Answer:
A. Yes, the calculated density in agreement with the tabulated value.
B. 19.28 mL of volume of cyclohexane should be used.
C. 742.20 is the mass of the sphere of lead.
Explanation:
A.
Volume of the liquid = V = 25.0 mL
Mass of the liquid = m = 21.95 g
Density of the liquid = d
[tex]d=\frac{m}{V}[/tex]
[tex]=\frac{21.95 g}{25.0 mL}=0.878 g/mL[/tex]
Density mentioned in the report book = d' = 0.878 g/mL
d' = d = 0.878 g/mL
Yes, the calculated density in agreement with the tabulated value.
B.
Volume of the liquid cyclohexane= V = ?
Mass of the liquid cyclohexane= m = 15.0 g
Density of the liquid cyclohexane = d = 0.7781 g/mL
[tex]d=\frac{m}{V}[/tex]
[tex]V=\frac{15.0 g}{0.7781 g/mL}=19.28 mL[/tex]
19.28 mL of volume of cyclohexane should be used.
C.
Diameter of the ball = d = 5.0 cm
Radius of the ball = r = 0.5 × d = 2.5 cm
Volume of sphere ,V= [tex]\frac{4}{3}\pi r^3[/tex]
[tex]V = \frac{4}{3}\times 3.14\times (0.25 cm)^3=65.45 cm^3[/tex]
Volume of the spherical lead ball = V
Mass of the spherical lead ball= m = ?
Density of the spherical lead ball = d = [tex]11.34 g/cm^3[/tex]
[tex]d=\frac{m}{V}[/tex]
[tex]m=d\times V=11.34 g/cm^3\times 65.45 cm^3=742.20 g[/tex]
742.20 is the mass of the sphere of lead.
A. The tabulated density of benzene [tex]15^0 C[/tex] is 0.878 g/mL. B. 19.28 mL of volume of cyclohexane should be used. C. The mass of the lead sphere is approximately 743.5 g.
A. To determine if the calculated density of the liquid matches the tabulated value for benzene, we need to calculate the density using the provided data and then compare it to the handbook value. The density is calculated by dividing the mass of the liquid by its volume.
Given:
Mass of the liquid = 21.95 g
The volume of the liquid = 25.0 mL
Calculated density = [tex]Mass / Volume = 21.95 g / 25.0 mL = 0.878 g/mL[/tex]
The tabulated density of benzene [tex]15^0 C[/tex] is 0.878 g/mL. Since the calculated density matches the tabulated value, the liquid in the bottle is likely benzene.
B. To find the volume of cyclohexane required for the experiment, we can use the formula: Volume = Mass / Density
Given:
Mass of cyclohexane needed = 15.0 g
Density of cyclohexane at [tex]25^0C[/tex] = 0.7781 g/mL
Volume of cyclohexane = 15.0 g / 0.7781 g/mL = 19.28 mL
Therefore, 19.28 mL of cyclohexane should be used for the experiment.
C. The mass of the lead sphere can be calculated using the formula for the volume of a sphere and the density of lead:
The volume of a sphere = [tex](4/3)\pi r^3[/tex]
Density = Mass / Volume
Given:
The diameter of the lead sphere = 5.0 cm
Radius of the lead sphere = Diameter / 2 = 5.0 cm / 2 = 2.5 cm
Density of lead = [tex]11.34 g/cm^3[/tex]
The volume of the lead sphere =[tex](4/3)\pi (2.5 cm)^3 =20.94375\pi cm^3[/tex]
Mass of the lead sphere = Density — Volume = [tex]11.34 g/cm^3 - 20.94375\pi cm^3= 11.34 g/cm^3- 65.45902 cm^3= 743.5 g[/tex]
Therefore, the mass of the lead sphere is approximately 743.5 g.
Given a sorted list of 1024 elements, what is the runtime for binary search if the search key is less than all elements in the list?
Answer: 1024
Linear search's runtime is proportional to the number of list elements.
A hot metal plate at 150°C has been placed in air at room temperature. Which event would most likely take place over the next few minutes?
A. Molecules in both the metal and the surrounding air will start moving at lower speeds.
B. Molecules in both the metal and the surrounding air will start moving at higher speeds.
C. The air molecules that are surrounding the metal will slow down, and the molecules in the metal will speed up.
D. The air molecules that are surrounding the metal will speed up, and the molecules in the metal will s
Answer:
The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down.
Explanation:
The hot metal at 150 °C loses heat energy by conduction to the surrounding air molecukes and as such cools down, the cooler metal consists of lesser enery to power the movement of the molecules of the metal hence the metal molecules slows down in their movement as seen in the equation of heat and temperature
ΔH = m×C×ΔT where ΔH is the change in heat energy (heat loss of the metal, C is the heat capacity and ΔT is the temperature change
For the surrounding air that experiences increase in temperature the same process follows leading to increase in the kinetic energy of the air molecules and decrease in kinetic energy of the metal molecules as shown in the formula
K = [tex]\frac{3}{2}[/tex]×[tex]\frac{R}{N_{A} }[/tex]×T where K = Kinetic Energy, R = gas constant (8.314J/mol×K) and [tex]N_{A}[/tex] = Avogadros number (6.022×[tex]10^{23}[/tex] atoms/mol)
Answer:
d
Explanation:
Why does the rate of a reaction generally increase with increased temperature? (Choose all that apply)
A.
at higher temperatures, molecules are more likely to collide with each other in the correct orientation.
B.
at higher temperatures, molecules will collide more often.
C.
at higher temperatures, the potential energy difference between reactants and products is smaller.
D.
at higher temperatures, more of the collisions will have the required activation energy.
Answer: Option (A) is the correct answer.
Explanation:
When we increase the temperature of a chemical reaction then molecules of the given reaction will gain kinetic energy. As a result, they will come into motion and then collide frequently with each other.
For a reaction to increase its rate it is important that the particles must collide in a correct orientation.
Thus, we can conclude that the rate of a reaction generally increase with increased temperature because at higher temperatures, molecules are more likely to collide with each other in the correct orientation.
Final answer:
The rate of a reaction increases with temperature due to more frequent collisions, collisions with greater energy exceeding the activation energy, and a higher likelihood of correct molecular orientation during these collisions.
So, correct options are A, B, and D.
Explanation:
The rate of a reaction generally increases with increased temperature for multiple reasons. At higher temperatures, molecules move more quickly which leads to an increase in the frequency of collisions between them. Furthermore, with the rise in kinetic energy, there is a greater chance that these collisions will have enough activation energy to result in a reaction. Also, as molecules move faster, they are more likely to collide in the correct orientation to instigate a reaction.
Options A and B are correct because molecules are more likely to collide and do so with the correct orientation at higher temperatures. Option D is also correct because the increase in kinetic energy means more collisions will have the required activation energy. However, Option C is not accurate because the potential energy difference between reactants and products does not depend on temperature.
The volume of a gas is reduced from 4 L to 0.5 L while the temperature is held constant. How does the gas pressure change?
It increases by a factor of two.
It increases by a factor of eight.
It decreases by a factor of eight.
It increases by a factor of four.
Answer:
It increases by a factor of eight
Explanation:
When temperature is held constant, gas pressure changes according the volume, in undirectly proportion.
Volume increases → Pressure decreases
Volume decreases → Pressure increases
As volume gas, was reducted from 4L to 0.5L, it was reduced by 1/8, so the pressure gas was increased by a factor of eight.
Answer:
It increases by a factor of eight.
Explanation:
A 2.12-g sample of scandium metal is reacted with excess hydrochloric acid to produce 0.1415 g hydrogen gas. What is the formula of the scandium chloride produced in the reaction?
Answer:
The formula of the scandium chloride produced in the reaction is ScCL₃
Explanation:
With the given data, you can know the molar relationship between Sc and H₂ (molar ratio) to determine the reaction stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction).
Then: [tex]molar ratio=\frac{moles Sc}{moles H_{2} }[/tex]
Knowing that:
mass Sc= 2.12 gmolar mass of Sc= 44.956 g/molmass H₂= 0.1415 gmolar mass of H₂= 2 g/moland knowing that the number of moles (n) of a compound can be calculated as: [tex]n=\frac{mass}{molar mass}[/tex]
Then:
[tex]n (Sc)=\frac{2.12g}{44.956\frac{g}{mol} }[/tex] then [tex]n (Sc)=0.047 moles[/tex][tex]n(H_{2} )=\frac{0.1415g}{2\frac{g}{mol} }[/tex] then [tex]n(H_{2} )=0.071 moles[/tex]So:
[tex]molarratio=\frac{0.047}{0.071}[/tex]
Then it is possible to say that the molar ratio is approximately equal to [tex]\frac{2}{3}[/tex]. This indicates that by stoichiometry 2 moles of Sc are needed to produce 3 moles of H₂.
So:
2 Sc + HCl → ScCL₃ + 3 H₂
The law of conservation of matter states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reagents has to be equal to the number of atoms present in the products.
Then, balancing the equation so that the same amount of moles of each element on each side of the equation is obtained:
2 Sc + 6 HCl → 2 ScCL₃ + 3 H₂
The formula of the scandium chloride produced in the reaction is ScCL₃
We expect the enthalpy of cumbustion of two isomers to be _________. The molecular formulas of two molecules are _____, so the balanced chemical equation for the two combustion reactions are________, in calculation of combustion enthalpy from _____ of products and reactants, the difference will be in the ______of the two_____
Different, Isomers, standard enthalpies of formation, combustion products, standard enthalpies of combustion, the same, very similar
The rod-shaped n-pentane has ________ possible________ than the almost________ neopentane
Tetrahedral, less, vibrational and rotational motions, more, translational motion, spherical
Final answer:
The enthalpy of combustion of two isomers with the same molecular formula will be different due to their different structural arrangements. The rod-shaped n-pentane has less possible vibrational and rotational motions compared to the almost spherical neopentane.
Explanation:
The enthalpy of combustion of two isomers with the same molecular formula will be different. Isomers are compounds that have the same molecular formula but different structural arrangements. For example, the isomers of C4H10 are n-butane and isobutane. The balanced chemical equations for their combustion reactions are:
n-Butane: C4H10 + 6.5O2 → 4CO2 + 5H2O
Isobutane: C4H10 + 6.5O2 → 4CO2 + 5H2O
When calculating the combustion enthalpy from the standard enthalpies of formation of products and reactants, the difference in enthalpy will be due to the different standard enthalpies of combustion of the two isomers.
The rod-shaped n-pentane has less possible vibrational and rotational motions than the almost spherical neopentane. This is because neopentane is a more compact, three-dimensional molecule, while n-pentane is a linear molecule. Linear molecules have fewer degrees of freedom and therefore have fewer potential vibrational and rotational motions compared to more complex, three-dimensional molecules like neopentane.
The enthalpies of combustion for isomers are different due to structural variations. N-pentane has more vibrational and rotational motions compared to neopentane.
'We expect the enthalpy of combustion of two isomers to be different. The molecular formulas of two molecules are isomers, so the balanced chemical equations for the two combustion reactions are very similar. In calculation of combustion enthalpy from standard enthalpies of formation of products and reactants, the difference will be in the standard enthalpies of combustion of the two molecules.'
The rod-shaped n-pentane has more possible vibrational and rotational motions than the almost spherical neopentane.
When DNA is replicated, it is necessary for the two strands to "unzip" temporarily. Choose which bonding type is most appropriate for holding the strands together in this way.
A) ionic bonding
B) hydrogen bonding
C) non-polar covalent bonding D) polar covalent bonding
Answer: B) hydrogen bonding
Explanation:
A) An ionic bond is formed when an element completely transfers its valence electron to another element.
B) Hydrogen bonds are special type of dipole dipole forces which are formed when hydrogen bonds with an electronegative element.
DNA strands are held together by hydrogen bonds between the bases on one strand and those on the other. Adenine and thymine have two hydrogen bonds between them while cytosine and guanine have three hydrogen bonds between them.
C) Non-polar covalent bond is defined as the bond which is formed when there is no difference of electronegativities between the atoms.
D) A polar covalent bond is defined as the bond which is formed when there is a difference of electronegativities between the atoms.
When DNA is replicated, the two strands unzip temporarily. The strands are held together by hydrogen bonding.
Explanation:When DNA is replicated, the two strands temporarily unzip or separate from each other. The type of bonding that holds the strands together is hydrogen bonding. Hydrogen bonding occurs between the nitrogenous bases of the DNA strands.
These bases, adenine (A), thymine (T), cytosine (C), and guanine (G), form complementary base pairs with each other, with adenine pairing with thymine and cytosine pairing with guanine.
The hydrogen bonds between these base pairs keep the two DNA strands connected.
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The density of water at 3.98°C is 1.00000 g/mL. What is the mass in pounds of 16.743 L of water?
Answer:
The answer to your question is 36.9 pounds
Explanation:
Data
density = 1 g/ml
mass = ?
volume = 16.743 L
- To solve this problem use the formula of density.
density = mass / volume
- Solve for mass
Mass = density x volume
- Convert volume to ml
1000 ml --------------- 1 L
x ---------------- 16.743 L
x = (16.743 x 1000) / 1
x = 16743 ml
- Substitution
Mass = (1 g/ml)(16743 ml)
Simplification and result
Mass = 16743 g
- Convert mass to pounds
1 pound ------------------ 453.58 g
x ------------------ 16743 g
x = (16743 x 1) / 453.58
x = 36.9 pounds
The mass of 16.743 L of water in pounds can be calculated as approximately 36.904 pounds.
Explanation:To calculate the mass of water in pounds, we first determine the mass in grams using the given water volume and density.
Since the density of water is 1.00000 g/mL, and we know that 1 L is equal to 1000 mL, we multiply the volume in liters by the density in g/mL and by 1000 to get mass in grams.
This gives us 16.743 L * 1.00000 g/mL * 1000 = 16743 grams.
Converting grams to pounds, we know 1 pound is approximately 453.592 grams
So, the mass of water is then 16743 g / 453.592 g/lbs = 36.904 lbs. Therefore, the mass of 16.743 L of water is approximately 36.904 pounds.
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A 222 N force is required to keep an object moving along a circular path with a diameter of 11.6 m at a speed of 7.60 m/s. What is the mass of the object?
Answer:
[tex]m=22.3\ kg[/tex]
Explanation:
The expression for centrifugal force is shown below as:-
[tex]F=\frac{m\times v^2}{r}[/tex]
Where, m is the mass of the object = ?
v is the velocity of the object = 7.60 m/s
r is the radius
F is the force = 222 N
Diamter = 11.6 m
Radius = d/2 = 11.6 m/2 = 5.8 m
Applying the values in the above expression as:-
[tex]222=\frac{m\times {7.60}^2}{5.8}[/tex]
[tex]57.76m=1287.6[/tex]
[tex]m=22.3\ kg[/tex]
Which of the following statements about acids/bases is correct: Group of answer choices
1. molecule with a pH of 9 has a higherA concentration of H than water
2. A molecule with a pH of 10 has a higher concentration of OH- than water
3. A molecule with a pH of 12 has a lower concentration of OH- than water
4. A molecule with a pH of 4 has a higher concentration of OH- than water
Answer:
2
Explanation:
To answer this correctly, we need to make a reference to the pH scale. Here we know that pH 1-6 stands for acidity where the smaller the value, the higher the concentration of H3O+ and the higher the acidity. Also, pH 7 stands for neutral, there is an interpreted balance here, where the concentrations are averagely equal. Also , we have pH 8 to 14 where we have the alkalinity, the higher the value the higher the concentration of OH-.
Now let’s solve the question at hand.
1 is wrong
Water is termed neutral with an average balance of the two ions. pH 9 means higher OH- which stipulates lesser H+ than water.
2 is correct
The pH is termed alkaline and has more hydroxide concentration than water.
3 is wrong
It should have a higher concentration of OH- than water
4 is wrong
It should have a higher acidity and lesser alkalinity. This translates to higher H+ and lower OH- relative to water
Some prokaryotes, such as the blue-green ____________ , are photosynthetic and contain ____________ where chlorophyll and other pigments absorb energy from the sun to produce carbohydrates via photosynthesis.
Blue-green algae are prokaryotes that carry out photosynthesis in chloroplasts, converting light energy into carbohydrates by using chlorophyll as a pigment for energy absorption.
Explanation:Some prokaryotes, such as blue-green algae, are photosynthetic and contain chloroplasts where chlorophyll and other pigments absorb energy from the sun to produce carbohydrates via photosynthesis. The algae, specifically, are among the groups of prokaryotes that are capable of photosynthesis, much like plants. The chloroplasts in these cells function as the site where light energy captured by chlorophyll is converted into chemical energy, which is then used to create carbohydrates from carbon dioxide and water.
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If a piece of ice weighs 36.0 NN at the surface of the earth, what is its mass on the earth's surface?
Answer:
3.67 kilograms is the mass of a given ice on the earth's surface.
Explanation:
Weight of an object is the force acting on an object by virtue of its mass.
Weight(W)=[tex]mass(m)\times acceleration(a)[/tex]
We have :
Mass of an ice = m =?
Acceleration due to gravity ,a = [tex]9.8m/s^2[/tex]
Weight of an ice , W= 36.0 N
Putting in the values we get,
[tex]36.0 N=m\times 9.8 g/m^2[/tex]
[tex]m=\frac{36.0 N}{9.8 m/s^2}=3.67 kg[/tex]
3.67 kilograms is the mass of a given ice on the earth's surface.
In aerobic cellular respiration, if four molecules of pyruvic acid enter steps two, the formation of acetyl CoA and three, the Krebs cycle, how many molecules of ATP, NADH, and FADH2 will be formed?
Answer:
The aerobic cellular respiration of the glucose where glucose is converted to energy via four steps as follows
1. Glycolysis (glucose break down to pyruvic acid)
2. Link reaction
3. Krebs cycle
4. Electron transport chain, or ETC
The four pyruvic acid produces Four ATP, twenty NADH, and four [tex]FADH_{2}[/tex] molecules
Explanation:
When four pyruvic acid enters step two of the aerobic cellular respiration, they are converted by Oxidative decarboxylation into acetyl-CoA, four molecules of NADH and four molecule of CO2 are formed. This process is otherwise called the link reaction or transition step, because it connects or links the Krebs cycle and glycolysis.
From the chemical reactions involved in cellular respiration of one glucose molecule, from two pyruvic acid molecules we have 2 ATP molecules, 10 NADH molecules, and 2 FADH2 molecules
Hence from four pyruvic acid molecules we have that the acetyl-CoA produced from the four pyruvic acid enters the the Krebs cycle and forms four ATP molecules, twenty NADH molecules, and four [tex]FADH_{2}[/tex] molecules.
In aerobic cellular respiration, four molecules of pyruvic acid will generate a total of four molecules of ATP, sixteen molecules of NADH (four from the conversion to Acetyl CoA, and twelve from the Krebs cycle) and four molecules of FADH2
Explanation:In aerobic cellular respiration, pyruvic acid is converted into acetyl CoA. One molecule of pyruvic acid generates one molecule of NADH during this conversion so four molecules of pyruvic acid will yield four molecules of NADH. Acetyl CoA then enters the Krebs cycle, for each molecule of Acetyl CoA that goes through the Krebs cycle, three molecules of NADH, one molecule of FADH2, and one molecule of ATP is formed. Therefore, the four molecules of pyruvic acid would end up generating four molecules of ATP, twelve molecules of NADH and four molecules of FADH2 during the Krebs cycle (not including the NADH generated during the conversion to Acetyl CoA).
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The freezing temperatures for water for Celsius and Fahrenheit scales are 0ºC and 32ºF. The boiling temperatures for water are 100 ºC and 212 ºF. Let C denote the temperature in Celsius and F in Fahrenheit. Write the conversion function from Celsius to Fahrenheit. Use the function to convert 25 ºC into ºF.
Answer:
Therefore the required function is
[tex]C= \frac{5}{9} (F-32)[/tex]
Therefore 25°C=57°F
Explanation:
F denotes temperature of Fahrenheit and C denotes temperature of Celsius.
[tex]\frac{C-0}{100-0} =\frac{F-32}{212-32}[/tex]
[tex]\Rightarrow C= \frac{100}{180} (F-32)[/tex]
[tex]\Rightarrow C= \frac{5}{9} (F-32)[/tex]
Therefore the required function is
[tex]C= \frac{5}{9} (F-32)[/tex]
Putting C=25°C in above equation
[tex]25=\frac{5}{9} (F-32)[/tex]
[tex]\Rightarrow 25 \times \frac{9}{5} = F-32[/tex]
⇒45 =F-32
⇒F=32+45
⇒F=57
Therefore 25°C=57°F
The conversion function from Celsius to Fahrenheit is F = (C × 9/5) + 32. Using this, 25°C is converted to 77°F.
The subject of this question is temperature conversion between Celsius and Fahrenheit scales. To create a conversion function from Celsius to Fahrenheit, we use the relationship that a change of 1°C is equivalent to a change of 1.8°F. Since the freezing point of water is 0°C and 32°F, we can derive the following function for converting temperature from Celsius to Fahrenheit:
F = (C × 9/5) + 32.
Using this function, we can convert 25°C into Fahrenheit as follows:
Multiply the Celsius temperature by 9/5: 25 × 9/5 = 45.Add 32 to the result: 45 + 32 = 77.Therefore, 25°C is equal to 77°F.
Air is 78.1% nitrogen, 20.9% oxygen, and 0.934% argon by moles. What is the density of air at 22 °C and 760 torr? Assume ideal behavior.
Answer:
The density of air at 22 °C and 760 torr is 1.195 KG/m³
Explanation:
The solution to the above question is arrived t by considering the given variables and calculating the number of moles in a 1 m³ sample of air by plugging values into the universal gas equation from which the number of moles of the constituent gases can be calculated by Dalton's law of partial pressure, then their masses and lastly the density of air is calculated using the formula, Density = mass/volume
The given variables are
Percentage Nitrogen = 78.1% by volume
Percentage oxygen = 20.9% by volume
Percentage argon = 0.934% by volume
The molar mass of nitrogen = 14.006g/mol
The molar mass of oxygen = 15.999g/mol
The molar mass of argon = 39.948 g/mol hence Considering a unit volume of air of one cubic meter (1m^3) we have
0.781 m³ of nitrogen, 0.209 m³ of oxygen and 0.00934 m³ of argon
The number of moles in 1 m³ of gas at 22 °C and 760 torr is given by
PV = nRT or n = [tex]\frac{PV}{RT}[/tex] = where 760 torr = 101325Pa we have n = [tex]\frac{(101325)(0.001)}{(8.314)(295.15)}[/tex] = 0.00413 mols per liter or 41.29 moles/m³
thus we have number of moles of nitrogen = 42.129 × 78.1% = 32.25 moles and the mass of nitrogen = 32.25×28.02 = 903.6 g
number of moles of oxygen= 42.129 × 20.1% = 8.63 moles and the mass of nitrogen = 8.63×32 = 276.16 g
number of moles of argon= 42.129 × 0.934% = 0.386 moles and the mass of nitrogen = 0.386×40 = 15.43 g
Therefore, mass of one cubic meter of air (1 m³), has a mass of
903.6 g + 276.16 g + 15.43 g = 1195.2 g or 1.195 KG Hence the density of air
is given by Density = [tex]\frac{mass}{volume}[/tex] =[tex]\frac{1.195 KG}{1 m^{3} }[/tex] = 1.195 KG/m³
The density of air at 22 °C and 760 Torr is 1.19 g/L.
Air is 78.1% nitrogen, 20.9% oxygen, and 0.934% argon by moles. We will calculate the average molar mass of the air (M) as a weighted average of the molar masses of its constituents.
[tex]M = 78.1\% \times M(N_2) + 20.9\% \times M(O_2) + 0.934\% \times M(Ar)\\\\M = 78.1\% \times 28.00g/mol + 20.9\% \times 32.00g/mol + 0.934\% \times 39.95 g/mol = 28.93 g/mol[/tex]
Then, we will convert 22 °C to Kelvin using the following expression.
[tex]K = \° C + 273.15 = 22\° C + 273.15 = 295 K[/tex]
Assuming ideal behavior, we can calculate the density (ρ) of the air at 295 K (T) and 760 Torr (P) using the following expression.
[tex]\rho = \frac{P \times M }{R \times T} = \frac{760 Torr \times 28.93g/mol }{(62.4mmHg.L/mol.K) \times 295K} = 1.19 g/L[/tex]
where,
R: ideal gas constantThe density of air at 22 °C and 760 Torr is 1.19 g/L.
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Classify these substances? More than one answer may apply in each case.
N 2 solution heterogeneous mixture homogeneous mixture element compound pure substance
O 2 pure substance homogeneous mixture solution heterogeneous mixture element compound
N 2 O compound heterogeneous mixture element homogeneous mixture solution pure substance Air
(mostly N 2 and O 2 )
homogeneous mixture heterogeneous mixture solution pure substance element compound
Answer:
N2 element, pure substance
O2 element, pure substance
N2O Compound, pure substance
Air (mostly N2 and O2 ) homogeneous mixture
Explanation:
N2, Nitrogen is known as the chemical element that is characterized by having atomic number 7 and that is symbolized by the letter N, in its molecular version, it is recognized as N2.
O2, Oxygen is the chemical element of atomic number 8, this molecular form is composed of two atoms of this element.
A chemical element is a type of matter, consisting of atoms of the same class.
N2O, Nitrous oxide is formed by the union of two molecules of nitrogen and one of oxygen, which is considered a chemical compound since it is a substance formed by the chemical combination of two different elements of the periodic table.
A pure substance is one that cannot change state or divide into other substances, except for a chemical reaction.
Air (mostly N2 and O2 ), it is a homogeneous mixture of gases that constitutes the earth's atmosphere. A homogeneous mixture is a type of mixture in which its components are not distinguished and in which the composition is uniform and each part of the solution has the same properties.
Substances can be classified based on their composition and uniformity: N2 and O2 are elements and pure substances, N2O is a compound and a pure substance, and air is a homogeneous mixture or solution.
Explanation:When classifying substances, we take into account their composition and uniformity. Here are the classifications for the mentioned substances:
N2 (Nitrogen) is an element and a pure substance since it is composed of only one type of atom.
O2 (Oxygen) is also an element and a pure substance, with two oxygen atoms bonded together.
N2O (Nitrous Oxide) is a compound as it is made up of two different elements, nitrogen and oxygen, in a fixed ratio and a pure substance due to its uniform composition.
Air, which is mostly made up of Nitrogen (N2) and Oxygen (O2), is a homogeneous mixture or solution because the composition is uniform throughout, and it is a mixture of multiple gases.
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Manganese sulfate forms a pale pink hydrate with the formula MnSO 4 ⋅ n H 2 O ( s ) . If this hydrate is heated to a high enough temperature, H 2 O ( g ) can be driven off, leaving the grey‑white anhydrous salt MnSO 4 ( s ) . A 16.260 g sample of the hydrate was heated to 300 ∘ C . The resulting MnSO 4 ( s ) had a mass of 14.527 g . Calculate the value of n in MnSO 4 ⋅ n H 2 O ( s ) .
Answer:
Value of n in MnSO₄.nH₂O is one.
Explanation:
The n represents the number of moles of water attached to the formula unit manganese sulfate. These moles (n) can be determined by taking the ratio of the moles of anhydrous salt and the moles of water. The moles of water can be determined by taking the difference of final and initial mass of the salt. This difference is equal to the mass of the water, mathematically it can be represented as,
Mass of H₂O = initial mass of the salt (g) - final mass of the salt (g)
Mass of H₂O = 16.260 g - 14.527 g
Mass of H₂O = 1.733 g
moles of H₂O = (1.733 g) ÷ (18.015 g/mole)
moles of H₂O = 0.0962
For the moles of anhydrous salt:
moles of MnSO₄ = mass of MnSO₄ ÷ molar mass of MnSO₄
moles of MnSO₄ = 14.5277 ÷ 151.001
moles of MnSO₄= 0.0962
Now for n:
n = moles of water ÷ moles of MnSO₄
n = 0.0962 ÷ 0.0962
n = 1
The above calculations show that one mole of H₂O is attached to the one formula unit of MnSO₄
Final answer:
To calculate the value of n in the hydrate formula MnSO₄ .nH₂O, the mass of water driven off by heating is found to be 1.733 g. This corresponds to 0.0962 mol of water. Since the ratio of water to MnSO₄is 1:1, n is determined to be 1.
Explanation:
To calculate the value of n in the hydrate formula MnSO₄ . H₂O, we need to find the number of moles of water lost upon heating. We subtract the mass of the anhydrous salt (14.527 g) from the original mass of the hydrate (16.260 g) to get the mass of water lost:
Mass of H₂O = 16.260 g - 14.527 g = 1.733 g
Next, we calculate the number of moles of water using its molar mass (18.015 g/mol):
Number of moles of H₂O = 1.733 g / 18.015 g/mol ≈ 0.0962 mol
To find the number of moles of MnSO₄ in the anhydrous sample, we need its molar mass, which is approximately 151.00 g/mol for MnSO₄. Using the mass of the anhydrous salt:
Number of moles of MnSO₄ = 14.527 g / 151.00 g/mol ≈ 0.0962 mol
This indicates that the mole ratio of H₂O to MnSO₄ is 1:1. Therefore, the value of n is 1, and the hydrate is MnSO₄.H₂O.
A 36.2 g object has a heat capacity of 12.5 J/ oC. How much energy (in J) is required to raise the temperature by 5.73 oC?
Answer:
[tex]\Delta H=2592.825\ J[/tex]
Explanation:
The expression for the calculation of the enthalpy change of a process is shown below as:-
[tex]\Delta H=m\times C\times \Delta T[/tex]
Where,
[tex]\Delta H[/tex] is the enthalpy change
m is the mass
C is the specific heat capacity
[tex]\Delta T[/tex] is the temperature change
Thus, given that:-
Mass of object = 36.2 g
Specific heat = 12.5 J/g°C
[tex]\Delta T=5.73\ ^0C[/tex]
So,
[tex]\Delta H=36.2\times 12.5\times 5.73\ J=2592.825\ J[/tex]
[tex]\Delta H=2592.825\ J[/tex]
Answer:71.625J
Explanation:Energy(Q)=mCpT
But recall that they said heat capacity (C), not specific heat capacity(cp)....Heat capacity=Mass *cp,but the heat capacity was clearly given so the mass is irrelevant in this case.
Energy (Q)=Heat capacity*Temperature change
Energy=12.5*5.73=71.625J
Some friends are trying to make wine in their basement. They've added yeast to a sweet grape juice mixture and have allowed the yeast to grow. After several days they find that sugar levels in the grape juice have dropped, but there's no alcohol in the mixture. The most likely explanation is that ______.
Answer:
The yeast respired aerobically
Explanation:
The expectation here was that the yeast was going to put the sugar in the grape juice through fermentation, which is an anaerobic process that results in alcohol as one of the products. However since no alcohol was found in this particular example, it is fair to assume that maybe there was oxygen in teh muxture and therefore the yeast respired aerobically, producing water and carbon dioxide as products.
27. How much energy is required to change 150.0 g of water from 10.0°C to 45.0°C? (Cwater =
4.18 J/g °C)
A. 125.43
B. 627.03
C. 219003
D. 28200 3
Answer:
The answer to your question is letter C. 21900.3. My result was 21945 J, letter C is close to this result.
Explanation:
Data
mass = 150 g
temperature 1 = T1 = 10°C
temperature 2 = T2 = 45°C
Cw = 4.18 J/g°C
Formula
Q = mCΔT = mC(T2 - T1)
Substitution
Q = (150)(4.18)(45 - 10)
Simplification
Q = (150)(4.18)(35)
Result
Q = 21945 J
One isotope of carbon (C) has exactly the same mass number and atomic mass since it was used as the definition of the atomic mass unit (amu). Which isotope is it and what is its atomic mass?
Answer:The isotope is Carbon-12 and its atomic mass is 12.
Explanation:
Mass number is the total number of protons and neutrons in a nucleus.
Atomic number is the number of protons in the nucleus of an atom.
An isotope of a chemical element is an atom that has a different mass number but the same atomic number as the element. The difference in mass number is from the number of neutrons (that is, a greater or lesser atomic mass) than the standard for that element.
Carbon-12 is an isotope of carbon it has 6 neutrons and 6 protons, giving it a mass number of 12 and atomic number of 6. Carbon-12 is a stable isotope of carbon, it has the same mass number and atomic number as carbon.
The isotope of carbon that has exactly the same mass and atomic number as used in the definition of the atomic mass unit is;
Isotope Carbon-12.
In chemistry, we know that;
Mass number is defined as the sum of the protons and the neutrons present in the nucleus of an atom.Meanwhile, Atomic number is defined the number of protons in the nucleus or number of electrons around the nucleus of an atom.
While an isotope is defined as an atom of an element that has the same atomic number but different mass number.
The isotope Carbon-12 is an isotope of carbon that has 6 protons and 6 neutrons.This means from the definition of mass number; Carbon-12 will have a mass number = 6 + 6 = 12Number of protons = number of electrons. Thus, number of electrons = 6 and therefore, atomic number = 6. From periodic table, the element Carbon has the same mass number and atomic number as its' isotope carbon-12.
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Determine the percent composition of CH2O.
Answer:
The given chemical compound has 2 atoms of hydrogen and one atom of oxygen for each atom of carbon. The mass of CH2O is 12 + 2*1 + 16 = 30. The molecular weight of the compound is 180.18 which is approximately 180. This gives the molecular formula of the chemical compound as C6H12O6.
Explanation:
To calculate the percent composition of CH2O, determine the molar mass of each element and the total molar mass of the compound. The percent composition is approximately 40.0% Carbon, 6.7% Hydrogen, and 53.3% Oxygen.
Explanation:To calculate the percent composition of CH2O, we will need to determine the molar mass of each element in the compound and the total molar mass of the compound. The molar masses from the periodic table are approximately 12.01 g/mol for Carbon (C), 1.01 g/mol for Hydrogen (H), and 16.00 g/mol for Oxygen (O).
First, let's calculate the total molar mass of CH2O: (1 × 12.01) + (2 × 1.01) + (1 × 16.00) = 12.01 + 2.02 + 16.00 = 30.03 g/mol.
Now, let's find the percent composition for each element:
Carbon: (12.01 g/mol ÷ 30.03 g/mol) × 100% = 40.0%Hydrogen: (2.02 g/mol ÷ 30.03 g/mol) × 100% = 6.7%Oxygen: (16.00 g/mol ÷ 30.03 g/mol) × 100% = 53.3%The percent composition of CH2O is therefore approximately 40.0% Carbon, 6.7% Hydrogen, and 53.3% Oxygen.
During a period of discharge of a lead-acid battery, 405 g of Pb from the anode is converted into PbSO4 (s).
What mass of PbO2 (s) is reduced at the cathode during this same period?
and
How many coulombs of electrical charge are transferred from Pb to PbO2?
Answer:
The answers to the question are as follows
First part
The mass of PbO2 (s) reduced at the cathode during the period is = 467.55_g
Second part
The electrical charge are transferred from Pb to PbO2 is 377186.86_C or 3.909 F
Explanation:
To solve this, we write the equation for the discharge of the lead acid battery as
H₂SO₄ → H⁺ + HSO₄⁻
Pb (s) + HSO⁻₄ → PbSO₄ + H⁺ + 2e⁻
at the cathode we have
PbO₂ + 3H⁺ + HSO⁻₄ + 2e⁻ → PbSO₄ + 2H₂O
Summing the two equation or the total equation for discharge is
Pb (s) + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O
From the above one mole of lead and one mole of PbO₂ are consumed simultaneously hence
Number of moles of lead contained in 405 g of Pb with molar mass = 207.2 g/mole = (405 g)/ (207.2 g/mole) = 1.95 mole of Pb
Hence number of moles of PbO₂ reduced at the cathode = 1.95 mole
mass of PbO₂ reduced at the cathode = (number of moles)×(molar mass)
= 1.95 mole × 239.2 g/mol = 467.55 g of Lead (IV) Oxide is reduced at the cathode
Part B
Each mole of Pb transfers 2e⁻ or 2 electrons, therefore 1.95 moles of Pb will transfer 2 × 1.95 = 3.909 moles of electrons transferred
Each electron carries a charge equal to -1.602 × 10⁻¹⁹ C or one mole of electrons carry a charge equal to 96,485.33 coulombs
hence 3.909 moles carries a charge = 3.909 × 96,485.33 coulombs =377186.86 Coulombs of electrical charge
or transferred electrical charge = 377186.86 C or 3.909 Faraday
Answer:
Mass of [tex]PbO_2[/tex] reduced = [tex]467g[/tex][tex]3.75*10^5C[/tex] of electrical charge is neededExplanation:
A) Moles Pb = [tex]\frac{405 g}{207.2 g/mol}[/tex]
[tex]= 1.95[/tex]
Moles Pb = moles [tex]PbO_2[/tex] reduced
Molar mass [tex]PbO_2 = 239.19 g/mol[/tex]
Grams [tex]PbO_2 = 1.95 mol * 239.19 g/mol[/tex]
[tex]= 466.42g[/tex]
B) 1 mol [tex]PbO_2[/tex] -------------------> 2 F electricity
1 .95 mol [tex]PbO_2[/tex] --------------> 2 * 1.95 F electricity = [tex]3.9F[/tex]
number of coulombs = [tex]3.9 * 96485C[/tex]
[tex]= 3.76*10^5C[/tex]
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The presence of many C-C and C-H bonds causes fats to be ... The presence of many C-C and C-H bonds causes fats to be ... (a) rich in energy. (b) insoluble in water. (c) low in energy. Both (a) and (b). Both (b) and (c).
Answer: Both (a) and (b)
Explanation:
Lipids are heterogeneous group of compounds of biochemical importance. Lipids may be defined as compounds which are relatively insoluble in water and are concentrated of energy source.
Fatty acids are aliphatic carboxylic acids and have the general formula, R-COOH, where COOH is the functional group and R group are hydrocarbon chain.
The structure of fat contains lot of C-C and C-H bonds and there are lot of calories, and therefore energy is packed into thier chemical structure.
Despite fat contains glycerol polar group, the long chains of hydrocarbon which are non polar makes fats insoluble in water.
Answer:
Both (a) and (b)
(a) rich in energy.
(b) insoluble in water.
Explanation:
Fats are stored as triesters (triglycerides), which when hydrolyzed form the three alcohol molecules (triglycerol) and three fatty acids. The acids that are liberated usually have long carbon chains that contain anywhere from 4 to 18 carbons. The C-C and C-H have high electron molecules present hence whey they are good sources of energy.
However, the bonding between carbon (C-C) and hydrogen (C-H) are not polar. This is because the electrons in covalent bonds are shared equally between the carbon and the hydrogen (due to their similar electronegative values) and there are no partial charges. Thus, long chains of C-C and C-H bonds form fats.
A railroad diesel engine weighs four times as much as a freight car. The diesel engine coasts at 5 km/h into a freight car that is initially at rest. Use the conservation of momentum to show that after they couple together, the engine + car coast at 4 km/h.
Please show work!!!! step by step
Explanation:
Conservation of momentum :
[tex]m_1u_1+m_2u_2=m_1v_1+m_1v_2[/tex]
Where :
[tex]m_1, m_2[/tex] = masses of object collided
[tex]u_1,u_2[/tex] = initial velocity before collision
[tex]v_1,v_2[/tex] = final velocity after collision
We have :
Mass of an engine = [tex]m_1=4M[/tex]
Mass of an car= [tex]m_2=M[/tex]
Initial velocity of railroad engine [tex]m_1=u_1=5 km/h[/tex]
Initial velocity of car [tex]m_2=u_2=0 km/h[/tex] (rest)
Final velocity of railroad engine [tex]m_1=v_1=v[/tex] (same direction )
Final velocity of car [tex]m_2=v_2=v[/tex] (same direction)
[tex]4M\times 5km/h+M(0 km/h)=4Mv+Mv[/tex]
[tex]4\times 5 km/h=5M[/tex]
v = 4 km/h
The speed of the engine and car after they coupled together is 4 km/h.
Using the principle of conservation of momentum, it's shown that the diesel engine and freight car will coast together at 4 km/h after coupling. This is calculated by equating the initial and final momentums of the system.
Explanation:The conservation of momentum can be used to solve this problem. The principle states that the total linear momentum of a closed system remains constant, regardless of any internal changes. Here, the system consists of the diesel engine and the freight car.
Let's denote the weight of the freight car as 'm'. Given that the diesel engine weighs four times as much as a freight car, the weight of the engine would be '4m'.
If the diesel engine is coasting at 5 km/hr, the initial momentum of the system is the momentum of the engine, because the freight car is at rest. Therefore, the initial momentum (Pi) is the weight of the engine times its velocity, which is 4m*5 km/hr = 20m km/hr.
After they couple together, there's no external force, so the total momentum should remain the same (the conservation of momentum principle). Let's denote the final velocity of the engine + car (now moving together) as 'v'. The final momentum (Pf) = (m + 4m) * v = 20m km/hr.
Therefore, we can establish the equation: Pi = Pf, meaning 20m km/hr = 5m * v. Solving for v, we find that v = 4 km/h. Therefore, the engine + car coast together at 4 km/h after coupling, demonstrating the conservation of momentum.
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Describe a sigma bond.a. orbital overlapping with the side of a p orbitalb. overlap of two d orbitalsend to end overlap of p orbitalc. sp orbital overlapping with an f orbitald. side by side overlap of d orbitals
A sigma bond b. overlap of two d orbitals end-to-end overlap of p orbital.
What is a sigma bond?A sigma bond is a strong bond that is made up of overlapping orbitals. In fact, these bonds are the strongest known bonds in chemical reactions. The overlapping orbitals are in the form of covalent bonding and this gives us the idea that sigma bonds are strong in nature.
In addition, sigma bonds are mostly common with diatomic elements and compounds. There are three orbitals where the sigma bond can be found and they are the p-p, s-p, and the s-s orbitals. These are known for forming symmetry groups.
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what events and experiences lead bruno to gradually give up some of his innocence and see things differently
Answer:
The events of Bruno seeing the young boy in the concentration camp, slowly seeing his father’s evil side come out, and having his sister explain to him what was really happening; all lead Bruno to gradually see what was really going on.
Explanation:
The gas in a 600. mL balloon has a pressure of 1.20 atm. If the temperature remains constant, what will be the pressure of the gas in the balloon when it is compressed to 400. mL?
Answer:
1.8 atm
Explanation:
As temperature and moles of a gas reamin both constant, we can say that:
Pressure₁ . Volume₁ = Pressure₂ . Volume₂
1.20 atm . 600 mL = Pressure₂ . 400 mL
(1.20 atm . 600 mL) / 400 mL = Pressure ₂ → 1.8 atm
What is the sound intensity level in decibels of ultrasound of intensity 7.53 ✕ 104 W/m2, used to pulverize tissue during surgery?
Answer : The intensity of sound level in decibels is, 169 dB
Explanation:
The expression used for the intensity of sound level is given by the equation:
[tex]P=10\times \log \frac{I}{I_o}[/tex]
where,
I = final intensity = [tex]7.53\times 10^4W/m^2[/tex]
[tex]I_o[/tex] = initial intensity = threshold intensity of human hearing = [tex]10^{-12}W/m^2[/tex]
Now put all the given values in the above formula, we get:
[tex]P=10\times \log \frac{7.53\times 10^4W/m^2}{10^{-12}W/m^2}[/tex]
[tex]P=168.767dB\approx 169dB[/tex]
Thus, the intensity of sound level in decibels is, 169 dB
Final answer:
The sound intensity level in decibels of the ultrasound used to pulverize tissue during surgery is 160 dB.
Explanation:
The sound intensity level in decibels (dB) of ultrasound with an intensity of 7.53 × 10^4 W/m² used to pulverize tissue during surgery can be calculated using the formula:
IdB = 10 log10 (I/I0)
where I is the given intensity (7.53 × 10^4 W/m²) and I0 is the reference intensity (10^-12 W/m²).
Using the formula:
IdB = 10 log10 (7.53 × 10^4 / 10^-12)
We can calculate:
IdB = 10 log10 (7.53 × 10^16)
IdB = 10 × 16 (using log10 (ab) = b × log10 (a))
IdB = 160 dB
Therefore, the sound intensity level in decibels of the given ultrasound is 160 dB.
Doc Inmaking is thinking about the density of liquid water (d = 0.99823 g/mL at 20 °C) compared to the density of ice (0.9168 g/mL at 0 °C). His favorite water bottle has a total volume of 300 mL. He fills it with exactly 288 mL of water at 20 °C. He tightens the lid and puts the bottle in the freezer. What mass of water did Doc placed in the bottle? (Three significant digits, unit of g.)
Answer:
The mass of water did Doc placed in the bottle is 288 grams.
Explanation:
Mass of water filled in water bottle = M
Volume of the water filled in water bottle = V = 288 mL
Density of the water at 20°C , d= 0.99823 g/ml
[tex]D=\frac{M}{V}[/tex]
[tex]M=D\times V=0.99823 g/ml\times 288 mL=287.49024 g\approx 288 g[/tex]
The mass of water did Doc placed in the bottle is 288 grams.
Final answer:
The mass of water at 20 °C with a density of 0.99823 g/mL that Doc placed in the bottle is 287 g when rounded to three significant digits.
Explanation:
The mass of water Doc placed in the bottle can be calculated using the density of liquid water at 20 °C, which is 0.99823 g/mL. Given he filled it with exactly 288 mL of water, the mass of the water is calculated by multiplying the volume by the density: The mass of water at 20 °C with a density of 0.99823 g/mL that Doc placed in the bottle is 287 g when rounded to three significant digits.
Mass = volume × density
Mass = 288 mL × 0.99823 g/mL
Mass = 287.63064 g
Since we are asked to provide the mass to three significant digits, the mass of water Doc placed in the bottle is 287 g.