Answer:
B:COTYLEDONS
Explanation:
Cotyledon is defined as;it is the part of embryo present whithin the seed of a plant and is often referred as "seed leaf"
The number of cotyledons is a defining characteristic of angiosperms.Cotyledons are present in the embryo of the angiosperms.On the basis of the number of cotyledons ;angiosperms are divided or distinguished into two classes which are termed as;Monocotyledonae (Monocots) and Dicotyledonae (Dicots).
MONOCOTS
The species which have one cotyledon in their seeds are monocots.corn,wheat,barley,rice.
DICOTS
The species having two cotyledons are called dicots.peas,beans,peanuts.
Yup. It's B, Cotyledons. I got it right on edge 2020
The octane rating of gasoline is a measure of the resistance to premature ignition, also known as engine knock. It is based on comparison to a mixture of 2,2,4-trimethylpentane, commonly known as isooctane (rating = 100), and heptane (rating = 0). An octane rating of 90 would compare to a mixture that is 90% isooctane and 10% heptane. Isooctane is a highly branched alkane with the condensed structure (CH3)3CCH2CH(CH3)2. Draw the complete struckture and skeletal struckture.
Answer: From the attached document
1. Complete structure
2. Skeletal structure
The complete structural formula for isooctane (2,2,4-trimethylpentane) is (CH3)3CCH2CH(CH3)2, while the skeletal formula is CH3-C-(CH3)2-CH2-CH(CH3)2. These represent its molecular structure in detail and a simplified version, respectively.
The complete structural formula and skeletal formula for 2,2,4-trimethylpentane, commonly known as isooctane, are as follows:
Complete Structural Formula:
In the complete structural formula, we represent each atom in the molecule explicitly. Isooctane has the following structure:
```
H
|
H - C - (CH3)2
| |
H - C - CH3
|
CH3
```
In this representation, each line represents a chemical bond, and we've labeled carbon (C) and hydrogen (H) atoms to show their positions in the molecule. Isooctane is highly branched, containing multiple methyl (CH3) groups attached to a central carbon atom.
Skeletal Formula:
The skeletal formula is a simplified way of representing the molecular structure. It focuses on the carbon framework, and hydrogen atoms bonded to carbon are usually omitted. Isooctane's skeletal formula is as follows:
```
CH3
|
CH3-C-(CH3)2
|
CH3
```
In this skeletal formula, each vertex represents a carbon atom, and the lines between them represent carbon-carbon bonds. The methyl (CH3) groups are indicated as branches off the carbon atoms.
Isooctane is an important component in the determination of gasoline's octane rating because it has a high resistance to engine knocking, making it a reference standard with an octane rating of 100. Gasoline blends with a higher percentage of isooctane are less prone to premature ignition and have higher octane ratings, indicating their suitability for high-performance engines.
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If the rate of formation of ammonia is 0.345 M/s, what is the rate of disappearance of N2?
Answer: The rate of disappearance of [tex]N_2[/tex] is 0.172 M/s
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
[tex]N_2(g)+3H_2(g)\rightarrow 2NH_3(g)[/tex]
The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
Rate in terms of disappearance of [tex]N_2[/tex] = [tex]-\frac{1d[N_2]}{dt}[/tex]
Rate in terms of disappearance of [tex]H_2[/tex] = [tex]-\frac{1d[H_2]}{3dt}[/tex]
Rate in terms of appearance of [tex]NH_3[/tex] = [tex]\frac{1d[NH_3]}{2dt}[/tex]
The rate of formation of ammonia is = 0.345 M/s
[tex]\frac{1d[NH_3]}{dt}=0.345 M/s[/tex]
The rate of disappearance of [tex]N_2[/tex] is one by three times the rate of appearance of [tex]NH_3[/tex]
[tex]-\frac{1d[N_2]}{dt}=\frac{d[NH_3]}{2dt}[/tex]
[tex]-\frac{1d[N_2]}{dt}=\frac{0.345M/s}{2}=0.172M/s[/tex]
Thus the rate of disappearance of [tex]N_2[/tex] is 0.172 M/s
If the rate of formation of ammonia is 0.345 M/s, the rate of disappearance of nitrogen is 0.173 M/s.
Let's consider the balanced equation for the formation of ammonia.
N₂(g) + 3 H₂(g) ⇒ 2 NH₃(g)
The molar ratio of N₂ to NH₃ is 1:2. If the rate of formation of ammonia is 0.345 M/s, the rate of disappearance of nitrogen is:
[tex]\frac{0.345molNH_3}{L.s} \times \frac{1molN_2}{2molNH_3} = \frac{0.173molN_2}{L.s} = 0.173 M/s[/tex]
If the rate of formation of ammonia is 0.345 M/s, the rate of disappearance of nitrogen is 0.173 M/s.
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In a 4.00 L pressure cooker, water is brought to a boil. If the final temperature is 115 °C at 3.10 atm, how many moles of steam are in the cooker?
Answer: 0.39mol of the steam are in the cooker.
Explanation:Please see attachment for explanation
Given:
Volume, V = 4 LPressure, P = 3.10 atmTemperature, T = 115°C= 115+273
= 388 K
We know the relation,
→ [tex]PV = nRT[/tex]
or,
→ [tex]n = \frac{PV}{RT}[/tex]
By substituting the values, we get
[tex]= \frac{4\times 3.1}{0.082\times 388}[/tex]
[tex]= 0.39 \ mol[/tex]
[tex]= 0.39 \ mol[/tex]
Thus the above answer is correct.
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What are two important quantum-mechanical concepts associated with the Bohr model of the atom?
Explanation:
The important quantum-mechanical concepts associated with the Bohr model of atom are :
1. Electrons are nothing but particles that revolve around the nucleus in discrete orbitals.
2. Energy is associated with each orbital is quantised. Meaning electron in each shell will have energy in multiple of a fixed quanta.
Final answer:
The Bohr model introduced two key concepts of quantum mechanics: the quantization of energy levels and the quantization of angular momentum, both crucial for understanding atomic structure and the emission spectra of atoms.
Explanation:
The Bohr model of the atom represents an early understanding of quantum mechanics, specifically applicable to the hydrogen atom. It introduced two fundamental quantum-mechanical concepts that underpin the modern understanding of atomic structure and behavior, despite its ultimate replacement by more comprehensive theories.
Quantized Energy Levels
The first important concept is the quantization of energy levels of electrons within an atom. According to the Bohr model, electrons can only occupy certain allowed energy levels, and transitions between these levels result in the absorption or emission of photons with specific energies. This concept matched the observed discrete spectra of hydrogen, although the model assumed circular orbits for electron paths, which was later found to be incorrect.
Quantized Angular Momentum
The second concept is the quantization of angular momentum. Bohr proposed that the angular momentum of an electron in orbit is quantized and can only take on certain values, each associated with a different orbit. Although Bohr's assumptions about specific, circular orbits were eventually supplanted by the probabilistic nature of electron positioning in quantum mechanics, the idea that angular momentum is quantized remains a fundamental aspect of quantum theory.
Ammonium carbamate (NH2COONH4) is a salt of carbamic acid that is found in the blood and urine of mammals. At 202°C, Kc = 0.008 for the following equilibrium: If 6.82 g of NH2COONH4 is put into a 0.5-L evacuated container, what is the total pressure (in atm) at equilibrium? Enter a number to 2 decimal places.
Answer:
P(total) = 46.08 atm
Explanation:
Given data:
Mass of NH₂COONH₄ = 6.8 g
Kc = 0.008
Temperature = 202°C = 202+273 = 475 K
Total pressure at equilibrium = ?
Solution:
Equilibrium equation:
NH₂COONH₄ ⇄ 2NH₃ + CO₂
Formula:
Kp = Kc(RT)³
Kp = 0.008 (0.0821 atm.L/mol.K )³(475K)³
Kp = 471.56
NH₃ CO₂
Initial concentration 0 0
Change in concentration 2x x
equilibrium concentration 2x x
AS
471.56 = 2x (x)
471.56 = 2x²
x² = 471.56 /2
x² = 235.78
x = 15.36
Pressure of ammonia = 2x = 2(15.36) = 30.72 atm
Pressure of carbon dioxide = x = 15.36 atm
Total pressure:
P(total) = P(NH₃) + P(CO₂)
P(total) = 30.72 atm + 15.36 atm
P(total) = 46.08 atm
2-19 Compare and contrast metallic and covalent primary bonds in terms of
(a) the nature of the bond,
(b) the valence of the atoms involved, and
(c) the ductility of the materials bonded in these ways.
While metallic bonds have strong attractions and ionic bonds involve the transfer and acceptance of electrons from the valence shell, covalent bonds involve sharing electrons.
What are chemical bonds?Chemical bonds are defined as a link made by two surfaces or items that have been brought together, often by heat, pressure, or an adhesive agent. Chemical bonds can have many different types, but covalent and ionic bonds are the most well-known. When one atom has less energy, the other has enough thanks to these bonds. Atoms are held together by the force of attraction, which enables the electrons to unite in a bond.
In contrast to covalent bonding, which involves atoms sharing their additional electrons locally, metallic bonding involves all atoms giving off their extra electrons and forming a sea of them. The ability of an atom to stick together and fill the orbits of its outermost electrons in order to arrange itself in a form that is the most stable.
Thus, while metallic bonds have strong attractions and ionic bonds involve the transfer and acceptance of electrons from the valence shell, covalent bonds involve sharing electrons.
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Calculate the amount of heat needed to boil 183.g of ethanol ( CH3CH2OH ), beginning from a temperature of 33.9°C . Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol.
To calculate the amount of heat needed to boil ethanol, use the formula Q = m * c * ΔT. Convert the temperature to Kelvin, calculate the change in temperature, and then apply the formula to find the amount of heat. The amount of heat needed to boil 183.g of ethanol is 20 kJ.
Explanation:To calculate the amount of heat needed to boil ethanol, we can use the equation:
Q = m * c * ΔT
Where Q is the amount of heat, m is the mass of ethanol, c is the specific heat capacity of ethanol, and ΔT is the change in temperature.
First, we need to calculate the change in temperature:
Convert the temperature from Celsius to Kelvin: 33.9°C + 273.15 = 306.05 KNext, subtract the boiling point of ethanol (78.4°C = 351.55 K) from the initial temperature: ΔT = 351.55 K - 306.05 K = 45.50 KNow we can calculate the amount of heat:
Find the specific heat capacity of ethanol: 2.44 J/g°CApply the formula: Q = 183.g * 2.44 J/g°C * 45.50 KFinally, round the answer to 3 significant digits: Q = 20000 J = 20 kJTherefore, the amount of heat needed to boil 183.g of ethanol is 20 kJ.
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The frequency of stretching vibrations is correlated to the strength and stiffness of the bond between two atoms. This can be thought of as a ball-and-spring model. Using this knowledge, rank the following bonds in each part of the question by increasing frequency.
(a) Alkyne
(b) Alkane
(c) alkene
Answer:
a > c > b
Explanation:
As higher is the strength and stiffness of the bond between two atoms, more stable it is, and more difficult is to these bonds vibrate. So, the stretching vibration decreases when the strength and stiffness increases.
As more bonds are done between the atoms, more strength, and stiffness they have. So, the order of increase is:
simple bond > double bond > triple bond
And the increased frequency of vibration is:
triple bond > double bond > simple bond
An alkane is a hydrocarbon that has only simple bonds between carbons, an alkene is a hydrocarbon with one double bond between carbon, and an alkyne is a hydrocarbon with one triple bond. So, the increase in vibration of them is:
alkyne (a) > alkene (c) > alkane (b)
Answer: (A) < (C) < (B)
Ranking in order of increasing frequency.
Explanation:
Consider a general reaction
a) The ΔG° of the reaction is -4.140 kJ mol. Calculate the equilibrium constant for the reaction. (Assume a temperature of 25°C.)
b) What is ΔG at body temperature (37.0°C) if the concentration of A is 1.6 M and the concentration of B is 0.45 M?
Answer:
a) K = 5.3175
b) ΔG = 3.2694
Explanation:
a) ΔG° = - RT Ln K
∴ T = 25°C ≅ 298 K
∴ R = 8.314 E-3 KJ/K.mol
∴ ΔG° = - 4.140 KJ/mol
⇒ Ln K = - ( ΔG° ) / RT
⇒ Ln K = - ( -4.140 KJ/mol ) / (( 8.314 E-3 KJ/K.mol )( 298 K ))
⇒ Ln K = 1.671
⇒ K = 5.3175
b) A → B
∴ T = 37°C = 310 K
∴ [A] = 1.6 M
∴ [B] = 0.45 M
∴ K = [B] / [A]
⇒ K = (0.45 M)/(1.6 M)
⇒ K = 0.28125
⇒ Ln K = - 1.2685
∴ ΔG = - RT Ln K
⇒ ΔG = - ( 8.314 E-3 KJ/K.mol )( 310 K )( - 1.2685 )
⇒ ΔG = 3.2694
At the beginning of the industrial revolution in 1850, the CO2concentration was 280 ppm. Today, it is 410 ppm.
1. How much extra radiative forcing is the Earth’s surface receiving today, relative to 1850?
2. What is the equivalent temperature change? ΔF (Wm^(-2)) = α ln(C/C0), ΔT(K) = λ*ΔF, α = 5.35, λ = 0.8 per (Wm^(-2)).
Answer:
1. 2.04 W/m²
2. 1.63°C
Explanation:
The radiative force that the Earth receives comes from the Sun. When the Sun rays come to the surface, some of them are absorbed and then it is reflected in the space. The greenhouse gases (like CO2) blocks some of these rays, and then the surface stays warm. The excessive amount of these gases makes the surface warmer, which unbalance the climate on Earth.
1. The variation of the radiative forcing can be calculated based on the concentration of the CO2 by the equation:
ΔF = 5.35*ln(C/C0)
Where C is the final concentration, and C0 is the initial concentration.
ΔF = 5.35*ln(410/280)
ΔF = 2.04 W/m²
2. The temperature change in the Earth's surface caused by the variation of the radiative forcing can be calculated by:
ΔT = 0.8*ΔF
ΔT = 0.8*2.04
ΔT = 1.63 K = 1.63°C
The increase in CO2 concentration from 280 ppm to 410 ppm today results in an extra radiative forcing of 2.2 W/m² on Earth's surface. The equivalent temperature change is approximately 1.7°C.
Explanation:The increase in CO2 concentration from 280 ppm to 410 ppm today has resulted in an extra radiative forcing on Earth's surface. The radiative forcing is calculated using the formula ΔF (Wm^(-2)) = α ln(C/C0), where α = 5.35. By plugging in the values, the radiative forcing is approximately 2.2 W/m².
The equivalent temperature change can be calculated using the formula ΔT(K) = λ*ΔF, where λ = 0.8 per (Wm^(-2)). By multiplying the radiative forcing by the climate sensitivity, we get a temperature increase of about 1.7°C.
It is important to note that the temperature increase may catch up to a new equilibrium once oceans warm and ice melts, resulting in a higher temperature increase in the future.
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what is the kinetic energy of a particle of mass m moving through space with velocity v?
Answer:
Kinetic Energy of the particle is half the mass of the particle multiplied by the square of the velocity of the particle
Explanation:
Kinetic Energy is given by 1/2mv^2
Where, m is the mass of the object and v is the velocity of the object
Answer: kinetic energy is equal to 1/2mv^2
A mineral sample is obtained from a region of the country that has high arsenic contamination. An elemental analysis yields the following elemental composition:
Element Atomic Weight (g/mol) Percent Composition
Ca 40.078 22.3%
As 74.9216 41.6%
O 15.9994 35.6%
H 1.00794 60%
What is the empirical formula of this mineral?
Answer:
CaAsHO₄
Explanation:
The data has a mistake in one of the values there. I believe the mistake is on the hydrogen. So, I'm going to assume the value of Hydrogen is 0.6%, so the total percent composition would be 100.1% (Something better). All you have to do is replace the correct value of H (or the value with the mistaken option) and do the same procedure.
Now, to calculate the empirical formula, we can do this in three steps.
Step 1. Calculate the amount in moles of each element.
In these case, we just divide the percent composition with the molar mass of each one of them:
Ca: 22.3 / 40.078 = 0.5564
As: 41.6 / 74.9216 = 0.5552
O: 35.6 / 15.9994 = 2.2251
H: 0.6 / 1.00794 = 0.5953
Now that we have done this, let's calculate the ratio of mole of each of them. This is doing dividing the smallest number of mole between each of the moles there. In this case, the moles of As are the smallest so:
Ca: 0.5564/0.5552 = 1.0022
As: 0.5552/0.5552 = 1
O: 2.2251/0.5552 = 4.0077
H: 0.5953/0.5552 = 1.0722
Now, we round those numbers, and that will give us the number of atoms of each element in the empirical formula
Step 3. Write the empirical formula with the rounded numbers obtained
In this case we will have:
Ca: 1
As: 1
O: 4
H: 1
The empirical formula would have to be:
CaAsHO₄
write out the acid dissociation reaction for hydrochloric acid. b) calculate the pH of a solution of 5.0 x 10^-4 M HCl. c) write out the acid dissociation reaction for sodium hydroxide. d) calculate the pH of a solution of 7.0 x 10^-5 M NaOH.
Answer:
The answer to your question is below
Explanation:
a) Acid dissociation reaction for HCl
HCl(g) + H₂O(l) ⇒ H⁺¹ (aq) + Cl⁻¹ (aq)
b) pH = -log [H⁺¹]
Concentration = 5 x 10⁻⁴
pH = -log [5 x 10⁻⁴]
pH = 3.3
c) Acid dissociation reaction
NaOH(s) + H₂O ⇒ Na⁺¹(aq) + OH⁻¹(aq)
d) pH
pOH = -log [7 x 10⁻⁵]
pOH = 4.2
pH = 14 - 4.2
pH = 9.8
The acid dissociation and pH for HCl and NaOH were detailed. For HCl, the reaction and solution pH were HCl -> H+ + Cl- and 3.3 respectively. For NaOH, the dissociation and solution pH were NaOH -> Na+ + OH- and 9.8 respectively.
Explanation:a) The acid dissociation reaction for hydrochloric acid (HCl) is as follows:
HCl → H+ + Cl-
b) The pH of a solution of hydrochloric acid can be calculated using the formula:
pH = -log[H+].
As HCl is a strong acid, it will completely ionize in water, meaning the concentration of H+ ions equals the concentration HCl. Therefore, the pH for a 5.0 x 10^-4 M HCl solution is -log(5.0 x 10^-4) = 3.3.
c) Sodium hydroxide is a base, not an acid, so its ionization does not involve releasing protons into the solution, but rather attracting them. It has the following dissociation reaction:
NaOH → Na+ + OH-.
d) As for the pH of a solution of 7.0 x 10^-5 M NaOH, we first need to find the pOH using the formula pOH = -log[OH-], since NaOH being a strong base completely ionizes. This equals -log(7.0 x 10^-5) = 4.2. Given that pH and pOH are related by the expression pH + pOH = 14 at 25°C, we can solve for the pH of the NaOH solution = 14 - pOH = 14 - 4.2 = 9.8.
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Consider a mixture of four proteins with various molecular weights. A histone molecule weighs 15 kDa, 15 kDa, a p53 molecule weighs 53 kDa, 53 kDa, an actin molecule weighs 42 kDa, 42 kDa, and an IgG molecule weighs 150 kDa. 150 kDa. Arrange the molecules in order of their elution from a gel filtration column.
Answer:
IgGp53ActinHistoneExplanation:
In gel filtration chromatography a porous matrix is used as the stationary phase. The smaller molecules are able to diffuse into this matrix while the larger ones are not. For this reason the larger molecules will elute first, followed by the smaller ones in decreasing order of size.
In other words the largest molecule will elute first and the smallest molecule will elute last:
IgGp53ActinHistoneCalculate the radius of a platinum atom in cm, given that Pt has an FCC crystal structure, a density of 21.4 g/cm3, and an atomic weight of 195.08 g/mol.
Answer:
Radius of platinum atom is [tex]1,38\times10^{-8} \ cm.[/tex]
Explanation:
We know, Density in solid state is also given by , [tex]\rho=\dfrac{Z\times M}{N_A\times a^3}[/tex]
Here, Z= total no of atom or molecule per unit cell. ( Z = 4 for FCC structure )
M = atomic weight = 195.08 gm/mol.
N_A= Avogadro's number = [tex]6.023\times 10^{23} .[/tex]
a= size of side of cube.
Also for, FCC crystal [tex]a=\sqrt2\times 2r[/tex] ....1
Since,[tex]\rho=\dfrac{Z\times M}{N_A\times a^3}\\[/tex]
Therefore, [tex]a^3=\dfrac{Z\times M}{N_A\times \rho}[/tex]
Putting all values , [tex]a=3.9\times 10^{-8} \ cm.[/tex]
Putting value of a in equation 1 we get,
[tex]r=\dfrac{a}{\sqrt2\times2}=1.38\times 10^{-8} \ cm.[/tex]
Hence, this is the required solution.
The radius of a platinum atom can be found using the density, atomic weight, and the nature of the crystal structure. The calculation involves finding the volume of an atom from the atomic weight, density and Avogadro's number and converting this volume to a radius by considering the atom as a sphere.
Explanation:The radius of a platinum atom can be calculated using the density, atomic weight, and the information about the crystal structure (in this case, face-centered cubic or FCC). Platinum (Pt) has a density (d) of 21.4 g/cm3 and an atomic weight (AW) of 195.08 g/mol.
First, let's calculate the number of atoms (n) in the unit cell of an FCC structure. An FCC unit cell has 4 atoms. Avogadro's number (Na) is 6.022 x 1023 mol-1.
The volume (V) of an atom can be calculated as follows: V = AW / (d * n * Na).Now we will calculate the radius (r) from the volume, considering the atom as a sphere: r = ((3*V) / (4*π))1/3.Keep in mind that the values have to be converted to cm in the end.
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Construct a simulated 1H NMR spectrum for ethyl acetate by dragging and dropping the appropriate splitting patterns into the boxes on the chemical shift baseline, and by dragging integration values into the small box above each signal. Items may be used more than once. Peak heights do not represent integration.
Answer:
Triplet at 1.15 ppm
Singlet at 2.02 ppm
Quartet at 4.10 ppm
Explanation:
In ethyl acetate there are three different sets of protons having different surrounding as compared to each other. The predicted ¹H-NMR of ethyl acetate is attached below and following are the signals shown by this compound,
(i) A singlet around 2.02 ppm is a characteristic peak shown by protons at carbon a. This is because there is no proton available on the adjacent carbon so there is no coupling.
(ii) A triplet around 1.15 ppm is a peak shown by protons at carbon c. This is because there are two proton present on the adjacent carbon so, according to (n+1) rule it will give a triplet peak.
(iii) A quartet around 4.10 ppm is a peak shown by protons at carbon b. This is because there are three proton present on the adjacent carbon so, according to (n+1) rule it will give a quartet peak. Also, as it is directly attached to oxygen atom hence, the chemical shift is downfielded or deshielded.
Typically, neutron stars are about 20 km in diameter and have around the same mass as our sun. What is a typical neutron star density in g/cm3?
Answer:
4.7 × 10¹⁴ g/cm³
Explanation:
The mass of the sun (and of the neutron star) is 1.98 × 10³⁰ kg = 1.98 × 10³³ g.
The diameter of the neutron star is 20 km and its radius is 1/2 × 20 km = 10 km = 10⁴ m = 10⁶ cm
If we consider the neutron star to be approximately spherical, we can calculate its volume using the following expression.
V = 4/3 × π × r³
V = 4/3 × π × (10⁶ cm)³
V = 4.19 × 10¹⁸ cm³
The density (ρ) is equal to the quotient between the mass and the volume.
ρ = 1.98 × 10³³ g / 4.19 × 10¹⁸ cm³ = 4.73 × 10¹⁴ g/cm³ ≈ 4.7 × 10¹⁴ g/cm³
In the following list, only __________ is not an example of a chemical reaction.
A) burning a plastic water bottle
B) the production of hydrogen gas from water
C) the tarnishing of a copper penny
D) chopping a log into sawdust
E) charging a cellular phone
Answer: Thus chopping a log into sawdust is not an example of a chemical reaction.
Explanation:
Physical reaction is a reaction in which there is no rearrangement of atoms and thus no new substance is formed. There is only change in physical state of the substance.
Example: chopping a log into sawdust as the shape and size of wood changes.
Chemical reaction is a reaction in which there is rearrangement of atoms and thus new substance is formed. There may or may not be a change in physical state.
Example: burning a plastic water bottle : as the plastic would combine with oxygen to give oxidised products.
the production of hydrogen gas from water: the chemical composition changes
the tarnishing of a copper penny: oxidation of copper takes place to give copper oxide.
charging a cellular phone: the electrical energy is used for a chemical reaction
Chopping a log into sawdust is an example of a physical change because it changes the size not the composition of the substance.
What are chemical reactions?Chemical reactions involve such kind change that leads to the formation of a new product (compound).
For example- the production of hydrogen gas from water, the burning of plastic produce gases.
Physical reactions:These reactions refer to the rearrangements of the atoms in a substance but do not to the formation of a new product such as temperature, phase, size, etc.
For example- chopping a log into sawdust
Therefore, chopping a log into sawdust is an example of a physical change because it changes the size not the composition of the substance.
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CaCl_2 is often used to melt ice on sidewalks.
a) Could CaCl_2 melt ice at -33°C? Yes or no? Assume that the solubility of CaCl_2 at this temperature is 70.1 g CaCI_2/100.0 g of H2O and that the van't Hoff factor for a saturated solution of CaCl_2 is i = 2.5.
b) What is the minimum amount of CaCl_2 that would be required to melt ice on sidewalks at the above temperature? Use 100.0 g of water as the amount of solvent.
Answer:
a) No, it couldn't
b) 78.81 g
Explanation:
When a nonvolatile solute is added to a pure solvent, the melting point of the solvent must decrease, which is called cryoscopy. The temperature variation (melting point of pure solvent - the melting point of solution) can be calculated by:
ΔT = Kc.W.i
Where Kc is the cryoscopy constant (for water it's 1.86 °C/mol.kg), W is the molality, and i is the van't Hoff factor of the solute.
W =m1/m2*M1
Where m1 is the mass of the solute (in g), m2 is the mass of the solvent (in kg), and M1 is the molar mass of the solute (CaCl2 = 111.0 g/mol). The van't Hoff factor indicates how much of the solute ionizes in the solution.
a) The melting point of water is 0°C, so let's calculate the new melting point with the given information:
m1 = 70.1 g
m2 = 100 g = 0.1 kg
W = 70.1/(0.1*111) = 6.31 mol/kg
ΔT = 1.86*6.31*2.5
ΔT = 29.34°C
0 - T = 29.34
T = -29.34°C
Thus, at -33°C, the ice will not melt yet.
b) Let's then found out the value of m1 in this case:
ΔT = Kc.W.i
0 - (-33) = 1.86*W*2.5
4.65W = 33
W = 7.1 mol/kg
W = m1/M*m2
7.1 = m1/(111*0.1)
m1 = 78.81 g
Calcium chloride cannot melt the ice at -33°C. Cryoscopy is the method of determining a decrease in melting point due to dissolved substances.
What is cryoscopy?The decrease in the melting point of a pure solvent when a non-volatile solute is added to the solution.
[tex]\Delta T = K_c. W.i[/tex]
Where
[tex]K_c[/tex] - cryoscopy constant= 1.86 °C/mol.kg for water
[tex]W[/tex] - molality = 6.31 mol/kg
[tex]i[/tex] - Van't Hoff factor of the solute = 2.5
Put the values in the formula,
[tex]\Delta T = 1.86\times 6.31\times 2.5\\\\\Delta T = \rm \ 29.34^oC[/tex]
Therefore, Calcium chloride cannot melt the ice at -33°C.
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Phosphorus crystallizes in several different forms, one of which is a simple cube with an edge length of 238 pm. What is the density of phosphorus in the simple cubic crystal form?
Answer: The density of phosphorus is [tex]3.81g/cm^3[/tex]
Explanation:
To calculate the density of phosphorus, we use the equation:
[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex]
where,
[tex]\rho[/tex] = density
Z = number of atom in unit cell = 1 (CCP)
M = atomic mass of phosphorus = 31 g/mol
[tex]N_{A}[/tex] = Avogadro's number = [tex]6.022\times 10^{23}[/tex]
a = edge length of unit cell = [tex]238pm=238\times 10^{-10}cm[/tex] (Conversion factor: [tex]1cm=10^{10}pm[/tex] )
Putting values in above equation, we get:
[tex]\rho=\frac{1\times 31}{6.022\times 10^{23}\times (238\times 10^{-10})^3}\\\\\rho=3.81g/cm^3[/tex]
Hence, the density of phosphorus is [tex]3.81g/cm^3[/tex]
A 27.40 g sample of liquid mercury is initially at 158.30°C. If the sample is heated at constant pressure ( = 1 atm), kJ of energy are needed to raise the temperature of the sample to 376.20°C. How much energy in kJ is required?
Answer:
energy required is 0.247kJ
Explanation:
The formula to use is Energy = nRdT;
Where n is number of mole
R is the molar gas constant
dT is the change in temperature
n = reacting mass of mercury / molar mass of mercury = 27.4/200.59 = 0.137
dT = final temperature - initial temperature = 376.20 - 158.30 = 217.90K
R = 8.314Jper mol per Kelvin
Energy = 0.137 x 8.314 x 217.90 = 247.12J
Energy in kJ= 247.12/1000= 0.247kJ
Calculate the density, in g / mL , of a liquid that has a mass of 0.125 g and a volume of 0.000285 L
Answer:
ρ = 0.439 g/mL
Explanation:
Given data
Mass of the liquid: 0.125 gramsVolume of the liquid: 0.285 milliliters0.000285 L × (1000 mL/ 1 L) = 0.285 mL
The density (ρ) of a liquid is equal to its mass (m) divided by its volume (V).
ρ = m/V
ρ = 0.125 g/0.285 mL
ρ = 0.439 g/mL
The density of the liquid is 0.439 g/mL.
The density of a liquid, defined as mass per unit volume, with given mass of 0.125 grams and volume of 0.000285 liters, can be calculated using the formula: Density = mass/volume. After converting liters to milliliters, the density is approximately 0.439 g/mL.
Explanation:The question is asking for the calculation of the density of a liquid given its mass and volume. Density is a property of matter defined as mass per unit volume. It can be calculated using the formula: Density = mass/volume. In this case, the mass of the liquid is 0.125 grams and the volume is 0.000285 liters. But please note that the density should be presented in g/mL. Therefore, we need to convert the volume from liters to milliliters. 1 liter equals 1000 milliliters, so 0.000285 liters equals 0.285 milliliters. Using the formula, Density = 0.125 g / 0.285 mL which gives us a density of approximately 0.439 g/mL.
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A standard solution of 86% (w/w) H3PO4 (specific density = 1.71 g/mL) is to be diluted to get a1200 mL H3PO4 solution at 4.00 M. How many mL of the standard H3PO4 solution are required?
Answer: 319 ml
Explanation:
Given : 86 g of [tex]H_3PO_4[/tex] is dissolved in 100 g of solution.
Density of solution = 1.71 g/ml
Volume of solution=[tex]\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.71g/ml}=58.5ml[/tex]
moles of [tex]H_3PO_4=\frac{\text {given mass}}{\text {Molar mass}}=\frac{86g}{98g/mol}=0.88mol[/tex]
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution in L}}[/tex] .....(1)
Molarity of standard [tex]H_3PO_4[/tex] solution =[tex] \frac{0.88\times 1000}{58.5ml}=15.04M[/tex]
To calculate the volume of acid, we use the equation given by neutralization reaction:
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of standard acid which is [tex]H_3PO_4[/tex]
[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of diluted acid which is [tex]H_3PO_4[/tex]
We are given:
[tex]M_1=15.04M\\V_1=?mL\\\\M_2=4.00M\\V_2=1200mL[/tex]
Putting values in above equation, we get:
[tex]15.04\times V_1=4.00\times 1200\\\\V_1=319mL[/tex]
Thus 319 ml of the standard [tex]H_3PO_4[/tex] solution are required.
To prepare a 1200 mL of 4.00 M H3PO4 solution from an 86% w/w H3PO4 solution with a specific density of 1.71 g/mL, 320 mL of the standard solution is required.
To determine how many milliliters of a standard 86% H3PO4 solution (with a specific density of 1.71 g/mL) are required to prepare a 1200 mL 4.00 M H3PO4 solution, we can use the concept of molarity and the equation M1V1 = M2V2, where M1 and V1 are the molarity and volume of the concentrated solution, and M2 and V2 are the molarity and volume of the diluted solution, respectively.
First, we need to calculate the molarity of the stock solution (86% H3PO4). We know that 86% (w/w) means there are 86 grams of H3PO4 in 100 grams of solution. As the density of the solution is 1.71 g/mL, we can find the volume that 100 grams of solution will occupy:
Volume = mass / density = 100 g / 1.71 g/mL = 58.48 mL
Now, we calculate the mols of H3PO4 in 86 grams:
Moles H3PO4 = mass (g) / molar mass (g/mol) = 86 g / 97.99 g/mol ≈ 0.877 mol
The molarity (M1) of the stock solution is:
Molarity = moles of solute / volume of solution (L) ≈ 0.877 mol / 0.05848 L ≈ 15.00 M
Next, using the dilution equation M1V1 = M2V2, where M2 = 4.00 M and V2 = 1200 mL (1.2 L), we can solve for V1:
V1 = (M2V2) / M1 = (4.00 M * 1.2 L) / 15.00 M = 0.32 L
Since we need the volume in milliliters:
V1 = 0.32 L * 1000 mL/L = 320 mL
Therefore, 320 mL of the standard 86% H3PO4 solution is required to prepare a 1200 mL of 4.00 M H3PO4 solution.
Which of the following statements is true? I. Carbon forms covalent rather than ionic bonds. II. Carbon has an electronegativity value that is midway between that of the most metallic element and the most nonmetallic element. III. Carbon forms four bonds in virtually all its compounds. IV. A Si-Si bond is much stronger than a C-C bond. V. Relatively little heat is released when a C chain reacts and one bond replaces the other. Group of answer choices I, II, II, V I only II only III only IV only V only I and II I and III I and IV I and V II and III II and IV II and V III and IV III and V
Explanation:
Atomic number of carbon is 6 and its electronic distribution is 2, 4. As there are 4 valence electrons present in a carbon atom. Hence, it belongs to group 14 of the periodic table.
So, being a non-metal carbon atom tends to form covalent bonds. Also, the electronegativity value of carbon atom is mid-way between those of metals and non-metals.
In order to complete its octet, carbon atom tends to gain or share its 4 valence electrons. Therefore, it forms four bonds in all its compounds.
The Si-Si bond will be less stronger than C-C bond due to the larger size of silicon atom. As a result, there will be less overlapping between the silicon atoms due to which the bond formed will be weak in nature.
Whereas C-C bond is more stronger as a carbon atom is smaller in size as compared to silicon atom. As a result, more will be the overlapping between the carbon atoms. Hence, more stronger will be C-C bond than Si-Si bond.
As there will be covalent bonds present in a carbon chain and these bonds are weak as compared to ionic bonds. Hence, relatively little heat is released when a C chain reacts and one bond replaces the other.
Thus, we can conclude that out of the given options statements of I, II, III and V are true.
Study this chemical reaction: FeSO4 (aq) + Zn (s) --> Fe (s) + ZnSO4 (aq) Then, write balanced half-reactions describing the oxidation and reduction that happen in this reaction.
oxidation:
reduction:
Answer: The oxidation and reduction half reactions are written below.
Explanation:
Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.
[tex]X\rightarrow X^{n+}+ne^-[/tex]
Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.
[tex]X^{n+}+ne^-\rightarrow X[/tex]
For the given chemical reaction:
[tex]FeSO_4(aq.)+Zn(s)\rightarrow Fe(s)+ZnSO_4(aq.)[/tex]
The half reactions for the given reaction follows:
Oxidation half reaction: [tex]Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-[/tex]
Reduction half reaction: [tex]Fe^{2+}(aq.)+2e^-\rightarrow Fe(s)[/tex]
Hence, the oxidation and reduction half reactions are written above.
The balanced half-reaction for oxidation in the reaction is Zn (s) -> Zn2+ (aq) + 2e-, signifying Zinc is being oxidised. The reduction half-reaction is Fe2+ (aq) + 2e- -> Fe (s) demonstrating Iron being reduced.
Explanation:In this chemical reaction:FeSO4 (aq) + Zn (s) --> Fe (s) + ZnSO4 (aq) two processes occur simultaneously; oxidation and reduction. Oxidation involves loss of electrons, and in this case, it is the Zinc metal that gets oxidized. It unites with Sulphate forming ZnSO4 and in the process gains a positive charge by losing 2 electrons.
The balanced half-reaction for oxidation is: Zn (s) --> Zn2+ (aq) + 2e-
Reduction on the other hand involves gain of electrons. Iron in this reaction gains electrons to form metallic iron (Fe). This can be represented by the following balanced half-reaction:
Fe2+ (aq) + 2e- --> Fe (s).
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A chemist makes of magnesium fluoride working solution by adding distilled water to of a stock solution of magnesium fluoride in water. Calculate the concentration of the chemist's working solution. Round your answer to significant digits.
The question is incomplete, here is the complete question:
A chemist makes 600. mL of magnesium fluoride working solution by adding distilled water to 230. mL of a stock solution of 0.00154 mol/L magnesium fluoride in water. Calculate the concentration of the chemist's working solution. Round your answer to 3 significant digits.
Answer: The concentration of chemist's working solution is [tex]5.90\times 10^{-4}M[/tex]
Explanation:
To calculate the molarity of the diluted solution (chemist's working solution), we use the equation:
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of the stock magnesium fluoride solution
[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of chemist's magnesium fluoride solution
We are given:
[tex]M_1=0.00154M\\V_1=230mL\\M_2=?M\\V_2=600mL[/tex]
Putting values in above equation, we get:
[tex]0.00154\times 230=M_2\times 600\\\\M_2=\frac{0.00154\times 230}{600}=5.90\times 10^{-4}M[/tex]
Hence, the concentration of chemist's working solution is [tex]5.90\times 10^{-4}M[/tex]
An acid that dissociates to the extent of 92% in water would be termed a strong acid.
True or Fasle?
Answer: False
Explanation:
Strong acid is defined as the acid which completely dissociates when dissolved in water. They have low pH. These releases ions in their aqueous states.
[tex]HNO_3(aq.)\rightarrow H^+(aq.)+NO_3^-(aq.)[/tex]
Weak acid is defined as the acid which does not completely dissociates when dissolved in water. They have high pH. These releases ions in their aqueous states.
[tex]CH_3COOH\rightleftharpoons CH_3COO^-+H^+[/tex]
Thus as the given acid is only 92% dissociated and cannot be completely dissociated, it is termed as a weak acid.
Final answer:
The given statement," An acid that dissociates to the extent of 92% in water would be termed a strong acid." is false.
Explanation:
In chemistry, the degree to which an acid dissociates in water classifies it as either strong or weak. An acid that dissociates 100% into ions when dissolved in water is deemed a strong acid. In contrast, acids that do not reach 100% dissociation are termed weak acids. An example of a weak acid is acetic acid (CH3COOH), which has a dissociation level significantly lower than 100% when dissolved in water.
Considering the given information, an acid that dissociates to the extent of 92% in water would not meet the criteria for a strong acid, as it does not fully dissociate. Thus, even with a high dissociation percentage such as 92%, this acid would still be classified as a weak acid. It is important to note that in practical terms a 92% dissociation indicates a relatively strong acid amongst weak acids, but it is not categorized as a strong acid in the technical sense.
Which of the following compounds would have the highest boiling point?
a. 1-propanol,
b. CH3CH2CH2OH ethanol,
c. CH3CH2OH dimethyl ether,
d. CH3OCH3 ethyl methyl ether,
e. CH3OCH2CH3
Answer:a. 1-propanol,
Explanation:
Boiling point depends on the relative molecular mass of the compound and the kind of intermolecular interactions present in the substance. 1-propanol has a relative molecular mass of 60 and posses strong intermolecular hydrogen bonding. The presence of hydrogen bonding is known to lead to an increase in boiling point. As a result of this, it is obvious that 1-propanol has the highest boiling point among other compounds in the list.
Final answer:
1-propanol has the highest boiling point among the mentioned compounds due to its ability to form hydrogen bonds, which significantly elevate its boiling point compared to the other options.
Explanation:
The question pertains to determining which of the listed compounds would have the highest boiling point. When comparing boiling points, factors such as molecular mass, polarity, and the presence of hydrogen bonding play crucial roles. Among the options, 1-propanol (C3H7OH) stands out because it is capable of hydrogen bonding, which significantly increases its boiling point compared to the other compounds listed, such as dimethyl ether (CH3OCH3) and ethyl methyl ether (CH3OCH2CH3), which primarily exhibit dipole-dipole interactions and London dispersion forces. Ethanol (CH3CH2OH) also exhibits hydrogen bonding but has a shorter carbon chain than 1-propanol, leading to a slightly lower boiling point.
Osmium has a density of 22.6 g/cm3. What volume (in cm3) would be occupied by a 21.8 g sample of osmium? 1.04 2.03 × 103 493 0.965 2.03 × 10-3
Answer:
volume =0.9645cm³
Explanation:
density refers to the ratio of mass to volume of a substance. It gives us a clue about how heavy an object is.
density = mass/volume
volume = mass/density
mass=21.8g
density=22.6g
volume =21.8/22.6
volume =0.9645cm³
The volume occupied by a 21.8 g sample of osmium with a density of 22.6 g/cm³ is approximately 0.9646 cm³, calculated by dividing the mass by the density.
To find the volume (in cm³) occupied by a 21.8 g sample of osmium, you can use the formula:
Volume = Mass / Density
Given:
- Mass (m) = 21.8 g
- Density (D) = 22.6 g/cm³
Now, plug these values into the formula:
Volume = 21.8 g / 22.6 g/cm³
Volume ≈ 0.9646 cm³
So, the volume occupied by the 21.8 g sample of osmium is approximately 0.9646 cm³.
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Barium has a body-centered cubic structure. If the atomic radius of barium is 222 pm, calculate the density of solid barium?
Answer: density =3.377g/cm³
Explanation:
Density =( molecular weight × effective number of atoms per unit cell) / (volume of unit cell × avogadro constant)
D= (M ×n) /(V×A)
M= 137g/mol
n= 2 (For BCC)
V=a³ , where a= 4r/√3
a= (4×222)/√3
a=512.69pm
a= 512.69×10^-10cm
V= ( 512.69×10^-10)^3
V= 1.3476×10^-22cm³
D= (137×2)/(1.3476×10^-22 × 6.02^23)
D= 3.377g/cm³
Therefore the density of barium is 3.377g/cm³
Given:
Radius of barium,
R = 222 pmWe know,
→ [tex]a = \frac{4r}{\sqrt{3} }[/tex]
[tex]= \frac{4\times 222}{\sqrt{3} }[/tex]
[tex]= 512.69 \ pm[/tex]
[tex]= 512.69\times 10^{-10} \ cm[/tex]
Now,
→ [tex]V = a^3[/tex]
[tex]= (512.69\times 10^{-10})^3[/tex]
[tex]= 1.3476\times 10^{-22} \ cm^3[/tex]
hence,
The density will be:
→ [tex]D = \frac{M\times n}{V\times A}[/tex]
[tex]= \frac{137\times 2}{1.3476\times 10^{-22}\times 6.02\times 10^{23}}[/tex]
[tex]= 3.377 \ g/cm^3[/tex]
Thus the response above is right.
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