The mass of nitrogen dissolved in the 92.0 L home aquarium at room temperature is 778.48 grams.
The mass of gases refers to the amount of matter or substance present in a gaseous state. It represents the total mass of all the gas particles within a given volume.
In the study of gases, the mass of gases is often expressed in terms of molar mass, which is the mass of one mole of the gas. Molar mass is typically measured in grams per mole (g/mol).
The mass of gases can be calculated using the ideal gas law, which relates the pressure, volume, temperature, and number of moles of a gas. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
n = X × P × V / (R × T)
Given :
P = 1.0 atm
V = 92.0 L
X = 0.78 (mole fraction of nitrogen)
T = 25 + 273.15 = 298.15 K.
n = 0.78 × 1.0 atm × 92.0 L / (0.0821 L.atm/mol.K × 298.15 K)
n = 27.86 moles
Mass of nitrogen = n × molar mass
Mass of nitrogen = 27.86 moles × 28.0 g/mol
Mass of nitrogen = 778.48 g
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6) Write the name of each of the following polyatomic ions. a) CO 3
2
b) PO 4
2
c) C 2
H 1
d) Cr 2
O 7
2
7) Write the formula, including charge, for each of the following. a) Hydroxide b) Nitrite c) Perchlorate d) Sulfate
7) The formula including charge for Hydroxide ion, Nitrite ion, Perchlorate ion and Sulfate ion are OH−, NO2−, ClO4−, and SO42− respectively
correct question:
6) Write the name of each of the following polyatomic ions.
a) CO₃2
b) PO 42
c) C ₂H ₁
d) Cr₂O ₇2
7) Write the formula, including charge, for each of the following.
a) Hydroxide
b) Nitrite
c) Perchlorate
d) Sulfate
answer:
6) Write the name of each of the following polyatomic ions.
a) CO32− (Carbonate ion)
b) PO42− (Phosphate ion)
c) C2H3O2− (Acetate ion)
d) Cr2O72− (Dichromate ion)
7) Write the formula, including charge, for each of the following.
a) Hydroxide ion - OH−
b) Nitrite ion - NO2−
c) Perchlorate ion - ClO4−
d) Sulfate ion - SO42−Polyatomic ions are made up of more than one atom.
These atoms are joined together with covalent bonds and have an overall electrical charge.
Polyatomic ions are very common in nature and they play a crucial role in various chemical processes.
Carbonate ion, Phosphate ion, Acetate ion and Dichromate ion are the names of CO32−, PO42−, C2H3O2−, and Cr2O72− respectively.
Hydroxide ion (OH−) has a charge of 1−, Nitrite ion (NO2−) has a charge of 1−,
Perchlorate ion (ClO4−) has a charge of 1−, and Sulfate ion (SO42−) has a charge of 2−.
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What is the magnitude of the charge of the electrons in 6.70 mol of neutral molecular-hydrogen gas (h2)?
The magnitude of the charge of the electrons in 6.70 moles of neutral molecular hydrogen gas, H₂ is 1.29×10⁶ C
How do i determine the magnitude of the charge of the electrons?The following data were obtained from the question:
Number of mole (n) = 6.70 molesAvogadro's constant (N) = 6.022×10²³ Number of molecules = n × N = 6.70 × 6.022×10²³ = 4.035×10²⁴ moleculesElementary charge (C) = 1.6×10⁻¹⁹ CMagnitude of charge =?The magnitude of the charge on the electron of the neutral hydrogen can be obtained as illustrated below:
Magnitude of charge = 2 × Number of molecules × C
= 2 × 4.035×10²⁴ × 1.6×10⁻¹⁹
= 1.29×10⁶ C
Thus, we can conclude from the above calculation that the magnitude of the charge is 1.29×10⁶ C
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17. the binding of the amino acid in aminoacyl-trna is a (n) a. amide c. hemiacetal b. ester d. ether
The binding of the amino acid in aminoacyl-tRNA involves the formation of an ester bond. Option b
Aminoacyl-tRNA is a complex molecule that plays a crucial role in protein synthesis. It consists of a tRNA molecule covalently linked to an amino acid. The amino acid is attached to the 3' end of the tRNA molecule through an ester bond.
An ester bond is formed between the carboxyl group (-COOH) of the amino acid and the hydroxyl group (-OH) of the ribose sugar at the 3' end of the tRNA molecule. This ester bond is also referred to as an ester linkage. The formation of the ester bond is catalyzed by the enzyme aminoacyl-tRNA synthetase.
The ester bond in aminoacyl-tRNA is essential for protein synthesis. During translation, the aminoacyl-tRNA molecule carries the specific amino acid to the ribosome, where it is incorporated into the growing polypeptide chain. The ester bond is later hydrolyzed, releasing the amino acid for further use in protein synthesis.
In summary, the binding of the amino acid in aminoacyl-tRNA involves the formation of an ester bond between the carboxyl group of the amino acid and the hydroxyl group of the ribose sugar in the tRNA molecule.
Option b
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water molecule bond to each other via hydrogen bonds. these bonds form between the slight negative charge of , and the slight positive charge of on adjacent water molecules
Water molecular bond to each other via hydrogen bonds. These bonds form between the slight negative charge of oxygen, and the slight positive charge of hydrogen, on adjacent water molecules.
Water molecules are made up of two hydrogen atoms bonded to one oxygen atom. Oxygen is more electronegative than hydrogen, which means it attracts electrons more strongly. As a result, the oxygen atom in a water molecule carries a slight negative charge, while the hydrogen atoms carry slight positive charges. These partial charges create an electrostatic attraction between the oxygen of one water molecule and the hydrogen of another water molecule. This attraction is called a hydrogen bond.
Hydrogen bonds are relatively weak compared to covalent bonds, but they are strong enough to give water its unique properties. The hydrogen bonding between water molecules contributes to high surface tension, which allows insects to walk on water. Additionally, hydrogen bonds allow water to absorb a large amount of heat, which helps regulate temperature in organisms and stabilize Earth's climate.
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Describe about protecting groups in organic synthesis
(more than 4).
Protecting groups are temporary functional groups that are added to a molecule to block certain reactive sites in the molecule while allowing other reactions to proceed unhindered. These groups are used in organic synthesis to protect certain functional groups from undesired reactions and to ensure selective reactivity. They are widely used in organic synthesis to enable a specific bond to be formed without unwanted side reactions.
The protecting groups in organic synthesis are mainly divided into two types, those that protect primary alcohols and those that protect carbonyl groups. The most common protecting groups used in organic synthesis include tert-butyldimethylsilyl (TBDMS), methoxymethyl (MOM), trimethylsilyl (TMS), and tert-butoxycarbonyl (Boc).For example, the TBDMS group is commonly used to protect primary alcohols and can be removed under mild conditions using fluoride ion. The MOM group is used to protect primary alcohols and can be removed using acidic conditions. The TMS group is used to protect carbonyl groups and can be removed using fluoride ion. The Boc group is used to protect amines and can be removed using acidic conditions.
Protecting groups are essential for the synthesis of complex organic molecules, and the development of new protecting groups is an ongoing area of research in organic chemistry. In summary, protecting groups are a crucial tool in organic synthesis that help ensure the desired product is obtained and unwanted side reactions are minimized.
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A certain liquid X has a normal boiling point of 121.70 "C and a boiling point elovation constant K i
=131 "C kg mol 1 a solution is prepared by. dissolving some olycine (C 2
H 3
NO 2
) in 800 . g of X. This solution boils at 123.7 C. Calculate the mass of C 2
H 5
NO that was dissolved. Round your answer to 2 significant digits.
The Mass of glycine = Molality × Molar mass × Mass of solvent=0.0153 × 75 × 800=91.35 g ≈ 91 g Mass of C2H5NO2 that was dissolved is approximately 91 g.
Hence, the correct option is 91.
Given,Normal boiling point of liquid X, T1
= 121.7°C Boiling point elevation constant, Kb
= 131°C kg mol-1 Mass of solvent, w2
= 800 g Boiling point of solution, T2
= 123.7°C
Now, Boiling point elevation is given by, ΔTb
= T2 - T1ΔTb
= 123.7 - 121.7ΔTb
= 2°C The boiling point elevation is given by,ΔTb
= Kb × bΔTb/Kb
= b Molality (b) of solution, b
= ΔTb/Kb
=2/131
=0.0153 mol/kg
The molality of the solution is 0.0153 mol/kg.Molar mass of glycine, C2H3NO2
= 75 g/mol
We need to calculate the mass of glycine dissolved in the solution.
The Mass of glycine
= Molality × Molar mass × Mass of solvent
=0.0153 × 75 × 800
=91.35 g ≈ 91 g Mass of C2H5NO2 that was dissolved is approximately 91 g.
Hence, the correct option is 91.
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A calibration curve was created to determine the quantity of protein in a solution. The calibration curve has the form of a straight line with the equation A=0.0182x+0.007 where A is the corrected absorbance of the solution and x is quantity of protein in the solution in units of micrograms ( μg ). Determine the quantity of protein in a solution that has an absorbance of 0.338. A blank solution has an absorbance of 0.055. quantity of protein:
The x = (0.283 - 0.007)/0.0182≈ 14.53 μg.
Therefore, the quantity of protein in the solution that has an absorbance of 0.338 is 14.53 μg.
The problem is to determine the quantity of protein in a solution that has an absorbance of 0.338. A blank solution has an absorbance of 0.055. The calibration curve has the form of a straight line with the equation
A=0.0182x+0.007
where A is the corrected absorbance of the solution and x is the quantity of protein in the solution in units of micrograms (μg).
To solve the problem, we need to use the equation of the calibration curve given above. But first, we need to find out the corrected absorbance of the sample solution.
The corrected absorbance is the difference between the absorbance of the sample solution and the absorbance of the blank solution.
This is given by:Corrected absorbance
= Absorbance of sample solution - Absorbance of blank solution
= 0.338 - 0.055
= 0.283 μg
Now, we can use the equation of the calibration curve to find out the quantity of protein in the solution.
This is given by:x
= (A - b)/m
Where,x
= quantity of protein in the solution A
= corrected absorbance of the sample solutionb
= intercept of the calibration curve
= 0.007m = slope of the calibration curve
= 0.0182. The x
= (0.283 - 0.007)/0.0182≈ 14.53 μg.
Therefore, the quantity of protein in the solution that has an absorbance of 0.338 is 14.53 μg.
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5.if 75.0 ml of 0.250 m hno3 and 75.0 ml of 0.250 m koh are mixed, what is the molarity of the salt in the resulting solution?
The molarity of the salt in the resulting solution will be 0.125 M.
What is molarity ?When HNO3 and KOH are mixed, they react to form KNO3, a salt. The balanced equation for the reaction is:
HNO3 + KOH → KNO3 + H2O
The molarity of a solution is the number of moles of solute per liter of solution. In this case, the solute is KNO3.
The total volume of the solution is 75.0 mL + 75.0 mL = 150.0 mL.
The number of moles of KNO3 formed is:
0.250 M * 75.0 mL = 18.75 mmol
The molarity of KNO3 is:
18.75 mmol / 150.0 mL = 0.125 M
Therefore, the molarity of the salt in the resulting solution is 0.125 M.
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What is the pH of a 0.03M solution of Acetic Acid?
Assume for Acetic acid, Ka=1.74x10-5. Give
your answer to 2 decimal places
Therefore, the pH of a 0.03 M solution of acetic acid is 2.88. The answer should be provided in two decimal places.
Acetic acid is a weak acid, meaning it only partially dissociates in water.
To find the pH of a 0.03 M solution of acetic acid, we first need to calculate the concentration of hydrogen ions (H+) produced when the acid partially dissociates.
We can do this by using the acid dissociation constant (Ka) of acetic acid.
Ka = [H+][C2H3O2−] / [HC2H3O2]
Ka = 1.74 × 10^-5 (given)
[H+] = x[C2H3O2−]
[H+] = x[HC2H3O2]
[H+] = 0.03 - x
Substituting these values in the expression for Ka gives:
1.74 × 10^-5 = x(0.03 - x) / 0.03
Solving for x: x = 0.00132 M
[H+] = 0.00132 M
pH = -log[H+]
pH = -log(0.00132)
pH = 2.88
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For the following reaction, 3.17 grams of hydrochloric acid are mixed with excess barium hydroxide. The reaction yields 7.02 grams of barium chloride.
hydrochloric acid (aq) + barium hydroxide (aq) barium chloride (aq) + water (l)
What is the theoretical yield of barium chloride ? grams
What is the percent yield of barium chloride ? %
The percent yield of BaCl2 is 58.8%.To find out the theoretical yield of barium chloride (BaCl2), we will use the given information to calculate the amount of barium chloride that should have been produced.
Given;3.17 grams of hydrochloric acid (HCl)7.02 grams of barium chloride (BaCl2)When we write the balanced chemical equation of the given reaction;
HCl (aq) + Ba(OH)2 (aq) → BaCl2 (aq) + 2H2O (l)
We find that for 1 mole of HCl, 1 mole of BaCl2 should be produced.
So, we can find the amount of BaCl2 that should have been produced from the amount of HCl that was used.
We can start by finding the number of moles of HCl that was used.
Molar mass of HCl = 36.5 g/mol
Moles of HCl = mass ÷ molar mass = 3.17 ÷ 36.5 = 0.087 moles
From the balanced equation, we know that 1 mole of HCl produces 1 mole of BaCl2.
Therefore, 0.087 moles of HCl should produce 0.087 moles of BaCl2.
Molar mass of BaCl2 = 137.3 g/mol
Theoretical yield of BaCl2 = moles of BaCl2 × molar mass
= 0.087 × 137.3= 11.94 grams
Therefore, the theoretical yield of BaCl2 is 11.94 grams.
Percent yield of BaCl2 is calculated using the formula;
Percent yield = (actual yield ÷ theoretical yield) × 100%
We are given that the actual yield of BaCl2 is 7.02 grams (which was obtained from the reaction).
So, Percent yield = (7.02 ÷ 11.94) × 100%
Percent yield = 58.8%.
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