The density of the object is 65.447 g/mL.
Density is a physical property of a substance that represents the amount of mass per unit volume. It is a measure of how closely packed the particles or molecules of a substance are. The formula for density is:
Density = Mass / Volume
In this formula, mass refers to the amount of matter in the substance, typically measured in grams (g) or kilograms (kg), and volume refers to the amount of space occupied by the substance, typically measured in cubic centimeters (cm³) or liters (L).
Density provides information about the compactness or concentration of a substance. Substances with a higher density have more mass packed into a given volume, while substances with a lower density have less mass packed into the same volume.
Given:
Mass of the object = 149.8 g
Initial volume of water = 0.001210 L
Final volume of water and object = 0.003498 L
Volume of the object = Final volume - Initial volume
Volume of the object = 0.003498 L - 0.001210 L
Volume of the object = 0.002288 L
Density = Mass / Volume
Density = 149.8 g / 0.002288 L
Density = 65.447 g/mL
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2. Write out: a. The formula for sodium hyponitrite. b. Determine the molecular weight (molar mass) of sodium hyponitrite. c. Calculate the number of moles of sodium hyponitrite in 56.89 grams. d. Calculate the total number of atoms of oxygen in 56.89 grams of sodium hyponitrite.
a) The formula for sodium hyponitrite is Na2N2O2.
b) Molar mass of sodium hyponitrite is 105.99 g/mol.
c) To calculate the number of moles of sodium hyponitrite in 56.89 grams,
we will use the formula:
n = m/Mn = 56.89 g / 105.99 g/moln = 0.536 mol
Thus, the number of moles of sodium hyponitrite in 56.89 grams is 0.536 mol.
d) We will first find the total number of atoms in 1 molecule of sodium hyponitrite which is equal to 7 atoms
(2 sodium, 2 nitrogen, and 3 oxygen).
Next, we will find the number of molecules present in 56.89 grams using the formula:
Number of moles = Mass / Molar mass
Thus, number of moles = 56.89 g / 105.99 g/mol= 0.536 mol
Number of molecules present = Number of moles x Avogadro’s number
Number of molecules present = 0.536 mol x 6.022 x 10²³ molecules/mole
= 3.227 x 10²³ molecules
Total number of atoms of oxygen present in 56.89 grams of sodium hyponitrite
= 3 x number of molecules present
= 3 x 3.227 x 10²³ atoms
= 9.681 x 10²³ atoms of oxygen.
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Answer:
a. The formula for sodium hyponitrite is NaNO2.
b. The molecular weight (molar mass) of sodium hyponitrite is 69.004 grams per mole.
c. The number of moles of sodium hyponitrite in 56.89 grams is 0.82 moles.
d. The total number of atoms of oxygen in 56.89 grams of sodium hyponitrite is 4.
Explanation:
Here are the calculations for each answer:
a. The formula for sodium hyponitrite is NaNO2. Sodium has a charge of +1, nitrogen has a charge of -3, and oxygen has a charge of -2. The charges must add up to zero, so there must be one sodium atom, one nitrogen atom, and two oxygen atoms in the formula.
b. The molecular weight of sodium hyponitrite can be calculated by adding the atomic weights of the atoms in the formula. The atomic weight of sodium is 22.990, the atomic weight of nitrogen is 14.007, and the atomic weight of oxygen is 15.999. The molecular weight of sodium hyponitrite is therefore 22.990 + 14.007 + 2(15.999) = 69.004 grams per mole.
c. The number of moles of sodium hyponitrite in 56.89 grams can be calculated by dividing the mass by the molecular weight. 56.89 / 69.004 = 0.82 moles
d. The total number of atoms of oxygen in 56.89 grams of sodium hyponitrite is 4. There are 2 oxygen atoms in each molecule of sodium hyponitrite, and there are 56.89 / 69.004 = 0.82 moles of sodium hyponitrite in 56.89 grams. 0.82 x 2 = 4.
magnesium metal is produced by the electrolysis of molten magnesium chloride using inert electrodes a. sketch the cell. labe the anode and cathode, indicate the sign of the electrodes, and show the direction of electron and ion flow.
The electrolysis of molten magnesium chloride produces magnesium metal at the cathode and chlorine gas at the anode.
The electrolysis of molten magnesium chloride is a process that uses electricity to split the molten salt into its component elements. The molten salt is placed in an electrolysis cell, which has two electrodes: a cathode and an anode. The cathode is connected to the negative terminal of the power supply, and the anode is connected to the positive terminal of the power supply.
When the power supply is turned on, electrons flow from the negative terminal of the power supply to the cathode, and protons flow from the positive terminal of the power supply to the anode. The electrons at the cathode react with the chloride ions in the molten salt to form chlorine gas. The protons at the anode react with the magnesium ions in the molten salt to form magnesium metal.
The overall reaction is:
MgCl₂ → Mg + Cl₂
The magnesium metal is produced at the cathode and rises to the top of the cell, where it can be collected. The chlorine gas is produced at the anode and bubbles out of the cell.
The electrolysis of molten magnesium chloride is a very efficient way to produce magnesium metal. The process is also relatively inexpensive, and it does not produce any harmful pollutants.
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what did bohr’s model of the atom include that rutherford’s model did not have? a nucleus energy levels electron clouds smaller particles
Bohr's model of the atom included the energy levels that Rutherford's model of the atom did not have.
The Bohr's model is a model of atomic structure proposed by Niels Bohr in 1913.
This model has since been replaced by the quantum mechanical model, but it still helps to clarify certain concepts and predict certain outcomes.
It depicted the electrons in the atom as being situated in specific layers, or energy levels, surrounding the nucleus.
The Rutherford's model of the atom did not include the concept of energy levels, but Bohr's model did.
According to this model, electrons traveled around the nucleus in specific, discrete energy levels.
Electrons could also move from one energy level to another by either absorbing or releasing energy.
As a result, Bohr's model is frequently referred to as the planetary model of the atom.
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A+compound+that+contains+only+carbon,+hydrogen,+and+oxygen+is+68.5%+c+and+8.63%+h+by+mass.+what+is+the+empirical+formula+of+this+substance?
A compound that contains only 68.5% carbon and 8.63% hydrogen by mass has empirical formula [tex]\rm C_4H_6O[/tex]. Option C is the correct answer.
The empirical formula of a compound is the simplest whole-number ratio of atoms in the compound. Empirical formulas are useful in determining the composition of a compound when the exact molecular formula is not known.
To find the empirical formula of the compound that contains only carbon, hydrogen, and oxygen, we need to assume that we have 100 g of the compound.
From the given information, we know that the compound is 68.5% carbon and 8.63% hydrogen by mass.
Therefore, the mass of carbon in 100 g of the compound is 68.5 g, and the mass of hydrogen is 8.63 g.
To find the mass of oxygen in the compound, we can use the fact that the compound contains only carbon, hydrogen, and oxygen, so the sum of the masses of these elements must add up to 100 g. Therefore, the mass of oxygen in 100 g of the compound is:
Mass of oxygen = 100 g - (68.5 g + 8.63 g) = 22.87 g
Now, we can convert the masses of each element to moles by dividing by their respective atomic masses:
Moles of carbon = 68.5 g / 12.01 g/mol = 5.71 mol
Moles of hydrogen = 8.63 g / 1.01 g/mol = 8.54 mol
Moles of oxygen = 22.87 g / 16.00 g/mol = 1.43 mol
Next, we need to find the simplest whole number ratio of the atoms in the compound by dividing each of the mole values by the smallest mole value:
Moles of carbon / smallest mole value = 5.71 mol / 1.43 mol = 4
Moles of hydrogen / smallest mole value = 8.54 mol / 1.43 mol = 6
Moles of oxygen / smallest mole value = 1.43 mol / 1.43 mol = 1
Therefore, the empirical formula of the compound
containing 56.0 g nitrogen and 32.0 g oxygen is [tex]\rm C_4H_6O[/tex] . The correct option is C.
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The question is incomplete. The complete question is:
A compound that contains only carbon, hydrogen, and oxygen is 68.5% C and 8.63% H by mass. What is the empirical formula of this substance?
A) [tex]C_3H_5O[/tex]
B) [tex]C_6H_4[/tex]
C) [tex]C_4H_6O[/tex]
D)[tex]C_4H_8O_2.[/tex]
An expanding gas does 147 j of work on its surroundings at a constant pressure of 1.04 atm. if the gas initially occupied 68.0 ml, what is the final volume of the gas? l
The final volume of the gas is 209.35 mL.
To find the final volume, we can use the formula W = P * ΔV, where W is the work done, P is the pressure, and ΔV is the change in volume.
We can rearrange the formula W = P * ΔV to solve for ΔV:
ΔV = W / P.
Substituting the given values,
ΔV = 147 J / (1.04 atm) = 141.35 mL.
To find the final volume, we add the change in volume to the initial volume:
Final volume = Initial volume + ΔV = 68.0 mL + 141.35 mL = 209.35 mL.
So, the final volume of the gas is 209.35 mL.
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A certain liquid has a vapor pressure of 92.0 Torr at 23.0 ∘
C and 299.0 Torr at 45.0 ∘
C. Calculate the value of ΔH vap
for this liquid. ΔH vap
= Calculate the normal boiling point of this liquid. boiling point:
The boiling point of the given liquid is 89.3°C.
Given data: Vapor pressure at 23°C = 92.0 Torr
Vapor pressure at 45°C = 299.0 Torr
We have to find the value of ΔHvap for this liquid.
ΔHvap= -R*(T2-T1) /ln(P2/P1)
Where, R is the gas constant
T1 = 23°C = 23+273 = 296K
T2 = 45°C = 45+273 = 318K
P1 = 92.0 Torr
P2 = 299.0 Torr
Plugging these values into the above formula,
ΔHvap= -8.314*(318-296) /ln(299/92)
= 29.44 kJ/mol (approx)
The normal boiling point of the liquid can be calculated as follows:
ln(P2/P1) = ΔHvap/R * (1/T1 - 1/T2)
Rearranging the above equation we get,
T2 = ΔHvap / R * (ln(P2/P1)) + 1/T1
T2 = 29.44 kJ/mol / (8.314 J/mol K) * ln(299/92) + 1/296
T2 = 362.3 K or 89.3°C
Therefore, the boiling point of the given liquid is 89.3°C.
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what was the general trend from 1985 to 2014 in the size of the area with ph levels less than or equal to 4.5 (below which the environment is considered to be acidic)?
From 1985 to 2014, the general trend in the size of the area with pH levels less than or equal to 4.5 (indicating acidic environment) showed an increase. This trend suggests that over this time period, the acidic areas expanded or became more prevalent.
Several factors contribute to this trend. One major factor is human activities, particularly the burning of fossil fuels and industrial emissions. These activities release pollutants such as sulfur dioxide and nitrogen oxides into the atmosphere, which can then react with water and form acids, leading to acid rain. Acid rain can contribute to the acidification of soil and water bodies, leading to a decrease in pH levels.
Another contributing factor is deforestation. When forests are cleared, the soil is exposed to rainfall, which can lead to increased leaching of acidic compounds into the soil and water systems. Additionally, without the buffering capacity of trees, the pH of the soil can become more susceptible to fluctuations.
Climate change can also impact pH levels. Rising temperatures can affect the carbon dioxide levels in the atmosphere, leading to increased absorption of CO2 by the oceans. This can result in ocean acidification, where the pH of seawater decreases, negatively impacting marine life.
It is important to note that the trend may vary in different regions due to specific local factors and mitigation efforts. However, overall, the general trend from 1985 to 2014 indicates an increase in the size of areas with pH levels less than or equal to 4.5, signaling a growing concern for acidic environments.
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Which is true for a calibration curve? It is directly related to Beer's Law. It is used to determine the lambda max of a chemical substance. It is synonymous with the term "absorption spectrum." It is a plot of absorbance vs. wavelength.
The calibration curve is a plot of absorbance versus concentration. It is used to determine the relationship between concentration and absorbance for a given substance. The curve is constructed by measuring the absorbance of a series of solutions with known concentrations and plotting the data.
The curve is typically linear, and the equation for the line is used to calculate the concentration of an unknown solution based on its absorbance value.Calibration curves are used in various analytical techniques, including spectrophotometry and chromatography. The curve is not synonymous with the term "absorption spectrum," which refers to the range of wavelengths absorbed by a substance. It is also not used to determine the lambda max of a chemical substance. The lambda max is the wavelength at which a substance absorbs the most light and is typically determined by measuring the absorbance of a solution at different wavelengths.
Therefore, the correct statement for the calibration curve is that it is directly related to Beer's Law, and it is a plot of absorbance vs. concentration.
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2. (2 pts) Make a series of five 1/17 serial dilutions each between 100 and 1000 mL with one (not the final product) being exactly 444.44 mL.
Therefore, the series of five 1/17 serial dilutions each between 100 and 1000 mL with one (not the final product) being exactly 444.44 mL is as follows: 1000 mL, 58.82 mL, 3.46 mL, 0.20 mL, and 0.01 mL.
A series of five 1/17 serial dilutions each between 100 and 1000 mL with one (not the final product) being exactly 444.44 mL can be made as follows:
To make a series of five 1/17 serial dilutions each between 100 and 1000 mL with one (not the final product) being exactly 444.44 mL, follow the steps below:
Step 1:
Start with 1000 mL of the solution
Step 2: Take 1/17 of the solution and mix it with 16/17 of water, giving you a 58.82 mL solution.
This is a 1:17 dilution.
Step 3: Repeat Step 2 using 58.82 mL of the previous dilution and 941.18 mL of water to get a 1/17 dilution of 1000/17 or 58.82 mL.
Step 4: Repeat the previous step, using 58.82 mL of the previous dilution and 941.18 mL of water to get a 1/17 dilution of 1000/17² or 3.46 mL.
Step 5: Repeat Step 4, using 3.46 mL of the previous dilution and 96.54 mL of water to get a 1/17 dilution of 1000/17³ or 0.20 mL.
Step 6: Finally, repeat Step 5, using 0.20 mL of the previous dilution and 99.80 mL of water to get a 1/17 dilution of 1000/17⁴ or 0.01 mL.
Note: The volumes can be scaled to any range as long as the same 1/17 ratio is used between them.
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So 0.5mL of the supplement was used and diluted to 50mL in a volumetric flask. How would I set that up for question 4?
Also in question 3, when you take the new concentration found in mg/L and are trying to get the molarity in mol/L. When you take the concentration and divide it by the molar mass, why are you then dividing it by 1000?
Please help solve and answer #5: The molar mass of methylcobalamin is 1344.41 g/mol. Assuming one mole of cobalt per mole of methylcobalamin, find the number of mg of vitamin B12 in a 1 mL dose of the supplement. "Note that 1 g = 1000 mg
The molarity of the diluted solution = 0.005M. The number of mg of vitamin B12 in a 1 mL dose of the supplement is 6.72 mg.
To calculate the molarity of the diluted solution, use the formula:
Initial volume × Initial molarity = Final volume × Final molarity
Where,Initial volume = 0.5mL = 0.0005L
Initial molarity = molarity of the solution before dilution = ?
Final volume = 50mL = 0.050L
Final molarity = molarity of the solution after dilution = 0.005M
Substitute the values in the formula and calculate the initial molarity.
Initial molarity = (Final volume × Final molarity) / Initial volume
= (0.050 × 0.005) / 0.0005 = 0.5 M
The molarity of the diluted solution = 0.005 M.
When you take the new concentration found in mg/L and are trying to get the molarity in mol/L.
When you take the concentration and divide it by the molar mass, you are then dividing it by 1000 because there are 1000 milligrams in a gram.
Therefore, to convert milligrams per liter (mg/L) to moles per liter (mol/L), you need to divide by the molar mass and then by 1000.
This will give you the concentration in mol/L.
Molar mass of methylcobalamin = 1344.41 g/mol
The number of mg of vitamin B12 in a 1 mL dose of the supplement
= (Concentration of vitamin B12 in the supplement × Molar mass of methylcobalamin × Volume of 1 dose in liters) / 1 g
= (5000 × 1344.41 × 0.001) / 1 g
= 6.72 mg
Therefore, the number of mg of vitamin B12 in a 1 mL dose of the supplement is 6.72 mg.
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Finish those equations (the most struggling part of my HW) some cannot react
45. Mg(OH)2 + NaCl → …………… + ……………
46. H2SO4 + ……………. → BaSO4↓ + HCl
47. AgNO3 + Fe(NO3)2 → …………… + ……………
48. MgCl2 + NaOH → …………… + ……………
49. NaOH + Fe2(SO4)3 → …………… + ……………
50. Pb(NO3)2 + …………… → PbCl2↓ + 2KNO3
51. MgSO4 + BaCl2 → …………… + ……………
52. Cu(NO3)2 + KOH → …………… + ……………
53. Al(OH)3 + NaOH → …………… + ……………
54. BaCO3 + H2SO4 → …………… + ……………
55. ZnCl2 + …………… → AgCl + ZnSO4
56. BaSO4 + KOH → …………… + ……………
57. Fe(NO3)3 + KOH → …………… + ……………
58. NaHCO3 + KHCO3 → ……………
59. ¬Ba(OH)2 + Na2CO3 → …………… + ……………
60. Al + H2SO4 → …………… + H2
The following chemical equations have been balanced using the appropriate coefficients.
Mg(OH)2 + NaCl → MgCl2 + 2NaOH
H2SO4 + BaCl2 → BaSO4↓ + 2HCl
AgNO3 + Fe(NO3)2 → AgFeO2↓ + 2NO3
How to balance chemical equationsBalancing chemical equations in chemistry involves adjusting the coefficients of the reactants and products in a chemical reaction so that the number of atoms of each element is equal on both sides of the equation.
Note: Never change the subscripts in a chemical formula to balance an equation. Only coefficients can be changed.
Mg(OH)2 + NaCl → MgCl2 + 2NaOH
H2SO4 + BaCl2 → BaSO4↓ + 2HCl
AgNO3 + Fe(NO3)2 → AgFeO2↓ + 2NO3
MgCl2 + 2NaOH → Mg(OH)2↓ + 2NaCl
3NaOH + Fe2(SO4)3 → Fe(OH)3↓ + 3Na2SO4
Pb(NO3)2 + 2KCl → PbCl2↓ + 2KNO3
MgSO4 + BaCl2 → MgCl2 + BaSO4↓
Cu(NO3)2 + 2KOH → Cu(OH)2↓ + 2KNO3
Al(OH)3 + NaOH → NaAlO2 + 2H2O
BaCO3 + H2SO4 → BaSO4↓ + CO2↑ + H2O
ZnCl2 + 2AgNO3 → 2AgCl↓ + Zn(NO3)2
BaSO4 + 2KOH → Ba(OH)2↓ + K2SO4
Fe(NO3)3 + 3KOH → Fe(OH)3↓ + 3KNO3
NaHCO3 + KHCO3 → NaKHCO3 + H2O + CO2↑
Ba(OH)2 + Na2CO3 → BaCO3↓ + 2NaOH
Al + H2SO4 → Al2(SO4)3 + H2↑
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Which conversion factor should be used to convert an area from in^2 to cm^2 if 1 in = 2.54 cm?
The conversion factor that should be used to convert an area from square inches ([tex]\text in^2[/tex]) to square centimeters ([tex]\text cm^2[/tex]) is [tex]\rm (2.54 cm)^2[/tex] / [tex]\rm (1 in)^2[/tex] .
A conversion factor is a numerical ratio that expresses the relationship between two different units of measurement. It is used to convert a quantity from one unit to another unit.
In this case, conversion factor between inches (in) and centimeters (cm) is 1 in = 2.54 cm, so the conversion factor between square inches and square centimeters is:
[tex]\rm (1 in)^2 = (2.54 cm)^2[/tex]
Squaring both sides of the equation gives:
[tex]\rm 1 in^2 = (2.54 cm)^2[/tex]
Therefore, to convert an area from square inches to square centimeters, we can multiply the area in square inches by the conversion factor:
area in [tex]\rm cm^2[/tex] = area in [tex]\rm in^2 \times (2.54 cm)^2 / (1 in)^2[/tex]
Simplifying the expression gives:
area in [tex]\rm cm^2[/tex] = area in[tex]\rm in^2 \times 6.4516[/tex]
So, to convert an area from square inches to square centimeters, we should multiply the area in square inches by 6.4516.
Therefore, The conversion factor between square inches and square centimeters is[tex]\rm (2.54 cm)^2[/tex] / [tex]\rm(1 in)^2[/tex], and to convert an area from square inches to square centimeters, we should multiply the area in square inches by 6.4516.
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the energy of an electrostatic interaction between two charged atoms is dependent on the charges on the atoms, the distance between them, and the dielectric constant of the solvent. for example, the strength of a weak acid ( Ka , acid dissociation constant) depends on the strength of the electrostatic interaction between a negatively charged carboxylic acid group and a proton. the solvent dielectric constant has a large influence on the pKa for weak acids.
The energy of an electrostatic interaction between two charged atoms depends on the charges on the atoms, the distance between them, and the dielectric constant of the solvent.
The strength of a weak acid, represented by its acid dissociation constant (Ka), relies on the strength of the electrostatic interaction between a negatively charged carboxylic acid group and a proton. The dielectric constant of the solvent also plays a significant role in determining the pKa value for weak acids.
The energy of an electrostatic interaction between two charged atoms is influenced by several factors. Firstly, the charges on the atoms themselves play a crucial role in determining the strength of the interaction. If the charges on the atoms are higher, the electrostatic interaction will be stronger. Secondly, the distance between the charged atoms also affects the energy of the interaction. As the distance between the atoms decreases, the electrostatic interaction becomes stronger. Lastly, the dielectric constant of the solvent has a significant influence on the energy of the electrostatic interaction. The dielectric constant of the solvent affects this interaction by either enhancing or reducing the electrostatic forces involved. Consequently, the pKa value for weak acids is greatly influenced by the dielectric constant of the solvent.
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what is the shortest wavelength in the molecule’s absorption spectrum? express your answer in nanometers.
The shortest wavelength in the molecule's absorption spectrum is 121.6 nm.
The energy of a photon is given by the equation:
E = hν
where h is Planck's constant and ν is the frequency of the photon. The frequency of a photon is related to its wavelength by the equation:
ν = c/λ
where c is the speed of light and λ is the wavelength of the photon.
In the molecule's absorption spectrum, the shortest wavelength corresponds to the highest energy photon. The highest energy photon will have a wavelength of:
λ = hc/E
Substituting the values for h, c, and E, we get:
λ = (6.626 × 10⁻³⁴ J⋅s)(3 × 10⁸ m/s)/(1.36 × 10⁻¹⁹ J) = 121.6 nm
Therefore, the shortest wavelength in the molecule's absorption spectrum is 121.6 nm.
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1- List the desired properties of precipitate formed in gravimetric methods of analysis. 2- The addition of dimethylglyoxime, H 2
C 4
H 6
O 2
N 2
to a solution of nickel (II) ion give rise to a precipitate. Ni 2+
+2H 2
C 4
H 6
O 2
N 2
>Ni(H 2
C 4
H 6
O 2
N 2
) 2
+2H +
If 0.15 g nickel alloy is treated with dimethylglyoxime and 175mg nickel dimethylglyoxime is collected. Determine the mass and the percent of nickel in the alloy.
The mass of Ni in the alloy is 0.0323 g, and the percent of nickel in the alloy is 21.53%.
1. The desired properties of precipitates that are formed in gravimetric methods of analysis include:
i. Be completely pure and easily filterable
ii. The precipitate should be of known composition
iii. Large and easily recognizable crystalline structure
iv. The precipitate should be free from impurities
v. The precipitate should be completely insoluble in the mother liquor.
2. Given: Amount of Ni alloy = 0.15 gAmount of Ni-dimethylglyoxime obtained = 175 mgTo determine: The mass and percent of nickel in the alloySolution: The mass of Ni present in Ni-dimethylglyoxime can be calculated by using the molecular weight of Ni-dimethylglyoxime.To calculate the molecular weight of Ni-dimethylglyoxime, multiply the atomic weight of each element by its subscript and then sum the values.
Molecular weight of Ni(H2C4H6O2N2)2 = (58.69 + 2(1.01 × 6) + 2(12.01 × 4) + 2(16 × 2) + 2(14.01 × 2)) g/mol = 318.12 g/mol
Mass of Ni in 175 mg of Ni-dimethylglyoxime = (58.69 g Ni / 318.12 g Ni(H2C4H6O2N2)2) × 0.175 g = 0.0323 g
Percent of Ni in the alloy = (Mass of Ni / Mass of alloy) × 100% = (0.0323 g / 0.15 g) × 100% = 21.53% (rounded off to two decimal places)Therefore, the mass of Ni in the alloy is 0.0323 g, and the percent of nickel in the alloy is 21.53%.
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t-aminocaproic acid with n -pentylamine reaction
This reaction is used in the preparation of polyamides, which are widely used in the manufacture of fibers, films, and plastics.
The reaction equation for t-aminocaproic acid and n-pentylamine is:
C6H13NH2 + HOOC(CH2)4CH(NH2)COOH → C6H13NHCO(CH2)4CH(NH2)COOH + H2O
The reaction of t-aminocaproic acid with n -pentylamine yields an amide which has a 100-word content loaded answer. Here's the detailed information regarding the t-aminocaproic acid with n -pentylamine reaction:
The reaction between t-aminocaproic acid and n-pentylamine results in the formation of amides. During the reaction, the amine group of n-pentylamine replaces the -OH group in t-aminocaproic acid via a condensation reaction.
As a result of this reaction, the molecule loses a molecule of water (H2O), forming an amide bond.
N-pentyl t-aminocaproamide is the chemical name for the product produced by the reaction between t-aminocaproic acid and n-pentylamine. The amide group contains the nitrogen atom (-NH2) and carbonyl carbon (-C=O), which is connected via a single covalent bond.
The reaction between t-aminocaproic acid and n-pentylamine is an example of an acid-amine condensation reaction. This reaction is used in the preparation of polyamides, which are widely used in the manufacture of fibers, films, and plastics.
The reaction equation for t-aminocaproic acid and n-pentylamine is:
C6H13NH2 + HOOC(CH2)4CH(NH2)COOH → C6H13NHCO(CH2)4CH(NH2)COOH + H2O
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Assuming that you have benzoic acid in your unknown. -Is it more efficient if you use 3.0MCH 3
COONa (sodium acetate) as your base in place of 3.0MNaOH for your extractions? - Draw out the acid-base equilibrium between benzoic acid- NaOH and benzoic acid-acetate, Calculate the equilibrium constant of each reaction. Use appropriate equilibrium arrow to show the direction of each equilibrium.
Yes, it is more efficient if you use 3.0 M CH3COONa (sodium acetate) as your base instead of 3.0 M NaOH for your extractions. This is because benzoic acid can be more efficiently converted to its salt form by a weak base (CH3COO- ion is a weak base) than by a strong base (OH- ion is a strong base).
In the presence of NaOH, benzoic acid is ionized to form a benzoate ion (C6H5COO-) and a proton (H+):
C6H5COOH + NaOH → C6H5COO- Na+ + H2O
In the presence of CH3COONa (sodium acetate), benzoic acid is converted to its salt form, i.e., the benzoate ion: C6H5COOH + NaOAc → C6H5COO- Na+ + H2O.
The benzoate ion is negatively charged and hence can be easily extracted into the organic layer. Thus, sodium acetate is more efficient in extraction than sodium hydroxide.
Here's the equilibrium equation for benzoic acid-NaOH:
C6H5COOH + NaOH → C6H5COO- Na+ + H2O
Here's the equilibrium equation for benzoic acid-acetate:
C6H5COOH + NaOAc → C6H5COO- Na+ + H2O
The equilibrium constant for each reaction is calculated using the equation:
K = [C6H5COO- Na+][H2O]/[C6H5COOH][NaOH]
K1 = [C6H5COO- Na+][H2O]/[C6H5COOH][NaOH]
K2 = [C6H5COO- Na+][H2O]/[C6H5COOH][NaOAc]
Where K1 and K2 are the equilibrium constants for the benzoic acid-NaOH and benzoic acid-acetate reactions, respectively.The direction of each equilibrium is shown by the appropriate equilibrium arrow.
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complete the mechanism for the reaction between acetophenone in acid and chlorine by adding the missing bonds, atoms, charges, nonbonding electrons, and curved arrows (forward reaction only). the mechanism for steps 1 to 3 are pre‑drawn for you.
The mechanisms include:
Chlorine adds to the carbonyl carbon of acetophenone, forming a chlorinated intermediate.The chlorinated intermediate loses a proton, forming a carbocation.A chloride ion attacks the carbocation, forming chloroacetophenone.The water molecule abstracts a proton from chloroacetophenone, forming the final product, 2-chloro-1-phenylethanol.How to create the mechanism?Step 1: The chlorine atom in Cl2 is electrophilic, meaning that it has a partial positive charge. The carbonyl carbon of acetophenone is nucleophilic, meaning that it has a partial negative charge. The chlorine atom and the carbonyl carbon form a new covalent bond, and a chloride ion is formed.
Step 2: The chlorinated intermediate is unstable because it has a positive charge on a carbon atom. The carbon atom loses a proton, forming a carbocation.
Step 3: A chloride ion attacks the carbocation, forming chloroacetophenone. The chloride ion is a good nucleophile because it has a lone pair of electrons. The lone pair of electrons on the chloride ion attacks the carbocation, forming a new covalent bond.
Step 4: The water molecule abstracts a proton from chloroacetophenone, forming the final product, 2-chloro-1-phenylethanol. The water molecule is a good base because it has a lone pair of electrons. The lone pair of electrons on the water molecule attacks the proton on chloroacetophenone, forming a new covalent bond and a hydroxide ion. The hydroxide ion then leaves, taking the proton with it.
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The Ksp of lead(II) carbonate, PbCO3, is 7.40×10−14.
Calculate the concentration of lead(II) ions in a saturated
solution of PbCO3.
The concentration of lead(II) ions in a saturated solution of PbCO3 is 8.60 × 10⁻⁸ M.
Solubility products are calculated with the help of solubility constants or Ksp values. These constants reflect the product of the concentrations of ions in a saturated solution at equilibrium. Therefore, solubility constants are a kind of equilibrium constant.In a saturated solution of PbCO3, the lead(II) ion concentration is to be calculated. The solubility product constant (Ksp) of lead(II) carbonate is 7.40 × 10⁻¹⁴.
The reaction for the dissolution of lead(II) carbonate in water can be written as:
PbCO₃(s) ⟶ Pb²⁺(aq) + CO₃²⁻(aq)
The solubility equilibrium equation can be expressed as:
Ksp = [Pb²⁺][CO₃²⁻]
We are given the Ksp for lead(II) carbonate, therefore, we can substitute its value:
7.40 × 10⁻¹⁴ = [Pb²⁺][CO₃²⁻]
The molar concentration of Pb²⁺ ion is equal to the concentration of CO₃²⁻ ion in the saturated solution because one mole of each ion is formed in the dissolution of one mole of PbCO3.Therefore, [Pb²⁺] = [CO₃²⁻]Let the concentration of lead(II) ion be x, therefore:[
Pb²⁺] = x
Therefore, the solubility product expression becomes:
Ksp = x²7.40 × 10⁻¹⁴ =
x²x = √(7.40 × 10⁻¹⁴)
x = 8.60 × 10⁻⁸
Therefore, the concentration of lead(II) ions in a saturated solution of PbCO3 is 8.60 × 10⁻⁸ M.
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the following procedure was used to determine the volume of a flask. the flask was weighed dry and then filled with water. if the masses of the empty flask and filled flask were and , respectively, and the density of water is , calculate the volume of the flask in . be sure your answer has the correct number of significant digits.
The volume of flask is equal to the volume of water filled which is: 31.345 cm³
How to find the Volume of the Flask?The parameters are given as:
Mass of the empty flask: m = 56.12 g
Mass of the flask filled with water: M = 87.39 g
Density of the water: ρ = 0.9976 g/cm³
Thus, we can calculate as follows:
Mass of water filled in the flask is gotten as:
Mw = M - m
Mw = 87.39 - 56.12
Mw = 31.27 g
Formula for density is expressed as:
Density = mass / volume
Therefore, for water we can say that:
0.9976 = 31.27 / volume
Volume of water = 31.345 cm³
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Complete question is:
The following procedure was used to determine the volume of a flask. The flask was weighed dry and then filled with water. If the masses of the empty flask and filled flask were 56.12 g and 87.39 g, respectively, and the density of water is 0.9976 g/cm³, calculate the volume of the flask in cubic centimeters.
When performing absorption spectrometry, Logger Pro should display a plot of: Molar absorptivity vs. Wavelength. Absorbance vs. Wavelength. Absorbance vs. Concentration. Molar absorptivity vs. Concentration.
The correct plot that Logger Pro should display while performing absorption spectrometry is Absorbance vs. Wavelength. This is because Absorbance vs. Wavelength is a graph of how much light of each wavelength is absorbed by the sample.
Absorbance is the logarithm of the ratio of the incident light intensity to the transmitted light intensity, and it is proportional to the concentration of the analyte in the solution.The relationship between absorbance and concentration is given by Beer-Lambert Law. The Beer-Lambert Law, also known as the Beer-Lambert-Bouguer Law, describes the absorption of light by a material as a function of its concentration. The law states that absorbance is proportional to the path length, the concentration of the absorbing species, and the molar absorptivity, which is a property of the absorbing species and the wavelength of light used.
Logger Pro is software that is used to collect and analyze data from various experiments, including absorption spectrometry. It is useful for plotting graphs of data collected from an experiment. In summary, the plot that Logger Pro should display during absorption spectrometry is Absorbance vs. Wavelength, which helps determine the concentration of an analyte in a solution using the Beer-Lambert Law.
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the atomic models after dalton’s time included ideas about the atomic structure. which atomic model that shows the atomic structure is missing from this set? bohr’s model schrödinger’s model rutherford’s model thomson’s model
The atomic models after Dalton's time that included ideas about the atomic structure are Rutherford's model, Thomson's model, Bohr's model, and Schrödinger's model.
Thomson's atomic model is missing from this set.
Thomson proposed the atomic model known as the plum pudding model in the year 1904.
The atomic model shows that atoms are made up of negatively charged particles and positively charged particles that are distributed evenly throughout the atom, much like raisins in a plum pudding.
In 1911, Ernest Rutherford discovered the atomic nucleus, which is made up of positively charged particles, through his gold foil experiment.
Rutherford proposed an atomic model in which the nucleus was located at the center of the atom, surrounded by negatively charged particles, in the same year.
This model is also known as the planetary model of the atom.
In 1913, Niels Bohr proposed a new atomic model based on Rutherford's planetary model that included electron orbits.
Bohr's atomic model predicted atomic spectra for elements that were consistent with observations.
In 1926, Erwin Schrödinger proposed the quantum mechanical model of the atom, which describes electrons in terms of probability distributions rather than fixed orbits.
The electrons in this model are described as wavefunctions.
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Show how to compute an energy level diagram for the first 50 and 500 states and an energy level diagram as a histogram for a particle in a box as:
a. one-dimensional particle in a box with L=1nm
b- three-dimensional particle in a box with Lx=Ly=Lz=1nm
In quantum mechanics, the particle in a box is a critical model of a particle in a potential well. It is also referred to as the infinite potential well. A particle in a box, also known as a quantum dot, is a term used to describe a solitary electron or an atom confined to a region in space by forces such as electric, magnetic, or electrostatic fields.
For the first 50 states of an energy level diagram for a particle in a box, the equation for determining the energy of a particle confined in a one-dimensional box of width L can be used as follows: $$E_n=\frac{n^2 h^2}{8mL^2}.$$Where E is the energy, n is the quantum number of the energy level, h is Planck's constant, m is the mass of the particle, and L is the length of the box. For the first 500 states of an energy level diagram for a particle in a box, the same equation as that of the first 50 states can be used to determine the energy level of the system. For a 1D particle in a box with L=1nm, the graph of the energy levels will be a straight line, where the energy levels increase as the quantum number n increases.
The energy values are plotted on the y-axis and the quantum numbers are plotted on the x-axis. For a 3D particle in a box with Lx=Ly=Lz=1nm, the energy levels will be distributed as a histogram. The histogram will show a higher concentration of states near the ground state energy, which is equal to 3/2 times the energy of a particle in a 1D box of the same length.
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There are two isotopes of an unknown elesnent, x−19 and x−21. The abundsnce of x−19 is 13.898, A weghted average uses the percentages of each isotope to scale their contribution to the total mass. Each isotope's contribution is the percentage (in decimal formi rultipied by the mass of the isotope. What is the comribution ( n arnu) to the weighted average from the X-19 iscoope, which has a mass of 19.00 amu? of Pont Earned- 313 utarets hemaning What is the cercentage of the X-2t woloce? 3f3 Muenest Almaking 917histeriad 3/3 kikntintakeseded
The contribution to the weighted average for the x-21 isotope = (0.86102) × (21.00) = 18.08642amu
Therefore, the contribution of x-19 isotope to the weighted average is 2.62482amu and the percentage of the x-21 isotope is 86.102%.
Let's solve the given problem. The unknown element has two isotopes; the abundances are given below:x-19; abundance is 13.898%x-21; abundance is 100 - 13.898
= 86.102%
The contribution to the weighted average from the x-19 isotope with a mass of 19.00 amu can be calculated as follows:
Mass of x-19 isotope
= 19.00 amu Abundance of x-19 isotope
= 13.898% or 0.13898Contributions to the weighted average for x-19 isotope
= (0.13898) × (19.00)
= 2.62482amu
The percentage of the x-21 isotope can be calculated as follows:
Mass of x-21 isotope
= 21.00 amu Abundance of x-21 isotope
= 100 - 13.898
= 86.102% or 0.86102.
The contribution to the weighted average for the x-21 isotope
= (0.86102) × (21.00)
= 18.08642amu
Therefore, the contribution of x-19 isotope to the weighted average is 2.62482amu and the percentage of the x-21 isotope is 86.102%.
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What volume of groundwater containing 2 mg l−1 dissolved oxygen (DO) would be required to oxidize 23 l of octane that leaked out of an underground storage tank (UST) and into an unconfined aquifer?
List five factors that could affect reaction rate, and create a schematic diagram to show potential pathways of octane as it leaks out of the UST. (1 sentence per factor)
List and briefly describe three potential remediation strategies. (1 sentence per strategy)
Potential pathways of octane as it leaks out of the UST are via volatilization into the air, absorption into the soil, dissolution into the groundwater, and biodegradation by microorganisms.
Three potential remediation strategies include in-situ bioremediation, air sparging, and chemical oxidation.
The volume of groundwater containing 2 mg l−1 dissolved oxygen (DO) required to oxidize 23 l of octane that leaked out of an underground storage tank (UST) and into an unconfined aquifer would be 9,200 liters of water.
Factors that could affect reaction rate include temperature, pH, presence of catalysts, surface area, and pressure.
Potential pathways of octane as it leaks out of the UST are via volatilization into the air, absorption into the soil, dissolution into the groundwater, and biodegradation by microorganisms.
Three potential remediation strategies include in-situ bioremediation, air sparging, and chemical oxidation.
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Did you use the same apparatus for all your measurements? if not, how could it affect your data analysis outcome?
No, using different apparatus for measurements can affect the data analysis outcome.
Using different apparatus for measurements can introduce variability in the data. Each apparatus may have slight variations in calibration, sensitivity, or precision, leading to inconsistent measurements. This can result in inaccurate or unreliable data. In data analysis, consistency and precision are crucial for making reliable conclusions. If different apparatus were used, it is important to account for these variations during data analysis.
This can be done by conducting appropriate calibration or standardization procedures to ensure the measurements are comparable. Additionally, statistical methods like averaging or calculating standard deviations can help mitigate the impact of variability introduced by different apparatus. Overall, using the same apparatus for all measurements ensures consistency and reduces the potential for systematic errors in data analysis.
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Find the ratio of the electrostatic to gravitational force between two electrons.
The ratio of the electrostatic to the gravitational force between two electrons is: Fe/Fg = 4.17 * 10⁴²
What is the ratio of the forces between the electrons?The electrostatic force between two electrons is given by the formula:
F_e = kq²/r²
where:
k is electrostatic force constant = 9 * 10⁹ N.m²/C²
q is charge on electron = 1.6 * 10⁻¹⁹ C
r is the separation between the electrons.
The gravitational force between two electrons is given by the formula:
G_e = Gm_e²/r²
where:
G is the universal gravitational constant = 6.67 * 10⁻¹¹ N.m²/kg²
m_e is mass of electron = 9.1 * 10⁻³¹ kg
Thus, the ratio of the electrostatic force to gravitational force between two electrons is:
Fe/Fg = kq²/(Gm_e²)
Fe/Fg = (9 * 10⁹ * (1.6 * 10⁻¹⁹)²)/(6.67 * 10⁻¹¹ * (9.1 * 10⁻³¹)²)
Fe/Fg = 4.17 * 10⁴²
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Gold has a density of 19.3g/cm^3. what is the mass, in kg, of a cube of gold measuring 2.0 inches on each side? use unit analysis, not the density formula.
The mass of the cube of gold measuring 2.0 inches on each side is 2.56 kg.
Given that gold has a density of 19.3 g/cm³ and the cube of gold has sides of 2.0 inches, we need to calculate the mass of the cube in kg without using the density formula. We can convert the length of each side of the cube from inches to cm by using the conversion factor 1 inch = 2.54 cm.
Therefore, each side of the cube measures:2.0 inches × 2.54 cm/inch = 5.08 cm. The volume of the cube can be calculated by raising the length of one side to the third power, i.e.,
V = (side)³ = (5.08 cm)³ = 132.5 cm³
The mass of the cube can be calculated by using the density of gold and the volume of the cube. However, we are not supposed to use the density formula, so we can use unit analysis to determine the mass in kg. We have:
19.3 g/cm³ = 19.3 g/cm³ × 1 kg/1000 g = 0.0193 kg/cm³
Multiplying this by the volume of the cube in cm³, we get:
0.0193 kg/cm³ × 132.5 cm³ = 2.56 kg
Therefore, the mass of the cube of gold measuring 2.0 inches on each side is 2.56 kg.
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The total concentration of a phosphate buffer containing NaH 2
PO 4
and Na 2
HPO 4
is 15mM and pH of the solution is 7.3. What would be the concentration of each phosphate species in the buffer. Given that pK a
=7.2
The concentration of each phosphate species in the buffer are H2PO4- = 5.34 mM and HPO42- = 9.66 mM.
The total concentration of a phosphate buffer containing NaH2PO4 and Na2HPO4 is 15 mM and pH of the solution is 7.3. We need to calculate the concentration of each phosphate species in the buffer.
Given that pKa= 7.2.The reaction equation for the phosphate buffer is as follows;
H2PO4- ⇔ HPO42- + H+
The pKa of the acid is 7.2.
Hence, pH = pKa + log([HPO42-]/[H2PO4-])
or, log([HPO42-]/[H2PO4-]) = pH - pKa
or, log([HPO42-]/[H2PO4-]) = 7.3 - 7.2
or, log([HPO42-]/[H2PO4-]) = 0.1
or, [HPO42-]/[H2PO4-] = 10^(0.1)
or, [HPO42-]/[H2PO4-] = 1.258
or, [H2PO4-] = 15/(1 + 1.258)
= 5.34 m
M [HPO42-] = 15 - 5.34
= 9.66 mM
Therefore, the concentration of each phosphate species in the buffer are as follows:
H2PO4- = 5.34 mM HPO42- = 9.66 mM
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For the air as a closed system, the _____ terms can be neglected in the energy balance.
The negligible terms in the energy balance refer to the specific energy change associated with potential energy and kinetic energy.
Which terms can be neglected?For the air as a closed system, the negligible terms in the energy balance typically refer to the specific energy change associated with potential energy and kinetic energy. In many cases, these terms can be neglected or assumed to be constant if the changes in elevation and velocity of the air are small and do not significantly affect the overall energy balance.
The energy balance equation for the closed system of air usually focuses on the conservation of internal energy, which includes terms related to heat transfer and work. Neglecting the potential energy and kinetic energy terms simplifies the analysis by assuming that any changes in elevation and velocity are insignificant compared to the other energy transfers involved.
It is important to note that neglecting these terms is an assumption made for simplification purposes and may not be applicable in certain scenarios where elevation changes or air velocity have a significant impact on the energy balance.
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