For the following reaction, calculate how many grams of each product are formed when 4.05 g of water is used.
2 H20 -----> 2 H2 + O2

Answers

Answer 1

Answer: The mass of hydrogen gas and nitrogen gas produced is 0.45 g and 3.15 g respectively.

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 water = 4.05 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of water}=\frac{4.05g}{18g/mol}=0.225mol[/tex]

For the given chemical reaction:

[tex]2H_2O\rightarrow 2H_2+O_2[/tex]

For hydrogen:

By Stoichiometry of the reaction:

2 moles of water is producing 2 moles of hydrogen gas

So, 0.225 moles of water will produce = [tex]\frac{2}{2}\times 0.225=0.225mol[/tex] of hydrogen gas.

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

Moles of hydrogen gas = 0.225 mol

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

[tex]0.225mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Mass of hydrogen gas}=0.45g[/tex]

For nitrogen:

By Stoichiometry of the reaction:

2 moles of water is producing 1 mole of nitrogen gas

So, 0.225 moles of water will produce = [tex]\frac{1}{2}\times 0.225=0.1125mol[/tex] of nitrogen gas.

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

Moles of nitrogen gas = 0.1125 mol

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

[tex]0.1125mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=3.15g[/tex]

Hence, the mass of hydrogen gas and nitrogen gas produced is 0.45 g and 3.15 g respectively.


Related Questions

Calculate the mass of 12 mL of a substance whose density is 1.1 g/mL. Express your answer in grams using the correct number of significant figures. Do not enter your answer using scientific notation.

Answers

Final answer:

To find the mass of the substance, multiply the given density (1.1 g/mL) by the given volume (12 mL) to get a mass of 13.2 grams, with the answer rounded to two significant figures.

Explanation:

To calculate the mass of a substance, you can use the formula mass = density × volume. Given a density of 1.1 g/mL and a volume of 12 mL, you simply multiply these two values together to find the mass.

Mass = 1.1 g/mL × 12 mL = 13.2 g

The density has two significant figures and the volume has two as well, so our final answer will be reported with two significant figures, hence 13.2 grams is the mass.

A worker is exposed to the following airborne concentrations of ethylbenzene: 8:00 – 10:00 125 ppm 10:15 – 12:15 88 ppm 12:45 – 3:15 112 ppm 3:30 – 5:00 70 ppm What is the 8-hour TWA exposure for the employee (calculate the exposure)? Does it exceed the OSHA PEL?

Answers

Answer:

The 8-hour TWA exposure for the employee is 101 ppm and it exceeds the OSHA PEL of 100 ppm for ethylbenzene.

Explanation:

The TWA for 8 hours is calculated by the sum of airbone concentrations multiplied by the time it has been exposed to that period. The total is divided by 8 which refers to the 8 hours total the employee has been exposed.

TWA = (125x2+88x2+112x2.5+70x1.5)/8.

The OSHA PEL is a known number for every compound and it can be find in PEL tables. In the case of ethylbenezene, it is 100 ppm.

How many grams of solid Ca(OH)2 (74.1 g/mol) are required to make 500 ml of a 3 M solution?

Answers

Answer:

111.15 g are required to prepare 500 ml of a 3 M solution

Explanation:

In a 3 M solution of Ca(OH)₂ there are 3 moles of Ca(OH)₂ per liter solution. In 500 ml of this solution, there will be (3 mol/2) 1.5 mol Ca(OH)₂.

Since 1 mol of Ca(OH)₂ has a mass of 74.1 g, 1.5 mol will have a mass of

(1.5 mol Ca(OH)₂ *(74.1 g / 1 mol)) 111.15 g. This mass of Ca(OH)₂ is required to prepare the 500 ml 3 M solution.

How are hydrogen bonds formed between water molecules? Use a drawing to illustrate.

Answers

Answer:

They hydrogen bonds exist at the covalents molecules, as water where the 0 is an atom electronegative. The hydrogen bonds are formed because the hydrogen  gives its electron to oxygen (d-), that's why we say, there is a dipole at the molecule. As the dipole has been formed, the hydrogen who gave the electron to oxygen will try to get another electro from other molecules. That is  how the hydrogen bonds are formed.

Explanation:

A 200 ml sample of 0.1015 Miric acid is mbred with 2300 ml of water. What is the molar concentration of nitric acid in the final solution 0 406 M 127M 325 x 10 M 5.08 x 10 M 8.12 x 10 M water Backspace lu 'o Pin

Answers

Answer: Molar concentration of nitric acid in the final solution is [tex]8.12\times 10^{-3}M[/tex]

Explanation:

According to the dilution law,

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1[/tex] = molarity of stock [tex]HNO_3[/tex] solution = 0.1015 M

[tex]V_1[/tex] = volume of stock [tex]HNO_3[/tex]solution = 200 ml

[tex]M_2[/tex] = molarity of dilute [tex]HNO_3[/tex] solution = ?

[tex]V_2[/tex] = volume of  dilute [tex]HNO_3[/tex]  solution = (2300 +200 )ml = 2500 ml

Putting in the values we get:  

[tex]0.1015M\times 200=M_2\times 2500[/tex]

[tex]M_2=8.12\times 10^{-3}M[/tex]

Thus the molar concentration of nitric acid in the final solution is [tex]8.12\times 10^{-3}M[/tex]

Calculate the concentration of A bottle of wine contains 12.9% ethanol by volume. The density of ethanol (CH3OH) is 0.789 g/cm ethanol in wine in terms of mass percent and molality Mass percent Molality =

Answers

Answer:

The mass percentage of the solution is 10.46%.

The molality of the solution is 2.5403 mol/kg.

Explanation:

A bottle of wine contains 12.9% ethanol by volume.

This means that in 100 mL of solution 12.9  L of alcohol is present.

Volume of alcohol = v = 12.9 L

Mass of the ethanol = m

Density of the ethanol ,d= [tex]0.789 g/cm^3=0.789 g/mL[/tex]

[tex]1 cm^3=1 mL[/tex]

[tex]m=d\times v=0.798 g/ml\times 12.9 mL = 10.1781 g[/tex]

Mass of water = M

Volume of water ,V= 100 mL - 12.9 mL = 87.1 mL

Density of water = D=1.00 g/mL

[tex]M=D\times V=1.00 g/ml\times 87.1 mL =87.1 g[/tex]

Mass percent

[tex](w/w)\%=\frac{m}{m+M}\times 100[/tex]

[tex]\frac{10.1781 g}{10.1781 g+87.1 g}\times 100=10.46\%[/tex]

Molality :

[tex]m=\frac{m}{\text{molar mass of ethanol}\times M(kg)}[/tex]

M = 87.1 g = 0.0871 kg (1 kg =1000 g)

[tex]=\frac{10.1781 g}{46 g/mol\times 0.0871 kg}[/tex]

[tex]m=2.5403 mol/kg[/tex]

Final answer:

To calculate the concentration of ethanol in wine, multiply the volume percent of ethanol by the density of ethanol. Then divide by the mass of the solution and multiply by 100 to get the mass percent.

Explanation:

To calculate the concentration of ethanol in wine, we can use the mass percent formula. Mass percent is calculated by dividing the mass of the solute (ethanol) by the mass of the solution (wine), and multiplying by 100. The mass of ethanol can be found by multiplying the volume percent of ethanol (12.9%) by the density of ethanol (0.789 g/cm³). The density of wine is typically close to 1 g/cm³. So, to find the concentration of ethanol in terms of mass percent, we can follow these steps:

Assuming we have 100 mL of wine, we can calculate the volume of ethanol in the wine by multiplying 100 mL by 12.9% (0.129).Next, we can convert the volume of ethanol to mass by multiplying it by the density of ethanol. This gives us 0.129 mL * 0.789 g/cm³ = 0.101961 g.Since the total mass of the wine is equal to the mass of ethanol (since ethanol is the only solute), the concentration of ethanol in terms of mass percent is 0.101961 g / 100 mL * 100% = 0.101961%.

Learn more about Concentration of Ethanol in Wine here:

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a reasonable pka for a weak base is:

10.3

0.9

5.8

7.4

13.1

Answers

Answer:

0.9

Explanation:

The pka represents the force by which the molecules need to dissociate for the acids ,

Hence , lower the pka stronger will be the acid and that therefore will  dissociate completely and vice versa , for a weak acid higher the pka .

And in case of a base , it will be completely reversed , lower pKa , weaker base ,

and higher pKa , stronger base .

From the data of the question ,

0.9 is the lowest value of the pKa , hence , weakest base .

What is the theoretical Van’t Hoff factor for the following substances: HCL & ethanol (CH3CH2OH)

Answers

Answer:

i=2 for HCl and i=1 for ethanol

Explanation:

Hello,

Since the hydrochloric acid is composed by sodium and chloride ions which are completely dissociated, two types of ions are present, that's why i becomes 2.

On the other hand, as long as the ethanol doesn't present dissociation in aqueous solution, the Van't Hoff factor becomes 1.

Best regards.

Many elements combine with hydrogen(valence = 1) to give compounds called hydrides. Use the position of an element in the periodic table to deduce its valence. Write formulas for these hydrides without using subscripts, for example XH3. If no hydride forms, write "none". What is the formula of the hydride formed by sulfur? What is the formula of the hydride formed by potassium ? ||

Answers

Answer:

Explanation:

Use the position of an element in the periodic table to deduce its valence.

The periodic table is a table that groups elements based on their periodic functions. A group is a vertical arrangement of elements. The group number shows the number of elements in the outer shell of the atoms.

The groups runs from 1 to 8. On the periodic table, elements that has more than 4 electrons in their outermost shell will have a valency of 8 less than the number of outermost electrons.

Write formulas for these hydrides without using subscripts, for example XH3. If no hydride forms, write "none".

for group I : NaH, KH,

    group II : MgH2, CaH2,

    group III: BH3, AlH3

  Group 8 do not form hydrides because of their inertness.

What is the formula of the hydride formed by sulfur?

the hydride of sulfur is H₂S

What is the formula of the hydride formed by potassium ?

the hydride of potassium is KH

Final answer:

The formula of the hydride formed by sulfur is H2S, while the formula for the hydride formed by potassium is KH, based on the valence of sulfur and potassium corresponding to their positions in the periodic table.

Explanation:

When determining the formulas for hydrides, the valence of the non-hydrogen element is crucial. Sulfur, which has a valence of 2, forms a hydride by combining with two hydrogen atoms, giving us the formula H2S. In contrast, potassium belongs to the alkali metals with a valence of 1, thus it combines with one hydrogen atom to form a hydride, resulting in the formula KH.

The periodic table helps us understand these valences due to an element's group number. Sulfur, located in group 16, typically forms compounds where it has two bonding sites, as seen with its hydride, hydrogen sulfide. Potassium, found in group 1, forms compounds by donating a single electron, resulting in a 1:1 ratio with hydrogen in potassium hydride.

The heat capacity of chloroform (trichloromethane,CHCl3)
in the range 240K to 330K is given
byCpm/(JK-1mol-1) = 91.47
+7.5x10-2(T/K). In a particular experiment,
1.0molCHCl3 is heated from 273K to 300K. Calculate the
changein molar entropy of the sample.

Answers

Answer : The change in molar entropy of the sample is 10.651 J/K.mol

Explanation :

To calculate the change in molar entropy we use the formula:

[tex]\Delta S=n\int\limits^{T_f}_{T_i}{\frac{C_{p,m}dT}{T}[/tex]

where,

[tex]\Delta S[/tex] = change in molar entropy

n = number of moles = 1.0 mol

[tex]T_f[/tex] = final temperature = 300 K

[tex]T_i[/tex] = initial temperature = 273 K

[tex]C_{p,m}[/tex] = heat capacity of chloroform = [tex]91.47+7.5\times 10^{-2}(T/K)[/tex]

Now put all the given values in the above formula, we get:

[tex]\Delta S=1.0\int\limits^{300}_{273}{\frac{(91.47+7.5\times 10^{-2}(T/K))dT}{T}[/tex]

[tex]\Delta S=1.0\times [91.47\ln T+7.5\times 10^{-2}T]^{300}_{273}[/tex]

[tex]\Delta S=1.0\times 91.47\ln (\frac{T_f}{T_i})+7.5\times 10^{-2}(T_f-T_i)[/tex]

[tex]\Delta S=1.0\times 91.47\ln (\frac{300}{273})+7.5\times 10^{-2}(300-273)[/tex]

[tex]\Delta S=8.626+2.025[/tex]

[tex]\Delta S=10.651J/K.mol[/tex]

Therefore, the change in molar entropy of the sample is 10.651 J/K.mol

a 50-g sample contains 32.1 g of oxygen 2 g of hydrogen and 15.9 g of carbon

Answers

Answer:

i need more to solve this

Explanation:

Options for full question:

(A) % oxygen is 32.1%

(B) % hydrogen is 4%

(C) total percent composition of all elements is approximately 100%

(D) % carbon is 15.9%

Answer:

Options B and C

Explanation:

Information Given;

Mass of sample - 50g

Mass of Oxygen - 32.1g

Mass of Hydrogen - 2g

Mass of Carbon - 15.9g

The percentage composition of an element in a compound is the mass percentage of the element present in the compound. It tells the mass percentage of each element present in a compound.

The formular is given as;

Percentage composition = (Mass of element / Mass of compound) * 100

For oxygen;

Percentage Composition = (32.1 / 50) * 100 = 0.642 * 100= 64.2%

For Hydrogen;

Percentage Composition = (2 / 50) * 100 = 0.04 * 100= 4%

For Carbon;

Percentage Composition = (15.9 / 50) * 100 = 0.318 * 100= 31.8%

Total percentage composition of all elements = 4% + 31.8% + 64.2% = 100%

Based on this,

Option A is incorrect

Option B is Correct.

Option C is Correct.

Option D is incorrrect

What is the overall (X-fold) dilution of 4 serial 2-fold dilutions?

Answers

Answer:

16-fold dilution.

Explanation:

A serial dilution is any dilution where the concentration decrease by the same quantity in each successive step. So, for a 2-fold dilution, the concentration decrease 1/2, it means that if we have a sample with 10 M of concentration, after a 2-fold dilution it will be 5 M. For the next step it will be 1/2 of 5= 2.5 M, and successively.

Then, we just multiply the factor for each dilution. After 4 serial dilutions:

1/2 x 1/2 x 1/2 x 1/2 = 1/16

So, it would be a 16-fold dilution in the end.

Order the following elements based upon their electronegativity, highest to lowest: carbon, flourine, hydrogen, nitrogen, oxygen, and sodium.

Answers

Answer:

The correct order is: Fluorine; Oxygen; Nitrogen; Carbon; Hydrogen; Sodium.

Explanation:

The electronegativity in the periodic table increases to the right in a period and up in a group. We can figure out the electronegativity of each element according to its electron configuration:

Carbon: [He] [tex]2s^{2} 2p^{2}[/tex]

Fluorine: [He] [tex]2s^{2} 2p^{5}[/tex]

Hydrogen: [tex]1s^{1}[/tex]

Nitrogen: [He] [tex]2s^{2} 2p^{3}[/tex]

Oxygen: [He] [tex]2s^{2} 2p^{4}[/tex]

Sodium: [Ne] [tex]3s^{1} [/tex]

The period of a chemical element is given by the last energy level of electron configuration.

The group of a chemical element is given by the amount of electrons in the last energy level of electron configuration.

Therefore,

Carbon: Period 2 Group 4

Fluorine: Period 2 Group 7

Hydrogen: Period 1 Group 1

Nitrogen: Period 2 Group 5

Oxygen: Period 2 Group 6

Sodium: Period 3 Group 1

Electronegativity increases when the number of the group increases.

Electronegativity increases when the number of the period decreases.

In conclusion, Fluorine has the greatest number of group and Sodium has the lowest number of group (also it has a greater number of period than hydrogen, so it is less electronegative than hydrogen)

F>O>N>C>H>Na

If you have 16 g of manganese (II) nitrate tetrahydrate, how much water is required to prepare 0.16 M solution from this amount of salt? OL 0 0.01 L the task is impossible, as manganese nitrate is insoluble in water O 0.10 L 398 ml

Answers

Answer: The volume of water required is 398 mL

Explanation:

To calculate the molarity of solution, we use the equation:

[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 (manganese (II) nitrate tetrahydrate) = 16 g

Molar mass of manganese (II) nitrate tetrahydrate = 251 g/mol

Molarity of solution = 0.16 M

Putting values in above equation, we get:

[tex]0.16M=\frac{16g\times 1000}{251g/mol\times \text{Volume of solution}}\\\\\text{Volume of solution}=398mL[/tex]

Hence, the volume of water required is 398 mL

why diamond has a very high melting point than aluminium??

Answers

Each carbon atom is covalently bonded to four other carbon atoms. A lot of energy is needed to separate the atoms in diamond. This is because covalent bonds are strong, and diamond contains very many covalent bonds. This makes diamond's melting point and boiling point very high.

If 12.5 mL of 0.200 M NaOH was used to reach the equivalence point for in the titration of 25.0 mL of unknown weak acid, HA, the concentration of HA is a. 0.050M b. 0.100M C.0.200M d.0.400M

Answers

Answer:

b. 0.100M

Explanation:

The balanced chemical reaction is: NaOH + HA ⇒ H₂O + NaA

The NaOH and HA react in a 1:1 molar ratio, so at the equivalence point, the amount of NaOH added equals the amount of HA that was present in the solution.

The amount of NaOH that was added can be calculated and set equal to the amount of HA that must have been present to react with it.

n = CV = (0.200 mol/L)(12.5 mL) = 2.50 mmol NaOH = 2.50 mmol HA

Thus, there were 2.50 mmol of HA in 25.0 mL. The concentration can be calculated as follow:

C = n/V = (2.50 mmol)/(25.0mL) = 0.100 M

The unit conversion between rankine and kelvin is (linear, exponential, quadratic, none of the above)?

The unit conversion between degrees Celsius and degrees Fahrenheit is (linear, exponential, quadratic, none of the above)?

Answers

Answer:

Unit conversion between Rankine and Kelvin is linear.

Unit conversion between degree Celsius and degree Fahrenheit is linear.

Explanation:

Relation between rankine and Kelvin is

[tex]R\ =\ \dfrac{9}{5}\ K[/tex]

So, the plot between Rankine and Kelvin is a straight line with zero intercept and has a slope having value [tex]\dfrac{9}{5}[/tex].

Relation between degrees Celsius and degrees Fahrenheit is given by

[tex]^{\circ}C\ =\ \dfrac{5}{9}\ (^{\circ}F\ -\ 32)[/tex]

So, the plot between degrees Celsius and degrees Fahrenheit is a straight line with slope [tex]\dfrac{5}{9}[/tex] and negative intercept of [tex]\dfrac{160}{9}[/tex].

A 0.1375 g sample of solid magnesium is burned in
aconstant-volume bomb calorimeter that has a heat capacity of3024
J/ oC. The temperature increases by 1.126oC.
Calculate the heat given off by burning Mg, inkJ/g and in
kJ/mol.

Answers

Answer:

24.76 kJ/g

844.8 kJ/mol

Explanation:

The heat produced by the burning of the solid magnesium is equal to the heat absorbed by the calorimeter, which causes its temperature to increase.

First, we calculate the heat absorbed by the calorimeter, where C is the heat capacity and Δt is the temperature change.

Q = CΔt = (3024 J·°C⁻¹)(1.126 °C) = 3405 J

This is the same amount of heat that was produced by burning the magnesium

Now we can calculate the heat produced per gram of magnesium:

(3405 J)(1 kJ/1000 J) / (0.1375 g) = 24.76 kJ/g

We can convert grams to moles using the atomic weight of Mg (24.305 g/mol).

(24.76 kJ/g)(24.305 g/mol) = 844.8 kJ/mol

What is the Damkohler number? b) What is the significance of a system with a low Damkohler number?

Answers

Explanation:

(a)  A number which is dimensionless and provides us an estimate of the degree of conversion which can be achieved in CSTR, that is, continuous stirred tank reactor is known as Damkohler number.

This number is denoted as Da.

Mathematically,     Da = [tex]\frac{\text{reaction rate}}{\text{convection rate}}[/tex]

             Da = [tex]\frac{-rA \times V}{Fa_{o}}[/tex]

Now, for first order system, Da = [tex]\frac{-rA \times V}{Fa_{o}}[/tex]

                        = [tex]\frac{k \times CA_{o} \times V}{v \times CA_{o}}[/tex] = Tk

where,      rA = rate of reaction

                V = volume of reactor

                [tex]Fa_{o}[/tex] = molar flow rate of component A

                 k = rate constant

               [tex]CA_{o}[/tex] = initial concentration of A

                  v = volumetric flow rate of A

                  T = residence time

(b)   Since, from a given Damkohler number we can figure out the possible conversion of CSTR, that is, continuous stirred tank reactor.

So, if we have a low Damkohler number then the system will give us a less conversion formula. As the conversion is as follows.

                       X = [tex]\frac{Da}{Da + 1}[/tex]

Hence, we can conclude that [tex]Da \leq 0.1[/tex] will give less than 10% conversion as calculated by using above formula.

Final answer:

The Damkohler number is a dimensionless number used in chemical engineering to characterize the importance of reaction rates relative to a system's residence time. A system with a low Damkohler number indicates that the chemical reaction is much faster compared to the transport of material. This means that the reaction can be considered essentially instantaneous, and the reactants are fully converted into products within the system before they have a chance to be transported out.

Explanation:

The Damkohler number is a dimensionless number used in chemical engineering to characterize the importance of reaction rates relative to a system's residence time. It is defined as the ratio of the characteristic time for a chemical reaction to occur to the characteristic time for a system to transport material through itself.

A system with a low Damkohler number indicates that the chemical reaction is much faster compared to the transport of material. In practical terms, this means that the reaction can be considered essentially instantaneous, and the reactants are fully converted into products within the system before they have a chance to be transported out.

For example, if we have a packed bed reactor (a reactor where reactants flow through a bed of solid catalyst particles), a low Damkohler number implies that the reaction is so fast that the reactants are fully converted into products even before they can flow through the bed, resulting in high conversion levels.

A liquid mixture contains water (H2O, MW = 18.0), ethanol (C2H5OH, MW = 46.0) and methanol (CH3OH, MW = 32.0). Using two different analytical techniques to analyze the mixture, it was determined that the water mole fraction was 0.250 while the water mass fraction was 0.134. Determine the mole fraction ethanol (C2H5OH) and the mole fraction methanol (CH3OH) in the solution. Report the values to the correct number of significant figures.

Answers

Answer:

Mole fraction of ethanol is 0.363.

Mole fraction of methanol is 0.387.

Explanation:

Mole fraction of water =[tex]\chi_1=0.250[/tex]

Mole fraction of ethanol =[tex]\chi_2[/tex]

Mole fraction of methanol = [tex]\chi_3[/tex]

[tex]\chi_1+\chi_2+\chi_3=1[/tex]

[tex]\chi_2+\chi_3=1-\chi_1=0.750[/tex]

[tex]\chi_2+\chi_3=0.750[/tex]

[tex]\chi_1=\frac{n_1}{n_1+n_2+n_3}[/tex]

[tex]\chi_2=\frac{n_2}{n_1+n_2+n_3}[/tex]

[tex]\chi_3=\frac{n_3}{n_1+n_2+n_3}[/tex]

[tex]n_1+n_2+n_3=1[/tex]

Moles of water = [tex]n_1=0.250 mol[/tex]

Moles of ethanol= [tex]n_2[/tex]

Moles of methanol= [tex]n_3[/tex]

[tex]n_2+n_3=0.750 mol[/tex]

Mass of the mixture = M

Mass of water, [tex]m_1[/tex]

[tex]=n_1\times 18.0 g/mol=0.250 mol\times 18.0 g/mol=4.5 g[/tex]

Fraction of water by mass = 0.134

[tex]\frac{n_1\times 18.0 g/mol}{M}=0.134[/tex]

[tex]M=\frac{0.250 mol\times 18.0 g/mol}{0.134}=33.58 g[/tex]

Mass of ethanol = [tex]m_2[/tex]

Mass of methanol = [tex]m_3[/tex]

[tex]m_1+m_2+m_3=M[/tex]

[tex]4.5 g+m_2+m_3=33.58 g[/tex]

[tex]m_2+m_3=29.08 g[/tex]..[1]

[tex]\frac{m_2}{46.0 g/mol}+\frac{m_3}{32.0 g/mol}=0.750 mol[/tex]

[tex]16m_1+23m_3=552[/tex]..[2]

On solving [1] and [2]:

[tex]m_2 = 16.70, m_3= 12.38 g[/tex]

Mole fraction of ethanol =[tex]chi_2[/tex]

[tex]\chi_2=\frac{n_2}{n_1+n_2+n_3}[/tex]

[tex]=\frac{\frac{16.70 g}{46.0 g/mol}}{1 mol}= 0.363[/tex]

Mole fraction of methanol = [tex]chi_3[/tex]

[tex]\chi_3=\frac{n_3}{n_1+n_2+n_3}[/tex]

[tex]=\frac{\frac{12.38 g}{32.0 g/mol}}{1 mol}= 0.387[/tex]

A protein has a binding site for a single ligand with a A.G-16.7 KJ/mol at 298 K. What is Keg for this reaction?

Answers

Explanation:

The given data is as follows.

          [tex]\Delta_{r} G[/tex] = -16.7 kJ/mol = [tex]-16.7 \times 10^{3}[/tex],       T = 298 K

          R = 8.314 J/mol K,       [tex]K_{eq}[/tex] = ?

Relation between [tex]\Delta_{r} G[/tex] and [tex]K_{eq}[/tex] is as follows.

                [tex]\Delta_{r} G[/tex] = [tex]-RT ln K_{eq}[/tex]

Hence, putting the values into the above equation as follows.

                [tex]\Delta_{r} G[/tex] = [tex]-RT ln K_{eq}[/tex]

                        [tex]-16.7 \times 10^{3} J/mol[/tex] = [tex]-8.314 J/mol K \times 298 K ln K_{eq}[/tex]

                     [tex]ln K_{eq}[/tex] = [tex]\frac{-16.7 \times 10^{3} J/mol}{-8.314 J/mol K \times 298 K}[/tex]    

                                         = 6.740

                            [tex]K_{eq}[/tex] = antilog (6.740)

                                            = 846

Thus, we can conclude that [tex]K_{eq}[/tex] for given values is 846.

                                   

1. Make a 100 mL volume of 100 mM acetic acid (HAC) HAC (glacial, 100%, 17.4 M, 1.05 g/ml, MW 60.05) ddh,0 ML add to 100 mL Measure 250 ml beaker. Add mL of ddH20 using a 100-ml graduated cylinder and transfer this volume to a ul of 100 mM HAC to the water. Mix with a stir bar.

Answers

Final answer:

To prepare a 100 mM acetic acid solution, calculate the moles needed, measure the glacial acetic acid based on its density and volume, and dilute with distilled water to the final volume. Mix well for homogeneity.

Explanation:

The question involves preparing a solution of acetic acid and relates to the subjects of molarity and solution preparation in chemistry. The solution is to be made to a specified concentration using a direct dilution from a more concentrated stock.

Steps to Prepare 100 mM Acetic Acid Solution

First, we calculate the number of moles of acetic acid required for 100 mL of a 100 mM solution using the molar mass of acetic acid (MW 60.05). Second, we use the density of glacial acetic acid (1.05 g/mL) and the volume needed (from the calculated moles and molarity) to determine how much of the glacial acetic acid to measure out. Finally, we add this measured amount of acid to a beaker and dilute to the mark with distilled water (ddH2O) to achieve the desired final concentration. Mixing with a stir bar ensures a homogeneous solution.

To calculate the mole fraction of acetic acid in a solution, we need the number of moles of acetic acid and the number of moles of water in the solution. This can be accomplished by converting the mass of each component to moles using their respective molar masses.

What is the wavenumber of the radiation emitted when a hydrogen
atom makes a transition corresponding to a change in energy of
1.634 x 10-18 J?

Answers

Answer: Wavenumber of the radiation emitted  is [tex]0.08\times 10^{8}m^{-1}[/tex]

Explanation:

The relationship between wavelength and energy of the wave follows the equation:

[tex]E=\frac{hc}{\lambda}[/tex]

where,

E = energy of the radiation = [tex]1.634\times 10^{-18}J[/tex]

h = Planck's constant  = [tex]6.626\times 10^{-34}Js[/tex]

c = speed of light = [tex]3\times 10^8m/s[/tex]

[tex]\lambda[/tex] = wavelength of radiation = ?

Putting values in above equation, we get:

[tex]1.634\times 10^{-18}J=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{\lambda}\\\\\lambda=12.16\times 10^{-8}m[/tex]

[tex]\bar {\nu}=\frac{1}{\lambda}=\frac{1}{12.16\times 10^{-8}}=0.08\times 10^{8}m^{-1}[/tex]

Thus wavenumber of the radiation emitted  is [tex]0.08\times 10^{8}m^{-1}[/tex]

Khoisan salts is the number 1 producer of salts in SA for both local and international markets. 500 kg of KCl is dissolved in sufficient water to make a saturated solution at 370 K. At 370 Kthe solubility of KCl is 42 mass %. The solution is cooled to 320 K and the solubility is 31,5 mass % It is assumed that no water is evaporated. 2.1. Determine the amount of water is added to the 500 kg of KCl to produce the required saturated solution at 370 K. (3) 2.2. Determine the mass of KCl crystals formed after the cooling process to a temperature of 320 K. (Use the formula method)

Answers

Answer:

690 kg of water.125 kg of KCl crystals are formed after cooling process

Explanation:

1) A saturated solution is a chemical solution containing the maximum concentration of a solute dissolved in the solvent. The solubility of KCl at 370K is 42%mass -42kg of KCl in 100kg of water+KCl , thus, the amount of water added to 500kg of KCl is:

500 kg of KCl × [tex]\frac{100 kg Water+KCl}{42kgKCl}[/tex] = 1190 kg of water +KCl

1190 kg of water +KCl - 500 kg of KCl = 690 kg of water

2) The maximum amount that this solution could solubilize of KCl at 320K is:

1190 kg of water + KCl × [tex]\frac{31,5 kgKCl}{100kgWater+KCl}[/tex] = 375 kg

Thus, the mass of KCl crystals formed are:

500 kg of KCl - 375 Kg of KCl = 125 kg of KCl

I hope it helps!

Give the names for each of the elements between atomic numbers 1 and 86 that has a symbol starting with "C".

Identify the family each element belongs to. If the element does not belong to a family, identify it as a metal or non-metal.

2. In your own words, define the following terms. a. Element b. Atom c. Molecule d. Compound 3. How many oxygen atoms are in one molecule (or formula unit) of each of the following compounds? NO3 Al(OH)3 Ca(NO3)2 Ba(OCN)2

Answers

Explanation:

1)

Carbon with atomic number 6. The symbol of carbon is C.It belongs to non metals.Chlorine with atomic number of 17. The symbol of chlorine is Cl.It belongs to non metals.Calcium with atomic number 20. The symbol of calcium is Ca. It belongs to metals.Chromium with atomic number 24. The symbol of chromium is Cr. It belongs to metals.Cobalt with atomic number of 27. The symbol of cobalt is Co. It belongs to metals.Copper with atomic number of 29. The symbol of copper is Cu. It belongs to metals.Cadmium with atomic number of 48. The symbol of cadmium is Cd. It belongs to metals.Cesium with atomic umber of 55.The symbol of cesium is Cs. It belongs to metals.Cerium with atomic number of 58. the symbol for cerium in Ce. It belongs to metals.

2)

Element are those substance which can not be split into simpler substance.they are made up of single atom.

Atom is the structural unit of the matter and smallest component of an element.

Molecules are group of atoms bonded together. The atom boned can be of more than one type.

Compounds are group of atoms of different elements bonded together.

3) Oxygen atoms in one molecule of following compounds:

a) [tex]NO_3[/tex]

1 × 3 = 3

There 3 oxygen atoms in the 1 molecule of [tex]NO_3[/tex].

b) [tex]Al(OH)_3[/tex]

3 × 1 = 3

There 3 oxygen atoms in the 1 molecule of [tex]Al(OH)_3[/tex].

c) [tex]Ca(NO_3)_2[/tex]

2 × 3 = 6

There 6 oxygen atoms in the 1 molecule of [tex]Ca(NO_3)_2[/tex].

d) [tex]Ba(OCN)_2[/tex]

2 × 1 = 2

There 6 oxygen atoms in the 1 molecule of [tex]Ba(OCN)_2[/tex].

At 0.70 atmosphere pressure, a certain mass of a gas has a volume of 3.8 L. What will be the volume if the pressure is decreased to 0.409 atmosphere pressure?

Answers

Answer: The new volume of the gas will be 6.50 L

Explanation:

To calculate the new pressure, we use the equation given by Boyle's law. This law states that pressure is inversely proportional to the volume of the gas at constant temperature.

The equation given by this law is:

[tex]P_1V_1=P_2V_2[/tex]

where,

[tex]P_1\text{ and }V_1[/tex] are initial pressure and volume.

[tex]P_2\text{ and }V_2[/tex] are final pressure and volume.

We are given:

[tex]P_1=0.70atm\\V_1=3.8L\\P_2=0.409atm\\V_2=?L[/tex]

Putting values in above equation, we get:

[tex]0.70atm\times 3.8L=0.409atm\times V_2\\\\V_2=6.50L[/tex]

Hence, the new volume of the gas will be 6.50 L

. Does each of these describe a physical change or a chemical change? Explain a. The moth balls gradually vaporize in a closet. Hydrofluoric acid attacks class, and is used to etch calibrations marks on glass laboratory utensils A French chef making a sauce with brandy is able to burn of the alcohol from the brandy leaving just the brandy flavoring d. Chemistry majors sometimes et holes in the cotton jeans they wear to lab, because of acid spills A piece of egg boiled in water for 20 minutes

Answers

Answer:

a. The mothballs gradually vaporize in a closet. Physical change

b. Hydrofluoric acid attacks glass, and is used to etch calibrations marks on glass laboratory utensils. Chemical change

c. A French chef making a sauce with brandy is able to burn of the alcohol from the brandy leaving just the brandy flavoring. Chemical change

d. Chemistry majors sometimes et holes in the cotton jeans they wear to lab, because of acid spills. Chemical change

e. A piece of egg boiled in water for 20 minutes. Chemical change

Explanation:

In a physical change, there is no change in the chemical composition of the substance, the change is only on the physical properties. For example, state changes (a).

In a chemical change instead, usually there is a combination of two or more substances that combine to form a new one different from the originals. For example, chemical reactions like combustion (c) or protein denaturalization (e). The attacking of hydrofluoric acid to glass (b) or the acid pill attacking cotton are chemical reactions too (d).

Physical change or a chemical change are given as:

a. The moth balls gradually vaporize in a closet -physical change.

b. Hydrofluoric acid attacks glass and is used to etch calibration marks on glass laboratory utensils-chemical change.

c. A French chef making a sauce with brandy is able to burn off the alcohol from the brandy, leaving just the brandy flavoring- a chemical change.

d. Chemistry majors sometimes etch holes in the cotton jeans they wear to the lab because of acid spills- physical change.

A piece of egg boiled in water for 20 minutes-  chemical change.

a. In a closet, moth balls progressively vaporise: This represents a physical transformation. The moth balls sublimate, going straight from a solid state to a gaseous state, undergoing a phase change from a solid to a gas.

b. Glass is attacked by hydrofluoric acid, which is also used to etch calibration marks on glass laboratory utensils: This refers to a chemical transformation. The glass and hydrofluoric acid interact, causing a chemical reaction that etches the surface of the glass.

c. A French cook who uses brandy in a sauce can burn out the alcohol, leaving only the brandy flavouring: This explains a chemistry alteration. Alcohol is burned off through a chemical reaction, specifically alcohol combustion, which transforms ethanol (alcohol) into carbon dioxide and water vapour.

d. Because of acid spills, chemistry majors occasionally etch holes in the cotton pants they wear to the lab: This sentence depicts a physical alteration. The fibres in the jeans melt or break down as a result of acid spills, creating holes.

e. A chemical change is described by a piece of egg that was boiled in water for 20 minutes. The denaturation of proteins and the coagulation of the egg white are two distinct chemical reactions that occur inside the egg as a result of boiling.

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One kilogram of saturated steam at 373 K and 1.01325 bar is contained in a rigid walled vessel. It has a volume of 1.673 m3. It is cooled to a temperature at which the specific volume of water vapour is 1.789 m. The amount of water vapour condensed in kilograms is (a) 0.0 (b) 0.065 (c) 0.1 (d) 1.0

Answers

Answer: Option (b) is the correct answer.

Explanation:

The given data is as follows.

        Initial volume [tex](v_{1})[/tex] = 1.673 [tex]m^{3}[/tex]

          Final volume [tex](v_{2})[/tex] = 1.789 [tex]m^{3}[/tex]

As, the amount of water vapor condensed will be as follows.

                     [tex]\frac{(v_{2} - v_{1})}{v_{2}}[/tex]

                     = [tex]\frac{(1.789 m^{3} - 1.673 m^{3})}{1.789 m^{3}}[/tex]

                     = [tex]\frac{0.116 m^{3}}{1.789 m^{3}}[/tex]

                     = 0.065 kg

Hence, we can conclude that the amount of water vapour condensed in kilograms is 0.065 kg.

Given a water solution that contains 1.704 kg of HNO:/kg H:O and has a specific gravity of 1.382 at 20 °C, express the composition in the following ways: (a) Weight percent HNO (b) Pounds HNO3 per cubic foot of solution at 20 °c (c) Molarity (gmoles of HNOs per liter of solution at 20 °C)

Answers

Answer:

a) 63,0%

b) 54,4 Pounds HNO₃ per cubic foot of solution

c) 13,8 M

Explanation:

a) Weight percent is the ratio solute:solution times 100:

[tex]\frac{1,704 kg HNO_3}{2,704 kg Solution}[/tex] = 63,0%

b) Pounds HNO₃ per cubic foot of solution at 20 °c

Pounds HNO₃:

1,704 kg [tex]\frac{2,20462 pounds}{1 kg}[/tex] = 3,7567 pounds

Cubic foot:

2,704 kg [tex]\frac{1 L}{1,382 kg}[/tex]x[tex]\frac{1 cubic foot }{28,3168 L}[/tex] = 0,069 ft³

Thus:[tex]\frac{3,7567 pounds}{0,069ft^3} =[/tex] = 54,4 Pounds HNO₃ per cubic foot of solution

c) Moles of HNO₃:

1704 g HNO₃ [tex]\frac{1 mol HNO_3}{63,01 g }[/tex] = 27,04 moles

Liters of solution:

2,704 kg [tex]\frac{1 L}{1,382 kg}[/tex] = 1,96 L of solution

Molarity:

[tex]\frac{27,04 mol}{1,96 L}[/tex] = 13,8 M

Final answer:

The composition of the solution can be expressed as 63.02 wt% HNO3, 54.28 lb HNO3/ft3, and a molarity of 14.2 M at 20 °C.

Explanation:

To express the composition of a water solution with 1.704 kg of HNO3 per kg of H2O and a specific gravity of 1.382 at 20 °C, we can calculate the following:

Weight Percent HNO3

Weight percent (wt%) is calculated as the mass of the solute divided by the total mass of the solution, multiplied by 100. For 1.704 kg HNO3 in 1 kg of water, the total mass of the solution is 1.704 kg + 1 kg = 2.704 kg. The weight percent HNO3 is then ((1.704 kg) / (2.704 kg)) × 100 = 63.02 wt%.

Pounds HNO3 per Cubic Foot of Solution at 20 °C

Firstly, we need to convert the specific gravity to density in g/mL: 1.382 (Specific Gravity) × 1.000 g/mL (density of water) = 1.382 g/mL. To find the density in lb/ft3, we multiply by 62.43 lb/ft3 which is the conversion factor from g/mL to lb/ft3. Then, multiply the density of the solution by the weight percent of HNO3 to find pounds HNO3 per cubic foot of solution: (1.382 g/mL) × (62.43 lb/ft3) × (63.02%) = 54.28 lb HNO3/ft3.

Molarity of HNO3 at 20 °C

Using the density of the concentrated HNO3 solution, 1.42 g/mL, and the given formula for molarity which is Molarity = [(%) (d)/ (Mw)] × 10, we substitute the values: Molarity = [(63.02) (1.42 g/mL) / (63.01)] × 10 = 14.2 M.

Which radioactive emission has the smallest (least) mass? Question options:

A) Positron

B) Alpha

C) All have equal mass

D) Gamma

E) Beta

Answers

Answer:

The answer is D. gamma rays

Explanation:

A radioactive atom can have three different types of emission:

alpha particles (α) = they have a mass of 4 amu and they have a very low penetrating power.

Beta particles (β) = they have 5x[tex]10^{-4}[/tex] amu and they have an intermediate penetrating power

Gamma rays (γ) = they are not particles basically just energy so its mass is ≈ 0 and its penetrating power is higher

For this reason Gamma emissions (γ) has the smallest mass value.

Final answer:

Among the options provided, the gamma photon in radioactive emissions has the smallest mass (zero mass), which means that it's the correct answer to the question. Gamma radiation is energy emission without a corresponding mass.

Explanation:

In the context of radioactive emissions, each type has a different mass. The alpha particle is the heaviest and is comprised of two neutrons and two protons. Beta particles, which are electrons (or positrons in the case of beta-plus decay), have a smaller mass. Positrons have the same mass as electrons but with a positive charge.

However, the gamma photon, which is a type of electromagnetic radiation, has no rest mass at all. So, out of the given options, the gamma ray has the smallest (in this case, zero) mass. This is an example of energy being emitted without a corresponding mass in a radioactive process, which often occurs when the remaining nucleus is in an excited state post decay and moves to a lower energy level by emitting a gamma photon.

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