Answer:
inches of mercury (inHg or “Hg) or millibars.
A 250ml sample of sir at 373.15k is warmed at 473.15 at constant pressure. What is the volume of the air sample at the new temperature
Answer:
317 mL.
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 250 mL
Initial temperature (T₁) = 373.15 K
Final temperature (T₂) = 473.15 K
Pressure = Constant
Final volume (V₂) =?
The final volume of the air sample can be obtained by using the Charles' law equation as illustrated below:
V₁/T₁ = V₂/T₂
250 / 373.15 = V₂ / 473.15
Cross multiply
373.15 × V₂ = 250 × 473.15
373.15 × V₂ = 118287.5
Divide both side by 373.15
V₂ = 118287.5 / 373.15
V₂ = 317 mL
Therefore, the final volume of the air sample is 317 mL
3. Which of the following can be physically separated?
Answer:
mixture
Explanation:
an example of one is a salad you can separate the ingredients
The energy released by a nuclear fusion reaction is produced when
A.
energy is converted to mass
B.
mass is converted to energy
C.
heat is converted to temperature
D.
temperature is converted to heat
The nuclear fusion results in the liberation of large amount of energy by the conversion of mass to energy. Thus, option B is correct.
The nuclear fusion has been a type of nuclear reaction in which the reaction of the two nuclei results in the nuclei with mass smaller than the reactants. In nuclear reactions, there has been the liberation of large amount of energy.
The energy released in the nuclear fusion has been formed by the conversion of the mass. Since, the formed product has nuclei with the mass smaller than the reactants, the remaining mass of the reactants has been converted to the energy.
Thus, in nuclear fusion, the energy released has been produced from the conversion of mass to energy. Thus, option B is correct.
For more information about the nuclear fusion, refer to the link:
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How many 250 mg tablets of metronidazole are needed to make 150 mL of suspension containing
100 mg/mL?
a. 25
b. 30
c. 50
d. 60
Answer:
How many 250 mg tablets of metronidazole are needed to make 150 mL of suspension containing
100 mg/mL?
a. 25
b. 30
c.50
d. 60
Explanation:
if mali po I'm so sorry
About 60 tablets of 250 mg tablets of metronidazole are needed to make 150 mL of suspension containing 100 mg/mL.
To calculate the number of tablets needed, we can use the following formula:
Number of tablets = (Desired total mass of metronidazole) / (Mass of one tablet)
The desired total mass of metronidazole in the suspension is given as 100 mg/mL × 150 mL = 15000 mg.
Since each tablet contains 250 mg of metronidazole, we can calculate the number of tablets as:
Number of tablets = 15000 mg / 250 mg = 60 tablets
Therefore, the correct answer is d. 60 tablets.
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#SPJ 2
Discuss the any two applications of Beer’s law with suitable justification.
Answer: It's used in chemistry to measure the concentration of chemical solutions.
Explanation:
Beer's Law is used in chemistry to measure the concentration of chemical solutions, to analyze oxidation, and to measure polymer degradation.
The law also describes the attenuation of radiation through the Earth's atmosphere
Boric acid, B(OH)3, acts as an acid in water, but does not do so via ionization of a proton from boric acid. Use a balanced chemical equation to show why boric acid may be described as an acid in water.
Answer:
B(OH)₃ + H₂O = B(OH)₄⁻ + H⁺
Explanation:
Let's consider Arrhenius acid-base theory:
An acid is a substance that in aqueous media releases H⁺.A base is a substance that in aqueous media releases OH⁻.Boric acid, B(OH)₃ reacts with water according to the following equation.
B(OH)₃ + H₂O = B(OH)₄⁻ + H⁺
As we can see, boric acid releases H⁺ in aqueous media. Thus, it is an acid.
Which of the following has the highest pH?
A. 0.01 M HCI
B. 0.1 M HCI
C. 1 M HCI
D. 0.001 M HCI
Answer:
D
Explanation:
pH=-log(x)
x=0.001M,pH=3
x=0.01M,pH=2
x=0.1M,pH=1
x=1M,pH=0
Highest pH is for option D
Answer:
D.
Explanation:
The highest pH is D because
0.01 M HCL => 2
0.1 M HCL => 1
1 M HCL => 0
0.001 M HCL => 3
so the answer is D.
How many line pairs of electrons does a square planar molecule have?
2
0
6
4
Answer:
the answer is 4 hope this helps
When sulfur loses one electron, it becomes a particularly stable, half-filled p subshell. The removal of this first electron requires less energy than the removal of an electron from phosphorus, which is initially a half-filled p subshell. This signifies that __________.
The question is incomplete, the complete question is;
When sulfur loses one electron, it becomes a particularly stable, half-filled p subshell. The removal of this first electron requires less energy than the removal of an electron from phosphorus, which is initially a half-filled p subshell. This signifies that the first ionization energy of sulfur is larger than the first ionization energy of phosphorus. the first ionization energy of sulfur is smaller than the first ionization energy of phosphorus. the second ionization energy of sulfur is smaller than the first ionization energy of phosphorus. the second ionization energy of phosphorus is larger than the second ionization energy of sulfur.
Answer:
the first ionization energy of sulfur is smaller than the first ionization energy of phosphorus
Explanation:
Let us look back at the electronic configuration of each of the atoms;
Sulphur; [Ne] 3s² 3p⁴
Phosphorus; [Ne] 3s² 3p³
We can easily see that phosphorus has an exactly filled half filled 3p sublevel. This partially filled orbital has a great deal of stability associated with it.
On the other hand, sulphur can attain this stability that results from a half filled orbital by loosing one of its p electrons. The energy required for this process is much lower than the energy required to remove an electron from an already half filled 3p orbital of the phosphorus atom.
Hence, the first ionization energy of sulfur is smaller than the first ionization energy of phosphorus.
In the following structure, carbons (I),(2),(3) and (4) are classified respectively as
Answer:
Carbon (i) : quaternary carbon
Carbon (ii) : secondary carbon
Carbon (iii) : tertiary carbon
Carbon (iv) : secondary carbon
Explanation:
Carbons can be classified into 4 categories:
(1) Primary carbon [tex](1^o)[/tex]: These are the atoms where the carbon atom is attached to one other carbon atom.
(2) Secondary carbon [tex](2^o)[/tex]: These are the atoms where the carbon atom is attached to two other carbon atoms.
(3) Tertiary carbon [tex](3^o)[/tex]: These are the atoms where the carbon atom is attached to three other carbon atoms.
(4) Quaternary carbon [tex](4^o)[/tex]: These are the atoms where the carbon atom is attached to four other carbon atoms.
In the given structure:
Carbon (i) is attached to 4 further carbon atoms and hence, it is a quaternary carbon.
Carbon (ii) is attached to 2 further carbon atoms and hence, it is a secondary carbon.
Carbon (iii) is attached to 3 further carbon atoms and hence, it is a tertiary carbon.
Carbon (iv) is attached to 2 further carbon atoms and hence, it is a secondary carbon.
A buffer solution contains 0.475 M nitrous acid and 0.302 M sodium nitrite . If 0.0224 moles of potassium hydroxide are added to 150 mL of this buffer, what is the pH of the resulting solution
Answer: The pH of the resulting solution will be 3.001
Explanation:
Molarity is calculated by using the equation:
[tex]\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}[/tex] ......(1)
We are given:
Moles of NaOH = 0.0224 moles
Molarity of nitrous acid = 0.475 M
Molarity of sodium nitrite = 0.302 M
Volume of solution = 150 mL = 0.150 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:
[tex]\text{Moles of nitrous acid}=(0.475mol/L\times 0.150L)=0.07125mol[/tex]
[tex]\text{Moles of sodium nitrite}=(0.302mol/L\times 0.150L)=0.0453mol[/tex]
The chemical equation for the reaction of nitrous acid and NaOH follows:
[tex]HNO_2+NaOH\rightleftharpoons NaNO_2+H_2O[/tex]
I: 0.07125 0.0224 0.0453
C: -0.0224 -0.0224 +0.0224
E: 0.04885 - 0.0677
The power of the acid dissociation constant is the negative logarithm of the acid dissociation constant. The equation used is:
[tex]pK_a=-\log K_a[/tex] ......(2)
We know:
[tex]K_a[/tex] for nitrous acid = [tex]7.2\times 10^{-4}[/tex]
Using equation 2:
[tex]pK_a=-\log (7.2\times 10^{-4})=3.143[/tex]
To calculate the pH of the acidic buffer, the equation for Henderson-Hasselbalch is used:
[tex]pH=pK_a+ \log \frac{\text{[conjugate base]}}{\text{[acid]}}[/tex] .......(3)
Given values:
[tex][NaNO_2]=\frac{0.0677}{0.150}[/tex]
[tex][HNO_2]=\frac{0.04885}{0.150}[/tex]
[tex]pK_a=3.143[/tex]
Putting values in equation 3. we get:
[tex]pH=3.143-\log \frac{(0.0677/0.150)}{(0.04885/0.150)}\\\\pH=3.143-0.142\\\\pH=3.001[/tex]
Hence, the pH of the resulting solution will be 3.001
sino may kuyang palaging nambibira
The decomposition of SO2Cl2 is first order in SO2Cl2 and has a rate constant of 1.42 x 10-4
s
-1 at a
certain temperature.
a. If the reaction started with 75.6 g of SO2Cl2 dissolved in 1.25 L, what would be the
concentration of SO2Cl2 after 3.00 hours?
b. A student claims that the half life of SO2Cl2 changes as the reaction progresses. Do you agree
or disagree with the students claim? Explain your answer. Note: The experiment was
preformed at a constant temperature.
Answer:
a. 13.0g/L is the concentration of SO2Cl2 after 3.00h
b. FALSE
Explanation:
The first order reaction follows the equation:
ln[SO2Cl2] = -kt + ln[SO2Cl2]₀
Where [] is the concentration after time t, k is rate constant = 1.42x10⁻⁴s⁻¹
[]₀ is initial concentration:
a. []₀ = 75.6g/1.25L = 60.48g/L
t in seconds: 3h * (3600s / 1h) = 10800s
Replacing:
ln[SO2Cl2] = -1.42x10⁻⁴s⁻¹*10800s + ln[60.48g/L]₀
ln[SO2Cl2] = 2.5687
[SO2Cl2] = 13.0g/L is the concentration of SO2Cl2 after 3.00h
b. The rate constant of a reaction remains constant if temeprature remains constant.
How does the law of conservation of mass relate to the number of atoms of each element that are present before a reaction vs. the number of atoms of each element that are present after a chemical reaction?
Suppose that you add 29.2 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the added solute, you find that there is a freezing point depression of 2.78 oC compared to pure benzene. What is the molar mass (in g/mol) of the unknown compound
Answer:
Suppose that you add 29.2 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the added solute, you find that there is a freezing point depression of 2.78 oC compared to pure benzene. What is the molar mass (in g/mol) of the unknown compound
Explanation:
The mass of nonvolatile solute added is ---- 29.2g
The mass of solvent benzene is ---- 0.250kg = 250g
The Kf value of benzene is ---- 5.12^oC/m.
Depression in the freezing point of the solution is --- 2.78^oC.
What is the molar mass of the unknown solute?
[tex]The depression in freezing point = Kf * molality of the solution\\molality of the solution = \frac{mass of solute}{molar mass of solute}*\frac{1}{mass of solvent in kg}[/tex]
Substitute the given values in this formula to get the molar mass of unknown solvent:
[tex]molality=\frac{29.2g}{M} * \frac{1}{0.250kg} \\depression in freezing point:\\2.78^oC=5.12^oC/m * \frac{29.2g}{M} * \frac{1}{0.250kg} \\\\=>M=5.12^oC/m * \frac{29.2g}{2.78^oC} * \frac{1}{0.250g} \\\\\\=>M=215.1g/mol[/tex]
Hence, the molar mass of unknown solute is --- 215g/mol.
An atom has the electron configuration of 1s22s22p5 how many electrons are in that atom
Answer:
There are [tex]9[/tex] electrons in this atom.
Explanation:
Electron configuration of this atom: [tex]1s^2\, 2s^2\, 2p^5[/tex].
The electron orbitals of an atom are denoted as [tex]1s[/tex], [tex]2s[/tex], [tex]2p[/tex], [tex]3s[/tex], [tex]3p[/tex], etc. At any given time, an electron in this atom is located in exactly one orbital.
The electron configuration of an atom gives the number of electrons in each orbitals of this atom.
For example, in this atom, the superscript "[tex]2[/tex]" on the right of "[tex]1s[/tex]" means that there are two electrons in the [tex]1s\![/tex] orbital of this atom. Hence, [tex]1s^2\, 2s^2\, 2p^5[/tex] would translate to:
The [tex]1s[/tex] orbital of this atom contains [tex]2[/tex] electrons.The [tex]2s[/tex] orbital of this atom contains [tex]2[/tex] electrons.The [tex]2p[/tex] orbitals of this atom contain [tex]5[/tex] electrons.Hence, there would be [tex]2 + 2 + 5 = 9[/tex] electrons in total in this atom.
THE DIAMETER OF A TAMBOURINE IS 10 INCHES.FIND THE AREA OF ITS SURFACE.USE π=3.14.
1.What is ask in the problem?
2.What are the given facts?
3.What operation to be used?
4.What is the number sentence
5.What is the answer?
Answer:
22/7 × [tex]10^{2}[/tex]
Explanation:
The area of a circle can be found out using π[tex]r^{2}[/tex]. Since r is the radius so if they multiply, they will give you an area of a square then multiply by 22/7 since it is a circle.
The surface area of the Tambourine at the given diameter of 10 inches is determined as 78.54 square inches.
Area of the TambourineA Tambourine has circular shape, and the area of the Tambourine can be determined by applying formula for area of a circle as shown below;
A = πr²
where;
r is the radius of the circleRadius of the Tambouriner = ¹/₂D
r = ¹/₂ x 10 in
r = 5 in
A = π(5)²
A = 25π in²
A = 78.54 in²
Thus, the surface area of the Tambourine at the given diameter of 10 inches is determined as 78.54 square inches.
Learn more about surface area here: https://brainly.com/question/76387
Determine the electron geometry, molecular geometry, and idealized bond angles for each of the following molecules. CF4CF4 NF3NF3 OF2OF2 H2SH2S In which cases do you expect deviations from the idealized bond angle
Answer:
CF4
Molecular geometry- tetrahedral
Electron geometry- tetrahedral
NF3
-molecular geometry - trigonal pyramidal
Electron geometry - tetrahedral
OF2
Molecular geometry - bent
Molecular geometry - tetrahedral
H2S
Molecular geometry- bent
Electron geometry - tetrahedral
Explanation:
According to Valence Shell Electron Pair Repulsion Theory, the shape of a molecule depends on the number of electron pairs on the valence shell of the central atom in the molecule.
For all the compounds listed, the central atom has four points of electron density. This correspond to a tetrahedra electron pair geometry. The presence of lone pairs on the central atom of OF2,NF3 and H2S accounts for the departure of the observed molecular geometry from the geometry and idealized bond angle predicted on the basis of the VSEPR theory.
C8H18+O2=CO2+H2O. Balance this equation and identify the number H2O molecules formed when 6 molecules of C8H18 react with 75 molecules of oxygen
Answer: 54 molecules of water will be formed in the reaction.
Explanation:
A balanced chemical equation is one where all the individual atoms are equal on both sides of the reaction. It follows the law of conservation of mass.
For the given unbalanced chemical equation, the balanced equation follows:
[tex]2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O[/tex]
We are given:
Molecules of [tex]C_8H_{18}[/tex] = 6
Molecules of [tex]O_2[/tex] = 75
By the stoichiometry of the reaction:
If 2 molecules of [tex]C_8H_{18}[/tex] produces 18 molecules of water
So, 6 molecules of [tex]C_8H_{18}[/tex] will produce = [tex]\frac{18}{2}\times 6=54[/tex] molecules of water
Hence, 54 molecules of water will be formed in the reaction.
Calcula la concentración de H+ de una sustancia que tiene pH 8.8
Answer:
[tex]pH = - log[H {}^{ + } ] \\ 8.8 = - log[H {}^{ + } ] \\ \: [H {}^{ + } ] = {10}^{ - 8.8} \\ [H {}^{ + } ] = 1.585 \times {10}^{ - 9} \: mol {dm}^{ - 3} [/tex]
The information code that an organism inherits can best be referred to as its -
O A genotype
B. territory
C. species
D. kingdom
Answer:
it will be no.A genotype
A 575.4575.4 mL sample of carbon dioxide was heated to 377377 K. If the volume of the carbon dioxide sample at 377377 K is 824.7824.7 mL, what was its temperature at 575.4575.4 mL
Answer:
263 K
Explanation:
Assuming ideal behaviour and constant pressure, we can solve this problem by using Charles' law, which states that at constant pressure:
T₁V₂=T₂V₁In this case:
T₁ = ?V₂ = 824.7 mLT₂ = 377 KV₁ = 575.45 mLWe input the data:
T₁ * 824.7 mL = 377 K * 575.45 mLAnd solve for T₁:
T₁ = 263 KHydrogen bonds within liquid water are attractions between protons and hydroxide ions. are dipole-dipole attractions. are ion-induced dipole attractions. are attractions between protons and oxygen nuclei. are attractions between two hydrogen atoms.
Answer:
true because the bonds cannot be broken down
19. What is the molarity of a
solution that contains 6 liters
of solution and 2 moles of
solute?
a. 8 M
b. 0.3 mol/L
c. 12 mol/L
d. 3M
Answer:
b. 0.3 mol/L is the closest.
Explanation:
A molar solution of a substance contains 1 mole per liter of solution
So if we have 2 moles in 6 liters the there are 2/6 =1/3 of a mole in 1 liter.
The pressure on 150 mL of a gas is increased from 500 mm Hg to 700 mm Hg at constant temperature. What is the new volume of the gas?
A. 107 ml
B. 210 mL
C. 350 ml
D. 2333 mL
Answer:
Option A. 107 mL
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 150 mL
Initial pressure (P₁) = 500 mmHg
Final pressure (P₂) = 700 mmHg
Temperature = constant
Final volume (V₂) =?
The final volume of the gas can be obtained by using the Boyle's law equation as shown below:
P₁V₁ = P₂V₂
500 × 150 = 700 × V₂
75000 = 700 × V₂
Divide both side by 700
V₂ = 75000 / 700
V₂ = 107 mL
Therefore, the final volume of the gas is 107 mL.
171 g of sucrose ( MW of 342, melting point 186 oC, boiling point very high, and vapor pressure is negligible) is dissolved in one liter of water at 25 oC. At 25 oC the vapor pressure of water is 24 mmHg. Which value is closest to the vapor pressure (VP) of this solution at 25 oC?
The complete question is as follows: 171 g of sucrose ( MW of 342, melting point 186 oC, boiling point very high, and vapor pressure is negligible) is dissolved in one liter of water at 25 oC. At 25 oC the vapor pressure of water is 24 mmHg. Which value is closest to the vapor pressure (VP) of this solution at 25 oC?
a. 16mm Hg
b. 24mm Hg
c. 20mm Hg
d. 12mm Hg
Answer: The vapor pressure (VP) of this solution at [tex]25^{o}C[/tex] is closest to the value 24 mm Hg.
Explanation:
Given: Mass of sucrose = 171 g
Mass of water = 1 L = 1000 g
Vapor pressure of water = 24 mm Hg
As moles is the mass of substance divided by its molar mass. Hence, moles of water (molar mass = 18.02 g) is calculated as follows.
[tex]Moles = \frac{mass}{molar mass}\\= \frac{1000 g}{18.02 g/mol}\\= 55.49 mol[/tex]
Similarly, moles of sucrose (molar mass = 342 g/mol) is as follows.
[tex]Moles = \frac{mass}{molar mass}\\= \frac{171 g}{342 g/mol}\\= 0.5 mol[/tex]
Total moles = 55.49 + 0.5 mol = 55.99 mol
Mole fraction of water is as follows.
[tex]Mole fraction = \frac{moles of water}{total moles}\\= \frac{55.49}{55.99}\\= 0.99[/tex]
Formula used to calculate vapor pressure of the solution is as follows.
[tex]P_{i} = P^{o}_{i} \times \chi_{i}[/tex]
where,
[tex]P_{i}[/tex] = vapor pressure of component i over the solution
[tex]P^{o}_{i}[/tex] = vapor pressure of pure component i
[tex]\chi_{i}[/tex] = mole fraction of i
Substitute the values into above formula to calculate vapor pressure of water as follows.
[tex]P_{i} = P^{o}_{i} \times \chi_{i}\\= 24 mm Hg \times 0.99\\= 23.76 \\or 24 mm Hg\\[/tex]
Thus, we can conclude that the vapor pressure (VP) of this solution at [tex]25^{o}C[/tex] is closest to the value 24 mm Hg.
What is a combustion reaction
Answer:
Combustion, a chemical reaction between substances, usually including oxygen and usually accompanied by the generation of heat and light in the form of flame.
Explanation:
Hope this helps!! :))
Compound A has the formula C8H8. It reacts rapidly with acidic KMnO4 but reacts with only 1 equivalent of H2 over a palladium catalyst. On hydrogenation under conditions that reduce aromatic rings, A reacts with 4 equivalents of H2, and hydrocarbon B, C8H16, is produced. The reaction of A with KMnO4 gives CO2 and a carboxylic acid C, C7H6O2.
Required:
Draw the structure of compound B below.
Answer:
C8H16 (Ethylcyclohexane).
Explanation:
From the given information:
Compound A is an alkene because it interacts with 1 unit of hydrogen across a palladium catalyst.
Also, we are given another hint that:
Compound A needs 4 equivalence of H2 to hydrogenate under circumstances that decrease aromatic rings, indicating that it is a phenyl substituted alkene.
Compound A with formula C8H8 reacts instantly with KMnO4 to produce CO2, as well as carboxylic acid, points out that Compound acts as a terminal alkene.
Therefore, we can opine that compound A is a terminal phenyl substituted alkene whose formula = C8H8 (Styrene)
The diagrammatic expression of the compound can be seen below.
However, in the presence of the palladium catalyst, the reduction of Compound A with 4 units of hydrogen produces Compound B: C8H16 (Ethylcyclohexane).
Write the molecular formula for the compound that exhibits a molecular ion at M+ = 112.0499. Assume that C, H, N, and O might be present, and use the exact masses below:
Exact mass of carbon = 12.000
Exact mass of hydrogen = 1.0078
Exact mass of nitrogen = 14.003
Exact mass of oxygen = 15.995
The order of atoms should be carbon, then hydrogen, then the others in alphabetical order.
Answer:
C₆H₁₀NO
Explanation:
In order to arrive at a molecular formula we have to make some assumptions and they are
Assuming there is one ( 1 ) N and one ( 1 ) O that is present in the said molecule
Total mass = 29.998
next step: subtract the total mass from 112.0499 = 82.501
next : assume the presence of 6 carbon atoms in said molecule
Total mass = 6 * 12 = 72
Mass of Hydrogens = 82.501 - 72 = 10.501
∴ number of hydrogens = 10.501 / 1.0078 ≈ 10
Hence Total mass = 29.998 + 82.501 ≈ 112.0499
Finally Molecular formula = C₆H₁₀NO
On another planet, the isotopes of titanium have the given natural abundances.
Isotope Abundance Mass (u)
46Ti 77.100% 45.95263
48Ti 17.100% 47.94795
50Ti 5.800% 49.94479
What is the average atomic mass of titanium on that planet?
Answer:
46.525 u
Explanation:
From the question given above, the following data were obtained:
Isotope A (⁴⁶Ti)
Abundance (A%) = 77.100%
Mass of A = 45.95263 u
Isotope B (⁴⁸Ti):
Abundance (B%) = 17.100%
Mass of B = 47.94795 u
Isotope C (⁵⁰Ti):
Abundance (C%) = 5.800%
Mass of C = 49.94479 u
Average atomic mass =?
The average atomic mass of titanium can be obtained as follow:
Average = [(Mass of A × A%)/100] + [(Mass of B × B%)/100] + [(Mass of C × C%)/100]
= [(45.95263 × 77.1)/100] + [(47.94795 × 17.1)/100] + [(49.94479 × 5.8)/100]
= 35.429 + 8.199 + 2.897
= 46.525 u
Therefore, the average atomic mass of titanium is 46.525 u
