Explanation:
The required concentration of [tex]HNO_3[/tex] M1 =0.222 M.
The required volume of [tex]HNO_3[/tex] is V1 =225 mL.
The standard solution of [tex]HNO_3[/tex] is M2 =16 M.
The volume of standard solution required can be calculated as shown below:
Since the number of moles of solute does not change on dilution.
The number of moles [tex]n=molarity * volume[/tex]
[tex]M_1.V_1=M_2.V_2[/tex]
[tex]V2=\frac{M_1.V_1}{M_2} \\=0.222M x 225 mL / 16 M\\=3.12 mL[/tex]
Hence, 3.12 mL of 16 m nitric acid is required to prepare 0.222 M and 225 mL of nitric acid.
Another method for creating a buffer, in situ, is to add an appropriate amount of a strong base, e.g., NaOH, to a weak acid OR add an appropriate amount of a strong acid, e.g., HNO3, to a weak base. As an example, mixing 1.0 mol of acetic acid with 0.5 mol of NaOH will result in a buffer solution with 0.5 mol of acetic acid and 0.5 mol of acetate. The acetate is created by the reaction of acetic acid and the strong base, hydroxide. Given this information, which of the following, when mixed with the appropriate amount of HCl, would create a buffer solution?
a. HNO3
b. HClO2
c. LiCl
d. NH3
Answer:
As an example, mixing 1.0 mol of acetic acid with 0.5 mol of NaOH will result in a buffer solution with 0.5 mol of acetic acid and 0.5 mol of acetate. The acetate is created by the reaction of acetic acid and the strong base, hydroxide.
When HClO2 is mixed with the appropriate amount of HCl it would create a buffer solution. That is option B.
Methods used to form buffer solutionA buffer solution is the solution that resists a change in pH of a solution when acid or base is added because it is made up of weak acid and the conjugate base or weak base and the conjugate acid.
The methods that can be used to form a buffer solution include:
Adding a strong base to a weak acid: For example, mixing 1.0 mol of acetic acid with 0.5 mol of NaOH will result in a buffer solution with 0.5 mol of acetic acid and 0.5 mol of acetate.Adding a weak acid to a conjugate base: For example HCl is a strong acid which will react with a conjugate base such as HClO2.Although HCl is a strong acid, it can be converted to a weak acid through dilution with water. It is in this context that it can be used to form a buffer solution.
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Consider the following reaction:
Cr(NO3)3 (aq) + 2NaF (aq) --> 3NaNO3 (aq) + CrF3 (s)
If 21.0 grams of NaF are needed to precipitate all of the Cr+3 ions present in 0.125L of a solution of Cr(NO3)3, what is the molarity of the Cr(NO3)3 solution?
Your answer should be to 2 decimal places.
Answer:
2.01
Explanation:
First, let's convert grams to moles
(Na) 22.99 + (F) 18.998 = 41.988
Every mole of NaF is 41.988 grams
21/41.988 = 0.500143 moles of NaF
For every Cr+3, we will need 2 NaF, so Cr+3 will be half of NaF
0.500143/2 = 0.250071
molarity = moles/liters
0.250071/0.125 = 2.0057 M
1. Draw the condensed structural formula of sodium benzoate showing all charges, atoms including any lone pairs in the side chain functional group, and all sigma and pi bonds.
2. Draw the condensed structural formula of benzoic acid showing all atoms including any lone pairs in the side chain functional group, and all sigma and pi bonds. Indicate the acidic hydrogen.
3. Draw the condensed structural formula of tetrahydrofuran (THF) showing all heteroatoms plus their lone pairs and all sigma and pi bonds.
The structures are shown in the image attached.
A structural formula is the representation of the molecule in which all atoms and bonds in the molecule are shown.
Since the question requires that all the lone pairs, formal charges and sigma and pi bonds should be shown, then the simple condensed structural formula becomes insufficient in this case.
I have attached images of the structural formula of sodium benzoate (image 1), benzoic acid (image 2) and tetrahydrofuran (image 3).
All the formal charges, lone pairs as well as sigma and pi bonds are fully shown.
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When comparing Be2 and H2:
I. Be2 is more stable because it contains both bonding and antibonding valence electrons.
II. H2 has a higher bond order than Be2.
III. H2 is more stable because it only contains 1s electrons.
IV. H2 is more stable because it is diamagnetic, whereas Be2 is paramagnetic
a. II,III,IV
b.II,III
c.III only
d.I,II
e.III,IV.
Answer:
The answer is "Option b".
Explanation:
H2 does have bond energy of 1, while Be2 has a covalent bond of zero. Be2 has eight electrons, each of which dwells in a distinct orbital. As just a result, four of them are linked molecular orbitals and two are antibonding molecular orbitals, respectively. As just a result, this molecule is unstable. This chemical orbital, with a bond order of 1, has just two electrons. As a result, it is a very solid substance. H2's bond length is higher than Be2's. Since it only has one electron, H2 is more stable than that of other compounds.
what are the properety of covalent bond
Explanation:
1. boiling and melting point
2. electrical conductivity
3. Bond strength
4. bond length
A covalent bond consists of negative electrons that are shared in between atoms. Because of this bond, they possess and manifest physical abilities, including electrical pressure/conductivity and lower melting points compared to ionic compounds.
For an atoms electrons, how many energy sublevels are present in the principal energy level n = 4?
A. 4
B. 9
C. 10
D. 16
E. 32
Answer:
by the own's formula energy sublevels are 2 the power of n or principal quantum number this means 2 the power of 4 equal to 32
how many moles of oxygen are present in 16 g of oxygen gas
Answer:
Mole = molecular weight / molecular mass
Mole = 16/16
Mole= 1
Van der Waals forces hold molecules together by: A. moving electrons from one molecule to another. B. attracting a lone pair of electrons to the positive charge of a hydrogen. C. inducing temporary dipoles that attract each other. D. sharing electrons between atoms.
Van der Waals forces hold molecules together by inducing temporary dipoles that attract each other. That is option C
Van Der Waals forces are example of those intermolecular forces which are weaker than ionic and covalent bonds that exists between molecules.
Van Der Waals forces was postulated by a Dutch physicist known as Van Der Waals. He postulated the existence of weak, short-range forces of attraction, which are independent of normal bonding forces, between non-polar molecules. He came to this conclusion after studying the of real gases at low temperatures and high pressures that:
electrons in a non-polar molecule such as hydrogen are close to one nucleus as to the other, although momentary concentration at one end of the molecule may occur, this momentary concentration of electron cloud on one side create a temporary dipole in the hydrogen molecule, that is, one side of the molecule acquires a partial negative charge while the other side acquires a partial positive charge of equal magnitude, the temporary dipole induces a similar dipole in an adjacent behavior molecule, this results in a temporary dipole-induced dipole attraction between the positive and negative ends of the adjacent molecules.This is how weak Van Der Waals forces are set up. Therefore, option C is CORRECT
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What type of bonding is occuring in the compound below?
A. Covalent polar
B. Metallic
C. Ionic
D. Covalent nonpolar
Answer:
(B). it's metallic bonding
How do I do this? What are the answers to the 5 questions shown?
Answer:
1. C₃H₆O₃
2. C₆H₁₂
3. C₆H₂₄O₆
4. C₆H₆
5. N₂O₄
Explanation:
1. Determination of the molecular formula.
Empirical formula => CH₂O
Mass of compound = 90 g
Molecular formula =?
Molecular formula = n × Empirical formula = mass of compound
[CH₂O]ₙ = 90
[12 + (2×1) + 16]n = 90
[12 + 2 + 16]n = 90
30n = 90
Divide both side by 30
n = 90/30
n = 3
Molecular formula = [CH₂O]ₙ
Molecular formula = [CH₂O]₃
Molecular formula = C₃H₆O₃
2. Determination of the molecular formula.
Empirical formula => CH₂
Mass of compound = 84 g
Molecular formula =?
Molecular formula = n × Empirical formula = mass of compound
[CH₂]ₙ = 84
[12 + (2×1)]n = 84
[12 + 2]n = 84
14n = 84
Divide both side by 14
n = 84/14
n = 6
Molecular formula = [CH₂]ₙ
Molecular formula = [CH₂]₆
Molecular formula = C₆H₁₂
3. Determination of the molecular formula.
Empirical formula => CH₄O
Mass of compound = 192 g
Molecular formula =?
Molecular formula = n × Empirical formula = mass of compound
[CH₄O]ₙ = 192
[12 + (4×1) + 16]n = 192
[12 + 4 + 16]n = 192
32n = 192
Divide both side by 32
n = 192/32
n = 6
Molecular formula = [CH₄O]ₙ
Molecular formula = [CH₄O]₆
Molecular formula = C₆H₂₄O₆
4. Determination of the molecular formula.
Empirical formula => CH
Mass of compound = 78 g
Molecular formula =?
Molecular formula = n × Empirical formula = mass of compound
[CH]ₙ = 78
[12 + 1]n = 78
13n = 78
Divide both side by 13
n = 78/13
n = 6
Molecular formula = [CH]ₙ
Molecular formula = [CH]₆
Molecular formula = C₆H₆
5. Determination of the molecular formula.
Empirical formula => NO₂
Mass of compound = 92 g
Molecular formula =?
Molecular formula = n × Empirical formula = mass of compound
[NO₂]ₙ = 92
[14 + (2×16)]n = 92
[14 + 32]n = 92
46n = 92
Divide both side by 46
n = 92/46
n = 2
Molecular formula = [NO₂]ₙ
Molecular formula = [NO₂]₂
Molecular formula = N₂O₄
name a factor tht affects the value of electron affinity
Answer:
Atomic sizeNuclear chargesymmetry of the electronic configurationcalculate the volume of 20.5g of oxygen occupied at standard temperature and pressure.what the volume
Answer :
volume of a gas = weight * 22.4 l / gram molecular weight
volume of o2 = ?
weight given = 20.5 g
gram molecular weight of oxygen = 32 (because of 2 oxygen atoms )
volume of oxygen = 20.5 * 22.4 / 32
volume of oxygen = 14.35 liters
Explanation:
hope this helps you
if wrong just correct me
If the starting material has no stereogenic centers, when carbonyl compounds are reduced with a reagent such as LiAlH4 or NaBH4 and a new stereogenic center is formed, what will the composition of the product mixture be?
A) Forms a racemic mixture of the two possible enantiomers.
B) Forms more of one enantiomer than another because of steric reasons around the carbonyl.
C) Forms more of one enantiomer than another depending on the temperature of the reaction.
D) Forms different products depending on the solvent used.
Answer:
A) Forms a racemic mixture of the two possible enantiomers
When carbonyl compounds are reduced with a reagent such as LiAlH₄ or NaBH₄ and new stereogenic center is formed chemical change will lead to products that form a racemic mixture of the two possible enantiomers.
What is a chemical change?
Chemical changes are defined as changes which occur when a substance combines with another substance to form a new substance.Alternatively, when a substance breaks down or decomposes to give new substances it is also considered to be a chemical change.
There are several characteristics of chemical changes like change in color, change in state , change in odor and change in composition . During chemical change there is also formation of precipitate an insoluble mass of substance or even evolution of gases.
There are three types of chemical changes:
1) inorganic changes
2)organic changes
3) biochemical changes
During chemical changes atoms are rearranged and changes are accompanied by an energy change as new substances are formed.
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Chromium-51 is a radioisotope that is used to assess the lifetime of red blood cells The half-life of chromium-51 is 27.7 days. If you begin with 39.7 mg of this isotope, what mass remains after 48.2 days have passed?
Answer:
11.9g remains after 48.2 days
Explanation:
All isotope decay follows the equation:
ln [A] = -kt + ln [A]₀
Where [A] is actual amount of the isotope after time t, k is decay constant and [A]₀ the initial amount of the isotope
We can find k from half-life as follows:
k = ln 2 / Half-Life
k = ln2 / 27.7 days
k = 0.025 days⁻¹
t = 48.2 days
[A] = ?
[A]₀ = 39.7mg
ln [A] = -0.025 days⁻¹*48.2 days + ln [39.7mg]
ln[A] = 2.476
[A] = 11.9g remains after 48.2 days
what is the difference between 25ml and 25.00ml
Answer:
There is no difference between the two.
Explanation:
They both show the same volume. But, adding decimal places shows the least count of the instrument used and is more acceptable when recording values in scientific experiments
5. Calcule las concentraciones cuando se alcanza el equilibrio si partimos de unas concentraciones iniciales [A]=[B]=1M ; [C]=[D]=0M y una constante de equilibrio de 5.
Las concentraciones en el equilibrio para la reacción química presentada son:
[tex][A] = [B] = 1-x = 1-0.69 = 0.31 M\\[C] = [D] = x = 0.69 M[/tex]
Consideremos la siguiente reacción química genérica:
A + B ⇄ C + D
Para calcular las concentraciones en el equilibrio, debemos construir una Tabla ICE. Cada fila representa una instancia (Inicial, Cambio, Equilibrio) y la completamos con la concentración o cambio de concentración ("x" para concentraciones desconocidas). Como inicialmente no hay productos, la reacción se desplazará hacia la derecha para alcanzar el equilibrio.
A + B ⇄ C + D
I 1 1 0 0
C -x -x +x +x
E 1-x 1-x x x
La constante de equilibrio, Kc, es:
[tex]Kc = 5 = \frac{[C][D]}{[A][B]} = \frac{x^{2} }{(1-x)^{2} } \\\sqrt{5} = x/1-x\\x = 0.69[/tex]
Las concentraciones en el equilibrio son:
[tex][A] = [B] = 1-x = 1-0.69 = 0.31 M\\[C] = [D] = x = 0.69 M[/tex]
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Select the choice that best completes the following sentence: When cooled slowly, transformations near the melting temperature tend to yield ______ grains due to the formation of ______ nucleation sites followed by ______ grain growth.
Question Completion with Options:
O coarse...few...rapid
O fine...few...slow
O fine...multiple...rapid
O coarse...few...slow
O fine...multiple...slow
Answer:
The choice that best completes the sentence is:
O coarse...few...slow
Explanation:
Transformations near the melting temperature develop coarse grains because few nucleation sites are formed and the rate of the grain growth is usually slow. This is because of the process that starts with recrystallization, recovery, and nucleation before growth can occur. While recrystallization enables the grain to increase in size at high temperature, nucleation gives the grain the energy to irreversibly grow into larger-sized nucleus.
20. An oxide of osmium (symbol Os) is a pale yellow solid. If 2.89 g of the compound contains 2.16 g of osmium, what is its empirical formula?
The empirical formula is OsO₄ :
Explanation:
Osmium oxide contains osmium and oxygen only.
Thus, we shall determine the mass of oxygen in osmium oxide. This can be obtained as follow:
Mass of compound = 2.89 g
Mass of Os = 2.16 g
Mass of O =?Mass of O = (Mass of compound) – (Mass of Os)
Mass of O = 2.89 – 2.16
Mass of O = 0.73 g
Finally, we shall determine the empirical formula of the compound. This can be obtained as follow:
Mass of Os = 2.16 g
Mass of O = 0.73 g
Empirical formula =..?Os = 2.16 g
O = 0.73 g
Divide by their molar mass of
Os = 2.16 / 190 = 0.011
O = 0.73 / 16 = 0.046
Divide by the smallest
Os = 0.011 / 0.011 = 1
O = 0.046 / 0.011 = 4
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Which subshells are found in each of the following shells
electron subshell - M shell
Answer:
3
Explanation:
The electron shells are labelled as K,L,M,N,O,P, and Q or 1,2,3,4,5,6, and 7.
As we go from innermost shell outwards, this number denotes the number of subshell in the shell. Electrons in outer shells have higher average energy and travel farther from the nucleus than those in inner shells.
Hence, M shell contains s,p and d subshells.
For the titration of 50. mL of 0.10 M ammonia with 0.10 M HCl, calculate the pH. For ammonia, NH3, Kb
Answer:
11.12 → pH
Explanation:
This is a titration of a weak base and a strong acid.
In the first step we did not add any acid, so our solution is totally ammonia.
Equation of neutralization is:
NH₃ + HCl → NH₄Cl
Equilibrium for ammonia is:
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻ Kb = 1.8×10⁻⁵
Initially we have 50 mL . 0.10M = 5 mmoles of ammonia
Our molar concentration is 0.1 M
X amount has reacted.
In the equilibrium we have (0.1 - x) moles of ammonia and we produced x amount of ammonium and hydroxides.
Expression for Kb is : x² / (0.1 - x) = 1.8×10⁻⁵
As Kb is so small, we can avoid the x to solve a quadratic equation.
1.8×10⁻⁵ = x² / 0.1
1.8×10⁻⁵ . 0.1 = x²
1.8×10⁻⁶ = x²
√1.8×10⁻⁶ = x → 1.34×10⁻³
That's the value for [OH⁻] so:
1×10⁻¹⁴ = [OH⁻] . [H⁺]
1×10⁻¹⁴ / 1.34×10⁻³ = [H⁺] → 7.45×10⁻¹²
- log [H⁺] = pH
- log 7.45×10⁻¹² = 11.12 → pH
11 Explain how you would obtain solid lead carbonate from a mixture of lead carbonate and sodium chloride
Explanation:
Add water, Na2CO3 dissolves, filter, PbCO3 stays in the paper and dissolved Na2CO3 goes through as the solution. Dry the PbCO3 and you have the dry solid.
OR
Add water to dissolve then filter to obtain PbCo3 as you're residue and Na2Co3 as the filtrate. Dry the insoluble PbCo3 between filter papers and you obtain solid PbCo3
which of the following is indicated by the ph value of a solution?
a- it's hydrogen ion concentration
b- its ammonium ion concentration
c- ability to undergo chemical reaction
d- its ratio of solute amount to solvent volume
Answer:
c- ability to undergo chemical reaction
Calculate the percent error in the atomic weight if the mass of a Cu electrode increased by 0.4391 g and 6.238x10-3 moles of Cu was produced. Select the response with the correct Significant figures. You may assume the molar mass of elemental copper is 63.546 g/mol. Refer to Appendix D as a guide for this calculation.
Answer:
10.77%
Explanation:
Molar mass of Cu = mass deposited/number of moles of Cu
Molar mass of Cu = 0.4391 g/6.238x10^-3 moles
Molar mass of Cu = 70.391 g/mol
%error = 70.391 g/mol - 63.546 g/mol/63.546 g/mol × 100
%error = 10.77%
if an element has an atomic number of 9 what is the electronic structure of the same element
9 is the element Florine
Florine has 9 electrons as well as the 9 protons that determine its atomic number.
The ground state configuration is the lowest energy configuration.
How many grams of magnesium chloride can be produced from 2.30 moles of chlorine gas reacting w excess magnesium Mg(s)+Cl2(g)->MgCl2(s)
The mass of magnesium chloride produced from 2.30 moles of chlorine gas is 218.99 grams.
How to calculate moles in stoichiometry?Stoichiometry refers to the study and calculation of quantitative (measurable) relationships of the reactants and products in chemical reactions.
According to this question, magnesium reacts with chlorine gas to form magnesium chloride as follows:
Mg + Cl₂ → MgCl₂
Based on the above chemical equation, 1 mole of chlorine gas forms 1 mole of magnesium chloride.
This means that 2.30 moles of chlorine gas will 2.30 moles of magnesium chloride.
Next, we convert moles of magnesium chloride to mass as follows:
molar mass of magnesium chloride = 95.211g/mol
mass of magnesium chloride = 95.211 × 2.30 = 218.99 grams.
Therefore, 218.99 grams of magnesium chloride will be formed.
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tea contains approximately 2% caffeine by weight. assuming that you started with 18g of tea leaves, calculate your percent yield of extraced caffeine
4.
Ammonia gas occupies a volume of 450. mL at a pressure of 720 mm Hg. What volume in
liters will the gas occupy at standard atmospheric pressure?
Answer:
[tex]\boxed {\boxed {\sf 426 \ mL}}[/tex]
Explanation:
We are asked to find the volume of ammonia gas given a change in pressure. We will use Boyle's Law, which states the volume of a gas is inversely proportional to the pressure of a gas. The formula is:
[tex]P_1V_1= P_2V_2[/tex]
The ammonia gas originally occupies a volume of 450 milliliters at a pressure of 720 millimeters of mercury. Substitute the values into the formula.
[tex]450 \ mL * 720 \ mm \ Hg = P_2V_2[/tex]
The pressure is changed to standard atmospheric pressure, which is 760 millimeters of mercury. The new volume is unknown.
[tex]450 \ mL * 720 \ mm \ Hg = 760 \ mm \ Hg * V_2[/tex]
We are solving for the volume at standard pressure. We will need to isolate the variable V₂. It is being multiplied by 760 millimeters of mercury. The inverse of multiplication is division. Divide both sides of the equation by 760 mm Hg.
[tex]\frac {450 \ mL * 720 \ mm \ Hg }{760 \ mm \ Hg}= \frac{760 \ mm \ Hg * V_2}{760 \ mm \ Hg}[/tex]
[tex]\frac {450 \ mL * 720 \ mm \ Hg }{760 \ mm \ Hg}= V_2[/tex]
The units of millimeters of mercury (mm Hg) cancel.
[tex]\frac {450 \ mL * 720 }{760} = V_2[/tex]
[tex]\frac {324,000}{760} \ mL = V_2[/tex]
[tex]426.3157895 \ mL =V_2[/tex]
The original values of volume and pressure have 3 significant figures. Our answer must have the same. For the number we calculated, that is the ones place. The 3 in the tenths place tells us to leave the 6 in the ones place.
[tex]426 \ mL \approx V_2[/tex]
The volume at standard atmospheric pressure is approximately 426 milliliters.
What is the molarity of a solution that contains 0.75 mol Naci in 3.0 L of solution? Select one: O a. 4.0 M O b. 2.3 M O d. 3.8 M O d. 0.25 M Clear my choice
Answer:
[tex]\boxed {\boxed {\sf D. \ 0.25 \ M}}[/tex]
Explanation:
Molarity is a measure of concentration in moles per liter.
[tex]molarity= \frac{moles \ of \ solute}{ liters \ of \ solution}[/tex]
The solution contains 0.75 moles of sodium chloride and has a volume of 3.0 liters.
moles of solute = 0.75 mol NaCl liters of solution = 3.0 LSubstitute these values into the formula.
[tex]molarity= \frac{ 0.75 \ mol \ NaCl}{3.0 \ L}[/tex]
[tex]molarity= 0.25 \ mol \ NaCl/L[/tex]
Molarity has the molar (M) as its unit. 1 molar is equal to 1 mole per liter.
[tex]molarity= 0.25 \ M \[/tex]
The molarity of the solution is 0.25 Molar and Choice D is correct.
the density of oxygen 1.43 gm/liter at 0°c and pressure 1.0 atm. if a 20 liter cylinder is filled with oxygen at pressure of 25 atm and temperature of 27°c. what is the mass of oxygen in the cylinder
Answer:
640 g
Explanation:
Step 1: Given and required data
Volume of the cylinder (V): 20 LPressure of the oxygen (P): 25 atmTemperature (T): 27 °C (300 K)Ideal gas constant (R): 0.082 atm.L/mol.KStep 2: Calculate the moles of oxygen gas
We will use the ideal gas equation
P × V = n × R × T
n = P × V / R × T
n = 25 atm × 20 L / (0.082 atm.L/mol.K) × 300 K = 20 mol
Step 3: Calculate the mass corresponding to 20 moles of oxygen
The molar mass of oxygen is 32.00 g/mol.
20 mol × 32.00 g/mol = 640 g
Considering a fish breeder decided to breed small fishes which needs a pH between 6,0 to 7,0 to stay alive. He needs to adjust the water's pH that is 5,0 to a value of 6.5, having available only calcium carbonate. The mass in mg added to 5L of water is about:
A)2,5
B)5,5
C)6,5
D)7,5
E)9,5
