Answer:
2.7 × 10⁻⁷ M
Explanation:
Step 1: Write the reaction for the solution of PbCO₃
PbCO₃(s) ⇄ Pb²⁺(aq) + CO₃²⁻(aq)
Step 2: Make an ICE chart
To relate the solubility of PbCO₃ (S) with its solubility product constant (Ksp) we will make an ICE chart.
PbCO₃(s) ⇄ Pb²⁺(aq) + CO₃²⁻(aq)
I 0 0
C +S +S
E S S
The solubility product constant is:
Ksp = 7.4 × 10⁻¹⁴ = [Pb²⁺] × [CO₃²⁻] = S²
S = √7.4 × 10⁻¹⁴ = 2.7 × 10⁻⁷ M
Which solution below would have the greatest buffering capacity? The solution was prepared from a weak acid and the salt of its conjugate base. 0.160 M C6H5OHCOOH and 0.160 M C6H5OHCOONa, Ka = 1.05e-3 0.0892 M H2NCH2COOH and 0.0892 M H2NCH2COOK, Ka = 4.50e-3 0.0725 M H2NCH2COOH and 0.0725 M H2NCH2COONa, Ka = 4.50e-3 0.1360 M C6H5OHCOOH and 0.1360 M C6H5OHCOOK, Ka = 1.05e-3
Answer:
0.1360 M C₆H₅OHCOOH and 0.1360 M C₆H₅OHCOOK,
Ka = 1.05x 10⁻³ .
Explanation:
The first mixture
0.160 M C₆H₅OHCOOH and 0.160 M C₆H₅OHCOONa,
Ka = 1.05x 10⁻³ .
The second mixture
0.0892 M H₂NCH₂COOH and 0.0892 M H₂NCH₂COOK,
Ka = 4.5 x 10⁻³ .
The third mixture
.0725 M H₂NCH₂COOH and 0.0725 M H₂NCH₂COONa,
Ka = 4.5 x 10⁻³
fourth mixture
0.1360 M C₆H₅OHCOOH and 0.1360 M C₆H₅OHCOOK,
Ka = 1.05x 10⁻³ .
In all the mixtures the ratio of acid and its salt are same and equal to one so this ratio will not determine their relative buffering capacity .
Now we know that weak acid has more buffering capacity so mixture having acid of less Ka will have more buffering capacity .
Ka is less if
Ka = 1.05 x 10⁻³ .
Dilute acids have greater buffering capacity
So ultimate answer is
0.1360 M C₆H₅OHCOOH and 0.1360 M C₆H₅OHCOOK,
Ka = 1.05x 10⁻³ .
Which of the following molecules is achiral? A) (2R,3R)-2,3-Dichloropentane B) (2R,3S)-2,3-Dichloropentane C) (2S,3S)-2,3-Dichlorobutane D) (2R,3S)-2,3-Dichlorobutane E) None of these
Answer:
E) None of these
Explanation:
For this question, we must remember the definition of a chiral carbon. In the chiral carbons, we have four different groups around the carbon.
So, an achiral molecule would be a molecule that does not have chiral carbons. That is, in an achiral molecule in all carbons we have at least one repeating group.
In all the molecules, which we have in the question, we have absolute configuration. That is, we have chiral carbons. In molecule A we have two chiral carbons (R and R), in molecule b we have two chiral carbons (R and S), in molecule c we have two chiral carbons (S and S) and in molecule d we have two chiral carbons (R and S). Therefore in all molecules, we have chiral carbons.
Now, if we analyze molecules a and c, we have a plane of symmetry. If we have a plane of symmetry, even though there are chiral carbons, there will be no optical activity. That is, these molecules do not have the ability to deflect polarized light despite having chiral carbons. Therefore molecules a and c are meso compounds.
But optical activity is not related to chirality.
See figure 1 to further explanations
I hope it helps!
Gold (Au) has a density of 19.3 g/ml. A gold filling falls out of your teeth and you bring it to chemistry class. When you drop it into a graduated cylinder the water rises from 9.27 ml to 12.03 ml.
What is the mass of your gold filling?
Answer:
53.27 g.
Explanation:
From the question given above, the following data were obtained:
Density of gold = 19.3 g/mL
Volume of water = 9.27 mL
Volume of water + gold = 12.03 mL
Mass of gold =.?
Next, we shall determine the volume of the gold filling. This can be obtained as follow:
Volume of water = 9.27 mL
Volume of water + gold = 12.03 mL
Volume of gold =?
Volume of gold = (Volume of water + gold) – (Volume of water)
Volume of gold = 12.03 – 9.27
Volume of gold = 2.76 mL
Finally, we shall determine the mass of the gold filling as follow:
Density of gold = 19.3 g/mL
Volume of gold = 2.76 mL
Mass of gold =.?
Density = mass/volume
19.3 = mass of gold /2.76
Cross multiply
Mass of gold = 19.3 × 2.76
Mass of gold = 53.27 g
The, the mass of the gold filling is 53.27 g.
What is the Temperature of the suns most middle layer the chromosphere
The temperature in the chromosphere varies between about 4000 K at the bottom (the so-called temperature minimum) and 8000 K at the top (6700 and 14,000 degrees F, 3700 and 7700 degrees C), so in this layer (and higher layers) it actually gets hotter if you go further away from the Sun
1. For one atom of lithium -6 I will need_____ atom of lithium -7 for an average atomic mass of 6.944? 2. For one atom of lithium -6 needed _____ Adams of lithium -7 for an average atomic mass of 6.944.
Answer:
For one atom of lithium -6 I will need_16.85____ atom of lithium -7 for an average atomic mass of 6.944 .
Explanation:
1 ) Let required atom be m .
[tex]\frac{6\times 1 + 7\times m}{1 + m} = 6.944[/tex]
6 + 7m = 6.944 + 6.944m
7 m - 6.944 m = .944
.056 m = .944
m = 16.85
Here, we are required to find how many atoms of lithium -7 is to combine with 1 atom of lithium -6 for an average atomic mass of 6.944.
For one atom of lithium -6 I will need 16.86 atoms of lithium -7 for an average atomic mass of 6.944.
Let the total number of lithium -6 and lithium -7 required be; x.
Therefore, the number of lithium -7 required is;
(x-1)
Therefore;
{(1/x) × 6 + ((x-1)/x)× 7} = 6.944
Therefore, (6 + 7x - 7)/x = 6.944
And, 7x -1 = 6.944x
0.056x = 1
And, x = 1÷0.056 = 17.86 atoms.
Therefore, the number of lithium -7 atoms needed is;
x - 1 = 17.86 - 1 = 16.85 atoms of lithium -7.
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Calculate the wavelength of light produced if an electron moves from n=6 state to n=5 state of an electron in a hydrogen atom.
The questions asks us to calculate the wavelength of light produced if an electron moves from n = 6 to n = 5 state in a hydrogen atom. The answer would be:
λ = [tex]7.46 * 10^{-6}[/tex] m
m = [tex]7.46 * 10^{-4}[/tex] cm
Explanation:
Please see my graphic below:
The wavelength of light produced if an electron moves from n=6 state to n=5 state of an electron in a hydrogen atom is 7.59 × [tex]10^-6[/tex] m
Using the Rydberg formula;
1/λ = 1.09737 × [tex]10^-6[/tex] (1/[tex]nf^2[/tex] - 1/[tex]ni^2[/tex])
Where;
λ = wavelength
nf = final state of the electron
ni = initial state of the electron
Substituting values;
1/λ = 1.09737 × [tex]10^7[/tex] ([tex]1/5^2[/tex] - [tex]1/6^2[/tex])
1/λ = 1.09737 × [tex]10^7[/tex] (1/25 - 1/36)
1/λ = 1.09737 × [tex]10^7[/tex] (0.04 - 0.028)
1/λ = 1.09737 × [tex]10^7[/tex] (0.012)
λ = 7.59 × [tex]10^-6[/tex] m
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Which of the following is not one of the four groups of organic compounds
The following that is not one of the four groups of organic compounds is a monosaccharide. The correct option is D.
What are organic compounds?Any of a vast group of chemical compounds known as organic compounds contain one or more carbon atoms that are covalently connected to atoms. Organic compounds are made up of carbon and hydrogen, and oxygen.
Carbohydrates, lipids, proteins, nucleic acids, and nucleotides are organic compounds. Due to their presence of both carbon and hydrogen, these substances are referred to as organic.
Monosaccharides are the single unit of carbohydrates. They join to form polymers of carbohydrates. They are glucose, fructose, etc. It is a single unit of carbohydrates. It is not one of the organic compound.
Thus, the correct option is D) monosaccharides.
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The question is incomplete. Your most probably complete question is given below:
A) carbohydrates D) monosaccharides B) lipids E) proteins C) nucleic acids
Which of the following describes freezing? A)A substance changing from the liquid state to the solid state. B) A substance changing from the solid state to the liquid state. C) A substance changing from the liquid to the gaseous state. D) A substance changing from the gaseous state to the liquid state.
Answer:
A
Explanation:
A because water turns into ice
An iceberg has a volume of 783 m3. If the density of ice = 0.917 g/cm3, how much mass does the iceberg have in kg?
Answer:
7.18 × 10⁵ kg
Explanation:
Step 1: Given data
Volume of the iceberg (V): 783 m³
Density of ice (ρ): 0.917 g/cm³
Step 2: Convert "V" to cm³
We will use the relationship 1 m³ = 10⁶ cm³.
783 m³ × (10⁶ cm³/1 m³) = 7.83 × 10⁸ cm³
Step 3: Calculate the mass (m) of the iceberg
We will use the following expression.
ρ = m/V
m = ρ × V
m = 0.917 g/cm³ × 7.83 × 10⁸ cm³
m = 7.18 × 10⁸ g
Step 4: Convert "m" to kg
We will use the relationship 1 kg = 10³ g.
7.18 × 10⁸ g × (1 kg/10³ g) = 7.18 × 10⁵ kg
What is the volume of an object that has a mass of 18 g and a density of 0.5 g/cm3 ?
6) What is located in the nucleus?
Protons
Protons and neutrons
Protons and electrons
Neutrons
Answer:
Protons and Neutrons are found in the nucleus
Calculate the mole fraction of nitric acid of a(n) 15.7% (by mass) aqueous solution of nitric acid. Calculate the mole fraction of nitric acid of a(n) 15.7% (by mass) aqueous solution of nitric acid. 2.56×10−2 0.102 5.33×10−2 5.11×10−2 The density of the solution is needed to solve the problem.
Answer:
0.0505
Explanation:
Step 1: Calculate the masses of solute and solvent
We have a 15.7% by mass nitric acid solution, that is, there are 15.7 g of nitric acid (solute) per 100 g of solution. The mass of water (solvent) is:
m(solution) = m(solute) + m(solvent)
m(solvent) = m(solution) - m(solute)
m(solvent) = 100 g - 15.7 g = 84.3 g
Step 2: Calculate the moles of nitric acid
The molar mass of nitric acid is 63.01 g/mol.
15.7 g × (1 mol/63.01 g) = 0.249 mol
Step 3: Calculate the moles of water
The molar mass of water is 18.02 g/mol.
84.3 g × (1 mol/18.02 g) = 4.68 mol
Step 4: Calculate the mole fraction of nitric acid
[tex]X(HNO_3) = \frac{nHNO_3}{nHNO_3+nH_2O} = \frac{0.249mol}{0.249mol+4.68mol} = 0.0505[/tex]
For which reaction, carried out at standard conditions, wouldboth the enthalpy and entropy changes drive the reaction in thesame direction? Please EXPLAIN.
A. 2H2(g) + O2(g) --->2H2O(l) ΔH = -571.1 kJ
B. 2Na(s) + Cl2(g)---->2NaCl(s) ΔH = -822.0kJ
C. N2(g) + 2O2(g) ---->2NO2(g) ΔH= +67.7 kJ
D. 2NH3(g) ----> N2(g) +3H2(g) ΔH= +92.4 kJ
Answer:
2NH3(g) ----> N2(g) +3H2(g) ΔH= +92.4 kJ
Explanation:
Entropy increases with increase in the number of particles from left to right in a reaction. Hence, the reaction; 2NH3(g) ----> N2(g) +3H2(g) ΔH= +92.4 kJ is favoured by increase in entropy.
Similarly, the enthalpy change for the reaction is only +92.4 KJ. Hence for this reaction, both enthalpy and entropy changes drive the reaction in the same direction.
If u burn paper in the jar, is it the same mass as before u burn it or less or more. And why.
Answer:
the mass would not be the same because of the ashes in the jar
Explanation:
Answer: The total mass before and after burning is the same
Explanation: When the carbon and hydrogen of the hydrocarbon-containing substance (the paper) chemically combine with the oxygen, the remaining materials may appear as ash, the solid remains of a fire.
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what is the average speed in miles per hour of a sprinter who who does thehundred yard dash in 11.3 seconds
Answer:
18.1 mi/h
Explanation:
Step 1: Given data
Distance traveled (d): 100 yd
Time elapsed (t): 11.3 s
Step 2: Convert "d" to miles
We will use the relationship 1 mi = 1,760 yd.
100 yd × (1 mi/1,760 yd) = 0.0568 mi
Step 3: Convert "t" to hours
We will use the relationship 1 h = 3,600 s.
11.3 s × (1 h/3,600 s) = 3.14 × 10⁻³ h
Step 4: Calculate the average speed (s)
We will use the following expression.
s = d / t
s = 0.0568 mi / 3.14 × 10⁻³ h
s = 18.1 mi/h
What is the relationship between the bases displayed when the arrow is pointed to the left versus when it is pointed to the right?
Answer:
Options
Explanation:
Draw all of the constitutional isomers of the molecule with formula C3H5Br. Ignore geometric and stereoisomers.
Answer:
-) 3-bromoprop-1-ene
-) 2-bromoprop-1-ene
-) 1-bromoprop-1-ene
-) bromocyclopropane
Explanation:
In this question, we can start with the I.D.H (hydrogen deficiency index):
[tex]I.D.H~=~\frac{(2C)+2+(N)-(H)-(X)}{2}[/tex]
In the formula we have 3 carbons, 5 hydrogens, and 1 Br, so:
[tex]I.D.H~=~\frac{(2*3)+2+(0)-(5)-(1)}{2}~=~1[/tex]
We have an I.D.H value of one. This indicates that we can have a cyclic structure or a double bond.
We can start with a linear structure with 3 carbon with a double bond in the first carbon and the Br atom also in the first carbon (1-bromoprop-1-ene). In the second structure, we can move the Br atom to the second carbon (2-bromoprop-1-ene), in the third structure we can move the Br to carbon 3 (3-bromoprop-1-ene). Finally, we can have a cyclic structure with a Br atom (bromocyclopropane).
See figure 1
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Benzoic acid is in the process of being recrystallized. Your lab partner is in a hurry and after the impure benzoic acid is dissolved, they quickly submerge the flask in an ice bath and see crystals. Besides possibly breaking the flask, what would be the problem with this technique
Answer:
Impurities will be trapped in the crystals of the benzoic acid crystallized in this manner.
Explanation:
After benzoic acid is dissolved in hot water, it should have been allowed to cool gradually before it is transferred into an ice bath.
This gradual cooling will aid the separation of impurities so that when the vessel is now submerged in an ice bath, only pure benzoic acid is recrystalized.
If the vessel is immediately submerged into an ice bath, impurities will be trapped in the crystals of the benzoic acid.
Thinking about simple organic molecules made with only carbon and hydrogen and
no other elements (hydrocarbons), what are four different ways to make new
molecules? (Hint: Two versions of the same "way" only count as one thing!)
Answer:
Explanation:
Simple organic molecules having only carbon and hydrogen are alkanes, alkenes and alkynes.
Ways of making these molecules are referred to as synthesis
Synthesis of
(1) alkanes can be achieved by halogenation of an alkene or alkyne as shown in the reaction below
C₂H₄ + H₂ ⇒ C₂H₆
The reaction above shows an alkene (ethene) undergoing halogenation to form an alkane (ethane). This reaction occurs in the presence of a platinum or nickel catalyst
C₂H₂ + 2H₂ ⇒ C₂H₆
The reaction above shows an alkyne (ethyne) undergoing halogenation to form an alkane (ethane). This reaction also occurs in the presence of a nickel or platinum catalyst
(2) Alkanes can be achieved by Wurtz reaction; treating alkyl halides with sodium metal as shown below
CH₃CH₂Br + 2Na + BrCH₂CH₃ ⇒ CH₃CH₂CH₂CH₃ + 2NaBr
This reaction proceed in the presence of a dry ether.
(3) alkanes from Kolbe's electrolytic method; this involves the synthesis of alkanes through the electrolysis of sodium or potassium salt of a carboxylic acid
2CH₃COO⁻Na⁺ + 2H₂O ⇒ CH₃CH₃ + 2CO₂ + H₂ + 2NaOH
(4) alkenes from dehydration of alcohol, as shown in the reaction below
C₂H₅OH ⇒ C₂H₄ (-H₂O)
The reaction above proceeds by heating the alcohol (ethanol) at high temperature in the presence of strong acid such as sulphuric acid.
4. Which statement gives you enough information to say that the atom is electrically neutral?
O A. The atom has 15 neutrons and 15 electrons.
OB. The atom has 4 neutrons and 4 protons.
O C. The atom has 7 protons and 7 electrons.
O D. The atom has 19 electrons and 19 neutrons.
Answer:
O C. The atom has 7 protons and 7 electrons.
If the proton amd electron of atoms are equal it is said to be electrically nuetral
437 mm converted to m
Answer:
0.437
Explanation:
Divide the length value by 1000 to get the answer.
An unknown substance is measured. It has a mass of 0.221 g and a volume of 2.25 mL. What is its density?
Answer:
d ≈ 0.098 g/mL
Explanation:
The density of a substance can be found by dividing the mass by the volume.
d=m/v
The mass of the substance is 0.221 grams and the volume is 2.25 milliliters.
m= 0.221 g
v= 2.25 mL
Substitute the values into the formula.
d= 0.221 g / 2.25 mL
Divide
d= 0.098222222 g/mL
Let’s round to the nearest thousandth. The 2 in the ten thousandths tells us to keep the 8 in the thousandth place.
d ≈ 0.098 g/mL
The density of the substance is about 0.098 grams per milliliter.
Answer:
Explanation:5.70
10 grams of compound J is found to have a mass ratio of 8:2 of lithium (Li): Hydrogen (H). How many grams
of hydrogen would found in 42.0 grams of compound J?
Answer:
Mass of hydrogen in 42 gram of J = 8.4
Explanation:
Given:
Amount of compound J = 10 gram
Mass ratio[lithium (Li): Hydrogen (H)] = 8:2
Find:
Mass of hydrogen in 42 gram of J
Computation:
Mass of hydrogen in 10 gram of J = [2 / 10] 10
Mass of hydrogen in 10 gram of J = 2 gram
Mass of hydrogen in 42 gram of J = [2 / 10] 42
Mass of hydrogen in 42 gram of J = 8.4
How do the chemical characteristics of carbon affect the characteristics of organic molecules?
Answer:
The chemical characteristics of carbon affect the characteristics of organic molecules due to its tetravalent nature. It has four valence electrons in which it shares with other elements in order to form an octet configuration.
Carbon atoms are also capable of forming double and triple bonds with other atoms. These properties help determine the functional group present and gives us a knowledge of the chemical features such as polarity, melting and boiling present in the compound.
How is Oxygen -18 different from oxygen -16?
Answer:
because they are both 2 deferent type of oxygen levels
CHML 1046 general chemistryExperiment "Buffers and Buffer Capacity" 1. What (NH3/ NH4+) ratio is required for a buffer solution that has pH = 7.00? Кь = 1.8x10-5 for NH3 and Ka = 56x10-10 for NH4.2. Why is a mixture of NHs and NH.CI a poor choice for a buffer having pH 7?
Answer:
Explanation:
For pH of a buffer solution , the equation is
pH = pKa + log [ A⁻ ] / [ HA ]
Putting the values
7 = - log 56 x 10⁻¹⁰ + log [ NH₃ ] / [ NH₄⁺]
7 = 10 - log 56 + log [ NH₃ ] / [ NH₄⁺]
- 3 + 1.748 = log [ NH₃ ] / [ NH₄⁺]
log [ NH₃ ] / [ NH₄⁺] = - 1.25
[ NH₃ ] / [ NH₄⁺] = [tex]10^{-1.25}[/tex]
Required ratio = [tex]10^{-1.25}[/tex]
2 )
In the formation of NH₄⁺ , generally HCl acid is required . A little bit of excess HCl will change the pH required . So , this mixture is a poor choice for getting pH 7 for a buffer solution .
Write the symbol for every chemical element that has atomic number less than 5 and atomic mass greater than 8.3 u.
Atomic m. Element Atomic no. Symbol
1.0079u Hydrogen 1 H
4.0026u Helium 2 He
6.941u Lithium 3 Li
Additional information:-★ Atomic number : The number of protons in one atom of an element is known as atomic number.
Atomic number of an element = No. of protons in one atom of element.The atomic no. of element is denoted by the letter z .The symbol for every chemical element that has an atomic number less than 5 and atomic mass greater than 8.3 u are hydrogen H, helium He, lithium Li and atomic masses 1,4 and 6.94 respectively in the periodic table.
What is a periodic table?A periodic table is an arrangement of the elements in which they are classified on the basis of the number of electrons, neutrons, and protons present in their nucleus and shell of them.
Hydrogen has one electron in it and atomic mass is also one, helium comes in the second number with 4 amu and lithium comes in the third number containing 6.94 mass with it.
Therefore, hydrogen H, helium He, lithium Li and atomic masses 1,4 and 6.94 respectively in the periodic table contain symbols for every chemical element that has an atomic number less than 5 and atomic mass greater than 8.3 u are hydrogen H.
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Draw the structure of 2,4,4,5-tetramethyl-2-hexene.
Answer:
The answer to your question is given below
Explanation:
2,4,4,5-tetramethyl-2-hexene.
To draw the structure of the above compound, we must bear the following in mind:
1. The compound is an alkene i.e it contains carbon to carbon double bond (C=C)
2. The parent name of the compound is hexene i.e it contain 6 carbon atoms.
3. The double is located at carbon 2.
4. The substituent group attached to the compound is methly.
5. There are four methly group attached to the compound of which one is located at carbon 2, two at carbon 4 and one at 5.
With the above information, we can easily draw the structure of the compound.
Please see attached photo for the structure of the compound.
PLEASE HELP ASAP!!!!!!!!!!! WITH PAGE 2 FOR QUESTION # 3 WITH THE GRAPHING THANK YOU. THE QUESTION IS AS FOLLOWED: GRAPH THE VAPOR PRESSURE OF WATER USING THE FOLLOWING TEMPERATURES. Temperature (k) Vapor Pressure (Torr) (dependent variable) 273 4.6 293 17.5 313 55.0 333 149.2 353 355.5 373 760.0
Answer:
a
Explanation:
Which molecule does not exhibit hydrogen bonding?a. HFb. CH3NH2c. CH2F2d. HOCH2CH2OH
Answer:
(c) CH₂F₂
Explanation:
Hydrogen bonds are weak intermolecular forces. They are the strongest kind of intermolecular forces, although they are weaker than the covalent bonds.
Hydrogen bonds arise from molecules which contain a hydrogen atom which is bonded to one of the most electronegative elements such as N, O or F.
(a) HF, → has H-F bond
(b) CH₃NH₂, → has N-H bond
(c) CH₂F₂, → has no H-F bond ( F- C- F)
(d) HOCH₂CH₂OH, → has O-H bond
Therefore, only CH₂F₂ does not exhibit hydrogen bonding.