tex] × 1,492 g/mL + [tex]\frac{Y}{20,0 mL}[/tex] × 2,890 g/mL = 1,82 g/mL (2)

If you replace (1) in (2):

[tex]\frac{X}{20,0mL\\}[/tex] × 1,492 g/mL + [tex]\frac{20,0 mL - X}{20,0 mL}[/tex] × 2,890 g/mL = 1,82 g/mL

Solving:

X = 15,3 mL

Thus, using (1):

20,0 mL - 15,3 mL = Y = 4,7 mL

Thus, to prepare 20,0 mL of the liquid mixture you must mix 15,3 mL of CHCl₃ with 4,7 mL of CHBr₃.

I hope it helps!

Answer : The overall mixture flow rate is 11905.71 lb.mass/hour

Explanation :

As we are given that 16.0 weight percent methanol. That means, 16.0 g of methanol present in 100 g of mixture (methanol-methyl acetate).

Mass of methanol = 16.0 g

Mass of mixture = 100 g

Mass of methyl acetate = 100 - 16.0 = 84.0 g

Now converting flow rate of methyl acetate from lb.mols/hour to lb.mass/hour.

[tex]\text{lb.mols/hour}=(\text{lb.mols/hour})\times \text{Molar mass}=\text{lb.mass/hour}[/tex]

Molar mass of methyl acetate = 74.08 g/mols

So, [tex]135.0\text{ lb.mols/hour}=(135.0\text{ lb.mols/hour})\times 74.08g/mols=10000.8\text{ lb.mass/hour}[/tex]

Now we have to calculate the overall mixture flow rate.

As, 84 g of methyl acetate flow rate = 10000.8 lb.mass/hour

So, 100 g of mixture flow rate = [tex]\frac{100}{84}\times 10000.8\text{ lb.mass/hour}=11905.71\text{ lb.mass/hour}[/tex]

Therefore, the overall mixture flow rate is 11905.71 lb.mass/hour

Final answer:

to find the overall mixture flowrate in lb,mass/hour, we add the flowrates of methyl acetate and methanol: F = FMA + FM = 135.0 + 21.6 = 156.6 lbmols/hour.

Explanation:

To find the overall mixture flowrate in lb,mass/hour, we need to determine the flowrate of methyl acetate and its weight percent in the mixture.

Given the flowrate of methyl acetate is 135.0 lbmols/hour and the weight percent of methanol is 16.0%, we can find the flowrate of methanol in the mixture. Let's denote the overall mixture flowrate as F, the flowrate of methyl acetate as FMA, and the flowrate of methanol as FM.

Since the weight percent of methanol is 16.0%, the weight percent of methyl acetate is 100% - 16.0% = 84.0%. This means that the weight ratio of methyl acetate to methanol in the mixture is 84.0% / 16.0% = 5.25.

Therefore, we can write the equation FMA / FM = 5.25. Solving for FM, we get FM = F / (1 + 5.25) = F / 6.25. Substituting the given flowrate of methyl acetate (135.0 lbmols/hour) into the equation, we get FM = 135.0 / 6.25 = 21.6 lbmols/hour.

Finally, to find the overall mixture flowrate in lb,mass/hour, we add the flowrates of methyl acetate and methanol: F = FMA + FM = 135.0 + 21.6 = 156.6 lbmols/hour.

Answer: 0.102 Liters

Explanation

According to the ideal gas equation:

[tex]PV=nRT[/tex]

P = Pressure of the gas = [tex]3.75\times 10^5 Pa[/tex] = 3675 atm     (1 kPa= 0.0098 atm)

V= Volume of the gas = ?

T= Temperature of the gas = 23.6°C = 296.6 K    [tex]0^00C=273K[/tex]

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas = 45.6

[tex]V=\frac{nRT}{P}=\frac{45.6\times 0.0821\times 296.6}{3675}=0.302L[/tex]

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

[tex]P\propto \frac{1}{V}[/tex]     (At constant temperature and number of moles)

[tex]P_1V_1=P_2V_2[/tex]  

where,

[tex]P_1[/tex] = initial pressure of gas  = [tex]3.75\times 10^5 Pa[/tex]

[tex]P_2[/tex] = final pressure of gas  = [tex]5.67\times 10^5 Pa[/tex]

[tex]V_1[/tex] = initial volume of gas   = 0.302 L

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

[tex]3.75\times 10^5 \times 0.302=5.67\times 10^5\times V_2[/tex]  

[tex]V_2=0.199L[/tex]

The final volume has to be 0.199 L, thus (0.302-0.199) L= 0.102 L must  release into the atmosphere.

Therefore the answer is 0.102 L

Answer:

At a pressure of I atm, the boiling point of a mixture of aniline and water is b) 100°C.

Explanation:

Assuming that aniline is insoluble in water there will be no interaction between liquids. There will be no positive interactions that can increase the boiling point or negative interactions that can decrease the boiling point. Thus the boiling points of each substance will not be affected. Since the boiling point of water is 100ºC, the mixture will start to evaporate at 100ºC.

Answer:

B.)

Explanation:

Final answer:

The rate of disappearance of NO2 in the experiment at 300 °C is approximately 1.32 × 10^-5 M/s, calculated by the change in concentration over time.

Explanation:

The rate of disappearance of NO2 in the experiment at 300 °C is calculated by determining the change in concentration over the change in time. The initial concentration of NO2 is 0.0138 M and it drops to 0.00886 M over 374 s. The change in concentration (Δ[NO2]) is 0.0138 M - 0.00886 M = 0.00494 M. The rate of disappearance is then Δ[NO2]/Δt = 0.00494 M/374 s.

After calculating this, we get the rate of disappearance of NO2 to be approximately 1.32 × 10-5 M/s, which is one of the possible choices provided in the question.

Answer:

Hydrogen bond

Explanation:

Hydrogen bond is a type of dipole dipole interaction.

Hydrogen bond is present between the hydrogen atom attached to an electronegative atom and other electronegative atom. Compounds in which hydrogen is attached with N, F and O, show hydrogen bonding.

As N, F and O are more electronegative than H, so a positively charge is develop on H atom and negative charge is develop on electronegative atom.

Hydrogen bond is a weaker bond. It is weak as compared to ionic bond and covalent bond but stronger than van der Waals interaction.

Ionic bond is formed by the complete transfer and gain of electrons. It is a electrostatic attraction between positively charged and negatively charged species. Fox example bonding in NaCl

Covalent bond is formed by the sharing of electrons between two atoms. For example bonding in Cl_2.

Metallic bond is present between mobile electrons and positively charged kernels. It exist in metals only.

Answer:

d) Hydrogen bond

Explanation:

Hydrogen bonding is a special type of the dipole-dipole interaction and it occurs between hydrogen atom that is bonded to highly electronegative atom which is either fluorine, oxygen or nitrogen atom.

Partially positive end of the hydrogen atom is attracted to partially negative end of these atoms which is present in another molecule. It is strong force of attraction between the molecules.

Answer:

Total feed rate = 98.3 lbmol/h

methanol mole fraction = 0.315

water mole fraction = 0.685

Explanation:

First of all, it is needed to calculate the feed mass of methanol and water in kg/h.

For methanol:

[tex]m_{methanol} = m\%wt_{methanol}/100 = (1000kg/h)(45\%)/100[/tex]

[tex]m_{methanol} = 450kg/h[/tex]

For water:

[tex]m_{water} = m\%wt_{water}/100 = (1000kg/h)(55\%)/100[/tex]

[tex]m_{water} = 550 kg/h[/tex]

Now, change from mass units (kg/h) to moles units (kmol/h and lbmol/h) using simple conversion factors:

For methanol:

[tex]n_{methanol} = (450\frac{kg}{h})(\frac{1 kmol}{32 kg} )[/tex]

[tex]n_{methanol} = 14.1kmol/h[/tex]

For water:

[tex]n_{water} = (550\frac{kg}{h})(\frac{1 kmol}{18 kg} )[/tex]

[tex]n_{water} = 30.6kmol/h[/tex]

Change units from kmol/h to lbmol/h

For methanol:

[tex]n_{methanol} = (14.1\frac{kmol}{h})(\frac{1 lbmol}{0.454 kmol} )[/tex]

[tex]n_{methanol} = 31.0 lbmol/h[/tex]

For water:

[tex]n_{water} = (30.6\frac{kg}{h})(\frac{1 lbmol}{0.454 kmol} )[/tex]

[tex]n_{water} = 67.3 lbmol/h[/tex]

Sum moles of methanol and water in lbmol/h to compute the total feed rate:

[tex]n = 31.0 lbmol/h + 67.3 lbmol/h[/tex]

[tex]n = 98.3 lbmol/h[/tex]

Divide both methanol and water moles feed rates by total feed rate:

For methanol:

[tex]X_{methanol} = \frac{31.0 lbmol/h}{98.3 lbmol/h}[/tex]

[tex]X_{methanol} =  0.315[/tex]

For water:

[tex][X_{water} =  \frac{67.3 lbmol/h}{98.3 lbmol/h}[/tex]

[tex]X_{water} =  0.685[/tex]

End

To draw the seven constitutional isomers of the cycloalkane with the formula C6H12, we need to consider different arrangements of carbon atoms in a cyclic structure

To draw the seven constitutional isomers of the cycloalkane with the formula C6H12, we need to consider different arrangements of carbon atoms in a cyclic structure with the given molecular formula:

1. Cyclohexane:

H H H

| | |

H--C--C--C--H

| | |

H H H

2. Methylcyclopentane:

H H H

| | |

H--C--C--C--C--H

| | |

H H H

3. Dimethylcyclobutane:

H H H

| | |

H--C--C--C--C--H

| | | |

H H H H

4. Ethylcyclopentane:

H H H

| | |

H--C--C--C--C--H

| | |

H H H

5. 1,1-Dimethylcyclobutane:

H H H

| | |

H--C--C--C--C--H

| | | |

H H H H

6. 1,2-Dimethylcyclopentane:

H H H

| | |

H--C--C--C--C--C--H

| | |

H H H

7. 1,3-Dimethylcyclobutane:

H H H

| | |

H--C--C--C--C--H

| | |

H H H

For the two stereoisomers of 1,3-dibromocyclobutane, we can have:

1. Cis-1,3-dibromocyclobutane:

Br H

| |

H--C--C--C--Br

| |

H H

2. Trans-1,3-dibromocyclobutane:

Br Br

| |

H--C--C--C--H

| |

H H

These structures represent the seven constitutional isomers of cycloalkane C6H12 and the two stereoisomers of 1,3-dibromocyclobutane.

Final answer:

An interpretation of the results of many tests is called a theory. It is a well-substantiated explanation of an aspect of the natural world, based on extensive research and experimentation.

Explanation:

An interpretation of the results of many tests is called a theory. A theory is a well-substantiated explanation of an aspect of the natural world. It is a scientific explanation that has been repeatedly tested and supported by many experiments.

For example, the theory of evolution is a well-supported explanation of how species have evolved over time. It is based on extensive research and experimentation.

It is important to note that a theory in science is different from the everyday use of the word, which often refers to a guess or speculation. In science, a theory is a well-tested and supported explanation.

Explanation:

FTIR spectroscopy is method which is used to determine structures of the molecules with characteristic absorption of the infrared radiation by the molecule.  

When the molecules is exposed to the infrared radiation, the sample molecules absorb the radiation of wavelengths (specific to molecule) which causes change of the dipole moment of the sample molecules. The vibrational energy levels of the sample molecules consequently transfer from the ground state to the excited state.  

Frequency of absorption peak is determined by vibrational energy gap.

Intensity of absorption peaks is related to change of dipole moment and possibility of transition of the energy levels.  

Thus, by analyzing infrared spectrum,abundant structure information of the molecule can be known.  

Most of the molecules/ compounds are infrared active but for homo-nuclear molecules which are diatomic like [tex]H_2[/tex], [tex]N_2[/tex], [tex]Cl_2[/tex], [tex]O_2[/tex] are inactive due to zero dipole change in vibration and the rotation of such molecules.

Final answer:

Molecular hydrogen does not produce an FTIR spectrum because as a homonuclear diatomic molecule, it does not have a change in dipole moment during vibration, which is a requirement for IR activity.

Explanation:

Molecular hydrogen does not produce a Fourier Transform Infrared (FTIR) spectrum because it is a homonuclear diatomic molecule, meaning it consists of two identical atoms. These molecules lack a permanent electric dipole moment, which is necessary for interaction with infrared light in FTIR spectroscopy. Infrared radiation can be absorbed by a molecule if the radiation causes a change in the dipole moment of the molecule, leading to vibrations within the molecule. Since molecular hydrogen's atoms are identical, they share electrons equally and there is no net dipole moment to be altered by the infrared radiation.

The fundamental requirement for a molecule to be IR active is a change in the molecular dipole moment as it vibrates. Homonuclear diatomic molecules like [tex]H_{2}[/tex] do not exhibit this change in dipole moment because the atoms involved are of the same electronegativity and thus share electrons equally, ensuing no dipole moment is generated during vibration. This is why molecular hydrogen doesn't show up on an FTIR spectrum.

Answer:

i) amount of MSW generated:

Laguna: 19200 Kg/day

Makati: 38000 Kg/day

ii) number of trucks to collect twice weekly:

Laguna: 3 trucks

Makati: 5 trucks

iii) volume of MSW in tons that enter landfill/week:

Laguna: 147.84 ton/week

Makati: 292 ton/week

Explanation:

i) Laguna: 0.96 Kg person / day of MSW * 20000 = 19200 Kg MSW / day

⇒ Laguna: 19200 Kg/day * ( 7day/ week ) = 134400 Kg/week

Makati: 1.9 Kg person / day of MSW * 20000 = 38000 Kg MSW / day

⇒ Makati: 38000 Kg/day * ( 7day/week ) = 266000 Kg/week

ii)  truck capacity = 4.4 ton * ( Kg / 0.0011 ton ) = 4000 Kg

⇒ quote/day = 4000 Kg * 0.75 = 3000 Kg

⇒ loads/day = 2 * 3000 kg = 6000 Kg

⇒ operate/week = 5 * 6000 Kg = 30000 Kg

∴ Laguna:  number of trucks needed/week= 134400 / 30000  = 4.48 ≅ 5 trucks

⇒ number of trucks to collect twice weekly = 5 / 2 = 2.5 ≅ 3 trucks

∴ Makati : number of trucks needed/week = 266000 / 30000 = 8.86 ≅ 9 trucks

⇒ number of trucks to collect twice weekly = 9 / 2 = 4.5 ≅ 5 trucks  

iii) enter landfill/week:

Laguna: 134400Kg MSW/week * ( 0.0011 ton/Kg ) = 147.84 ton/week

Makati: 266000Kg MSW/week * ( 0.0011 ton/Kg ) = 292 ton/week

Answer:

You have to weigh 0.079 g of Na₂S₂0₃ and dissolve it in water at a final volume of 250 ml.

Explanation:

A 0.00200 M Na₂S₂O₃ solution has 0.002 mol of Na₂S₂O₃ per liter of solution. As we know that 1 L = 1000 ml, and that the molecular weight of Na₂S₂O₃ IS 158 g/mol, we can calculate the mass of Na₂S₂O₃ to weigh as follows:

mass= [tex]\frac{0.002 mol Na2S2O3}{1000 ml solution}[/tex] x [tex]\frac{158 g}{1 mol Na2S2O3}[/tex] x 250 ml solution

mass= 0.079 g

To prepare the solution, we have to weigh in a beaker with 0.079 g of Na₂S₂O₃ by employing an analytical balance. Then, we have to dissolve the mass in a volume of aproximately 230 ml water. For this, we measure the 230 ml of water in a graduated cylinder, we add the volume to the beaker with the mass and we agitate until total disolution. Finally, we tranfer the total amount of the solution in the glass to a volumetric flask with a capacity of 250 ml. We add water until we reach the capacity, and then we homogenize the solution.

Answer :

[tex]Cl_2[/tex] is reduced species.

[tex]KI[/tex] is oxidized species.

[tex]Cl_2[/tex] is oxidizing agent.

[tex]KI[/tex] is reducing agent.

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Reducing agent : It is defined as the agent which helps the other substance to reduce and itself gets oxidized. Thus, it will undergo oxidation reaction.

Oxidizing agent : It is defined as the agent which helps the other substance to oxidize and itself gets reduced. Thus, it will undergo reduction reaction.

The balanced redox reaction is :

[tex]Cl_2(aq)+2KI(aq)\rightarrow 2KCl(aq)+I_2(aq)[/tex]

The half oxidation-reduction reactions are:

Oxidation reaction : [tex]2I^-\rightarrow I_2+2e^-[/tex]

Reduction reaction : [tex]Cl_2^++2e^-\rightarrow 2Cl^-[/tex]

From this we conclude that the [tex]'KI'[/tex] is the reducing agent that loses an electron to another chemical species in a redox chemical reaction and itself gets oxidized and [tex]'Cl_2'[/tex] is the oxidizing agent that gain an electron to another chemical species in a redox chemical reaction and itself gets reduced.

Thus, [tex]Cl_2[/tex] is reduced species.

[tex]KI[/tex] is oxidized species.

[tex]Cl_2[/tex] is oxidizing agent.

[tex]KI[/tex] is reducing agent.

Iodine from potassium iodide (KI) is oxidized, and chlorine from chlorine gas (Cl2) is reduced; Cl2 is the oxidizing agent and KI is the reducing agent in the reaction.

In the reaction Cl2(aq) + 2 KI(aq) → 2 KCl(aq) + I2(aq), the oxidized species is iodine (I) from potassium iodide (KI), and the reduced species is chlorine (Cl) from chlorine gas (Cl2). During the reaction, I- loses electrons and is therefore oxidized to I2; its oxidation number changes from -1 to 0. Conversely, each Cl atom in Cl2 gains one electron to become Cl-, going from an oxidation state of 0 to -1. Thus, the oxidizing agent is Cl2 because it accepts electrons (is reduced), and the reducing agent is KI because it donates electrons (is oxidized).

a.) Asp and Lys

Asp will elute first from the column because it has less positively charged functional groups than Lys.

b.) Arg and Met

Met will elute first from the column because it has less positively charged functional groups than Lys.

c.) Glu and Val

Glu will elute first from the column because it has more negativity functional groups than Lys and will be not be much retained by the -SO₃⁻ groups from the ion-exchange coloumn.

d.) Gly and Val

 Gly will elute first from the column because Lys have a longer alkyl chain which will be attracted by the strongly hydrophobic backbone for the resin.

e.) Ser and Ala

Ser will be eluted first from the column because Ala alkyl chain will be more attracted by the strongly hydrophobic backbone for the resin. Ser have an -OH group which will decrease the hydrophobicity of the alkyl chain and will not be so much retained on the column.

Answer:

C H2SO4 = 9.79 M

Explanation:

∴ %w/w H2SO4 = 960% = g H2SO4 / g sln * 100

⇒ 9.6 = g H2SO4 / g sln

calculation base: 1000 g sln

⇒ g H2SO4 = 9600g

⇒ mol H2SO4 = 9600 g H2SO4 * ( mol H2SO4/ 980g H2SO4 ) = 9.796  mol H2SO4

⇒ V sln = 1000g sln / 1000g/L = 1 L sln

∴ ρ H20 ≅ 1000 Kg/m³ = 1000 g/L

⇒ C H2SO4 = 9.796 mol H2SO4 / 1 L sln

⇒ C H2SO4 = 9.796 M

Answer:

[tex]3.446\times 10^{-2}[/tex]

Explanation:

Scientific notation is the way of writing numbers which are either large or small. The number is written in the scientific notation when the number is between 1 and 10 and then multiplied by the power of 10.

The given number:

0.03445750 can be written as [tex]3.445750\times 10^{-2}[/tex]

Answer upto 4 significant digits = [tex]3.446\times 10^{-2}[/tex]

The distance travelled by charlotte is 7269ft/mi.

Given:

Charlotte's speed = 70.4 mi/h.

Time she takes her eyes off the road = 4.54 seconds.

To calculate the distance Charlotte has traveled in feet while looking at her phone, it is important to convert her speed from miles per hour (mi/h) to feet per second (ft/s), and then use that speed to calculate the distance.

Convertion of speed to feet per second:

1 mile = 5280 feet

1 hour = 3600 seconds

Speed in ft = (70.4 (mi/h) × 5280(ft/mi)) / (3600(s/h))

Speed in ft = 103.25ft.

Calculation of distance:

Distance =Speed / Time

Distance = 103.25 * 70.4

Distance = 7269 ft/mi

Therefore, the distance travelled by charlotte is 7269ft/mi.

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Charlotte traveled approximately 469.761 feet while looking at her phone for 4.54 seconds by driving at a speed of 70.4 mi/h.

Charlotte is driving at 70.4 mi/h. To find out how far she has traveled in feet while looking at her phone for 4.54 seconds, we need to convert the speed from miles per hour to feet per second and then multiply by the time in seconds.

First, we convert the speed from miles per hour to feet per second using the conversion factors 1 mile = 5280 feet and 1 hour = 3600 seconds:

70.4 mi/h × 5280 ft/mi × 1/3600 h/s = 103.46667 ft/s.

Now, we multiply the speed in feet per second by the time she is distracted:

103.46667 ft/s × 4.54 s = 469.761 ft.

Charlotte has therefore traveled approximately 469.761 feet while looking down at her phone for 4.54 second

Answer:

1) 0,081 ft/s

2) 0,746 lb/s

Explanation:

The relation between flow and velocity of a fluid is given by:

Q=Av

where:

1)

To convert our data to appropiate units, we use the following convertion factors:

1 ft=12 inches

1 ft3=7,48 gallons

1 minute=60 seconds

So,

[tex]Q=\frac{3,60 gallons}{1 min}*\frac{1min}{60 s}*\frac{1ft3}{7,48gallons}=0,00802 \frac{ft3}{s}[/tex]

As the pipe has a circular section, we use A=πd^2/4:

[tex]d=4,26 inch *\frac{1ft}{12 inch}=0,355ft\\  A=\pi \frac{0,355^{2} }{4}=0,0989ft2[/tex]

Finally:

Q=vA......................v=Q/A

[tex]v=\frac{0,00802ft^{3} /s}{0,0989ft^{2} }=0,081ft/s[/tex]

2)

The following formula is used to calculate the specific gravity of a material:

SG = ρ / ρW  

where:

then:

ρ = SG*ρW   = 1,49* 62,4 lb/ft3 = 93 lb/ft3

To calculate the mass flow, we just use the density of the chloroform in lb/ft3 to relate mass and volume:

[tex]0,00802 \frac{ft3}{s}*\frac{93lb}{1ft3}=0,746lb/s[/tex]

The reduction of permanganate ion to manganese ion in acidic solution is represented by the balanced half-reaction: MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l). It shows the permanganate ion gaining 5 electrons to become a manganese(II) ion, with the protons reacting with permanganate to form water.

The reduction of the permanganate ion (MnO⁴⁻) to the manganese(II) ion (Mn²⁺) in acidic solution can be illustrated with the following balanced half-reaction:

MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l)

This equation indicates that the permanganate ion gains 5 electrons (is reduced) in an acidic solution to result in a manganese(II) ion. The 8 protons react with permanganate to form 4 molecules of water, which is also reflected in the equation.

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The reduction half-reaction for permanganate to manganese ion in an acidic solution is MnO4- + 8H+ + 5e- → Mn2+ + 4H2O(l), where oxygen and hydrogen are balanced by adding water and hydrogen ions, respectively, and the charge is balanced by adding electrons.

To write a balanced half-reaction for the reduction of the permanganate ion (MnO4-) to manganese ion (Mn2+) in an acidic aqueous solution, we start with the half-reaction:

MnO4- + 8H+ + 5e- → Mn2+ + 4H2O(l)

In this balanced equation, the oxygen atoms are balanced by adding water molecules (H2O) to the right side, and the hydrogen atoms are balanced by adding hydrogen ions (H+) to the left side. Additionally, 5 electrons (5e^-) are added to the left side to ensure that the overall charge is balanced between the reactant and product sides of the equation.

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