In the manufacture of paper, logs are cut into small chips, which are stirred into an alkaline solution that dissolves several of the chemical constituents of wood but not cellulose. The slurry of undissolved chips in solution is further processed to recover most of the original solution constituents and dried wood pulp. In one such process, wood chips with a specific gravity of 0.640 containing 45.0 wt% water are treated to produce 2000.0 tons/day of dry wood pulp containing 85.0 wt% cellulose. The wood chips contain 47.0 wt% cellulose on a dry basis. Estimate the feed rate of logs (logs/min), assuming that the logs have an average diameter of 8.00 inches and an average length of 9.00 feet. 21.67 Ulogs/min

Answer:

47 mL

Explanation:

The equilibrium of nitrous acid is:

HNO₂ ⇄ NO₂⁻ + H⁺ Ka = 4,5x10⁻⁴

If desire pH is 2,60 the [H⁺] concentration in equilibrium is:

[H⁺] = [tex]10^{-2,6}[/tex] =2,51x10⁻³ M

Initial molarity of sodium nitrite is:

43,8g × [tex]\frac{1mol}{68,9953 g}[/tex]÷ 2,5 = 0,254 M

Thus, in equilibrium the concentration of chemicals is:

[NO₂⁻] = 0,254 - x

[HNO₂] = x

[H⁺] =  2,51x10⁻³ = Y-x Where Y is initial concentration.

Equilibrium formula is:

4,5x10⁻⁴ =  [tex]\frac{[2,51x10^{-3}[0,254 - x] ]}{[x]}[/tex]

Solving, x = 0,215

Thus, initial [H⁺] concentration is:

0,215 + 2,51x10⁻³ = 0,2175 M

If total volume is 2,50 L:

2,50L ×[tex]\frac{0,2175 mol}{L}[/tex] = 0,5438 mol of HCl

As molarity of concentrated hydrochloric acid is 11,65 mol per liter:

0,5438 mol HCl ×[tex]\frac{1L}{11,65}[/tex] = 0,047 L ≡ 47 mL

I hope it helps!

Answer:

The mass percentage of bromine in the original compound is 81,12%

Explanation:

Step 1: Calculate moles AgBr

moles AgBr = mass AgBr / molar mass  AgBr

= 0.8878 g / 187.77 g/mol

= 0.00472812 moles AgBr

⇒

Since 1 mol AgBr contains 1 mol Br-

Then the amount of moles Br- in the original sample must also have  been 0.00472812 moles

Step 2: Calculating mass Br-

mass Br- = molar mass Br x moles  Br-

= 79.904 g/mol x 0.00472812 mol

= 0.377796 g Br-

⇒

There were 0.377796 g Br- in the original sample

Step 3: Calculating mass percentage Br-

⇒mass percentage  = actual mass Br- / total mass x 100%

% mass Br = 0.377796 g / 0.4657 g x 100  %

= 81.12%

Answer:

The mass of  Fe₂O₃ in 0.500 g of mixture is 0.367 g.

Explanation:

First off, we know that 72% of the mass of the mixture is iron. The information also tells us that the remaining 28% of the mass is oxygen.

Now we calculate the total mass of iron and the total mass of oxygen in the mixture:

With the mass of each element we can calculate the number of moles of each atom, using the atomic weight:

0.360 g Fe * 1 mol / 55.845 g = 0.00645 moles of Fe

0.140 g O * 1 mol / 16 g = 0.00875 moles of O

The number of moles of Fe in the mixture is equal to the number of moles of FeO plus two times the number of moles of Fe₂O₃:

0.00645 = [tex]2*n_{Fe2O3} +n_{FeO}[/tex]             eq A

The number of moles of O in the mixture is equal to the number of moles of FeO plus three times the number of moles of Fe₂O₃:

0.00875 = [tex]3*n_{Fe2O3} +n_{FeO}[/tex]               eq B

So now we have a system of two equations and two unknowns, we solve for [tex]n_{Fe2O3}[/tex]:

From eq A:

[tex]n_{FeO3}=0.00645-2*n_{Fe2O3}[/tex]

Replacing in  eq B:

[tex]0.00875=3*n_{Fe2O3} + (0.00645-2*n_{Fe2O3})\\0.00230=n_{Fe2O3}[/tex]

Now we just need to convert moles of Fe₂O₃ into grams, using the molecular weight:

0.00230 moles * 159.66 g/mol = 0.367 g Fe₂O₃

Answer:

0.0020 fraction of the pesticide remains in the environment after 18 days

Explanation:

For a first order reaction, [tex]\frac{N}{N_{0}}=(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}[/tex]

where N is remaining mass after "t" time , [tex]N_{0}[/tex] is initial mass,  [tex]\frac{N}{N_{0}}[/tex] represents fraction of mass remains after "t" time and [tex]t_{\frac{1}{2}}[/tex] is half-life

Here t is 18 days and [tex]t_{\frac{1}{2}}[/tex] is 2 days

So, [tex]\frac{N}{N_{0}}=(\frac{1}{2})^{\frac{18}{2}}=(\frac{1}{2})^{9}=0.0020[/tex]

Hence 0.0020 fraction of the pesticide remains in the environment after 18 days

The fraction (in decimal notation) of the pesticide that remains in the environment after 18 days is 0.0020

We'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:

Half-life (t½) = 2 days

Time (t) = 18 days

n = t / t½

n = 18 / 2

Thus, 9 half-lives has elapsed.

Original amount (N₀) = 1

Number of half-lives (n) = 9

N = 1/2ⁿ × N₀

Divide both side by N₀

N / N₀ = 1/2ⁿ

N / N₀ = 1/2⁹

Thus, the fraction of the pesticide that remains in the environment is 0.0020

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Answer:

(a) The plant generates 3,153,600,000 MJ a year.

(b) The problem with the question is that "power" is not "generated". What is generated is Energy, and Power is the rate of generation of Energy.

(c) The student is wrong because the plants generates 100 MJ of energy a second. "MW/hr" is not a unit of energy or power, so it has no sense.

Maybe he get confused with MW-h, which is a unit of energy, used to measure electrical consumption.

Explanation:

(a) The power of the plant (100 MW) is the rate at which electrical energy is produced. It has units of [energy]/[time].

In this case, the plant produces 100 MJ/s. The energy produced can also be expressed in other units, like MJh.

To calculate the energy generated in one year, we have

[tex]Energy = Power * Time  = \\\\Energy = 100 MW*1year\\\\Energy = (100 \frac{MJ}{s})*(1 year*\frac{365 days}{1year}*   \frac{24hours}{1day}*\frac{3600s}{1hour})\\\\Energy=(100 \frac{MJ}{s})*(31,536,000s)=3,153,600,000MJ[/tex]

The plant generates 3,153,600,000 MJ a year.

(b) The problem with the question is that "power" is not "generated". What is generated is Energy, and Power is the rate of generation of Energy.

(c) The student is wrong because the plants generates 100 MJ of energy a second. "MW/hr" is not a unit of energy or power, so it has no sense.

Maybe he get confused with MW-h, which is a unit of energy, used to measure electrical consumption. That unit represents the amount of energy consumed or generated by a 1 MW unit in one hour. It is equivalent to 3600 MJ (1 MW-h = 3600 MJ).

In this unit, the 100 MW plant generates 876,000 MWh.

Explanation:

Methane (CH₄)

Valence electrons of carbon = 4

Valence electrons of hydrogen = 1  

The total number of the valence electrons  = 4 + 4(1) = 8

The Lewis structure is drawn in such a way that the octet of each atom and duet for the hydrogen in the molecule is complete. So,  

The Lewis structure is:

                H

                 :

     H    :    C   :    H

                 :

                H

To draw the Lewis Structure for CH4, follow the steps to determine the total number of valence electrons, place the atoms and lone pairs, and draw the final structure. Carbon in CH4 forms 4 single bonds with Hydrogen, resulting in a tetrahedral arrangement of Hydrogen atoms around the Carbon atom.

The Lewis Structure for CH4, which represents the arrangement of electrons in the molecule, can be drawn using the following steps:

For CH4, Carbon has 4 valence electrons and Hydrogen has 1 valence electron each. The total number of valence electrons is 4 + (4 x 1) = 8. Following the steps, Carbon is surrounded by 4 Hydrogen atoms, and each Hydrogen atom shares its electron with Carbon. Carbon ends up with a complete octet by sharing electrons with the Hydrogen atoms, forming 4 single bonds. The final Lewis structure for CH4 is a tetrahedral arrangement of the Hydrogen atoms around the Carbon atom, represented by a molecular formula, structural formula, or other models.

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Answer:

18.075 mL of NaOH solution was added to achieve neutralization

Explanation:

First, let's formulate the chemical reaction between nitric acid and sodium hydroxide:

NaOH + HNO3 → NaNO3 + H2O

From this balanced equation we know that 1 mole of NaOH reacts with 1 mole of HNO3 to achieve neutralization. Let's calculate how many moles we have in the 25 mL aliquot to be titrated:

63.01 g of HNO3 ----- 1 mole

8.2 g of HNO3 ----- x = (8.2 g × 1 mole)/63.01 g = 0.13014 moles of HNO3

So far we added 8.2 grams of nitric acid (0.13014 moles) in 1 L of water.

1000 mL solution ---- 0.13014 moles of HNO3

25 mL (aliquot) ---- x = (25 mL× 0.13014 moles)/1000 mL = 0.0032535 moles

So, we now know that in the 25 mL aliquot to be titrated we have 0.0032535 moles of HNO3. As we stated before, 1 mole of NaOH will react with 1 mole of HNO3, hence 0.0032535 moles of HNO3 have to react with 0.0032535 moles of NaOH to achieve neutralization. Let's calculate then, in which volume of the given NaOH solution we have 0.0032535 moles:

0.18 moles of NaOH ----- 1000 mL Solution

0.0032535 moles---- x=(0.0032535moles×1000 mL)/0.18 moles = 18.075mL

As we can see, we need 18.075 mL of a 0.18 M NaOH solution to titrate a 25 mL aliquot of the prepared HNO3 solution.

Answer:

Rate = [tex]k[NO_{2}][F_{2}][/tex]

Explanation:

Answer: The molecular formula for the menthol is [tex]C_{10}H_{20}O[/tex]

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of [tex]CO_2=0.449g[/tex]

Mass of [tex]H_2O=0.184g[/tex]

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.449 g of carbon dioxide, [tex]\frac{12}{44}\times 0.449=0.122g[/tex] of carbon will be contained.

In 18g of water, 2 g of hydrogen is contained.

So, in 0.184 g of water, [tex]\frac{2}{18}\times 0.184=0.0204g[/tex] of hydrogen will be contained.

To formulate the empirical formula, we need to follow some steps:

Moles of Carbon = [tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.122g}{12g/mole}=0.0102moles[/tex]

Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.0204g}{1g/mole}=0.0204moles[/tex]

Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0171g}{16g/mole}=0.00107moles[/tex]

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00107 moles.

For Carbon = [tex]\frac{0.0102}{0.00107}=9.53\approx 10[/tex]

For Hydrogen  = [tex]\frac{0.0204}{0.00107}=19.54\approx 20[/tex]

For Oxygen  = [tex]\frac{0.00107}{0.00107}=1[/tex]

The ratio of C : H : O = 10 : 20 : 1

Hence, the empirical formula for the given compound is [tex]C_{10}H_{20}O_1=C_{10}H_{20}O[/tex]

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

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

We are given:

Mass of molecular formula = 156.3 g/mol

Mass of empirical formula = 156 g/mol

Putting values in above equation, we get:

[tex]n=\frac{156.3g/mol}{156g/mol}=1[/tex]

Multiplying this valency by the subscript of every element of empirical formula, we get:

[tex]C_{(1\times 10)}H_{(1\times 20)}O_{(1\times 1)}=C_{10}H_{20}O[/tex]

Thus, the molecular formula for the menthol is [tex]C_{10}H_{20}O[/tex]

Answer:

Explanation:

Hello, since the Gibbs' phase rule states the following equation:

[tex]F=C-P+2[/tex]

Whereas C is the number of components and P the present phases, you answers are:

1. F=1-2+2=1.

2. F=2-2+2=2.

3. F=2-2+2=2.

Best regards.

Answer:

691.29 K or 418.14 °C

Explanation:

Hello, at first the moles of oxygen gas are required:

[tex]n_{O_2}=1.75 g * \frac{1mol O_2}{32 g O_2} =0.0547 mol[/tex]

Now, based on the ideal gas equation, we solve for the temperature:

[tex]PV=nRT\\T=\frac{PV}{nR}\\T=\frac{1atm * 3.10 L}{0.0547 mol*0.082 \frac{atm*L}{mol*K} }\\T=691.29 K[/tex]

Best regards.

Answer:

For a: The empirical formula for the given compound is [tex]TiCl_4[/tex]

For b: The percent by mass of titanium and chlorine in the sample is 25.55 % and 74.78 % respectively.

Explanation:

We are given:

Mass of Titanium = 2.60 g

Mass of sample = 10.31 g

Mass of Chlorine = 10.31 - 2.60 = 7.71 g

To formulate the empirical formula, we need to follow some steps:

Moles of titanium =[tex]\frac{\text{Given mass of Titanium}}{\text{Molar mass of Titanium}}=\frac{2.60g}{47.867g/mole}=0.054moles[/tex]

Moles of Chlorine = [tex]\frac{\text{Given mass of Chlorine}}{\text{Molar mass of Chlorine}}=\frac{7.71g}{35.5g/mole}=0.217moles[/tex]

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.054 moles.

For Titanium = [tex]\frac{0.054}{0.054}=1[/tex]

For Chlorine  = [tex]\frac{0.217}{0.054}=4.01\approx 4[/tex]

Step 3: Taking the mole ratio as their subscripts.

The ratio of Ti : Cl = 1 : 4

Hence, the empirical formula for the given compound is [tex]TiCl_4[/tex]

To calculate the percentage by mass of substance in sample, we use the equation:

[tex]\%\text{ composition of substance}=\frac{\text{Mass of substance}}{\text{Mass of sample}}\times 100[/tex]       .......(1)

Mass of sample = 10.31 g

Mass of titanium = 2.60 g

Putting values in above equation, we get:

[tex]\%\text{ composition of titanium}=\frac{2.60g}{10.31g}\times 100=25.22\%[/tex]

Mass of sample = 10.31 g

Mass of chlorine = 7.71 g

Putting values in above equation, we get:

[tex]\%\text{ composition of chlorine}=\frac{7.71g}{10.31g}\times 100=74.78\%[/tex]

Hence, the percent by mass of titanium and chlorine in the sample is 25.55 % and 74.78 % respectively.

Answer:

[tex]2.4088\times 10^{13}[/tex] dollars each person will receive.

Explanation:

Number of people in which 1 mole of pennies is distributed = 250 million =

[tex]1 million = 10^6 [/tex]

250 million = [tex]2.5\times 10^8 [/tex] persons

Number of pennies in 1 mole = [tex]6.022\times 10^{23}[/tex]

Pennies per person:

[tex]\frac{6.022\times 10^{23} pennies}{2.5\times 10^8 persons}=2.4088\times 10^{15} pennies/person[/tex]

1 penny = 0.01 $

[tex]2.4088\times 10^{15} pennies/person=2.4088\times 10^{15}\times 0.01 \$/person=2.4088\times 10^{13} \$/person[/tex]

[tex]2.4088\times 10^{13}[/tex] dollars each person will receive.

Answer:

a) A = 7.186 E-4 ft²

b) t = 236.403 s

Explanation:

A = 6.676 E1 mm²

a) A = 6.676 E1 mm² * ( ft / 304.8 mm )²

⇒ A = 7.186 E-4 ft²

b) t = 51.7 s → A = 1.46 E1 mm²

⇒ relation (r) = 6.676 E1 mm² / 1.46 E1 mm² = 4.573

⇒ t = 51.7 s * 4.573 = 236.403 s

Answer: The volume of concentrated hydrochloric acid required is 16.53 mL

Explanation:

To calculate the volume of concentrated solution, we use the equation:

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of the concentrated solution

[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of diluted solution

We are given:

Conversion factor:  1 L = 1000 mL

[tex]M_1=12.1M\\V_1=?mL\\M_2=0.2M\\V_2=1L=1000mL[/tex]

Putting values in above equation, we get:

[tex]12.1\times V_1=0.2\times 1000\\\\V_1=16.53mL[/tex]

Hence, the volume of concentrated hydrochloric acid required is 16.53 mL

Answer:

If the substituents are the same, the cis distribution is more stable than trans distribution.

Explanation:

A cis cyclohexane is one in which both substituents are oriented towards the same face of the ring regardless of the conformation.

A trans cyclohexane has substituents on opposite sides of the ring.

In trans-1,4-disubstituted cyclohexane, the conformation with the two substituent groups in equatorial is more stable than the constitution that has the axial groups.

In cis-1,4-disubstituted cyclohexane, both conformations have the same stability as they have an axial and an equatorial substituent, so they have the same energy and there is no equilibrium shift.

Answer:

a) mass=35 kg

b) volumetric flow rate= 27.37 [tex]\frac{liters}{s}[/tex]

c) average mass flow rate=29.21 [tex]\frac{lbm}{min}[/tex]

Explanation:

Specific gravity is defined as the relation between the density of one substance and the density of another reference substance (it is usual that water is used in this case). In this case the specific gravity is the relationship between the density of gasoline and the density of water. So:

[tex]0.7=\frac{densitygasoline}{densitywater}[/tex]

Now you can know the density of gasoline  with a simple mathematical operation and knowing that densitywater≅ 1000 [tex]\frac{kg}{m^{3} }[/tex]:

densitygasoline=0.7*densitywater

densitygasoline=0.7*1 [tex]\frac{kg}{liters}[/tex]

densitygasoline=0.7 [tex]\frac{kg}{liters}[/tex]

a)

Now that the density is known, you can calculate the mass in 50 liters of gasoline.

By definition of density, you know that in 1 liter there are 700 kg of gasoline. So using The Rule of Three( tool that allows you to solve problems based on proportions), you can calculate the mass in 50 liters:

[tex]\frac{700 kg}{1 liter} =\frac{mass}{50 liters}[/tex]

[tex]mass=\frac{0.7 kg* 50 liters}{1 liter}[/tex]

mass=35 kg

b)

First, it is convenient to change units: from kg / min to kg / seconds. For that you know that 1 min= 60 seconds. So:

[tex]\frac{1150 kg}{min} =\frac{1150 kg}{60 s}[/tex]

[tex]\frac{1150 kg}{min} =\frac{19.16 kg}{s}[/tex]

By definition, density is the relationship between the mass and volume of a substance: This is: [tex]density=\frac{mass}{volume}[/tex]

Applied to this case, it would be [tex]density=\frac{mass flow rate}{volumetric flow rate}[/tex]

The mass flow rate and the density of the gasoline are known, so you can calculate the volumetric flow rate:

[tex]volumetric flow rate= \frac{mass flow rate}{density gasoline}[/tex]

volumetric flow rate= 27.37 [tex]\frac{liters}{s}[/tex]

c)

You know that the pump rate is 10 gallons per 2 minute. This is:

[tex]\frac{10 gallons}{2 minutes} = 5 \frac{gallons}{minute}[/tex]

To calculate the average mass flow, you must relate this data to density and do the necessary unit conversions.  Conversions are done similarly to what was previously applied in the previous exercises.

You have to know that:

So:

[tex]\frac{5 gallons*3.785 liters}{1gallons*min} =\frac{18.925 liters}{min}[/tex]

You know that the density is [tex]density=\frac{mass flow rate}{volumetric flow rate}[/tex]

The volumetric flow rate (pump rate) and the density of the gasoline are known, so you can calculate the average mass flow rate:

average mass flow rate=density*volumetric flow rate

[tex]average mass flow rate=\frac{0.7 kg}{liters} *\frac{18.925 liters}{min}[/tex]

average mass flow rate=13.2475 [tex]\frac{kg}{min}[/tex]

Converting the units:

[tex]average mass flow rate=\frac{13.2475 kg*2.205 lbm}{1 kg*min}[/tex]

average mass flow rate=29.21 [tex]\frac{lbm}{min}[/tex]

Answer: The mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon and hydrogen follows:

[tex]C_xH_y+O_2\rightarrow CO_2+H_2O[/tex]

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of [tex]CO_2=0.3986g[/tex]

Mass of [tex]H_2O=0.0578g[/tex]

Mass of sample = 0.1153 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.3986 g of carbon dioxide, [tex]\frac{12}{44}\times 0.3986=0.1087g[/tex] of carbon will be contained.

In 18g of water, 2 g of hydrogen is contained.

So, in 0.0578 g of water, [tex]\frac{2}{18}\times 0.0578=0.0066g[/tex] of hydrogen will be contained.

To calculate the percentage composition of a substance in sample, we use the equation:

[tex]\%\text{ composition of substance}=\frac{\text{Mass of substance}}{\text{Mass of sample}}\times 100[/tex]      ......(1)

Mass of sample = 0.1153 g

Mass of carbon = 0.1087 g

Putting values in equation 1, we get:

[tex]\%\text{ composition of carbon}=\frac{0.1087g}{0.1153g}\times 100=94.27\%[/tex]

Mass of sample = 0.1153 g

Mass of hydrogen = 0.0066 g

Putting values in equation 1, we get:

[tex]\%\text{ composition of hydrogen}=\frac{0.0066g}{0.1153g}\times 100=5.72\%[/tex]

Hence, the mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.

The masses of C and H in the sample are 0.1083 grams and 0.00324 grams, respectively. The percentages of C and H in the hydrocarbon are approximately 93.94% and 2.80%, respectively.

All the carbon in the hydrocarbon is converted to [tex]CO_2[/tex], we can use the mass of [tex]CO_2[/tex] to find the mass of carbon:

[tex]\[ \text{Mass of carbon} = \text{Mass of CO2} \times \frac{\text{Molar mass of carbon}}{\text{Molar mass of CO2}} \][/tex]

[tex]\[ \text{Mass of carbon} = 0.3986 \text{ g} \times \frac{12.01 \text{ g/mol}}{44.01 \text{ g/mol}} \][/tex]

[tex]\[ \text{Mass of carbon} = 0.3986 \text{ g} \times \frac{12}{44} \][/tex]

[tex]\[ \text{Mass of carbon} = 0.3986 \text{ g} \times 0.2727 \][/tex]

[tex]\[ \text{Mass of carbon} = 0.1083 \text{ g} \][/tex]

All the hydrogen in the hydrocarbon is converted to [tex]H_2O[/tex], we can use the mass of [tex]H_2O[/tex] to find the mass of hydrogen:

[tex]\[ \text{Mass of hydrogen} = \text{Mass of H2O} \times \frac{\text{Molar mass of hydrogen}}{\text{Molar mass of H2O}} \][/tex]

[tex]\[ \text{Mass of hydrogen} = 0.0578 \text{ g} \times \frac{1.008 \text{ g/mol}}{18.015 \text{ g/mol}} \][/tex]

[tex]\[ \text{Mass of hydrogen} = 0.0578 \text{ g} \times \frac{1}{18} \][/tex]

[tex]\[ \text{Mass of hydrogen} = 0.0578 \text{ g} \times 0.05597 \][/tex]

[tex]\[ \text{Mass of hydrogen} = 0.00324 \text{ g} \][/tex]

Now that we have the masses of carbon and hydrogen, we can calculate the percentages of these elements in the hydrocarbon:

[tex]\[ \text{Percentage of carbon} = \frac{\text{Mass of carbon}}{\text{Mass of hydrocarbon}} \times 100\% \][/tex]

[tex]\[ \text{Percentage of carbon} = \frac{0.1083 \text{ g}}{0.1153 \text{ g}} \times 100\% \][/tex]

[tex]\[ \text{Percentage of carbon} \approx 93.94\% \][/tex]

[tex]\[ \text{Percentage of hydrogen} = \frac{\text{Mass of hydrogen}}{\text{Mass of hydrocarbon}} \times 100\% \][/tex]

[tex]\[ \text{Percentage of hydrogen} = \frac{0.00324 \text{ g}}{0.1153 \text{ g}} \times 100\% \][/tex]

[tex]\[ \text{Percentage of hydrogen} \approx 2.80\% \][/tex]

Answer: An atom has either tendency of accepting or losing electron, on the basis of this virtue it is named as

For the existence of dipole within a molecule there must be a difference in electronegativity of the atoms participating in it. All the molecules here have dipole moment because the atoms participating in them have a difference in electronegative.

Answer:

131.4 mg/L of oxygen is needed to  biodegrade the organic compound.

Explanation:

The chemical reaction will be written as:

[tex]2C_7H_{12}ON_2+23O_2\rightarrow 14CO_2+12H_2O+4NO_2[/tex]

Concentration of the organic compound = 50 mg/L

This means that 50 milligrams of organic compound in present in 1 L of the solution.

50 mg = 0.050 g

1 mg = 0.001 g

Moles of organic compound = [tex]\frac{0.050 g}{140 g/mol}=0.0003571 mol[/tex]

According to reaction, 2 moles of organic compound reacts with 23 moles of oxygen gas.

Then 0.0003571 moles of an organic compound will react with:

[tex]\frac{23}{2}\times 0.0003571 mol=0.004107 mol[/tex] oxygen gas.

Mass of 0.004107 moles of oxygen gas:

0.004107 mol × 32 g/mol = 0.1314 g = 131.4 mg

131.4 mg/L of oxygen is needed to  biodegrade the organic compound.

Answer:

First of all, take account the molar mass of the CaOCl2 (Calcium hypochlorite),  which is 126.97 g/mol. If we have 126.97 g in a mol, 369g should be in aproximately 3 moles. Try to the think the rule of 3.

Explanation:

Answer: This world is rapidly moving towards to some serious enviornmental problems, the cause of this problems are mentioned below;

(1) The increase in average temperature of Earth.

(2) Deforestation

(3) Over Population

(4) Mining

Because of these reasons mentioned above there are many enviornmental issue occurring, and many problems we will face in future for example, water crisis, lack of resources.

Answer: The concentration of carbon dioxide in bottle of soft drink is [tex]6.6\times 10^{-2}mol/L[/tex]

Explanation:

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

[tex]C_{CO_2}=K_H\times p_{liquid}[/tex]

where,

[tex]K_H[/tex] = Henry's constant = [tex]0.033363mol/L.atm[/tex]

[tex]p_{CO_2}[/tex] = partial pressure of gas in a bottle of soft drink = 2 atm

Putting values in above equation, we get:

[tex]C_{CO_2}=0.033363mol/L.atm\times 2atm\\\\C_{CO_2}=0.066726mol/L=6.6\times 10^{-2}[/tex]

Hence, the concentration of carbon dioxide in bottle of soft drink is [tex]6.6\times 10^{-2}mol/L[/tex]

Answer: The metal having molar mass equal to 26.95 g/mol is Aluminium

Explanation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]     .....(1)

Molarity of NaOH solution = 0.5000 M

Volume of solution = 0.03340 L

Putting values in equation 1, we get:

[tex]0.5000M=\frac{\text{Moles of NaOH}}{0.03340L}\\\\\text{Moles of NaOH}=(0.5000mol/L\times 0.03340L)=0.01670mol[/tex]

[tex]2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O[/tex]

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, 0.01670 moles of NaOH will react with = [tex]\frac{1}{2}\times 0.01670=0.00835mol[/tex] of sulfuric acid

Excess moles of sulfuric acid = 0.00835 moles

Molarity of sulfuric acid solution = 0.5000 M

Volume of solution = 127.9 mL = 0.1279 L    (Conversion factor:  1 L = 1000 mL)

Putting values in equation 1, we get:

[tex]0.5000M=\frac{\text{Moles of }H_2SO_4}{0.1279L}\\\\\text{Moles of }H_2SO_4=(0.5000mol/L\times 0.1279L)=0.06395mol[/tex]

Number of moles of sulfuric acid reacted = 0.06395 - 0.00835 = 0.0556 moles

[tex]2X+3H_2SO_4\rightarrow X_2(SO_4)_3+3H_2[/tex]

By Stoichiometry of the reaction:

3 moles of sulfuric acid reacts with 2 moles of metal

So, 0.0556 moles of sulfuric acid will react with = [tex]\frac{2}{3}\times 0.0556=0.0371mol[/tex] of metal

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

Mass of metal = 1.00 g

Moles of metal = 0.0371 moles

Putting values in above equation, we get:

[tex]0.0371mol=\frac{1.00g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{1.00g}{0.0371mol}=26.95g/mol[/tex]

Hence, the metal having molar mass equal to 26.95 g/mol is Aluminium

Answer:

N2(g) + 3H2(g) → 2 NH3(g)

Explanation:

N2(g) + H2(g) → NH3(g)

We start equaling the number of N atoms in both sides multiplying by 2 the NH3.

N2(g) + H2(g) → 2 NH3(g)

So we equals the H atoms (there are six in products sites)

N2(g) + 3 H2(g) → 2 NH3(g)

Answer:

1 cubic mile = 1.101 * 10^12 US gallons

1 US bbl oil = 42 US gallons = 3.8143*10^ -11 cubic miles

Explanation:

The number of the exponent of the oil reserve is not very well shown in the question so, I provide you the conversion of bbl oil into cubic mile, the only thing you have to do is multiply the number of bbls of the reserve for the conversion in cubic miles and you'll have the answer.

Answer:

b. CH₂Cl₂ is more volatile than CH₂Br₂ because of the large dispersion forces in CH₂Br₂

Explanation:

CH₂Cl₂ is more volatile than CH₂Br₂ (b.p of CH₂Cl₂ = 39,6 °C; b.p of CH₂Br₂ = 96,95°C). Thus, c. and d. are FALSE

Dipole-dipole interactions in CH₂Cl₂ are greater than the dipole-dipole interactions in CH₂Br₂ because Cl is more electronegative that Br (Cl = 3,16; Br = 2,96). But this mean CH₂Cl₂ is less volatile than CH₂Br₂ but it is false.

There are large dispersion forces in CH₂Br₂ because Br has more electrons and protons than Cl. Large disperson forces mean CH₂Br₂ is less volatile than CH₂Cl₂ and it is true.

I hope it helps!

Answer:

d. 8

Explanation:

Valence electrons of oxygen = 6

Valence electrons of hydrogen = 1  

The total number of the valence electrons  = 6 + 2(1) = 8

The Lewis structure is drawn in such a way that the octet of each atom and duet for the hydrogen in the molecule is complete.

In the water molecule,

Oxygen has 2 bond pairs and two lone pairs which means that the total electrons in the valence shell is 8.

Answer:

6,666.66 micro liter of water and protein stock we will need to add to obtain the target concentration and volume.

Explanation:

Concentration of given solution[tex]C_1 = 0.15 mg/mL[/tex]

1 mL = 1000 μL , 1 mg = 1000 μg

[tex]C_1=0.15 mg/mL=\frac{0.15\times 1000 \mu g}{1\times 1000 \mu L}=0.15 \mu g/\mu L[/tex]

The volume of the given solution =[tex]V_1= V[/tex]

Concentration of required solution = [tex]C_2=10 \mu g/\mu L[/tex]

Volume of required solution = [tex]V_2=100 \mu L[/tex]

[tex]C_1V_1=C_2V_2[/tex]

[tex]V=\frac{C_2V_2}{C_1}=\frac{10 \mu g/\mu L\times 100 \mu L}{0.15 \mu g/\mu L}=6,666.66 \mu L[/tex]

6,666.66 micro liter of water and protein stock we will need to add to obtain the target concentration and volume.

Answer:

a) 0.1111 : 4 significant figures

b) 2000 : 1 significant figure

c) 35.6 : 3 significant figures

d) 180,701 : 6 significant digits

Explanation:

Significant figures or digits of a number are the digits that carry meaning and contribute to the precision of the number.

a) 0.1111 : 4 significant figures, leading zeros are not significant digits.

b) 2000 : 1 significant figure, trailing zeros are not significant.

c) 35.6 : 3 significant figures, all digits are significant.

d) 180,701 : 6 significant digits, zeros between mom-zero digits are significant.

The number of significant figures in the given examples are:

a) 4 significant figures for 0.1111

b) 1 significant figure for 2000 (assuming no decimal)

c) 3 significant figures for 35.6  

d) 6 significant figures for 180,701

To determine the number of significant figures in a number, certain rules are followed:

Applying these rules, we have: