Answer:
a) [tex]CH_3CH_2CH_2CH_3[/tex]
Explanation:
In this question we have the following answer choices:
a) [tex]CH_3CH_2CH_2CH_3[/tex]
b) [tex]CH_3CH_2OH[/tex]
c) [tex]HF[/tex]
d) [tex]CH_3Cl[/tex]
e) [tex]HOCH_2CH_2OH[/tex]
We have to remember the relationship between intermolecular forces and vapor pressure. If we have stronger intermolecular forces we will have less vapor pressure because the molecules have more interactions between them, so, the molecules will prefer to stay in a liquid state rather than a gaseous state. Now, we have to check each molecule:
a) [tex]CH_3CH_2CH_2CH_3[/tex] (Van der waals interactions)
b) [tex]CH_3CH_2OH[/tex] (Hydrogen bonding)
c) [tex]HF[/tex] (Hydrogen bonding)
d) [tex]CH_3Cl[/tex] (Dipole-dipole interaction)
e) [tex]HOCH_2CH_2OH[/tex] (Hydrogen bonding)
For molecules b, c and e we have hydrogen bond to a heteroatom (O, N, S, or P). In this case oxygen, therefore we will have hydrogen bonding interactions (a very strong interaction). So, we can discard these ones.
In molecule e, we have "Cl" bond to a "C" therefore we will have the presence of a dipole (due to the electronegativity difference). If we have a dipole, we will have a dipole-dipole interaction (a strong interaction, less than hydrogen bonding but still is a strong interaction).
In molecule a, we have only Van der Waals interactions because in this molecule we have only carbon and hydrogen atoms bonded by single bonds. So, we will have a non-polar molecule. These interactions are the weakest interactions of all the molecules given. So, if we have weaker interactions the molecules can be converted to a gas state more easily and we have more vapor pressure.
Helium (He) is the lightest noble gas component of air, and xenon (Xe) is the heaviest. Perform the following calculations, using R = 8.314 J/(mol·K) and ℳ in kg/mol. (a) Find the rms speed of He in winter (0.°C) and in summer (30.°C). Enter your answers in scientific notation. × 10 m/s (winter) × 10 m/s (summer) (b) Find the ratio of the rms speed of He to that of Xe at 30.°C. (rate He)/(rate Xe) (c) Find the average kinetic energy per mole of He and of Xe at 30.°C. Enter your answers in scientific notation. × 10 J/mol for He × 10 J/mol for Xe (d) Find the average kinetic energy per molecule of He at 30.°C. Enter your answer in scientific notation. × 10 J/He atom
Answer:
Explanation:
Hello,
Data;
R = 8.314J/(mol.K)
Temp (winter) = 0°C = (0 + 273.15)K = 274.15K
Temp.(summer) = 30°C = (30 + 273.15)K = 303.15K
Molar mass of He = 4g/mol = 0.004kg/mol
Molar mass of Xe = 131.29g/mol = 0.131kg/mol
a) rms speed in winter and summer
Vrms = √(3RT/M)
R = gas constant
T = temperature of the gas
M = molar mass of the gas
In winter,
Vrms = √(3×8.314×273.15) / 0.004
Vrms = 1.30×10³m/s
In summer
Vrms = √(3×8.314×303.15) / 0.004
Vrms = 1.37×10³m/s
b) Vrms of Xe at 30°C
Vrms = √(3 × 8.314 × 303.15) / 0.131
Vrms of Xe = 240.24m/s = 2.40×10²m/s
Vrms of He = 1.37×10³m/s
Rate of He / rate of Xe = 1.37×10³ / 2.40×10²
Rate of He / rate of Xe = 5.7
c) K.E per mole
At 30°C
K.E of He = (3/2) × 8.314 × 303.15
K.E of He = 3.78×10³J/mol
K.E of Xe = (3/2) × 8.314 × 303.15
K.E of Xe = 3.78×10³J/mol
d) K.E per molecule = ½mv²
K.E of He molecule = ½ × 0.004 × 1.37×10³
K.E = 2.74J
The rms speed of He in winter and in summer are 1.30×10³m/s & 1.37×10³m/s respectively, ratio of the rms speed of He to that of Xe at 30 degree celsius is 5.7, average kinetic energy per mole of He and of Xe and average kinetic energy per molecule of He are discussed below.
How do we calculate the root mean square velocity?Root mean square velocity of the gases will be calculated as:
Vrms = √(3RT/M), where
R = universal gas constant = 8.314J/(mol.K)
M = molar mass of gas in kg/mol
T = temperature in K
Root mean square velocity of Helium at 0°C or 273.15K and 30°C or 303.15K will be calculated as:At 273.15K-
Vrms = √(3×8.314×273.15) / 0.004 = 1.30×10³m/s
At 303.15K-
Vrms = √(3×8.314×303.15) / 0.004 = 1.37×10³m/s
Vrms (winter×summer) = 1.30×10³m/s × 1.37×10³m/s = 1.78×10⁶m/s
Vrms of Xe at 303.15KVrms = √(3 × 8.314 × 303.15) / 0.131
Vrms of Xe = 240.24m/s = 2.40×10²m/s
Vrms of He = 1.37×10³m/s
Rate of He / rate of Xe = 1.37×10³ / 2.40×10²
Rate of He / rate of Xe = 5.7
Average kinetic energy will be calculated as:K.E = 3RT / 2N, where
N = avgadros number = 6.022×10²³ atoms/mole
Average kinetic energy per mole of He & Xe at 303.15K is same as:
K.E = 3/2(1.38 × 10⁻²³J/K)(303.15K) = 6.2 × 10⁻²⁵J/mol
Average kinetic energy per molecule of He will be calculated as:K.E of He molecule = ½ × 0.004 × 1.37×10³
K.E = 2.74J
Hence, calculations for the given points are described above.
To know more about root mean square velocity, visit the below link:
https://brainly.com/question/2020688
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Answer:
ok
Explanation:
The two reactions above, show routes for conversion of an alkene into an oxirane. If the starting alkene is cis-3-hexene the configurations of the oxirane products, A and B are Product A: _______ Product B: _______ Will either of these two oxirane products rotate the plane of polarization of plane polarized light? _____
Answer:
Product A: cis; no
Product B: cis: no
Explanation:
Two common methods of forming oxiranes from alkenes are:
Reaction with peroxyacids Formation of a halohydrin followed by reaction with base
1. Reaction with peroxyacids
(a) Stereochemistry
The reaction with a peroxyacid is a syn addition, so the product has the same stereochemistry as the alkene.
The starting alkene is cis, so the product is cis-2,3-diethyloxirane.
(b) Configuration
The product is optically inactive because it has an internal plane of symmetry.
It will not rotate the plane of polarized light.
2. Halohydrin formation
(a) Stereochemistry
The halogenation of the alkene proceeds via a cyclic halonium ion.
The backside displacement of halide ion by alkoxide is also stereospecific, so a cis alkene gives a cis epoxide.
The product is cis-2,3-diethyloxirane.
(b) Configuration
The cyclic halonium ion has an internal plane of symmetry, as does the product (meso).
The oxirane will not rotate the plane of polarized light.
g Which statement is incorrect regarding oxidation? Oxidation is a "gain" of electrons. Oxidation is the combination with O atoms. Oxidation is an increase in oxidation state. Oxidation is always accompanied by reduction. none of these
Answer:
The incorrect statement from the options is OXIDATION IS A "GAIN" OF ELECTRONS
Explanation:
Oxidation in a redox reaction is the loss of electrons. It is also the increase in the oxidation states of an atom or ion or atoms in a molecule. A redox reaction is a type of chemical reaction in which there is a transfer of electrons from an atom or ion to another resulting in a change in oxidation states of the substances involved. The reducing agent in the reaction is undergoes oxidation by losing electrons while the oxidating agent is reduced that is it gains electrons at the end of the reaction. The atom or ion from which electron is lost is said to be oxidized while the other atom or ion involved in the reaction is reduced.
Oxidation is also the combination with O atoms and it is always accompanied by reduction because oxidation forms a half of the whole redox reaction. A substance cannot be oxidized except it has reduced another substance by losing electrons to it.
What volume (mL) of a concentrated solution of sodium hydroxide (6.00 M) must be diluted to 187 mL to make a 1.53 M solution of sodium hydroxide
Answer:
47.68 mL
Explanation:
In this case, we have a dilution problem. So, we have to start with the dilution equation:
[tex]C_1*V_1=C_2*V_2[/tex]
We have to remember that in a dilution procedure we go from a higher concentration to a lower one. With this in mind, We have to identify the concentration values:
[tex]C_1~=~6.00~M[/tex]
[tex]C_2~=~1.53~M[/tex]
The higher concentration is C1 and the lower concentration is C2. Now, we can identify the volume values:
[tex]V_1~=~X[/tex]
[tex]V_2~=~187~mL[/tex]
The V2 value has "mL" units, so V1 would have "mL" units also. Now, we can include all the values into the equation and solve for "V1", so:
[tex]6.00~M*V_1=1.53~M*187~mL[/tex]
[tex]V_1=\frac{1.53~M*187~mL}{6.00~M}=47.68~mL[/tex]
So, we have to take 47.68 mL of the 6 M and add 139.31 mL of water (187-47.68) to obtain a solution with a final concentration of 1.53 M.
I hope it helps!
Question 4
2 pts
A careless chemistry student performed a chemical reaction where his theoretical yield of
Magnesium oxide was 57.82 grams, but he actually produced 12.89 grams. What is his percent yield
for this experiment? (include the number with 4 significant figures but no units)
Answer:
22.29%
Explanation:
Percent yield = experimental yield / theoretical yield * 100
= 12.89 / 57.82 * 100 = 22.29%
How many water molecules are in a block of ice containing 1.25 mol of water (H2O)
Answer:
Molecules = 7.5 × 10²³ molecules
Explanation:
Given:
Moles = 1.25 mol
Avogadro's No. = [tex]N_{A}[/tex] = 6.022 * 10²³
Required:
Molecules = ?
Formula:
Molecules = Moles × [tex]N_{A}[/tex]
Solution:
Molecules = 1.25 × 6.022 × 10²³
Molecules = 7.5 × 10²³ molecules
What energy transfer happens when wood is burning?
Answer:
Mechanical to Heat
explanation:
The wood itself can make mechanical energy but when it's on fire it makes heat energy
Answer: Chemical to heat and light
Explanation: The energy transforms from chemical energy to heat and light energy. Because when the candle burns a chemical reaction occurs and produces heat and light.
A 8.06 g piece of solid CO 2 is allowed to sublime in a balloon. The final volume of the balloon is 1.00 L at 300.0 K. What is the pressure of the gas?
Answer:
The pressure of the gas is 4.428 atm.
Explanation:
Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.
Considering a certain amount of ideal gas confined in a container where the pressure, volume and temperature can vary, but keeping the mass constant, that is, without altering the number of moles, the pressure, P, the temperature can be related, T and the volume, V, of an ideal gas using the ideal gas law:
P*V = n*R*T
where R is the ideal gas constant, and n is the number of moles of the gas.
In this case, to know n you must know the molar mass of the CO₂ compound. Being:
C: 12 g/moleO: 16 g/molethen, the molar mass of CO₂ is: 12 g/mole + 2*16 g/mole= 44 g/mole
Then you apply a rule of three: if 44 grams of CO₂ are present in 1 mole, 8.06 grams in how many moles are they?
[tex]moles of CO_{2} =\frac{8.06 grams*1 mole}{44 grams}[/tex]
moles of CO₂= 0.18 moles
Then, you know:
P: ?V: 1 Ln: 0.18 molesR= 0.082[tex]\frac{atm*L}{mol*K}[/tex]T= 300 KReplacing in the equation of the ideal gas law:
P* 1 L= 0.18 moles* 0.082[tex]\frac{atm*L}{mol*K}[/tex] * 300 K
Solving:
[tex]P=\frac{0.18 moles* 0.082 \frac{atm*L}{mol*K} *300 K}{1 L}[/tex]
P= 4.428 atm
The pressure of the gas is 4.428 atm.
As , P , S , Na which of these atomes has the highest ionization energy ?
Answer:
it is Na
Explanation:
because u need to pay attention i school and u wouldnt need to know these answer
If 175mL of oxygen is produced at STP, how many grams of hydrogen peroxide, H2O2
were decomposed? At STP, 1 mole of gas occupies 22.4L. Be sure to balance first.
2 H202 > H202 + O2
what's the
Mass of H2O2
Answer:
0.53g
Explanation:
We'll begin by converting 175mL to L. This is illustrated below:
1000mL = 1L
Therefore 175mL = 175/1000 = 0.175L
Next, we shall calculate the number of mole of O2 that occupy 0.175L. This is illustrated below:
1 mole of O2 occupy 22.4L at stp.
Therefore, Xmol of O2 will occupy 0.175L i.e
Xmol of O2 = 0.175/22.4
Xmol of O2 = 7.81×10¯³ mole
Therefore, 7.81×10¯³ mole of O2 occupy 175mL.
Next, we shall determine the number of mole of H2O2 that decomposed to produce 7.81×10¯³ mole of O2. This is illustrated below:
2H2O2 —> 2H2O + O2
From the balanced equation above,
2 moles of H2O2 decomposed to produce 1 mole of O2.
Therefore, Xmol of H2O2 will decompose to produce 7.81×10¯³ mole of O2 i.e
Xmol of H2O2 = 2 x 7.81×10¯³
Xmol of H2O2 = 1.562×10¯² mole
Therefore, 1.562×10¯² mole of H2O2 decomposed in the reaction.
Finally, we shall convert 1.562×10¯² mole of H2O2 to grams. This is illustrated below:
Molar mass of H2O2 = (2x1) + (16x2) = 34g/mol
Mole of H2O2 = 1.562×10¯² mole
Mass of H2O2 =..?
Mole = mass /Molar mass
1.562×10¯² = mass /34
Cross multiply
Mass of H2O2 = 1.562×10¯² x 34
Mass of H2O2 = 0.53g
Therefore, 0.53g of Hydrogen peroxide, H2O2 were decomposition in the reaction.
Identify the precipitation reaction in the set?
Answer:
The third reaction
(2NaOH + NiCL2 ---> 2NaCl + Ni(OH)2)
Explanation:
By definition, a precipitation reaction refers to the formation of an insoluble salt when two solutions containing soluble salts are combined.
(Source: lumenlearning)
From the 4 options, we can eliminate the first and second one immediately because there is no formation of an insoluble salt.
Then, the last one can also be eliminated because even though there is insoluble solid formed, but it is not a salt, and, the reactants are not solutions too. In fact, the last one is a displacement reaction. A more reactive metal displaces a less reactive metal to form an ion.
Since the third reaction matches the definition of precipitation reaction, this is the answer.
The following ions contain the same number of electrons. Rank them in order of decreasing ionic radii. Rank from largest to smallest radius. To rank items as equivalent, overlap them.K^+,Sc^3+,Cl^-,S^2-,P^3-
Answer:
P^3- > S^2- > Cl^- > K^+ > Sc^3+
Explanation:
Ionic radii is an example of physical properties of Periodicity.
The size of an atom's ion is difficult to estimate because of the electronic distribution and arrangement. This is due to the fact that the atom' ion have no definite outer boundary. In order to circumvent this problem, the atom's ion is estimated in a crystal lattice in terms of its ionic radii.
Ionic radii is taken as half the distance between atomic ions in a crystal lattice. Across a period in the periodic table , ionic radii decreases progressively from left to right.
Down the group, the ionic radius increases from top to bottom.
So the arrangement of the given elements from largest to smallest radius in the decreasing order of ionic radius will be:
P^3- > S^2- > Cl^- > K^+ > Sc^3+
The ranging from largest to smallest should be P^3- > S^2- > Cl^- > K^+ > Sc^3+.
What are ionic radii?It represents an example of the physical properties of Periodicity. It is considered to be half of the distance that lies between the atomic ions in a crystal lattice. It should be decreased progressively from left to right.
The ionic radius rises from top to bottom. It contains similar no of electrons, there should be the size difference because of the strength of the nuclear attraction.
Learn more about ions here: https://brainly.com/question/24599315
A sample of magnesium ribbon is ignited in a crucible to form magnesium oxide. Determine the empirical formula of magnesium oxide from the following data:
mass of crucible and cover + magnesium metal
33.741 g
mass of crucible and cover
33.500 g
mass of crucible and cover + magnesium oxide
33.899 g
Answer:
MgO
Explanation:
The following data were obtained from the question:
mass of crucible and cover + magnesium metal = 33.741 g
mass of crucible and cover = 33.5 g
mass of crucible and cover + magnesium oxide = 33.899 g
Next, we shall determine the mass of magnesium metal. This can be obtained as follow:
mass of crucible and cover + magnesium metal = 33.741 g
mass of crucible and cover = 33.5 g
Mass of magnesium metal =..?
Mass of magnesium metal = (mass of crucible and cover + magnesium metal) – (mass of crucible and cover)
Mass of magnesium metal = 33.741 – 33.5
Mass of magnesium metal = 0.241g
Next, we shall determine the mass of magnesium oxide. This can be obtained as follow:
mass of crucible and cover + magnesium oxide = 33.899 g
mass of crucible and cover = 33.5 g
Mass of magnesium oxide =?
Mass of magnesium oxide = (mass of crucible and cover + magnesium oxide) – (mass of crucible and cover)
Mass of magnesium oxide = 33.899 –. 33.5
Mass of magnesium oxide = 0.399g
Next, we shall determine the mass of oxygen. This can be obtained as follow:
Mass of magnesium oxide = 0.399g
Mass of magnesium metal = 0.241g
Mass of oxygen =..?
Mass of oxygen = (Mass of magnesium oxide) – (Mass of magnesium metal)
Mass of oxygen = 0.399 – 0241
Mass of oxygen = 0.158g
Now, we can obtain the empirical formula for the magnesium oxide as follow:
Mg = 0.241g
O = 0.158g
Divide by their molar mass
Mg = 0.241 / 24 = 0.01
O = 0.158 / 16 = 0.0099
Divide by the smallest
Mg = 0.01 / 0.0099 = 1
O = 0.0099 / 0.0099 = 1
Therefore, the empirical formula for the magnesium oxide is MgO
We are all familiar with the general principles of operation of an internal combustion engine: the combustion of fuel drives out the piston. It is possible to imagine engines that use reactions other than combustions, and we need to assess the work they can do. A chemical reaction takes place in a container of cross-sectional area 100.0 cm^2; the container has a piston at one end. As a result of the reaction, the piston is pushed out through 10 cm against a constant external pressure of 100 kPa. Calculate the work done by the system in Joules.
Answer:
The work done by the system is 100 J
Explanation:
Given details
The cross sectional area of the of the container is A = 100.0 cm^2 = 0.01m²
The total distance pushed by the piston is d = 10 cm = 0.10m
The total external pressure by which piston pushed is P = 100 kPa
From above data, the following relation can be used to determine the change in volume of the container
∆V = A * d
∆V = 0.01 * 0.10 = 0.001 m³
By using the following relation, the work done by the system is calculated as;
Work done W = P * ∆V
W = 100 * 0.001 = 0.1 kJ = 100 J
The work done by the system is 100 J
How much water would be needed to completely dissolve 1.52 L of the gas at a pressure of 730 torr and a temperature of 21 ∘C?
Answer:
The correct answer is 0.4 L.
Explanation:
The mentioned question can be solved by using the equation,
C = K × Pgas--------(i)
Here K is the Henry law constant whose value is 0.158 mol/L/atm, C is the concentration of the gas in liquid state, and Pgas is the partial pressure of the gas.
Now to find the volume of water, the formula to be used is,
PV = nRT-----------(ii)
Here P is the pressure of the gas, V is the volume, R is the universal gas constant whose value is 0.082 Latm/mol/K and T is the temperature.
PgasV = nRT
Pgas = nRT/Vgas
The value of Pgas is inserted in equation (i) we get,
C = K × nRT/Vgas
It is to be noted that C = n/V, here n is the no. of the moles and V is the volume of liquid.
n/Vliquid = K × nRT/Vgas
1/Vliquid = KRT/Vgas
Vliquid = Vgas/KRT--------------(iii)
Based on the given information, the volume of the gas is 1.52 L, the value of K is 0.158 mol/L/atm, the value of R is 0.082 Latm/mol/K and value of T is 21 degree C or 273 + 21 = 294 K.
Now putting the values in equation (iii) we get,
Vliquid = 1.52 L / 0.158 × 0.082 × 294
Vliquid = 1.52 / 3.809
Vliquid = 0.399 or 0.4 L
Hence, the volume of water required to dissolve 1.52 L of gas is 0.4 L.
Morphine, C 17H 19NO 3, is often used to control severe post-operative pain. What is the pH of the solution made by dissolving 25.0 mg of morphine in 100. mL of water? (For morphine, K b = 1.62 × 10 –6.)
Answer:
pH = 9.58
Explanation:
First of all, we need to determine the molarity of the solution.
We determine the molar mass of morphine:
12g/m . 17 + 1 g/m . 19 + 14 g/m + 16 g/m . 3 = 285.34 g/m
molar mass g/m, is the same as mg/mm
25 mg . 1 mmol / 285.34 mg = 0.0876 mmoles / 100 mL = 8.76×10⁻⁴ M
In diltuted solution, we must consider water.
Mass balance for morphine = [Morphine] + [Protonated Morphine]
8.76×10⁻⁴ M = [Morphine] + [Protonated Morphine]
As Kb is too small, I can skipped, the [Protonated Morphine]
8.76×10⁻⁴ M = [Morphine]
In the charge balance I will have:
[OH⁻] = [H⁺ morphine] + [H⁺]
Let's go to the Kb expression
Morphine + H₂O ⇄ MorphineH⁺ + OH⁻ Kb
Kb = [MorphineH⁺] [OH⁻] / [Morphine]
Kb = [MorphineH⁺] [OH⁻] / 8.76×10⁻⁴ M
So now, we need to clear [MorphineH⁺] to replace it in the charge balance
Kb . 8.76×10⁻⁴ M / [OH⁻] = [MorphineH⁺]
Now, the only unknown value is the [OH⁻]
[OH⁻] = Kb . 8.76×10⁻⁴ M / [OH⁻] + Kw/[OH⁻]
Remember that Kw = [H⁺] . [OH⁻]
[H⁺] = Kw/[OH⁻]
[OH⁻]² = 1.62×10⁻⁶ . 8.76×10⁻⁴ + 1×10⁻¹⁴
[OH⁻] = √(1.62×10⁻⁶ . 8.76×10⁻⁴ + 1×10⁻¹⁴)
[OH⁻] = 3.76×10⁻⁵ → - log [OH⁻] = pOH = 4.42
pH = 14 - pOH → 14 - 4.42 = 9.58
Two stereoisomers are obtained from the reaction of HBr with (S)-4-bromo-1-pentene. One of the stereoisomers is optically active, and the other is not. Draw the structure of the optically active stereoisomer.
Answer:
See explanation
Explanation:
In this case, we have an addition reaction. Additionally, this is a marknovnikov addition, therefore the "Br" atom would be added in the most substituted carbon (in this case carbon a). And we are going to have 2 enantiomers (2S,4S)-2,4-dibromopentane and (2R,4S)-2,4-dibromopentane. In the case of (2R,4S)-2,4-dibromopentane we will have a symmetry plane (a point in the molecule in which we can divide the molecule into two equal parts). When this happens we will have a mesocompound and we will not have optical activity.
See figure 1
I hope it helps!
The basic function of a carburetor of an automobile is to atomize the gasoline and mix it with air to promote rapid combustion. As an example, assume that 30 cm3 of gasoline is atomized into N spherical droplets, each with a radius of 2.0 × 10−5 m. What is the total surface area of these N spherical droplets? Answer: [A] m2.
Answer:
The total surface area of these N spherical droplets is 4.4929 m²
Explanation:
From the information given :
assuming that :
30 cm³ of gasoline is atomized into N spherical droplets &
each with a radius of 2.0 × 10−5 m
We are tasked to determine the total surface area of these N spherical droplets
We all known that:
[tex]1 \ cm^3 = 10 ^{-6} m^3[/tex]
Therefore
[tex]30 \ cm^3 = 30 * 10 ^{-6} m^3 = 3 *1 0^{-5} \ m^3[/tex]
For each droplet; there is a required volume which is = [tex]\dfrac{4}{3} \pi r ^3[/tex] since it assumes a sphere shape .
Thus;
replacing radius(r) with 2.0 × 10−5 m; we have:
[tex]= \dfrac{4}{3} \pi * (2.0 *10^{-5} m) ^3[/tex]
= [tex]3.35 * 10^{-14} \ m^3[/tex]
However; there are [tex]3*10^{-5} \ m^3[/tex] gasoline atomized into N spherical droplets with each with radius 2.0 × 10−5 m
For N ; we have ;
[tex]=\dfrac{3*10^{-5} \ m^3}{3.35 * 10^{-14} \ m^3/ droplet}[/tex]
= [tex]8.95*10^8 \ droplet s[/tex]
So; each droplet have a surface area = [tex]4 \pi r^2[/tex]
= [tex]4 \pi (2.0*10^{-5}m) ^2[/tex]
= [tex]5.02*10^{-9} \ m^2/droplets[/tex]
The surface area per droplet is equivalent to [tex]5.02*10^{-9} \ m^2/droplets[/tex]
Thus;
The total surface area of these N spherical droplets will be :
= [tex]8.95*10^8 \ droplet s * 5.02*10^{-9} \ m^2/ droplets[/tex]
= 4.4929 m²
The total surface area of these N spherical droplets is 4.4929 m²
Select the chemical equation that represents an acid base reaction??? Please help
Answer:
HCl + AgNO3 ------ AgCl + HNO3
Explanation:
HCl is an acid called hydrochloric acid, while AgNO3 is a base called Silver Nitrate.
Hope it helps.
20. Which of the following statements regarding subshell filling order for a neutral atom is/are CORRECT? 1. Electrons are assigned to the 4f subshell before they are assigned to the 6s subshell. 2. Electrons are assigned to the 5d subshell before they are assigned to the 6p subshell. 3. Electrons are assigned to the 4p subshell before they are assigned to the 4s subshell. a. 1 only b. 2 only c. 3 only d. 1 and 3 e. 1, 2, and 3
Answer:
2 only
Explanation:
Electrons are filled in atoms according to the Aufbau principle. Electrons are filled into lower energy orbital before the filling of higher energy orbitals and this sequence must be followed in filling electron orbitals.
The order of arrangement of energy levels may be shown as follows; 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p. This implies that 4f>6s, hence 6s is filled before 4f.
Also, the 6p level > 5d level hence this is the correct option. You must fill the 5d level before you feel the 6p level.
At a certain temperature this reaction follows second-order kinetics with a rate constant of 0.00317sâ1: 2N2O5(g) â2N2O4(g) + O29(g) Suppose a vessel contains SO3 at a concentration of 1.44M . Calculate the concentration of SO3 in the vessel 0.240 seconds later. You may assume no other reaction is important.Round your answer to 2 significant digits.
Answer:
[A] = 1.438M = 1.4M (Two s.f)
Explanation:
Rate constant, k = 0.00317
Initial Concentration, [A]o = 1.44M
Final Concentration, [A] = ?
Time, t = 0.240 s
Since this is a second order reaction, the formula for this is given as;
1 / [A] = 1 / [A]o + kt
1 / [A] = 1 / 1.44 + (0.00317 * 0.240)
1 / [A] = 0.6944 + 0.0007608
1 / [A] = 0.6952
[A] = 1.438M = 1.4M (Two s.f)
How many ng(nanogram) are in 3.3 mg(milligram)
Answer:
There are [tex]3.3(10^6)[/tex] nanograms in 3.3 milligrams
Explanation:
The conversion is 1 milligram is equal to [tex]1(10^6)[/tex] nanograms. Use the base 10 decimal system to help you.
Answer:
3 300 000 000 000 ng * nanogram
Explanation:
True
False
Question 7 (4 points)
In thermodynamics, the universe is defined as the system plus its surroundings.
True
False
Question 8 (4 points)
The law of conservation of energy states that energy is neither created nor destroyed.
Answer:
The question 7 answer is false
the question 8 answer is true
How many grams of LiNO3 must be added to 25.0 g of water to prepare a 5.00% (m/m) solution of LiNO3?
Answer:
1.25 g.
Explanation:
5% or 25 g
= 0.05 * 25
= 1.25 g (answer).
What is the rate constant of a reaction if rate = 1.5 (mol/L)/s, [A] is 1 M, [B] is
3 M, m = 2, and n = 1?
k=
rate
[A]" [B]"
A. 0.17
B. 13.5
C. 0.5
D. 4.5
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Actually Welcome to the Concept of the Rate Constant.
Here, the "K" is the Rate Constant.
so the ANSWER IS C.) 0.5
The rate of constant is 0.5.
The answer is option C.
How do find the rate of constant?To determine the fee regulation from a desk, you have to mathematically calculate how differences in molar concentrations of reactants affect the response charge to parent out the order of every reactant. Then, plug in values of the response charge and reactant concentrations to discover the particular rate constant.
What's the rate of constant?The particular rate constant is the proportionality steady touching on the rate of the reaction to the concentrations of reactants. The fee regulation and the unique charge constant for any chemical response need to be decided experimentally. The price of the fee regular is temperature-established.
Learn more about the rate of constant here: brainly.com/question/8813467
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(a) How many stereoisomers are possible for 4-methyl-1,2-cyclohexanediol? ___ (b) Name the stereoisomers formed by oxidation of (S)-4-methylcyclohexene with osmium tetroxide. If there is only one stereoisomer formed, leave the second space blank. Isomer #1: Isomer #2: (c) Is the product formed in step (b) optically active? _____
Answer:
See explanation
Explanation:
For the first part of the question, we have to check the chiral carbons in 4-methyl-1,2-cyclohexanediol. In this case carbons, 1 and 2 are chiral, if we have 2 chiral carbons we will have 4 isomers. We have to remember that formula 2^n in which "n" is the number of chiral carbons, so:
2^n = 2^2 = 4 isomers
And the isomers that we can have are:
1) (1R,2S)-4-methylcyclohexane-1,2-diol
2) (1S,2S)-4-methylcyclohexane-1,2-diol
3) (1S,2S)-4-methylcyclohexane-1,2-diol
4) (1S,2R)-4-methylcyclohexane-1,2-diol
See figure 1
For the second part of the question, we have to remember that the oxidation with [tex]OsO_4[/tex] is a syn addition. In other words, the "OHs" are added in the same plane. In this case, we have the methyl group with a wedge bond, so the "OH" groups will have a dashed bond due to the steric hindrance. Due to this we only can have 1 isomer ((1S,2R,4S)-4-methylcyclohexane-1,2-diol). Finally, on this molecule, we dont have any symmetry planes (this characteristic will cancel out the optical activity), so the product of this reaction has optical activity.
See figure 2
I hope it helps!
. Explain with examples following characteristics of chemical reactions:
a. Change of colour b. Evolution of gas c. Change of smell d. Change of state
Answer:
See explanation.
Explanation:
Hello,
a. In this case, the change of color is evident for instance when copper reacts with nitric acid to form hydrogen and copper (II) nitrate since copper orange-like and nitric acid is colorless, but copper (II) nitrate is green (dry) or blue (hydrated).
b. In this case, when we make react hydrochloric acid and magnesium, we notice a gas giving off while the magnesium chloride remains aqueous, due to the fact that magnesium displaces hydrogen which is given off as a gas.
c. In this case, we can consider an egg since when it is edible it has a tasty smell but when it decomposes to rotten egg, hydrogen sulfide is given off due to the action of specific bacteria, causing a change in smell to a quite stinky one.
d. In this case, a reaction by which a change of state is exhibited is for instance when aqueous lead (II) nitrate reacts with aqueous potassium iodide to yield potassium nitrate which remains aqueous whereas the lead (II) iodide precipitates out as a solid due to its tiny solubility as a yellow solid.
Best regards.
As lead chemist for a pharmaceutical manufacturing company, you need to inform the purchasing office of a supply order for the next batch of cisplatin, PtCl2(NH3)2. If you intend to make a 500kg batch, how many kg chlorine gas do you need to order?
Answer:
mass of chlorine gas required is 118 kg.
Explanation:
Total mass of the drug (Cisplatin) required = 500 kg
For the drug PtCl2(NH3)2, we first find the molar mass of the compound.
The molar mass of the drug is the total of all the molar mass of the elements in the drug
molar mass of Pt (platinum) in the drug = 195.078 g/mol
molar mass of chlorine (Cl) in the drug = 2 x (35.453 g/mol) = 70.908 g/mol
molar mass of ammonia (NH3) in the drug = 2 x (17.031 g/mol) = 34.062 g/mol
Total molar mass of the drug = 195.078 g/mol + 70.908 g/mol + 34.062 g/mol = 300.048 g/mol
fractional composition of chlorine in the drug = 70.908/300.048 = 0.236
mass of chlorine required for 500 kg of the drug = 0.236 x 500 = 118 kg
How many moles of aqueous magnesium ions and chloride ions are formed when 0.240 mol of magnesium chloride dissolves in water
Answer:
1. 0.240 mole of magnesium ion, Mg^2+
2. 0.48 mole of chloride ion, Cl^-
Explanation:
We'll begin by writing the balanced dissociation equation for magnesium chloride. This is given below:
MgCl2(aq) —> Mg^2+(aq) +2Cl^-(aq)
1. Determination of the number of mole of magnesium ion, Mg^2+ obtained from the dissolution 0.240 mol of magnesium chloride, MgCl2 in water. This is illustrated below:
From the balanced equation above,
1 mole of MgCl2 produced 1 mole of Mg^2+.
Therefore, 0.240 mole of MgCl2 will also produce 0.240 mole of Mg^2+.
Therefore, 0.240 mole of magnesium ion, Mg^2+ is produced.
2. Determination of the number of mole of chloride ion, Cl^- obtained from the dissolution 0.240 mol of magnesium chloride, MgCl2 in water. This is illustrated below:
From the balanced equation above,
1 mole of MgCl2 produced 2 moles of Cl^-.
Therefore, 0.240 mole of MgCl2 will produce = (0.240 x 2) = 0.48 mole of Cl^-.
Therefore, 0.48 mole of chloride ion, Cl^- is produced.