Answer:
The nswer to the question is
The maximum fraction of the air in the room that could be displaced by the gaseous nitrogen is 0.548 or 54.8 %
Explanation:
To solve the question we note that
The density of the liquid nitrogen = 0.808g/mL and the volume is 195 L tank (vaporised)
Therefore since density = mass/volume we have
mass = Density × volume = 0.808 g/mL × 195 L × 1000 ml/L =157560 g
In gaseous form the liquid nitrogen density =1.15 g/L
That is density = mass/volume and volume = mass/density = 157560 g/(1.15g/L) or
volume = 137008.69565 L
The dimension of the room = 10 m × 10 m × 2.5 m = 250 m³ and
1 m³ is equivalent to 1000 L, therefore 250 m³ = 250 m³ × 1000 L/m³ = 250000L
Therefore fraction of the volume occupied by the gaseous nitrogen =
137008.69565 L/250000 L = 0.548
Therefore the gaseous nitrogen occpies 54.8% of the room
To find the maximum fraction of air displaced by vaporized nitrogen in a room, calculate the mass of liquid nitrogen, convert it to gaseous volume, and compare it to the room's volume.
Explanation:When considering the accidental vaporization of 195 liters of liquid nitrogen in a closed room, we need to calculate the volume of the room and determine what fraction of the room's air could potentially be displaced by the nitrogen gas. To do this, we will use the given room dimensions and the densities of liquid and gaseous nitrogen. The room's volume is 10.00 m x 10.00 m x 2.50 m, which equals 250 cubic meters or 250,000 liters. The density of liquid nitrogen is 0.808 g/mL, and it converts to 0.808 kg/L since there are 1000 milliliters in a liter. Multiplying the density by the total volume of liquid nitrogen gives us the mass in kilograms, which is then converted to the volume of gaseous nitrogen at standard temperature and pressure using the given density of 1.15 g/L.
Now, converting the mass to volume for gaseous nitrogen, we can find out the fraction of room's air displaced by using the ratio of nitrogen gas volume over the room's total volume. This is an application of the ideal gas law where a given mass of gas occupies different volumes based on its phase (liquid or gas) and other conditions such as temperature and pressure.
State the number of solutions for Matrix B. a. No Solution b. One Solution c. Infinitely Many Solutions
Answer:
C. Infinitely Many Solutions
Explanation:
No solution case : This is the case when all given variables are not equal to any constant, for example: there is one row of zeros in matrix e.g 0=3. matrix B don't have any zero row. So, Not True.
One Solution case: This is the case when all variables are independent variables like if they are equal to some constant. e.g x=1,y=2,z=4 , Matrix B have more than one variable in first row due to which it made equation look like x+y=-5. so matrix B can't have only one solution. So, Not True.
Infinitely Many Solutions case: when there is one or more variables which is not equal to any constant and acting as linearly dependent variable, then that matrix have infinite solutions. Matrix B have that variable which is linearly dependent as show in the attachment solution. So, True.
The atomic mass of sulfur is 32.1 amu, and the atomic mass of oxygen is 16.0 amu. To the nearest tenth of a percent, what is the percent by mass of sulfur in sulfur trioxide (SO3)?
Answer:
[tex]\%\ Composition\ of\ sulfur=40.1\ \%[/tex]
Explanation:
Percent composition is percentage by the mass of element present in the compound.
Given , Mass of sulfur= 32.1 amu
Mass of oxygen = 16.0 amu
Mass of sulfur trioxide [tex]SO_3[/tex] = 32.1 amu + 3*16.0 amu = 80.1 amu
[tex]\%\ Composition\ of\ sulfur=\frac{Mass_{sulfur}}{Mass_{SO_3}}\times 100[/tex]
[tex]\%\ Composition\ of\ sulfur=\frac{32.1\ amu}{80.1\ amu}\times 100=40.1\ \%[/tex]
Final answer:
To determine the percent by mass of sulfur in sulfur trioxide (SO3), calculate the molar mass of the compound and divide the atomic mass of sulfur by the total molar mass. The percent by mass of sulfur in SO3 is found to be 40.0%.
Explanation:
To find the percent by mass of sulfur in sulfur trioxide (SO3), we first calculate the molar mass of the compound. We have one sulfur atom and three oxygen atoms in SO3.
Molar mass of S = 32.07 amuMolar mass of O = 16.00 amu (for one oxygen atom)Next, we calculate the molar mass of SO3:
Molar mass of SO3 = Molar mass of S + 3 × Molar mass of O
= 32.07 amu + 3 × 16.00 amu
= 32.07 amu + 48.00 amu
= 80.07 amu
To find the percent composition of sulfur, we divide the molar mass of sulfur by the total molar mass of the compound and multiply by 100:
Percent by mass of S in SO3 = (Molar mass of S / Molar mass of SO3) × 100%
= (32.07 amu / 80.07 amu) × 100%
= 40.0% (to the nearest tenth of a percent)
When a variable is determined by a factor outside of the function or model being evaluated, it is said to be A. statistically insignificant. B. exogenous. C. endogenous. D. unexplained.
Answer:
The correct answer is B. exogenous
Explanation:
Let us try to describe exogenous and endogenous variables an exogenous variable value is influenced only by factors outside a model or system and is forced onto the model, while a change in an exogenous variable is known as an exogenous change. Also an endogenous variable is one whose value is influenced only by the system or model under study.
Which of the following sites has volcanoes related to subduction? A is on an island arc, B is in the Andes, C is near Hawaii, D is along a mid-ocean ridge.
Answer:
A is on an island arc,
B is in the Andes
Explanation:
The subduction region is referred to as the subduction zone. When a plate slips underneath another, it forms a trench. Earthquakes resulting from the plate panelling often result in magma spilling across the underwater volcano trench. Island arcs are vast chains of active volcanoes with strong seismicity located along the converging tectonic plate borders. Much of the islands arcs emerge from the oceanic crust and also have originated from fall of the lithosphere into the mantle across the subduction zone. In the Andes mountain range, Under the South American plate the Nazca plate is subducting.
The sites has volcanoes related to subduction is on an island arc and is in the Andes.
A subduction zoneA subduction zone forms when continental crust and oceanic crust collide.
Subduction zones are located in the Pacific Ocean.
The subduction zone is at the western margin of South America (Andes).
Island arcs have active volcanoes found along convergent tectonic plate boundaries.
Thus option A and B are correct answer.
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The physicals properties of a substance can be used to identify the substance, because_____. Select all that apply.
1. at any given location, the physical properties of a substance do not change
2. when the substances are chemically reacted the physical properties remain the same
3. each substance has a unique set of physical properties
4. no two substances have any of the same properties
Answer:
1. at any given location, the physical properties of a substance do not change.
2. Each substance has a unique set of physical properties
Explanation:
The physical property of a substance is a feature of a substance that can be noticed or measured without changing the key identity of the substance. Physical properties include density, hardness, melting point, colour etc...
Answer:
1. at any given location, the physical properties of a substance do not change
3. each substance has a unique set of physical properties
Explanation:
1. The physical properties of a substance do not change provided that the physical conditions in the niche environment remain same.
3. Every substance on Earth has a unique set and arrangement of atoms and molecules that give them their own unique properties.
A catalyst can increase the rate of a reaction ________. Select one: a. by increasing the overall activation energy (Ea) of the reaction b. by providing an alternative pathway with a lower activation energy c. by lowering the activation energy of the reverse reaction d. by changing the value of the frequency factor (A) e. All of these are ways that a catalyst might act to increase the rate of reaction.
Catalysts increase the reaction rate by providing an alternative pathway with lower activation energy. They don't alter other energy factors.
Explanation:In a chemical reaction, a catalyst increases the rate of the reaction by providing an alternative pathway with a lower activation energy. This option 'b' is correct. The activation energy is the minimum amount of energy required for a reaction to occur. By reducing this energy, catalysts speed up the reaction process without being consumed in the reaction itself. They don't alter the overall energy (option 'a'), they don't lower the energy of the reverse reaction (option 'c'), and they do not change the value of the frequency factor ('option 'd'). Hence, answer 'e' is also not correct.
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At Mountain X, you could boil water in less than 3 minutes. At Mountain Y, it takes you almost 7 minutes to boil the same amount of water. What can be said about the elevations of both mountains?
A. The elevation of both mountains is the same.
B. Mountain X has a higher elevation than Mountain Y.
C. Mountain X has a lower elevation than Mountain Y.
D. Cannot determine from the information given
Answer:
The answer to your question is letter B.
Explanation:
Water boils at sea level at 100°C at higher heights the boiling point diminishes. That means that at higher heights the time needed to boil water will be lower than at sea level.
In this problem at mountain B is only 3 minutes needed to boil water that means that the height is higher here.
Answer:
The correct answer is B. Mountain X has a higher elevation than Mountain Y
Explanation:
The boiling point of a liquid reduces as the altitude where the boling takes place increases. With every 500 ft increase in altitude the boiling point of water devreses by 0.5 °C as such the boiling point of water at 8000 ft is just 92 °C. To compenasate cooking at altidude require the use of a pressure cooker as it is hard to boil items such as potatoes at very high altitudes of thousands of feet
Neils Bohr published his model of the atomic structure in 1913. He introduced the theory of electrons traveling in orbits around the atom's nucleus, with electrons orbiting only in certain orbits called shells. Each shell can contain a fixed number of electrons.
According to this model of the atom, the chemical properties of an element are mainly determined by ________.
According to this model of the atom, the chemical properties of an element are mainly determined by arrangement and distribution of electrons.
According to Niels Bohr's atomic theory, the distribution and arrangement of an element's electrons in its outermost shell, commonly known as the valence shell, largely determines its chemical characteristics.
An element's capacity to form bonds and interact with other elements to form compounds depends on the number of electrons in its valence shell and how quickly they may be obtained, lost, or shared during chemical processes.
Thus, key factor in determining an element's reactivity and place in the periodic table, as well as its chemical behavior and properties, are the valence electrons.
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A 8.6 g rock sample is added to a graduated cylinder filled with water, causing the level to increase from 12.8 mL to 13.6 mL. What is the density of the rock sample?
The density of the rock sample is 10.75 g/cm³.
Explanation:The density of an object can be calculated using the formula:
Density = mass/volume
In this case, we have the mass of the rock sample (8.6 g) and the change in volume of the water in the graduated cylinder (13.6 mL - 12.8 mL = 0.8 mL). To calculate the density, we need to convert the volume from mL to cm³, since the mass is in grams.
1 mL = 1 cm³, so the volume of the rock sample in cm³ is 0.8 cm³.
Now, we can calculate the density:
Density = mass/volume
Density = 8.6 g / 0.8 cm³
Density = 10.75 g/cm³
Therefore, the density of the rock sample is 10.75 g/cm³.
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The density of the rock sample is found to be 10.75 g/mL by dividing its mass, 8.6 g, by the volume of water it displaced, which is 0.8 mL.
To determine the density of the rock sample, we use the formula density = mass/volume. The mass of the rock is given as 8.6 g. The volume displaced by the rock is the change in the water level in the graduated cylinder: 13.6 mL - 12.8 mL = 0.8 mL.
Now, we can calculate the rock's density:
Density = Mass / Volume
Density = 8.6 g / 0.8 mL
Density = 10.75 g/mL
Therefore, the density of the rock sample is 10.75 g/mL.
When aqueous solutions of lead(II) nitrate and potassium iodide are mixed, what is the formula for the insoluble solid (precipitate) that forms?
Answer:
The formula for the precipitate is PbI₂
Explanation:
The chemical reaction is:
Pb(NO₃)₂ (aq) + 2KI (aq) → 2KNO₃ (aq) + PbI₂ (s)↓
All the iodides, with these cations (Ag⁺, Pb²⁺, Cu⁺, Hg²⁺, Bi⁺³ and Sn⁺⁴) produce insolubles solid.
Which states of matter can flow?
1) gas and liquid
2) gas
3) gas, liquid, and solids
4) liquid
Answer:
1) Gas and Liquid
Explanation:
hope it helps
How many unhybridized p atomic orbitals are present when a central atom exhibits trigonal planar geometry?
Answer:
1
Explanation:
Trigonal planar geometry is shown by the compounds where hybridization of central atom is [tex]sp^2[/tex].
In [tex]sp^2[/tex] hybridization, three hybrid orbitals are equally spaced at an angle of 120°.
Some of the compounds having [tex]sp^2[/tex] hybridization are [tex]CO_3^{2-}[/tex], [tex]SO_3[/tex], etc
In [tex]sp^2[/tex] hybridizationm, one s-orbital and 2 p-orbitals are involved.
Total no. of orbitals present in p-subshell is 3.
As 2 is involved in [tex]sp^2[/tex] hybridization, therefore no. of unhybridized orbital in [tex]sp^2[/tex] hybridization is 1.
You conduct an experiment in which you measure the temperature (T) and volume (V) of a mysterious sphere of gas at several different temperatures. The sphere can expand or shrink to any volume without changing the pressure of the gas inside. As a result of your experiment, you find that the volume (V) of the gas divided by its temperature (T) is always equal to 1.75. What would the volume (V) of the gas be at a temperature of 300K? (assume the unit for volume is liters) Show your work and answer the question with a clear statement. Show each step of your calculations for full points. After showing your calculations, answer the question with a clear statement.
Volume of the gas at 300K will be 525l itres.
Explanation:According to Charles' law, the volume of a gas is inversely proportional to its absolute temperature when the pressure of the gas is constant.
Here the V/T is given to be constant and equal to 1.75.
The temperature of the gas = T = 300K.
Let the volume of the gas be V liters.
So according to the question,
V / T = 1.75.
So,V / 300 = 1.75.
So, V = 300 × 1.75.
Or, V = 525 litres.
So, the volume of the gas at 300K will be 525 litres.
The Volume of the gas at 300K will be 525l liters.
Charles law:As per Charles' law, the volume of gas should be inversely proportional to the absolute temperature at the time when the pressure of the gas is constant.
Calculation of the volume of the gas:Here the V/T should be constant and equivalent to 1.75.
The temperature of the gas = T = 300K.
Also, we assume that the volume of the gas is V liters.
So,
V / T = 1.75.
V / 300 = 1.75.
V = 300 × 1.75.
V = 525 litres.
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What is the osmotic pressure of a solution formed by dissolving 44.3 mg of aspirin (C9H8O4) in 0.358 L of water at 25 ∘C?
The osmotic pressure of a solution formed by dissolving 44.3 mg of aspirin in 0.358 L of water at 25 ∘C is 0.01633 atm. This is calculated using the formula for osmotic pressure 'Pi = n/V RT', inserting the required values including the mole of aspirin, the volume, the gas constant, and temperature.
Explanation:To calculate the osmotic pressure, we need to use the formula Pi = n/V RT, where 'n' is the number of moles of the solute, 'V' is the volume in liters, 'R' is the gas constant (0.08206 L atm/mol K), and 'T' is the temperature in Kelvin. Firstly, we need to find the molar mass of aspirin (C9H8O4) which is approximately 180.16 g/mol. Consequently, we can determine the mole of aspirin used as 44.3 mg / 180.16 g/mol = 0.000246 moles. Now knowing all values, plug them into the formula: Pi = 0.000246 moles /0.358L * 0.08206 atm mol^-1K^-1 * 298.15K which equals 0.01633 atm, the osmotic pressure of the solution.
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For the osmotic pressure of an aspirin solution, first convert the mass of aspirin to moles, find the molarity, convert the temperature to Kelvin, and then apply the osmotic pressure formula. The osmotic pressure is found to be approximately 0.0168 atm.
The question asks to determine the osmotic pressure of a solution containing aspirin at a certain temperature. The osmotic pressure can be calculated using the formula:
\(\Pi = MRT\)
Where:
\(\Pi\) is the osmotic pressure,
\(M\) is the molarity of the solution (moles of solute per liter of solution),
\(R\) is the gas constant (0.0821 L atm K^{-1} mol^{-1}), and
\(T\) is the temperature in Kelvin.
First, convert the mass of aspirin to moles using the molecular weight of aspirin (C9H8O4), which is 180.16 g/mol:
\(44.3 mg = 0.0443 g\)
\(0.0443 g \times \dfrac{1 mol}{180.16 g} \approx 2.46 \times 10^{-4} mol\)
Next, find the molarity \(M\) by dividing the number of moles by the volume in liters:
\(M = \dfrac{2.46 \times 10^{-4} mol}{0.358 L} \approx 6.87 \times 10^{-4} M\)
Now convert the temperature to Kelvin:
\(T = 25 \degree C + 273.15 = 298.15 K\)
Finally, calculate the osmotic pressure:
\(\Pi = (6.87 \times 10^{-4} M)(0.0821 L atm K^{-1} mol^{-1})(298.15 K)\)
\(\Pi \approx 0.0168 atm\)
Therefore, the osmotic pressure of the aspirin solution is approximately 0.0168 atm.
Which type of intermolecular attractive force operates between (part A) all molecules, (part B) polar molecules, (part C) the hydrogen atom of a polar bond and a nearby small electronegative atom
Answer:
PART A: The LDF occurs between all molecules. Dispersion forces result from shifting electron clouds, which cause weak, temporary dipole.
PART B: Dipole dipole operates only between polar molecules. This is when two polar molecules get near each other and the positively charged portion of the molecule is attracted to the negatively charged portion of another molecule.
PART C: Dipole dipole and in some cases hydrogen bonding operate between the hydrogen atom of a polar bond and a nearby small electronegative atom. Only if the atom bonded to it were F, O or N it would be hydrogen bonding. Otherwise it is dipole dipole.
All molecules - Dispersion force.
Polar molecules - Dipole–Dipole interaction.
The hydrogen atom of a polar bond and a nearby small electronegative atom - Dipole–Dipole and Hydrogen Bonding.
What are intermolecular attractive force?Intermolecular force is a force that connect molecules to each other.
There are four types of forces present
DispersionDipole–DipoleHydrogen Bonding Ion-DipoleThus, the bonds present are: All molecules - Dispersion force.
Polar molecules - Dipole–Dipole interaction.
The hydrogen atom of a polar bond and a nearby small electronegative atom - Dipole–Dipole and Hydrogen Bonding.
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Calculate the concentrations of all species in a 1.15 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants for sulfurous acid are K a 1 = 1.4 × 10 − 2 and K a 2 = 6.3 × 10 − 8 .
Final answer:
The concentration of Na+ in a 1.15 M Na2SO3 solution is 2.30 M, and the concentration of SO32- is 1.15 M, as Na2SO3 dissociates entirely into these ions in solution.
Explanation:
To calculate the concentration of all species in a 1.15 M Na2SO3 (sodium sulfite) solution, we need to consider the ionization of sulfurous acid (H2SO3), which isn't explicitly present but relates to the anion derived from Na2SO3. The ionization constants given are Ka1 = 1.4 × 10−2 for the first dissociation (H2SO3 to HSO3−) and Ka2 = 6.3 × 10−8 for the second dissociation (HSO3− to SO32−). Although, given the context, we aren't going through the complete step-by-step equilibrium calculation for Ka1 and Ka2, we understand that Na2SO3 dissociates completely in solution to Na+ and SO32−. Therefore, initially, the concentration of Na+ is 2.30 M (since each Na2SO3 unit yields two Na+ ions), and the concentration of SO32− is 1.15 M.
The concentrations in the 1.15 M Na2SO3 solution are approximately:
- [tex][H2SO3] = [HSO3^-] = 1.15 M,[/tex]
- [tex][SO3^2-] = 2.10 \times 10^-5 M,[/tex]
- [tex][Na^+] =1.15 M, and[/tex]
- [tex][OH^-] = 8.70 \times 10^-14 M.[/tex]
Given:
- Initial concentration of [tex]\(Na_2SO_3\) = 1.15 M[/tex]
- [tex]\(K_a1\)[/tex] for sulfurous acid [tex]= \(1.4 \times 10^{-2}\)[/tex]
- [tex]\(K_a2\)[/tex]for sulfurous acid [tex]= \(6.3 \times 10^{-8}\)[/tex]
We can assume that the ionization of [tex]\(Na_2SO_3\)[/tex] into [tex]\(HSO_3^-\)[/tex] and [tex]\(Na^+\)[/tex] is small due to the weak acid nature of sulfurous acid. So, the initial concentration of [tex]\(H_2SO_3\)[/tex] is approximately equal to 1.15 M.
Now, we approximate the concentration of [tex]\(H^+\)[/tex] ions to be approximately equal to the concentration of [tex]\(H_2SO_3\)[/tex], which is 1.15 M.
Using [tex]\(K_a2\)[/tex], we calculate the concentration of [tex]\(SO_3^{2-}\)[/tex] ions:
[tex]\[ [SO_3^{2-}] = \frac{[H^+]^2}{K_a2} \approx \frac{(1.15)^2}{6.3 \times 10^{-8}} \approx 2.10 \times 10^{-5} \, \text{M} \][/tex]
Since [tex]\(HSO_3^-\)[/tex] ions concentration is approximately equal to [tex]\(H^+\)[/tex] ions concentration and [tex]\(Na^+\)[/tex] concentration is the same as [tex]\(Na_2SO_3\)[/tex], we have:
- [tex]\([H_2SO_3] = [HSO_3^-] \approx 1.15 \, \text{M}\)[/tex]
- [tex]\([SO_3^{2-}] \approx 2.10 \times 10^{-5} \, \text{M}\)[/tex]
- [tex]\([Na^+] \approx 1.15 \, \text{M}\)[/tex]
The concentration of [tex]\(OH^-\)[/tex] ions can be calculated using the [tex]\(K_w\)[/tex]expression:
[tex]\[ [OH^-] = \frac{K_w}{[H^+]} \approx \frac{1.0 \times 10^{-14}}{1.15} \approx 8.70 \times 10^{-14} \, \text{M} \][/tex]
Which contains more molecules, a mole of water or a mole of glucose?
Answer:
Both, water and glucose contain the same molecules
Explanation:
Although one mole of glucose has more mass than one mole of water, if we refer to the amount of atoms (or molecules) it is known that 1 mole contains the number of Avogadro in particles. No matter if the molar mass is bigger or smaller, 1 mol of anything has always 6.02×10²³ particles
What is the percentage composition of soap (C12H35COON)?
Answer:
Carbon percentage 65.8% Hydrogen percentage 14.9% Oxygen percentage 13.4% Nitrogen percentage 5.9%
Explanation:
The molar mass of C12H35COON is 237.41g/mol
Carbon percentage composition= (12.01*13)/237.41 *100= 65.8%
Hydrogen percentage composition= (1.008*35)/237.41 *100= 14.9%
Oxygen percentage composition= (15.99*2)/237.41 *100= 13.4%
Nitrogen percentage composition= (14.007)/237.41 *100= 5.9%
The percent composition of soap is as follows carbon-65.82 %, hydrogen-14.76%,oxygen-13.50%,nitrogen- 5.90%.
What is percent composition?The Percent composition is defined as a convenient way to record concentration of solution.It is a expression which relates mass of each type of an element present in a compound. It is useful in analysis of compounds providing elemental composition.It is useful in even analysis of ores giving metal content information.
The molar mass of soap C₁₂H₃₅COON=237 g/mole, thus for carbon=12×13/237×100=65.82%, for hydrogen=1×35/237×100=14.76%., for oxygen= 16×2/237×100=13.50%, for nitrogen=14/237×100=5.90%.
Thus, the percent composition of soap is as follows carbon-65.82 %, hydrogen-14.76%,oxygen-13.50%,nitrogen- 5.90%.
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What mass of strontium nitrate (Sr(NO3)2) would be required to prepare 2.000 L of a 0.0150 M aqueous solution of this salt?
Final answer:
To prepare a 0.0150 M solution of strontium nitrate in 2.000 L, you would need to use 4.49 grams of strontium nitrate.
Explanation:
To calculate the mass of strontium nitrate required to prepare the solution, we can use the formula:
moles = concentration x volume
First, convert the concentration to moles per liter:
moles/liter = concentration = 0.015 M
Next, multiply the moles per liter by the volume in liters to find the moles of strontium nitrate:
moles = (0.015 M) x (2.000 L) = 0.030 mol
Finally, calculate the mass of strontium nitrate using the molar mass:
mass = moles x molar mass = 0.030 mol x 149.6 g/mol = 4.49 g
the temperature of a sample of copper increased by 23.0 C when 265 J of heat was applied. What is the mass of the sample?
To find the mass of the copper sample, we use the formula for heat transfer Q = mcΔT. By substituting the given values of specific heat capacity of copper, the amount of heat applied and the change in temperature into the formula and solving for mass, we find the mass of the copper sample to be 29.54g.
Explanation:The subject of this question is physics, specifically dealing with the concept of heat transfer. To determine the mass of the copper sample, we'll use the formula for heat transfer: Q = mcΔT, where 'Q' is the amount of heat transferred, 'm' is the mass of the substance, 'c' is the specific heat capacity, and 'ΔT' is the change in temperature.
In this case, we know that the specific heat capacity of copper is 0.39 J/g °C, the amount of heat applied (Q) is 265 J and the change in temperature ΔT is 23.0 °C. By substituting these values into the formula, we can solve for 'm' (mass). Rearranging the formula gives us m = Q / (cΔT). Substituting the given values, we find m = 265 J / (0.39 J/g°C * 23.0°C) gives us the mass of the copper sample to be 29.54 g.
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A sample may contain any or all of the following ions: Hg22+, Ba2+, and Mn2+.
(a) No precipitate formed when an aqueous solution of NaCl was added to the sample solution.
(b) No precipitate formed when an aqueous solution of Na2SO4 was added to the sample solution.
(c) A precipitate formed when the sample solution was made basic with NaOH.
Which ion or ions are present in the sample solution? (Select all that apply.)
Mn2+ ions
Ba2+ ions
Hg22+ ions
Answer:
The ion present in the sample solution is Mn2+
Explanation:
(a) No precipitate formed when an aqueous solution of NaCl was added to the sample solution.
Most chlorides are quite soluble in water except for he H g 2 C l 2, that is an insoluble salt. Since no precipitation formed upon addition of NaCl, H g 2 + was not present in the sample.
(b) No precipitate formed when an aqueous solution of Na2SO4 was added to the sample solution.
The B a S O 4 is an insoluble salt too. Since no precipitation formed upon addition of N a 2 S O 4 , the ion B a 2 + was not present in the sample.
(c) A precipitate formed when the sample solution was made basic with NaOH.
As the M n ( O H ) 2 is insoluble in this last solution, the precipitate formed corresponds to B a S O 4 and H g 2 C l 2 compounds , therefore it can be concluded that the present ion in the sample solution is Mn+2.
N2(g) + 3 H2(g) ⇌ 2 NH3(g) + Energy
For the reaction of nitrogen gas and hydrogen gas to make methane, what stresses would shift the equilibrium to the left toward the reactants?
Answer:
The answer to your question is below
Explanation:
Reaction
N₂ + 3H₂ ⇔ 2 NH₃ + energy
a) The concentration of NH₃, if the concentration of NH₃ increases, the reaction will move to the left.
b) Diminishing the temperature, no more energy will be released and the reaction will move to the reactants.
What is the balanced chemical equation for the word equation below?
Iron and oxygen react to form iron III oxide.
Answer:
4Fe + 3O₂ → 2Fe₂O₃
Explanation:
Fe → ²⁺
O → ²⁻
But Iron III is Fe³⁺
So we have Fe³⁺ and O²⁻, the formula for the oxide must be Fe₂O₃ so the equation can be:
4Fe + 3O₂ → 2Fe₂O₃
What was the significance to India of the Persian conquest of the Indus Valley and Kashmir under Emperor Darius in the sixth century B.C.E.?
Answer:
India had a greater contact and net work with other areas or territory.
Explanation:
This lead to the introduction of new ideas and they also learnt new techniques.
A 50 W engine generates 50 J of energy. How long did it run for?
A. 10 s
B. 1 s
C. 5 s
D. 100 s
Answer: A Explanation:
Calculate the fraction of atoms in a sample of argon gas at 400 K that have an energy of 10.0 kJ or greater.
Answer:
The answer to this can be arrived at by clculating the mole fraction of atoms higher than the activation energy of 10.0 kJ by pluging in the values given into the Arrhenius equation. The answer to this is 20.22 moles of Argon have energy equal to or greater than 10.0 kJ
Explanation:
From Arrhenius equation showing the temperature dependence of reaction rates.
[tex]K = Ae^{\frac{Ea}{RT} }[/tex] where
k = rate constant
A = Frequency or pre-exponential factor
Ea = energy of activation
R = The universal gas constant
T = Kelvin absolute temperature
we have
[tex]f = e^{\frac{Ea}{RT} }[/tex]
Where
f = fraction of collision with energy higher than the activation energy
Ea = activation energy = 10.0kJ = 10000J
R = universal gas constant = 8.31 J/mol.K
T = Absolute temperature in Kelvin = 400K
In the Arrhenius equation k = Ae^(-Ea/RT), the factor A is the frequency factor and the component e^(-Ea/RT) is the portion of possible collisions with high enough energy for a reaction to occur at the a specified temperature
Plugging in the values into the equation relating f to activation energy we get
[tex]f = e^{\frac{10000J}{(8.31J/((mol)(K)))(400K)} }[/tex] or f = [tex]e^{3.01}[/tex] = 20.22 moles of argon have an energy of 10.0 kJ or greater
The question asks for the fraction of argon atoms at 400 K with an energy of 10.0 kJ or higher, which would require integrating the Maxwell-Boltzmann distribution, but key information is missing for a definitive answer.
Explanation:To calculate the fraction of argon atoms in a gas sample at 400 K with an energy of 10.0 kJ or greater, the principles of statistical mechanics and the Maxwell-Boltzmann distribution are applied. The Maxwell-Boltzmann distribution gives the fraction of particles at a certain energy level within a system at thermal equilibrium. However, we are missing some key information like the Maxwell-Boltzmann distribution function for the particular conditions set by the question, which is essential for performing such a calculation.
Typically, the calculation would require integrating the partition function over all energy states equal to or greater than 10.0 kJ. Without the exact form of the energy distribution, it is not possible to provide an accurate answer. The question pertains to a high level of physical chemistry or statistical mechanics typically studied at the college or university level.
The biopsychosocial perspective, compared with the perspectives of Freud, Jung, and Rogers, __________.
Answer:
Is broader than other modelsExplanation:
George Engel conceptualised the biospychosocial model in 1977. He suggested apart from the biological condition a persons medical condition is also dependent on social and psychological factors.
This model is used in cases of chronic pain and considers the pain as a psychophysiological behaviour pattern that cannot be categorised into social, psychological and biological perspective.
There are suggestions that physiotherapy should be integrated into psychological treatment to address the various components that comprises the experience of pain.
When 2−bromo−3,3−dimethylbutane is treated with K⁺ −OC(CH₃)₃, a single product T having molecular formula C₆H₁₂. When 3,3−dimethyl−2−butanol is treated with H₂SO₄, the major product U has the same molecular formula. Given the following ¹H NMR data, what are the structures of T and U?
Answer:
T = 3,3-dimethylbut-1-ene
U = 2,3-dimethylbut-2-ene
Explanation:
The chemical reactions for the two reactions and the mechanisms of reaction are shown in the two attachments to this answer.
Hope it helps!
The osmotic pressure, Π, of a solution of glucose is 83.1 atm . Find the molarity of the solution at 298 K. Express the molarity to three significant figures and include the appropriate units.
The molarity of the glucose solution is approximately 3.38 M, found by substituting values of osmotic pressure and temperature into the formula for osmotic pressure and solving for molarity.
Explanation:The osmotic pressure, Π, of a solution is given by the equation Π = cRT, where c represents the molarity of the solution, R is the ideal gas constant (0.0821 L·atm/K·mol), and T is the temperature in Kelvin. Given that the osmotic pressure is 83.1 atm and the temperature is 298 K, you can solve for c (molarity) by rearranging the equation: c = Π/(RT). Substituting the given values into the equation gives you c = 83.1 / (0.0821 x 298) = 3.38 M. Therefore, the molarity of the glucose solution is approximately 3.38 M.
Learn more about Osmotic Pressure here:https://brainly.com/question/37985044
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Nitrogen gas and hydrogen gas are combined in a reaction to produce ammonia. If 3.0 moles of N2 and 12.0 moles of H2 are present in the mixture, which is the limiting reactant?
Answer:
The limiting reactant is the N₂
Explanation:
The reaction for production of ammonia is:
N₂ + 3H₂ → 2NH₃
Ratio is 1:3. Let's make a rule of three to solve this:
1 mol of nitrogen is needed to react with 3 moles of hydrogen.
3 moles of nitrogen may react with (3 . 3) /1 = 9 moles of H₂
It is ok because I have 12 moles, so the limiting reactant is the N₂ but let's confirm it.
3 moles of H₂ react with 1 mol of N₂
12 moles of H₂ would react with (12. 1) / 3 = 4 moles of N₂
It's ok to say the N₂ is the limiting because I don't have enough moles to react. I need 4 and I only have 3
Final answer:
To determine the limiting reactant in the reaction to produce ammonia, we compare the stoichiometric amounts needed according to the balanced equation. Nitrogen is the limiting reactant in this reaction because it requires more hydrogen than what is available to completely react.
Explanation:
To determine the limiting reactant between nitrogen gas (N2) and hydrogen gas (H2) in the synthesis of ammonia (NH3), we use the balanced chemical equation:
N2 + 3 H2 ->2 NH3
According to the equation, each mole of nitrogen reacts with three moles of hydrogen to produce two moles of ammonia. Given the reaction mixture of 3.0 moles of N2 and 12.0 moles of H2, we compare the stoichiometric amounts needed for the reaction. The nitrogen gas would require 9.0 moles of hydrogen gas to fully react (3 moles N2 imes 3 moles H2/1 mole N2 = 9 moles H2). Since there are 12.0 moles of hydrogen gas available, hydrogen is in excess, and nitrogen is the limiting reactant.