A 0.0447−mol sample of a nutrient substance, with a formula weight of 114 g/mol, is burned in a bomb calorimeter containing 6.19 × 102 g H2O. Given that the fuel value is 6.13 × 10−1 in nutritional Cal when the temperature of the water is increased by 5.05°C, what is the fuel value in kJ in scientific notation?

Answers

Answer 1

Answer:

The value is  [tex]x = 2.565 *10^{3} \ kJ/kg[/tex]

Explanation:

From the question we are told that

  The  no of moles of the sample is  n = 0.0447 mole

  The formula weight is  [tex]M = 114 \ g/mol[/tex]

   The  mass of water is  [tex]m = 6.19 *10^{2}\ g[/tex]

   The amount of the fuel is  [tex]f= 6.13*10^{-1} \ nutritional \ Cal[/tex]

   The temperature rise is  [tex]\Delta T = 5.05^o[/tex]

Generally

      [tex]1 \ nutritional \ Cal => 4.184*10^{3} \ kJ/kg[/tex]

=>  [tex]f= 6.13*10^{-1} \ nutritional \ Cal \to x[/tex]

=>     [tex]x = \frac{6.13 *10^{-1} * 4.184 *10^{3}}{1}[/tex]

=>     [tex]x = 2.565 *10^{3} \ kJ/kg[/tex]

   


Related Questions

The value of the equilibrium constant for a reaction is 2.65 x 10-6 at 35° Calculate the value of LaTeX: \DeltaΔG°rxn.

Answers

Answer:

The answer is "33.95 [tex]\bold{ \frac{kj}{mol}}\\[/tex]".

Explanation:

Formula:

[tex]\bigtriangleup G_0= -R \times T l_n \times K_{eq}\\[/tex]

where

[tex]K_{eq} = \text{equilibrium constant}\\[/tex]

Given value:

[tex]T =35^{\circ} C\\[/tex]

convert temperature celsius (°C) to Kelvin (K):

[tex]= (273+45) \ kelvin \\\\= 318 \ Kelvin \\\\= 318 \ K[/tex]

[tex]R = 8.314 \ \ \frac{J}{ mol \cdot K}[/tex]

[tex]\bigtriangleup G_0= -(8.314 ) \times 31.8 \times l_n (2.65\times 10^{-6})\\[/tex]

        [tex]=-(8.314 ) \times 31.8T l_n \times (2.65\times 10^{-6})\\\\[/tex]

After solving the value it will give:

        = 33.95  [tex]\bold{ \frac{kj}{mol}}\\[/tex]

PLEASE HELP WILL GIVE BRAINLIEST!!!!
Xenon-133 has a half life of 5.2 days. A 237 gram sample of Xenon is collected and stored. It is later measured to be 29.625 grams. How long has it been in storage?
A. 15.6 days
B. 39.9 days
C. 20.8 days
D. 8 days

Answers

Answer:

A. 15.6 days

Explanation:

We have a general formula used to solve for questions dealing with half life.

N(t) = No (1/2)^t/t½

Where

N(t) = Quantity of sample left after t days

No = Initial amount of sample

t½ = Half life of sample

t = Duration or time required for sample to decay

In the above question, we are asked to determine how long(time) that the sample has been in storage.

The formula for time (t) has been derived as:

t =[ t½ × In(Nt/No)] ÷ - In 2

Where

No = 237 grams

N(t) = 29.625 grams

t½ = 5.2 days

t = ????? Unknown

t = [ t½ × In(Nt/No)] ÷ - In 2

t = [ 5.2 × In(29.625/237)] ÷ - In 2

t = [5.2 × -2.0794415417] ÷ - In 2

t = -10.813096017 ÷ - ln 2

t = 15.6 days

Therefore, the sample has been in storage for 15.6 days.

A mixture containing only FeCl3 and AlClz weighs 5.95 g . The chlorides are converted to the hydrous oxides and ignited to Fe2O3 and Al2O3 . The oxide mixture weighs 2.62 g . Calculate the percent Fe and Al in the original mixture .​

Answers

Answer:

The answer is "Al= 9.8% and Fe=18.0%"

Explanation:

Given:

The weight of [tex]FeCl_3\ and \ ALCl_3[/tex] = 5.95g

[tex]gFeCl_3=gFe (\frac{Mw\ FeCL_3}{\text{atomic weight of Fe}})\\\\gAlCl_3=gAl (\frac{Mw\ AlCL_3}{\text{atomic weight of Al}})\\\\[/tex]

[tex]\to a = FeCl_3+AlCl_3\\\\\to a=x+y \\\\ \to a= 5.95[/tex]

[tex]\to a =x \ gFe (\frac{Mw\ FeCL_3}{\text{atomic weight of Fe}})+ y \ gAl (\frac{Mw\ AlCL_3}{\text{atomic weight of Al}})\\\\[/tex]

[tex]\to x (\frac{162.2}{55.85})+ y (\frac{133.34}{26.98})= 5.95\\\\\to 2.90x+4.94y=5.95\\\\\ similarly \ for \ oxidies:\\\\\to 143x+1.89y=2.62\\\\\to x= 1.07 \ \ \ and \ \ y= 0.58\\\\\to \ Al \% = \frac{0.58}{5.95} \times 100= \bold{9.8} \%\\\\\to \ Fe \% = \frac{1.07}{5.95} \times 100= \bold{18.0} \%[/tex]

The mass of the compound has been given as the mass of each element in the compound. The percent Fe in the mixture is 18%, and the percent Al in the mixture is 9.8%.

What is the percent composition?

The percent composition has been given as the mass of the element in the compound. The percent of iron and aluminum in their chlorides can be given as:

[tex]\rm Fe=Fe\;\times\;\dfrac{Mwt\;FeCl_3}{Mwt\;Fe}[/tex]

[tex]\rm Al=Al\;\times\;\dfrac{Mwt\;AlCl_3}{Mwt\;Al}[/tex]

The total mass of the compounds has been 5.95 grams.

Thus,

[tex]5.95=\rm FeCl_3+AlCl_3[/tex]

The mass of Fe and Al will be x and y of the mass in the compounds. Thus,

[tex]5.95=\rm x\;Fe\;\times\;\dfrac{Mwt\;FeCl_3}{Mwt\;Fe}\;+\;y\;Al\;\times\; \dfrac{Mwt\;AlCl_3}{Mwt\;Al}[/tex]

The oxides of the chemical has the sum of mass of 2.62 grams.

The mass of oxides can be given as:

[tex]2.62=\rm x\;Fe\;\times\;\dfrac{Mwt\;Fe2O_3}{Mwt\;Fe}\;+\;y\;Al\;\times\; \dfrac{Mwt\;Al_2O_3}{Mwt\;Al}[/tex]

Substituting the values of mass of Al, aluminum chloride, oxide, and Fe and its compounds.

[tex]\rm 2.90x+4.94y=5.95\\143x+1.89y=2.62[/tex]

Solving the equation, x = 1.07, and y = 0.58

The percent mass of the compounds has been given as the ratio of mass with compound mass.

Thus, the percent mass of Fe has been:

[tex]\rm Fe=\dfrac{1.07}{5.95}\;\times\;100\\ Fe=18\;\%[/tex]

The percent mass of Al has been:

[tex]\rm Al=\dfrac{0.58}{5.95}\;\times\;100\\ Al=9.8\;\%[/tex]

Learn more about percent mass, here:

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5
Select the correct answer.
Which of the following is a physical model of the Sun?
A.
an equation that describes the Sun's motion
B.
a chart that lists the temperatures of different parts of the Sun
O c.
a computer program that shows how the Sun changes over time
OD
a paragraph that describes the Sun's structure
O E.
a small yellow ball that represents the Sun
Reset
Next

Answers

Answer:

E. a small yellow ball that represents the Sun

Explanation:

key word on the question, 'physical'

Students in a science class placed ice cubes in a cup. They were studying variables that affected how long it takes the ice to melt. Select the variable that
would most likely NOT affect the time it takes the ice to melt.
O Size of cup
O Size of ice cube
O Number of students
O Temperature of the room

Answers

Answer:

Number of students-this has nothing to do with the question, its called a distractor.

The variable that would most likely not affect the time it takes the ice to melt is the Number of students. The correct option is C.

What are the variables in the experiments?

Variables are the different substances or objects that participate in the experiment directly or indirectly. They are substances that together form the experiment. These variables impact the result of the experiment.

There are two types of variables. They are Dependent and independent variables. Dependent variables can be controlled in the experiment, but independent variables can't be controlled.

Students are performing the experiment, so they cannot be the variables because their presence or absence would be to impact the experiment.

Thus, the correct option is C. Number of students.

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Monomer liquid and polymer powder nail enhancement are service

Answers

Answer:

Monomer liquid and polymer powder nail enhancement are lucrative service

Which of the following molecules has a characteristic broad stretch at 3300 cm-1?
A) (CH3)2CHCH2OH
B) (CH3)3CH
C) (CH3)2CHCCCH3
D) (CH3)2CHCH=CH2
E) (CH3)2CHCO2CH3

Answers

Answer:

The answer is "Option A"

Explanation:

At 3300 cm-1, the wide region is equal to OH. It is a possible group of alcoholic features that includes drugs. In the given choices option a is correct that's the structure can be defined in attachment file please find it, and the wrong choice can be defined as follows:

In choice B, It is wrong because it doesn't include OH.  In choice C, it uses the triple bond, that's why it is wrong. In choice D, it uses the two bonds, that's why it is wrong. In choice E, it uses the characteristic bond, that's why it is wrong.

Acetyl chloride undergoes nucleophilic substitution at a faster rate than methyl acetate because:_________
A) the ester is more sterically hindered than the acid chloride.
B) the acid chloride is more sterically hindered than the ester.
C) the methoxide is a better leaving group than chloride.
D) esters hydrolyze faster than acid chlorides.
E) chloride is a better leaving group than methoxide.

Answers

I’m pretty sure the correct answer is E

Give the effect on the melting point of the presence of a cis double bond in a fatty acid.

Answers

Answer:

The cis double bond present in unsaturated fatty acids acids results in lower melting point when compared to saturated fatty acids of the same chain length.

Explanation:

Melting point of a fatty acids are affected by the length and degree of unsaturation of the hydrocarbon chain.

At room temperature, saturated fatty acids with hydrocarbon chain lengths between 12-24 are waxy solids whereas unsaturated atty acids of the same chain length are liquids. This is due to the nature of the packing of the fatty acid molecules in the saturated and unsaturated compounds.

In the saturated compounds, the molecules are tightly packed side by side with minimal steric hindrance and maximal van der Waals forces of attraction between molecules. However, in unsaturated fatty acids, the cis double bond introduces a bend or kink in the molecules which then interferes with the tight packing of the molecules and reducing interaction between molecules. Therefore, less energy is required to cause a disorder in the arrangement of unsaturated fatty acids, leading to a lowering of melting point.  

Select the conjugate acid-base pair(s). a) HI, I b) HCHO2, SO4^2- c) CO3^2-, HCI d) PO4^3-, HPO4^2-

Answers

Answer:

d) PO4^3-, HPO4^2-

Explanation:

Basically, an acid and a base which differs only by the presence or absence of  proton (hydrogen ion) are called a conjugate acid-base pair.

a) HI, I

This is incorrect. For the acid, HI the conjugate base is I⁻ ion.

b) HCHO2, SO4^2-

This is incorrect, there's no relationship between both entities.

c) CO3^2-, HCI

This is incorrect, there's no relationship between both entities.

d) PO4^3-, HPO4^2-

This is correct. The difference between both entities is the Hydrogen ion. This is the conjugate acid-base pair

Which measurement is best supported in units of microliters ?
The mass of an atom of uranium
The volume of a water droplet in a cloud
The thickness of a strand of hair
The volume of the Atlantic Ocean

Answers

Answer:the volume of a water droplet in a cloud

Explanation:

Answer:

The volume of a water droplet in a cloud

Explanation:

just took the test

HCl(aq) + NaOH(aq) ––––> NaCl(aq) + H2O(l) What volume of 0.631 M HCl is required to react with 15.8 mL of 0.321 M NaOH?

Answers

Answer:

The correct answer is 8.04 mL

Explanation:

Given the neutralization reaction:

HCl(aq) + NaOH(aq) ––––> NaCl(aq) + H₂O(l)

1 mol of HCl reacts with 1 mol of NaOH.

The number of moles of NaOH there is in 15.8 mL is calculated as follows:

0.321 mol/L x 1 L/1000 mL x 15.8 mL = 5.07 x 10⁻³ mol NaOH

Thus, we need 5.07 x 10⁻³ mol of HCl to react with 5.07 x 10⁻³ mol NaOH. We have a 0.631 M HCl solution. We calculate the volume of HCl we need by considering that 1 mol NaOH reacts with 1 mol HCl and that there are 0.631 moles of HCl in 1 liter of solution (1000 mL):

5.07 x 10⁻³ mol NaOH x 1 mol HCl/1 mol NaOH x 1000 mL/0.631 mol HCl = 8.04 mL

1. Aluminum forms a layer of aluminum oxide when exposed to air which protects the bulk metal from further corrosion.
4Al(s) + 3O2(g) ? 2Al2O3(s)
Using the thermodynamic data provided below, calculate S° for this reaction.
S°(J/K.mol)
Al(s) 28.3
O2(g) 205.0
Al2O3(s) 50.99
2. Aluminum forms a layer of aluminum oxide when exposed to air which protects the bulk metal from further corrosion.
4Al(s) + 3O2(g) ? Al2O3(s)
Calculate G° for this reaction, given that ?G°f of aluminum oxide is –1576.4 kJ/mol.

Answers

Answer:

Al(s) 28.3

Explanation:

Aluminum forms a layer of aluminum oxide when exposed to air which protects the bulk metal from further corrosion.

Aluminum forms a layer of aluminum oxide when exposed to air which protects the bulk metal from further corrosion.4Al(s) + 3O2(g) ? Al2O3(s)

AnswerAluminum forms a layer of aluminum oxide when exposed to air which protects the bulk metal from further corrosion.

4Al(s) + 3O2(g) ? 2Al2O3(s)

Using the thermodynamic data provided below, calculate S° for this reaction.

S°(J/K.mol)

Al(s) 28.3

O2(g) 205.0

Al2O3(s) 50.99

2. Aluminum forms a layer of aluminum oxide when exposed to air which protects the bulk metal from further corrosion.

4Al(s) + 3O2(g) ? Al2O3(s)

Calculate G° for this reaction, given that ?G°f of aluminum oxide is –1576.4 kJ/mol.

Answer:

Explanation:

Explanation:

0.200 grams of HCl is dissolved in 0.801 grams water. The density of the solution formed is 1.10 g/mL. What is the molarity of the solution

Answers

Answer:

[tex]M=6.03M[/tex]

Explanation:

Hello,

In this case, since the molarity is computed by:

[tex]M=\frac{n_{solute}}{V_{solution}}[/tex]

Whereas the solute is the hydrochloric acid, we compute the corresponding moles with its molar mass (36.45 g/mol):

[tex]n_{solute}=0.200gHCl*\frac{1molHCl}{36.45gHCl} =0.00549molHCl[/tex]

Next, since the solution contains both HCl and water, we compute the volume in liters by using its density:

[tex]V_{solution}=(0.200+0.801)g*\frac{1mL}{1.10g} *\frac{1L}{1000mL} =9.1x10^{-4}L[/tex]

Therefore, the molarity turns out:

[tex]M=\frac{0.00549mol}{9.1x10^{-4}L}\\ \\M=6.03M[/tex]

Regards.

A student is designing an experiment about the factors that affect math
grades. Which statement best represents a possible hypothesis for this
experiment?
O A. I think practicing math problems is the most important factor in a
student's math grade.
OB. If students spend at least 1 hour per night studying math, then
they will get a grade of 85 or higher.
C. Students who study more get higher math grades.
O D. Steve studied math for 1 hour every night and got a grade of 87.

Answers

Answer:

B. If students spend at least 1 hour per night studying math, then  they will get a grade of 85 or higher.

Explanation:

A hypothesis is defined as a tentative or predictions or an educated guess of scientific questions. A hypothesis is always done before performing an experiment.

The correct hypothesis for the given experiment will be "If students spend at least 1 hour per night studying math, then  they will get a grade of 85 or higher" because it has an if-then statement which is a conditional statement where statement after "if represents hypothesis" and "then represents conclusion".

Hence, the correct answer is "B. If students spend at least 1 hour per night studying math, then  they will get a grade of 85 or higher."

Dinitrogen tetroxide and hydrazine (N2H4) undergo a redox reaction in which nitrogen and water are formed as products. What mass of nitrogen could be produced when 50.0 g of dinitrogen tetroxide and 45.0 g of hydrazine are combined?

Answers

Answer:

The mass of nitrogen molecule [tex]N_2[/tex] = 45.65 g

Explanation:

The equation for the redox reaction can be represented as follows:

[tex]\mathtt{2N_2H_4 +N_2O_4 \ \to \ 3N_2 + 4H_2O}[/tex]

We know that:

numbers of moles = mass/molar mass

For [tex]\mathtt{N_2O_4}[/tex] :

number of moles = 50g/92 g/mol

number of moles = 0.5435 mol

For [tex]\mathtt{N_2H_4}[/tex] :

number of moles = 45 g/ 32 g/mol

number of moles = 1.40625 mol

From the above equation;

number of moles of [tex]\mathtt{N_2O_4}[/tex] needed = 1/2 moles of [tex]\mathtt{N_2H_4}[/tex]  = 1/2 ×  1.40625 mol

= 0.703125 mol

The amount of [tex]\mathtt{N_2O_4}[/tex] present = 0.5435 moles which is less than the needed. As such [tex]\mathtt{N_2O_4}[/tex] is the limiting reagent

The number of moles of nitrogen molecule [tex]N_2[/tex] produced = 3 × ([tex]\mathtt{N_2O_4}[/tex])

= 3 × 0.5435

= 1.6305  mol

The mass of nitrogen molecule [tex]N_2[/tex] = number of moles of  [tex]N_2[/tex]  ×  molar mass of [tex]N_2[/tex]

The mass of nitrogen molecule [tex]N_2[/tex] =  1.6305  mol × 28 g/mol

The mass of nitrogen molecule [tex]N_2[/tex] = 45.654 g

The mass of nitrogen molecule [tex]N_2[/tex] = 45.65 g

Predict the reactivity of silicon in water relative to that of sodium, magnesium, and aluminium. Explain your answer. How does the reactivity of the halogens vary within its own group from top to bottom?

Answers

Answer:

See explanation

Explanation:

The reactivity of metals has a lot to do with their position in the electrochemical series. However, it is also known that metallic character decreases across the period. This implies that as we move from left to right along the periodic table. Sodium, magnesium, aluminum and silicon continues to decrease in metallic character. As a matter of fact, silicon is a metalloid and not a pure metal.

Sodium reacts with cold water to give a vigorous reaction,magnesium and aluminium reacts with steam at red heat.

Silicon does not react with water, even as steam, under normal conditions.

Reactivity can be given as the ability of the atom to lose or gain electrons. The reactivity can be silicon is least as compared to sodium and other metals in water.

What is the trend of reactivity in the periodic table?

The periodic table has the arrangement of the elements in groups and periods in the table.

On moving from left to right in the periodic table, the number of electrons in the same shell increase, thereby making the loss of electrons difficult. Thus, on moving from left to right in the periodic table, the metal reactivity decreases.

Hence, silicon present in the right most of the periodic table has the lower reactivity as compared to the sodium, magnesium, and aluminum on the left side of the periodic table.

In the periodic table, on moving from top to bottom, the number of shells increases, decreasing the force of attraction to the outermost electron. Thus, the metal reactivity increases down the group.

Learn more about reactivity, here:

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Calculate the mass percent of HOCH 2CH 2OH in a solution made by dissolving 3.2 g of HOCH 2CH 2OH in 43.5g of water.

Answers

Answer:

[tex]\%m/m=6.85\%[/tex]

Explanation:

Hello,

In this case, we are asked to compute the by mass percent for the given 3.2 g of ethylene glycol in 43.5 g of water. In such a way, since the by mass percent is computed as follows:

[tex]\%m/m=\frac{m_{solute}}{m_{solute}+m_{solvent}} *100\%[/tex]

Whereas the solute is the ethylene glycol and the solvent the water, therefore we obtain:

[tex]\%m/m=\frac{3.2g}{3.2g+43.5g} *100\%\\\\\%m/m=6.85\%[/tex]

Best regards.

Which of the following is included in nuclide symbols, but is not strictly necessary for the identification of the nuclide?
a. mass number
b. atomic number
c. isotope number
d. none of the above

Answers

Answer:

c. isotope number

Explanation:

Mass Number is the sum total of mass of protons and neutrons present in the nucleus of an atom.  Generally they are being used in distinguishing isotopes. E.g Carbon - 12, Carbon - 13

Atomic Number is the number of protons. Every single element has it's unique atomic number and can be used in identification purpose. E.g Carbon - 6, Hydrogen - 1.

The correct option is option C. This is the symbol that is not necessary for the identification of a nuclide.

Answer: B, atomic number

Which has the greatest mass?a. one atom of carbon, b. one atom of hydrogenc. one atom of lithium

Answers

Answer:

one atom of carbon

Explanation:

The relative atomic mass of elements can be used to determine their relative masses. If we look at all the options, we will notice that carbon has a relative atomic mass of 12, hydrogen has a relative atomic mass of 1 and lithium has a relative atomic mass of 7

We can now see that carbon has the greatest relative atomic mass among the options provided in the question.

Calculate the solubility of PbF2 in water at . You'll find data in the ALEKS Data tab. Round your answer to significant digits.

Answers

Answer:

0.5 g/L.

Explanation:

Hello,

In this case, for this solubility problem, we can write for the lead (II) fluoride:

[tex]PbF_2(s)\rightleftharpoons Pb^{2+}(aq)+2F^-(aq)[/tex]

And the equilibrium expression is:

[tex]Ksp=[Pb^{2+}][F^-]^2[/tex]

Whereas Ksp of lead (II) fluoride is 3.3x10⁻⁸. In such a way, we can write the equilibrium expression in terms of the molar solubility [tex]x[/tex] as follows:

[tex]Ksp=(x)(2x)^2=3.3x10^{-8}[/tex]

Hence, solving for [tex]x[/tex] we find:

[tex]x=\sqrt[3]{\frac{3.3x10^{-8}}{4} }\\\\x=2.02x10^{-3}M[/tex]

Moreover, since the molar mass of lead (II) fluoride is 245.2 g/mol, the solubility turns out:

[tex]2.02x10^{-3}\frac{molPbF_2}{L}*\frac{245.2gPbF_2}{1molPbF_2}\\ \\0.5\frac{g}{L}[/tex]

Best regards.

Calculate the pH of a 0.10 M HCN solution that is 0.0070% ionized.
A) 1.00
B) 0.00070
C) 3.15
D) 5.15
E) 7.00

Answers

Answer:

D) 5.15

Explanation:

Step 1: Write the equation for the dissociation of HCN

HCN(aq) ⇄ H⁺(aq) + CN⁻(aq)

Step 2: Calculate [H⁺] at equilibrium

The percent of ionization (α%) is equal to the concentration of one ion at the equilibrium divided by the initial concentration of the acid times 100%.

α% = [H⁺]eq / [HCN]₀ × 100%

[H⁺]eq = α%/100% × [HCN]₀

[H⁺]eq = 0.0070%/100% × 0.10 M

[H⁺]eq = 7.0 × 10⁻⁶ M

Step 3: Calculate the pH

pH = -log [H⁺] = -log 7.0 × 10⁻⁶ = 5.15

The pH of the 0.10 M HCN solution that is 0.0070% ionized is 5.15 and the correct option is option D.

pH is a measure of the acidity or alkalinity of a solution. It is a logarithmic scale that indicates the concentration of hydrogen ions (H+) in the solution.

The pH scale ranges from 0 to 14, where a pH of 7 is considered neutral, pH values below 7 indicate acidity, and pH values above 7 indicate alkalinity.

The pH of a solution can be calculated using the formula:

pH = -log[H+]

Where [H+] represents the concentration of hydrogen ions in moles per liter.

Since HCN is a weak acid and only 0.0070% of it is ionized, assuming that the concentration of H+ ions is equal to 0.0070% of 0.10 M HCN.

0.0070% of 0.10 M HCN = (0.0070/100) × 0.10

= 0.000007 M

pH = -log(0.000007)

= 5.15

Thus, the ideal selection is option D.

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A rigid tank contains 1.80 moles of nitrogen, which can be treated as an ideal gas, at a pressure of 25.5 atm. While the tank and gas maintain a constant volume and temperature, a number of moles are removed from the tank, reducing the pressure to 5.70 atm. How many moles are removed

Answers

Answer:

1.4 moles.

Explanation:

The following data were obtained from the question:

Initial mole (n1) = 1.8 moles

Initial pressure (P1) = 25.5 atm

Final pressure (P2) = 5.7 atm

Number of mole removed =?

Next, we shall obtain an expression relating pressure and number of mole together.

This is illustrated below:

From the ideal gas equation:

PV = nRT

Divide both side by V

P = nRT /V

Divide both side by n

P/n = RT/V

But volume (V) and temperature (T) are constant.

P/n = constant

P1/n1 = P2/n2

P1 is initial pressure.

P2 is final pressure.

n1 is the initial mole.

n2 is the final mole.

Next, we shall determine the final mole in tank as follow:

Initial mole (n1) = 1.8 moles

Initial pressure (P1) = 25.5 atm

Final pressure (P2) = 5.7 atm

Final mole (n2) =.?

P1/n1 = P2/n2

25.5/1.8 = 5.7/n2

Cross multiply

25.5 x n2 = 1.8 x 5.7

Divide both side by 25.5

n2 = (1.8 x 5.7) /25.5

n2 = 0.40 mole.

Therefore, the final mole in the tank is 0.40 mole.

Finally, we shall determine the number of mole that was removed as follow:

Initial mole (n1) = 1.8 moles

Final mole (n2) = 0.4 mole

Number of mole removed =?

The number of mole removed from the tank can be obtained by finding the difference between the initial mole and final mole as illustrated below:

Number of mole removed = initial mole – final mole

Number of mole removed = n1 – n2

Number of mole removed = 1.8 – 0.4

Number of mole removed = 1.4 moles

Therefore, 1.4 moles were removed from the tank.

In the measurement 0.502 L, which digit is the estimated digit? A. 5 B. 3 C. The 0 to the left of the decimal point D. The 0 immediately to the left of the 3

Answers

Answer:

no c

Explanation:

the distance between atomes is sometimes given in picometers,where 1pm is equivalent to 1×10^12m if the distance between the carbon atomes in diamond is 2.81×10^8,what is the distance in picometers?​

Answers

Answer:

Distance between the carbon atom = 491 pm

Explanation:

Given:

1 pm = 1 × 10⁻¹² m

Distance between the carbon atom = 2.81 × 10⁻⁸ cm

Find:

Distance in picometers

Computation:

Distance between the carbon atom = 2.81 × 10⁻⁸ cm = 491 X 10⁻¹² m

Distance between the carbon atom = 491 pm

Write the condensed nd structural formulas as well as the names for all isomers of C3H5Cl3.

Answers

Answer:

1) 1,1,1-trichloropropane

2) 1,1,2-trichloropropane

3) 1,2,2-trichloropropane

4) 1,2,3-trichloropropane

Explanation:

For this question, we must remember that isomers are molecules that have the same formula but different structures. For the formula [tex]C_3H_5Cl_3[/tex] we can draw a linear chain of three carbons and change the position of the chlorine atoms in the carbon chain.

With this in mind, if we put all the chlorine atoms on the same carbon we will get 1,1,1-trichloropropane. If we change an atom from chlorine to carbon 2 we will obtain 1,1,2-trichloropropane. If we move another chlorine atom to carbon two we will get 1,2,2-trichloropropane. Finally, if we put a chlorine atom in each carbon we will obtain 1,2,3-trichloropropane.

See figure one for further explanations

I hope it helps!

3. How many significant figures are in the number 805000? *

Answers

Answer:

3

Explanation:

Three significant figures

10.
17.0cm
23.0cm
4,00cm
A 38.6 kg marble slab is shown above. What is its density? Give your
answer to the nearest tenth.
g/cm3
Enter the answer
Check it

Answers

Answer:

Density, [tex]\rho=24.68\ g/cm^3[/tex]

Explanation:

Mass of a marble slab is 38.6 kg 38600 grams

The dimensions of the marble is 17 cm×23 cm×4 cm

We need to find its density. Mass per unit volume equals density. So,

[tex]\rho=\dfrac{m}{V}\\\\\rho=\dfrac{m}{lbh}\\\\\rho=\dfrac{38600\ g}{(17\times 23\times 4)\ cm^3}\\\\\rho=24.68\ g/cm^3[/tex]

So, the density of the marble slab is [tex]24.68\ g/cm^3[/tex]

find the value of y in the equation Зу – 20 = 7.​

Answers

[tex]3y-20=7\implies y=\frac{20+7}{3}=9[/tex].

Hope this helps.

y= 9

Explanation:

=> 3y - 20 = 7

=> 3y = 7+20

=> 3y = 27

=> y = 27/3

=> y = 9

HOPE OT HELPS. ◉‿◉

Peripherals are used to?

Answers

Answer:

Peripheral device, also known as peripheral, computer peripheral, input-output device, or input/output device, any of various devices (including sensors) used to enter information and instructions into a computer for storage or processing and to deliver the processed data to a human operator or, in some cases.

Explanation:

Hope it helps

Answer: Input and output data

Explanation:

Plato!!!!!!!!!

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