Answer: 0.75 moles of Phosphorus and 4.5 moles of Hydrogen are produced respectively from 3 moles of Phosporus.
Explanation:
4PH3 --------> P4 + 6H2
From the stochiometry of the reaction,
4 moles of Phosphine gives 1 mole of Phosphorus
3 moles of Phosphine will give (3×1)/4 moles of Phosphorus.
Therefore, 0.75 moles of Phosphorus is produced.
Similarly, 4 moles of Phosphine gives 6 moles of Hydrogen
3 moles of Phosphine will give (3×6)/4 moles of Hydrogen.
Therefore, 4.5 moles of Hydrogen is produced.
QED!
When 0.5000 grams of an unknown hydrocarbon, CxHy, is completely combusted with excess oxygen, 1.037 L CO2 gas and is produced at 98.3 °C and 1.000 atm. What is the empirical formula of the hydrocarbon? (R = 0.08206 L×atm/mol×K)
Answer: C₃H₈
Explanation:
From PV = nRT
Moles of CO2 = PV / RT =(1 x 1.037)/(0.08206 x ( 273+98.3)) = 0.0340 moles of CO2
CxHy + (x + (y/2))O₂ ----> xCO₂ + yH₂O
there is 1 mole of C in each mole of CO2 so moles of C in CO₂ = 0.0340 moles
mass of C in CO₂ = 0.0340 x 12 = 0.41 g
This means mass of C in the hydrocarbon = 0.41g
so mass of H in the hydrocarbon = 0.50 - 0.41= 0.09 g
moles of H in the hydrocarbon = mass/molar mass = 0.09/1 = 0.09 moles
molar ratio of C:H = 0.034 : 0.09 = 1 :2.67
or 3 :8
empirical formula is C3H8
To find the empirical formula of the hydrocarbon, calculate the moles of carbon dioxide produced using the ideal gas law and the moles of carbon and hydrogen in the hydrocarbon. The empirical formula is C12H1.
Explanation:In order to determine the empirical formula of the hydrocarbon, we need to find the mole ratio of carbon to hydrogen in the compound. To do this, we first calculate the moles of carbon dioxide (CO2) produced using the ideal gas law:
n = (PV)/(RT) = (1.000 atm)(1.037 L)/(0.08206 L.atm/mol.K)(98.3 + 273.15 K) = 0.0443 mol
Since the combustion of 1 mole of hydrocarbon produces 1 mole of CO2, we can conclude that the moles of hydrocarbon consumed is also 0.0443 mol. Next, we calculate the moles of carbon and hydrogen in the hydrocarbon:
moles of carbon = (mass of carbon in hydrocarbon) / (molar mass of carbon) = (12.01 g/mol)(0.5000 g) / (1 g) = 6.003 mol
moles of hydrogen = (mass of hydrogen in hydrocarbon) / (molar mass of hydrogen) = (1.008 g/mol)(0.5000 g) / (1 g) = 0.5008 mol
The empirical formula of the hydrocarbon is then C6H0.5008. We can simplify this ratio by multiplying all the subscripts by 2 (the smallest whole number ratio) to get the empirical formula: C12H1.
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When aluminum hydroxide (dissolved in water);is mixed with aqueous sulfuric acid (H2SO4) the products are aluminum sulfate (a precipitate) and liquid water
Answer:
The equation for this reaction is:
2Al(OH)3 (aq) + 3H2SO4 (l) ---> Al2(SO4)3 (s) + 6H2O (l)
Explanation:
Given a list of 10,000 elements, and if each comparison takes 2 µs, what is the fastest possible runtime for linear search?
Answer:
2 µs
Explanation:
Which sequence of events is required to form a limestone cave where you can walk around and observe cave formations, such as stalactites?
Answer:
Which sequence of events is required to form a limestone cave where you can walk around and observe cave formations, such as stalactites? (Note: stalactites hang from the ceiling - they have to hold on tight to the roof.)
A geological sequence of events as involving the lowering of the water table to expose cave structures where stalactites and stalagmites form which is described as follows,
Explanation:
1. Acidic percolated water formed cavities of solution beneath the natural water table known as phreatic zone
2. After the passage of time there is a drop in the water table dropped forming caves from cavities
3. These caves, which are air filled voids that contains adequate environment for forming stalactites and stalagmites and where they are found
Describe and explain how electrical conductivity occurs in mercury bromide and mercury, in both solid and molten states.
Answer:
HgBr2 conducts when molten because there are mobile ions in molten HgBr which allows flow of current when an electrical potential difference is introduced to the HgBr in molten state
However HgBr2 does not conduct in the solid state as the ions are fixed in the solid HgBr2 lattice structure
Mercury, which is a metal in its natural form conducts both in the solid and molten states as the delocalized electrons are able to move both in the solid and molten mercury states and as such current flows through mercury when there is an electrical potential difference placed across it
Explanation:
Electricity or electric current flow is the term used to describe the state of movement or flow of matter that carries an electrical charge
It is the steady movement of or flow of electrons. The moving electrons transfer electrical charge round an electrical circuit. In metals, there are freely shared electrons between individual atoms so as to efficiently conduct electricity and so when an electrical potential difference is placed across a piece of metallic object an electron is readily displaced by another electron entering from one end and exiting from the other end of the electrical potential difference
Mercury bromide and mercury exhibit electrical conductivity in both solid and molten states due to the presence of freely mobile, charged entities. In the solid state, ionic compounds like mercury bromide are not conductive, but they become conductive when molten.
Explanation:In both the solid and molten states, mercury bromide and mercury demonstrate electrical conductivity due to the presence of freely mobile, charged entities. In the solid state, ionic compounds like mercury bromide are not electrically conductive because their ions are unable to flow. However, in the molten state, the ions are able to move freely through the liquid, allowing for electrical conductivity.
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It took 38.33 mL of 0.0944 M HCl to titrate (react completely with) the ammonia. What is the concentration of the original ammonia solution?
The question is incomplete, here is the complete question:
A 50.60 mL sample of an ammonia solution is analyzed by titration with HCl. The reaction is given below.
[tex]NH_3(aq.)+H^+(aq.)\rightarrow NH_4^+(aq.)[/tex]
It took 38.33 mL of 0.0944 M HCl to titrate (react completely with) the ammonia. What is the concentration of the original ammonia solution?
Answer: The concentration of original ammonia solution is 0.0715 M
Explanation:
To calculate the concentration of base, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is HCl
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is [tex]NH_3[/tex]
We are given:
[tex]n_1=1\\M_1=0.0944M\\V_1=38.33mL\\n_2=1\\M_2=?M\\V_2=50.60mL[/tex]
Putting values in above equation, we get:
[tex]1\times 0.0944\times 38.33=1\times M_2\times 50.60\\\\M_2=\frac{1\times 0.0944\times 38.33}{1\times 50.60}=0.0715M[/tex]
Hence, the concentration of original ammonia solution is 0.0715 M
At what speed, in meters per second, does the Moon recoil after the perfectly inelastic collision? The mass of the Moon is 7.36 × 1022 kg?
Answer:The moon will recoil at 1.019×10^-6m/s
Explanation:To determine the recoil velocity of the moon after the inelastic collisions
Given:
Mass of moon=7.36×10^22kg=m1
Mass of asteroids = 5×10^12=m2
Initial velocity of moon =0
Initial velocity of asteroids =15km/s=15000m/s
The law of conservation of momentum is given as
M1v1 + m2v2 =(m1+m2)v'
O + m2v2= (m1 +m2)v'
V'=m2v2/(m1+m2)
V'=(5×10^22)×15000/(7.36×10^22)+(5×10^12)
V'=1.019×10^-6m/s
Answer:
The moon will recoil at the speed of 1.019×10^-6 m/s the whole mass
Explanation:
Mass of moon=7.36×10^22kg=m1
Mass of asteroids = 5×10^12=m2
Initial velocity of moon =0
Initial velocity of asteroids =15km/s=15000m/s
The law of conservation of momentum is given as
M1v1 + m2v2 =(m1+m2)v'
0 + m2v2= (m1 +m2)v'
V'=m2v2/(m1+m2)
V'=(5×10^22)×15000/(7.36×10^22)+(5×10^12)
V'=1.019×10^-6m/s
Please find the attached picture to get the full explanation.
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As a scuba diver descends under water, the pressure increases. At a total air pressure of 2.71 atm and a temperature of 25.0 C, what is the solubility of N2 in a diver's blood?
[Use the value of the Henry's law constant k calculated , 6.26 x 10^{-4} (mol/(L*atm).]
Assume that the composition of the air in the tank is the same as on land and that all of the dissolved nitrogen remains in the blood.
Express your answer with the appropriate units.
Answer:
The molar solubility of nitrogen gas is [tex]1.32\times 10^{-3}mol/L[/tex].
Explanation:
Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.
To calculate the molar solubility, we use the equation given by Henry's law, which is:
[tex]C_{N_2}=K_H\times p_{liquid}[/tex]
where,
[tex]K_H[/tex] = Henry's constant = [tex]6.26\times 10^{-4}mol/L.atm[/tex]
[tex]p_{N_2}[/tex] = partial pressure of nitrogen gas
Total air pressure = P = 2.71 atm
Percentage of nitrogen in air = 78.09%
Mole fraction of nitrogen ,[tex]\hi_{N_2}= 0.7809[/tex]
[tex]p_{N_2}=P\times \chi_{N_2}=2.71 atm\times 0.7809[/tex]
Putting values in above equation, we get:
[tex]C_{N_2}=6.26\times 10^{-4}mol/L.atm\times 2.71 atm\times 0.7809\\\\C_{N_2}=1.32\times 10^{-3}mol/L[/tex]
Hence, the molar solubility of nitrogen gas is [tex]1.32\times 10^{-3}mol/L[/tex].
The solubility of N₂ in a diver's blood at 2.71 atm pressure and 25.0 C temperature is calculated using Henry's law to be 1.323 x 10⁻³ mol/L, assuming a mole fraction of N₂ of 0.78.
The solubility of N₂ in a diver's blood at a total air pressure of 2.71 atm and a temperature of 25.0 C can be determined using Henry's law, which states that the solubility of a gas in a liquid is directly proportional to the partial pressure of that gas above the liquid. In this case, the Henry's law constant k for N₂ is given as 6.26 x 10⁻⁴ mol/(L·atm).
Assuming the composition of the air is the same as on land with a mole fraction of N₂ being 0.78 (from the given standard composition of dry air), the partial pressure of N₂ is 0.78 x 2.71 atm = 2.1138 atm. Applying Henry's law:
Solubility = k × (partial pressure of N₂)
Solubility = 6.26 x 10⁻⁴ mol/(L·atm) × 2.1138 atm
Solubility = 1.323 x 10-3 mol/L
Therefore, the solubility of N₂ in the diver's blood at the given conditions is 1.323 x 10⁻³ mol/L.
Cab someone pls help me? I don't understand this
1. From Part 2 of the lab activity, plot a graph of volume vs. temperature (in kelvins) with the two data points that resulted from your experiment. Draw a straight line connecting the two points, thus assuming a linear relationship between volume and temperature.
The x-intercept corresponds to where the volume would be zero and the temperature can be no lower, defined as absolute zero. Absolute zero is 0 K. Compare your results to those expected. How close to absolute zero was your intercept? Why might your value be different from absolute zero?
Answer:
The activity consists on:
1. Lab activity
During the lab you will measure the volume of a gas at two different temperatures and record the results pairing each volume with its temperature. For instance, in the form of ordered pairs (Temperature, Volume).Let's do an hypothetical case in which you measured an initial temperature of 23ºC and the respective volume of 2 liter: (23, 2) is the first point. Then, you heated the sample of gas up to 80ºC and measured a volume of 2.4 liters.Your lab activity ends.
2. Plot a graph volume vs temperature (in kelvins) with the two data points that resulted from your experiment:
a) First convert the temperatures in ºC to kelvins:
[tex]Kelvin=\ºC+273.15[/tex]
[tex]T_1=23\ºC+273.15=296.15[/tex]
[tex]T_2=80\ºC+273.15=353.15[/tex]
b) Your points, now, are: (296.15, 2) and (353.15, 2.4)
c) Prepare your graph paper
Label the vertical volume in liters (dependent variable)Label the horizonal axis as temperature in kelvins (independent variable)You will not have enough precision to mark the decimals for the temperature; so, use only the integer part of the temperature, i.e. 296 and 353.Select and adequate scale for the temperature, in kelvins. For instance, for the x-axys mark 50, 100, 150, 200, 250, 300, 350, and 400 (you will find that you cannot use marks for 10, 20, 30, 40, 50, 60, 70, 80, 80, ...., 360).Select and adequate scale for the volume in liters. For instance: 0.5, 1.0, 1.5, 2.0, 2.5, 3.00.d) Place the two points (296, 2) and (353, 2.4) at the best your graph paper lets you.
d) Draw a straight line that joins the two points and cross both the vertical and the horizontal axis.
e) Read the point at which the line intercepts the horizontal axis. This is the x-intercept, and is the where the volume would be zero. The temperature cannot be lower, because the volume cannot be negative.
3. Compare your results to those expected.
The point at which the temperature crosses the x-axys should be 0, because that is the absolute zero (0 K).How close to absolute zero was your intercept: sure there will be a difference, whose magnitude and sign depend on several points: i) experimental errors (in the measurements), ii) imprecisions graphing, and iii) real the gases are not perfect gases, thus they do not follow the graph of V vs T is not a perfect line.According to Charle's Law at a fixed pressure, the amount of the given gas will be undeviatingly proportional to the temperature.
In this situation when the pressure is a fixed number then it is said to isobaric condition.
Lab Activity:
Estimate the volume of the given gas at two various temperatures.Now write the recorded volume with its temperature.Let's take hypothetical data over here. For the temperature at 30°C the measured volume is 2 liter, this will be the first point.For the next reading, you heated the sample of gas up to 87 °C and the volume that you measured is 2.4 liters, this will be the second point.Plotting of the graph:
1. To plot the graph between the temperature (kelvins) and the volume, we need to convert the temperature from C to kelvins.
[tex]\text{Kelvin} & = \text{C}^{\text{o}} + 273.15[/tex]
[tex]\text{T1} & = \text{30}^{\text{o}} + 273.15 = 303.15[/tex]
[tex]\text{T2} & = \text{87}^{\text{o}} + 273.15 = 360.15[/tex]
2. The coordinates we get for graphs now are:
(303, 2) and (360, 2.4)
3. Plot the above data on the graph paper.
See the attached image for the graph below.
Mark the vertical column as the volume (liters) and the horizontal axis as temperature (kelvins)Roundoff the temperature up to the integer part for plotting the graph.Select the appropriate scales for temperature and volumes for the graph.Mark the point that you recorded during your experiment.Draw a straight line joining the two points.Mark the point that intercepts the horizontal axis as the intercept. This point will have 0 volume.Absolute 0 will be 0 K as the X intercept shows 0 volume and the temperature can no longer be lower.4. Comparing the recorded results with the expected data: Compare the data on basis of errors, and imprecisions.
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In nature, the element X consists of two naturally occurring isotopes. 107X with abundance 53.84% and isotopic mass 106.9051 amu and 109X with isotopic mass 108.9048 amu. Use the given information to calculate the atomic mass of the element X to an accuracy of .001% (Report your answer like this yyy.yyyy)
The atomic mass of element X is approximately 107.8682 amu with an accuracy of .001%.
To calculate the atomic mass of element X, we need to use the given information about the isotopes of element X.
We are given that element X consists of two naturally occurring isotopes: 107X with an abundance of 53.84% and an isotopic mass of 106.9051 amu, and 109X with an isotopic mass of 108.9048 amu.
To calculate the atomic mass of element X, we can use the formula:
Atomic mass = (abundance of isotope 1 x mass of isotope 1) + (abundance of isotope 2 x mass of isotope 2)
Plugging in the values for element X, we get:
Atomic mass = (0.5384 x 106.9051 amu) + (0.4616 x 108.9048 amu)
Atomic mass ≈ 107.8682 amu
Therefore, the atomic mass of element X is approximately 107.8682 amu to an accuracy of .001%.
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To calculate the atomic mass of element X, multiply the mass of each isotope by its abundance and sum them up. The atomic mass of element X is 106.22288 amu.
Explanation:To calculate the atomic mass of element X, we need to consider the abundance and isotopic mass of its two naturally occurring isotopes. The atomic mass is calculated by multiplying the mass of each isotope by its abundance and summing them up.
For isotope 107X with an abundance of 53.84%, we multiply its mass (106.9051 amu) by its abundance (0.5384). For isotope 109X with an abundance of (100% - 53.84% = 46.16%), we multiply its mass (108.9048 amu) by its abundance (0.4616).
Finally, we add these two values together to get the atomic mass of element X to an accuracy of .001%. The calculation is as follows:
(106.9051 amu * 0.5384) + (108.9048 amu * 0.4616) = 106.22288 amu
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The density of air under ordinary conditions at 25 degrees * C is 1.19g / L . How many kilograms of air are in a room that measures 9.0ft * 11.0ft and has a 10.0 ft ceiling?
Answer:
33.3 kg of air
Explanation:
This is a problem of conversion unit.
Density is mass / volume
Therefore we have to calculate the volume in the room, to be multiply by density. That answer will be the mass of air.
Volume of the room → 9 ft . 11 ft . 10 ft = 990 ft³
Density is in g/L, therefore we have to convert the ft³ to dm³ (1 dm³ = 1L)
990 ft³ . 28.3 dm³ / 1ft³ = 28017 dm³ → 28017 L
This is the volume of the room, if we replace it in the density formula we can know the mass of air in g.
1.19 g/L = Mass of air / 28017 L
Mass of air = 28017 L . 1.19 g/L → 33340 g of air
Finally, let's convert the mass in g to kg → 33340 g . 1kg / 1000 g = 33.3 kg
What happens when you add more solute to a saturated solution
Answer:
If you add more solute in a saturated solution, it will have no effect. Additional solute does not dissolve in a saturated solution.
Explanation:
If more solute is added and it does not dissolve, then the original solution was saturated. If the added solute dissolves, then the original solution was unsaturated. A solution that has been allowed to reach equilibrium but which has extra undissolved solute at the bottom of the container must be saturated.
The element gallium has an atomic weight of 69.7 and consists of two stable isotopes gallium -69 and gallium -71. The isotope gallium -69 has a mass of 68.9 amu and a percent natural abundance of 60.4 %. The isotope gallium -71 has a percent natural abundance of 39.6%. What is the mass of gallium -71?
Answer:
The answer to your question is Gallium-71 = 70.9202 amu
Explanation:
Gallium atomic weight = 69.7
Gallium-69 = 68.9 amu abundance = 60.4%
Gallium-71 = x abundance = 39.6%
To solve this problem just write an equation and solve it for the mass of gallium-71.
Equation
Gallium = Gallium-69(abundance) + Gallium-71(abundance)
Substitution
69.7 = (68.9)(0.604) + Gallium-71(0.396)
69.7 = 41.6156 + Gallium-71(0.396)
Gallium-71(0.396) = 69.7 - 41.6156
Gallium-71(0.396) = 28.0844
Gallium-71 = 28.0844/0.396
Gallium-71 = 70.9202 amu
If the rate of evaporation is equal to the rate of condensation, the system is in a state of dynamic equilibrium, which cannot be disturbed. TRUE FALSE
Answer: False
Explanation:
From definition dynamic equilibrium is a state of balance between continuing processes. When the rate of evaporation equals the rate of condensation the system has reached a dynamic equilibrium.
However, it is possible, to disturb a system that is in dynamics equilibrium by changing conditions of the system. In fact anything that changes the thermodynamic state of the system will disturb the system and it will no longer be at equilibrium.
An increase in temperature to the system will favour evaporation more than condensation. Molecules with higher kinetic energy will escape the system.
________ is defined as the number of protons plus the number of neutrons. ________ is defined as the number of protons plus the number of neutrons. Atomic number Isotopic number Ionic number Mass number Nuclear number
Answer:
Mass number is defined as the number of protons plus the number of neutrons.
Explanation:
Every atom of an element have proton(s), neutron(s) and electron(s). The proton number of an element is the atomic number of that element. For an electronically neutral atom the proton number is equal to the electron numbers. The neutron and the proton contributes to the mass of every atom. The electron is more active when atoms are bonding.
Mass number of an element is the number of proton plus the number of neutron.
Atomic number of an atom is the number of proton present, so it can never be atomic number.
Isotopy talks about same element having different number of neutron but same number of protons in each atom. Example is hydrogen that exist as protium, deuterium and tritium. It cannot be isotopic number.
Ionic number talks about elements that possess a charge. The elements have been ionized.
The answer is Mass number because the sum of proton number and neutron number is equals to mass number.
Mass number is defined as the number of protons plus the number of neutrons in an atom's nucleus. Atomic number is defined as the number of protons in an atom's nucleus.
The term "mass number" is defined as the number of protons plus the number of neutrons in the nucleus of an atom. It provides the total mass of the atomic nucleus and is represented by the symbol "A."On the other hand, the term "atomic number" (also known as the "proton number") is defined as the number of protons in the nucleus of an atom.It is a fundamental property of an element and determines its identity, chemical behavior, and its position in the periodic table. The atomic number is typically denoted by the symbol "Z."To clarify further:Atomic Number (Z): Defines the unique identity of an element, as each element has a distinct number of protons in its nucleus.Mass Number (A): Represents the total number of nucleons (protons and neutrons) in the atomic nucleus. It is used to differentiate between isotopes of the same element.The terms "isotopic number," "ionic number," and "nuclear number" are not standard terminology in chemistry or physics. The key concepts used to describe atomic structure are the atomic number (Z) and the mass number (A).For more such questions on protons
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A uniform, solid, 1100.0 kg sphere has a radius of 5.00 m. Find the gravitational force this sphere exerts on a 2.40 kgkg point mass placed at the following distances from the center of the sphere: (a) 5.01 mm , and (b) 2.50 mm .
Explanation:
(a) The given data is as follows.
M = 1100 kg, r = 5.01 mm = [tex]5.01 \times 10^{-3}[/tex] m
m = 2.40 kg
Now, formula for Newton's law of gravitation is as follows.
F = [tex]\frac{GMm}{r^{2}}[/tex]
Putting the given values into the above formula as follows.
F = [tex]\frac{GMm}{r^{2}}[/tex]
= [tex]\frac{6.673 \times 10^{-11} Nm^{2}/kg^{2} \times 1100 \times 2.40 kg}{(5.01 \times 10^{-3})^{2}}[/tex]
= [tex]7.02 \times 10^{-3}[/tex] N
Therefore, gravitational force exerted by the given sphere on 2.40 kg point mass is [tex]7.02 \times 10^{-3}[/tex] N.
(b) Now, mass of the sphere beyond r = 2.50 mm or [tex]2.50 \times 10^{-3} m[/tex] will not contribute any force on the point mass.
Hence, mass within the radius r < R that experiences force is,
[tex]m^{1} = M \frac{r^{3}}{R^{3}}[/tex]
According to Newton's law of gravitation,
F = [tex]\frac{Gm^{1}m}{r^{2}}[/tex]
= [tex]\frac{Gm(M \frac{r^{3}}{R^{3}})}{r^{2}}[/tex]
= [tex]G \frac{mMr}{R^{3}}[/tex]
Here, r is the radius of point mass and R is the radius of solid sphere.
Therefore, putting the given values into the above formula as follows.
F = [tex]G \frac{mMr}{R^{3}}[/tex]
= [tex]\frac{6.673 \times 10^{-11} Nm^{2}/kg^{2} \times 1100 kg \times 2.40 kg \times 2.50 \times 10^{-3}}{(5)^{3}}[/tex]
= [tex]3.52 \times 10^{-9}[/tex] N
Therefore, the gravitational force this sphere exerts on a 2.40 kg is [tex]3.52 \times 10^{-9}[/tex] N.
Boston, MA and Barcelona, Spain have very similar latitudes (they are both a similar distance from the equator.) Why is the climate of Boston so different from the climate of Barcelona?
Answer:
The climate of Boston is so different from the climate of Barcelona because regional climate does not depend only on latitude. The flow of ocean current and its temperature causes this different climates. East coast cities such as Boston have a continental climate because prevailing westerly winds continually bring them under the influence of continental air masses while westerlies bring maritime air masses and therefore a maritime climate to the coast cities such as Barcelona and others.
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Which of the following is true of greenhouse gases?
They exist in fixed quantities.
They reflect incoming solar radiation.
They trap energy in the atmosphere.
They are all naturally occurring.
Greenhouse gases trap energy in the atmosphere.
Explanation:
Greenhouses gases trap the radiation, especially long infrared wavelengths, that emanate from the earth's surface after solar radiation hits the earth’s surface. These infrared wavelengths are electromagnetic radiation that are significant in heat transfer. Mostly a higher percentage of this radiation from the earth’s surface escapes into space. Trapping this radiation in the atmosphere causes warming of the atmosphere - a phenomenon called greenhouse effect. This is why an increase in greenhouse gases causes an increase in global temperatures and subsequently drives climate change.
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An element has four naturally occurring isotopes with the masses and natural abundances given in the table below. Find the atomic mass of the element (express your answer to four significant figures and include the appropriate units).
Isotope Mass (amu) Abundance (%)
1 203.97304 1.390
2 205.97447 24.11
3 206.97590 22.09
4 207.97665 52.41
Identify the element, by spelling out its full name.
Answer:
Lead (Pb)
Explanation:
To find the atomic mass of the element, we need to take into consideration all the naturally occurring isotopes. We then find the atomic mass by multiplying the natural abundance of the isotopes by their mass for all of the isotopes and summing them together.
1. 1.390/100 * 203.97304 = 2.835225256
2. 24.11/100 * 205.97447 = 49.660444717
3. 22.09/100 * 206.97590 = 45.72097631
4. 52.41/100 * 207.97665 = 109.000562265
We then add all of these masses together:
109.000562265 + 45.72097631+ 49.660444717 + 2.835225256 = 207.217208548
To 4 sf = 207.2 amu
Element is Lead (Pb)
The following equation lnN - N₀ = -kt Of time t is plotted on x-axis and lnN is plotted on y-axis, what are the slopes and y-intercept algebraically? Hint: lnN = lnN - LnN₀
Answer:
-k slope
The term No is the y intercept(does not mean NO) y-intercept
Explanation:
For a straight line graph, the normal equation could be written as
y = mx + c
Where m is the slope and c is the y intercept.
Writing the equation is the form of y = mx + c gives the following;
lnN = -kt + No
As obtained from the question, t is plotted in the x-axis, while lnN is the plot on the y-axis.
Now, we are asked to obtain the slope and the y-intercept. It can be seen that the slope is the coefficient of the term x in this case t. Hence, our slope is -k.
The y-intercept is the other term which in this case is No
if you begin with 210.3 g of Cl2, how many grams of HCl will you end up with? CH4 + 3Cl2 → CHCl3 + 3HCl *
Answer:
108.04g
Explanation:
Looking at the balanced chemical equation theoretically, we can see that 3 moles of chlorine yielded 3 moles of HCl. This means that they have equal mole ratio.
Now let’s get the actual reactive moles. The number of moles is obtained by dividing the mass by the molar mass.
The molecular mass of chlorine gas is (2 * 35.5) = 72g/mol
The number of moles is thus 210.3/71 = 2.96 moles
Since the number of moles are equal, the number of moles of HCl produced too is 2.96 moles.
Now to get the mass of HCl produced, we multiply the number of moles by the molar mass of HCl. The molar mass of HCl is 36.5g/mol
The mass is thus 2.96 * 36.5 = 108.04g
Which of the following is true about a municipal bond with a put option? (A) An investor will exercise the option to put the bond if yields rise significantly. (B) An investor must have the issuer's permission to put the bond. (C) The market price of bonds is never affected by a put option feature. (D) Yields are usually higher for a new issue bond with a put option than for a new issue bond without a put option.
Answer:
(A) An investor will exercise the option to put the bond if yields rise significantly
Explanation:
A put option on the bond is a mechanism to allow the buyer of the bond the ability to compel the lender to repay the principal on the bond. The put option offers the buyer of the bond the ability to collect the principal of the bond anytime they choose until maturity for any purpose.
Recall that once the price drops (that is, the yield increases), put options are exercised. If the yield significantly increased, the put choice on a municipal bond is executed.
A person doing a chin-up weighs 535 N, exclusive of the arms. During the first 25.4 cm of the lift, each arm exerts an upward force of 394 N on the torso. The acceleration of gravity is 9.8 m/s 2 . If the upward movement starts from rest, what is the person’s velocity at this point? Answer in units of m/s.
Answer:
person’s velocity at this point = 1.535m/s
Explanation:
The detailed steps and appropriate substitution is as shown in the attachment.
A concrete pile from a waterfront pier was pulled from the harbor water at port hueneme california. It's dimensions we're 14in x 14 in x 15in. a. What is the surface area expressed in square feet of a single face of the pile
The surface area expressed in square feet of a single face of a pile is 1.45 square feet.
Explanation:
The dimensions of the pillar are given as 14 in [tex]\times[/tex] 14 in
(1 feet = 12 inch)
According to the pillar dimensions mentioned in the problem, the length of the pillar should be 15 inches, height by 14 inches and the width by 14 inches.
There will be two surface areas of the concrete pile, one implies side surface area and the other is end surface area.
For calculating the side surface area,
Area of the side surface = 14 in [tex]\times[/tex] 15 in
= [tex]14 in.[/tex] [tex]\times[/tex] [tex]\frac{1 ft.}{12 in.}[/tex]
= 1.16 [tex]\times[/tex] 1.25
= 1.45 square feet.
(In the given question, only a single face is asked so there is no need to find the surface area of the end face.)
The surface area of the side face of the given pile is 1.45 ft². The surface area is the area of all surfaces in a 3D object such as a pillar, ball, etc.
What is surface area?The surface area is the area of all surfaces in a 3D object such as a pillar, ball, etc.
A pillar has two types of faces, side face and end phase with dimensions 14x14 and 14x15.
The dimentions of the pile,
Length - 14 inches = 1.16 ft
Width - 14 inches = 1.16 ft
height - 15 inches = 1.25 ft.
So, the surface area of a side face will be
As = 1.16 ft x 1.25 ft
As = 1.45 ft²
Therefore, the surface area of the side face of the given pile is 1.45 ft².
Learn more about the Surface area:
https://brainly.com/question/24720806
If an oxygen nucleus consists of eight protons and eight neutrons, the charge on that nucleus is positive. Since even I learned in high school that like charges repel, such a nucleus would find all its positive protons repelling and quickly fall apart." How would you answer his argument?
Answer:
The binding energy present in the atomic nucleus that holds the protons and the neutrons together and its magnitude is one million times stronger than the electron binding energy in small atoms
Explanation:
The minimum required force to dismember an atomic nucleus into its constituent components, of protons and nucleus (collectively called nucleons) in known as the nuclear binding energy.
Energy is required in separating the nucleons hence the binding energy of a nucleus is always positive
According to Einstein's Energy and light relation E = mc², when a nucleus is formed from the number of free protons and neutrons, the sum of their individual masses is more than the mass of the formed atomic nucleus. The mass deficit of the neutron, also known as the 'missing mass' or mass defect indicates the amount of energy released in forming of the nucleus which therefore has different characteristics from its constituents as mentioned above
The amount of mass that is equivalent to the binding energy of the nucleus as shown in the Einstein's equation (E=mc²) is represented by the missing mass or mass defect of the formed nucleus or the difference in mass between the nuclear mass and that of the sum of the individual masses of its constituent protons and neutrons
Given the unbalanced equation: N2(g) + H2(g) → NH3(g)
When the equation is balanced using the smallest whole-number coefficients, the ratio of moles of hydrogen consumed to moles of ammonia produced is
1 ) 1:3
2) 2:3
3) 3:1
4) 3:2
Answer:
Option-4 (3:2) is the correct answer.
Explanation:
Following steps are taken to balance the given unbalanced chemical equation.
Step 1: Write the unbalanced chemical equation,
N₂ + H₂ → NH₃
Step 2: Balance Nitrogen Atoms;
There are 2 nitrogen atoms on left hand side and 1 nitrogen atoms on right hand site therefore, to balance them multiply NH₃ on right hand side by 2 i.e.
N₂ + H₂ → 2 NH₃
Step 3: Balance Hydrogen Atoms;
Now, there are 2 hydrogen atoms on left hand side and 6 hydrogen atom on right hand site therefore, to balance them multiply H₂ on left hand side by 3 i.e.
N₂ + 3 H₂ → 2 NH₃
Now, the equation is balanced.
Step 4: Finding out mole ratios:
From balanced chemical equation it can be concluded that 3 moles of H₂ are involved in producing 2 moles of NH₃ hence, the mole ratio of consumption of H₂ to production of NH₃ is 3:2.
Answer:
4) 3:2
Explanation:
Castle learning
Dont mind the tiny tab
Nitrogen dioxide (NO2) cannot be obtained in a pure form in the gas phase because it exists as a mixture of NO2 and N2O4. At 26°C and 0.80 atm, the density of this gas mixture is 2.2 g·L−1. What is the partial pressure of each gas?
Answer: The partial pressure of [tex]NO_2[/tex] is 0.426 atm and that of [tex]N_2O_4[/tex] is 0.374 atm
Explanation:
Assuming ideal gas behavior, the equation follows:
PV = nRT
We know that:
[tex]\text{Density}=\frac{\text{Mass}}{\text{Volume}}[/tex]
Rearranging the above equation:
[tex]M=\frac{dRT}{P}[/tex]
where,
d = density of gas mixture = 2.2 g/L
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature = [tex]26^oC=[26+273]K=299K[/tex]
P = pressure of the mixture = 0.80 atm
M = average molar mass of mixture
Putting values in above equation, we get:
[tex]M_{avg}=\frac{2.2g/L\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 299}{0.80atm}\\\\M_{avg}=67.5g/mol[/tex]
We know that:
Molar mass of [tex]NO_2[/tex] = 46 g/mol
Molar mass of [tex]N_2O_4[/tex] = 92 g/mol
Let the mole fraction of [tex]NO_2[/tex] be 'x' and that of [tex]N_2O_4[/tex] be '(1-x)'
For average molar mass calculation:
[tex]M_{avg}=M_{NO_2}\chi_{NO_2}+M_{N_2O_4}\chi_{N_2O_4}[/tex]
Putting values in above equation:
[tex]67.5=46x+92(1-x)\\\\x=0.533[/tex]
Mole fraction of [tex]N_2O_4[/tex] = (1 - x) = (1 - 0.533) = 0.467
To calculate the partial pressure, we use the equation given by Raoult's law, which is:
[tex]p_{A}=p_T\times \chi_{A}[/tex]
For [tex]NO_2[/tex] :We are given:
[tex]p_T=0.80atm\\\chi_{NO_2}=0.533[/tex]
Putting values in above equation, we get:
[tex]p_{NO_2}=0.80atm\times 0.533\\\\p_{NO_2}=0.426atm[/tex]
For [tex]N_2O_4[/tex] :We are given:
[tex]p_T=0.80atm\\\chi_{N_2O_4}=0.467[/tex]
Putting values in above equation, we get:
[tex]p_{N_2O_4}=0.80atm\times 0.467\\\\p_{N_2O_4}=0.374atm[/tex]
Hence, the partial pressure of [tex]NO_2[/tex] is 0.426 atm and that of [tex]N_2O_4[/tex] is 0.374 atm
Suppose a 1.30 g nugget of pure gold has zero net charge. What would be its net charge after it has 1.68% of its electrons removed?
Answer:
The net charge of 1.3 g nugget of pure gold after 1.68% of its electrons are removed is 559 C
Explanation:
When an atom gains electrons it becomes negatively charged. Conversely, when it looses electrons the atoms becomes positively charged thus
To solve this question, we rely on the relationship between the nmber of particles present in a given mass of an atom, Avogadro's number and number of moles, n
The given variables are
mass of pure gold nugget = 1.30 g
Quantity of electrons removed = 1.68% of electrons present in the gold sample
Molar mass of gold = 197 g/mol
Avogadro's number = 6.02 × 10²³ atoms/mole
qc = one electron charge = -1.06 × 10⁻¹⁹ C/electron
Electrical charge of gold nugget = 0 C
Number of electrons in one gold atom = 79 electrons
Solving for the number of prticles or gold atoms in 1.3 grams of gold we get
n mass/(molar mass) = 1.3/197 moles of gold = 0.0066 moles
number of particles in 0.0066 moles of gold N = n×[tex]N_{A}[/tex] = 0.0066 × 6.02 × 10²³ = 3.97 × 10²¹ atoms
since 79 electrons are present per particle we have
3.97 × 10²¹ × 79 = 3.14 × 10²³ electrons
quantity of elecrtrons removed = 1.68% of 3.14 × 10²³ electrons =1.68/100 × 3.14 × 10²³ electrons = 0.0168 × 3.14 × 10²³ electrons = 5.3 × 10²¹ electrons
The net charge of 5.3 × 10²¹ electrons = 5.3 × 10²¹ electrons × -1.06 × 10⁻¹⁹ C/electron =
5.59 × 10² C = 559 C
Final answer:
After 1.68% of its electrons are removed, a 1.30 g nugget of pure gold will have a net positive charge of approximately +842 C, calculated based on the initial number of electrons and the charge per electron.
Explanation:
To calculate the net charge of a 1.30 g nugget of pure gold after 1.68% of its electrons are removed, we first need to understand the composition and characteristics of gold. The atomic mass of gold (Au) is approximately 197 g/mol, and each atom has 79 protons and, when neutral, 79 electrons. The number of gold atoms in the nugget can be calculated using its mass and the atomic mass of gold.
First, calculate the number of moles of gold in the nugget:
Moles of gold =
(1.30 g) / (197 g/mol) = 0.0066 mol. The number of gold atoms = 0.0066 mol ×
(6.022 × 1023 atoms/mol) ≈ 3.97 × 1021 atoms. Since each gold atom has 79 electrons when neutral, the total number of electrons initially is approximately 3.13 × 1023 electrons.
To find out how many electrons 1.68% represents:
1.68% of the total electrons = 1.68 / 100 × 3.13 × 1023 ≈ 5.26 × 1021 electrons.
The charge of one electron is approximately -1.6 × 10-19 C. Therefore, removing 5.26 × 1021 electrons means the nugget would have a net positive charge equal to:
5.26 × 1021 × 1.6 × 10-19 C ≈ 8.42 × 102 C. Thus, the net charge of the gold nugget after removing 1.68% of its electrons would be approximately +842 C.
A. After the label fell off a bottle containing a clear liquid believed to be benzene, a chemist measured the density of the liquid to verify its identity. A 25.0-mL portion of the liquid had a mass of 21.95 gg. A chemistry handbook lists the density of benzene at 15∘C∘C as 0.878 g/mg/mL. Is the calculated density in agreement with the tabulated value?
B. An experiment requires 15.0 g of cyclohexane, whose density at 25oC is 0.7781 g/mL. What volume of cyclohexane should be used?
C. A spherical ball of lead has a diameter of 5.0 cm. What is the mass of the sphere if lead has a density of 11.34 g/cm3?
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.
At − 12.0 ∘ C , a common temperature for household freezers, what is the maximum mass of sorbitol (C6H14O6) you can add to 2.00 kg of pure water and still have the solution freeze? Assume that sorbitol is a molecular solid and does not ionize when it dissolves in water.
Answer:
2,347.8 grams
Explanation:
The freezing point depression Kf of water = 1.86° C / molal.
To still freeze at -12° C, then the molality of the solution 12/ 1.86 = 6.45 moles
The molecular weight of sorbitol (C6H14O6)is:
6 C = 6 ×12 = 72
14 H = 14 × 1 = 14
6 O = 6 × 16 = 96
...giving a total of 182
So one mole of sorbitol has a mass of 182 grams.
Since there are 2 kg of water, 2 × 6.45 moles = 12.9 moles can be added to the water to get the 12° C freezing point depression.
Therefore
grams = moles × molar mass
12.9 moles × 182 grams / mole = 2,347.8 grams of sorbitol can be added and still freeze