Answer:
3.9 seconds
Explanation:
Given:
Final velocity = 72 m/sDistance covered = 540 mTo find:
Time taken = ?Solution:
We can use the equation,
v = u + at
where,
v is the final velocityu is the initial velocitya is the acceleration t is the time takenWe know that the initial velocity is 0, so the equation becomes,
v = at
We can also use the equation,
s = ut + ½* at²
where,
s is the distance coveredu is the initial velocity a is the acceleration t is the time takenWe know that the distance covered is 540 m and the final velocity is 72 m/s, so we can substitute these values into the equation to solve for the time taken.
540 = 0 * t + ½ * a * t²
540 = ½ * a * t²
1080 = a * t²
[tex]t^2= \frac{1080}{a}[/tex]
[tex]t = \sqrt{ \frac{1080}{a}}[/tex]
We know that the acceleration is the change in velocity divided by the time taken, so we can substitute this value into the equation to solve for the time taken.
[tex]t = \sqrt{\frac{1080 }{ 72 m/s}}[/tex]
[tex]t = \sqrt{15} s[/tex]
t = 3.9 s
Therefore, the time taken is t = 3.9 seconds
59.6 ml of air is at 20.5 °C. What is its volume at 73.9 °C?
The volume of air at 73.9 °C would be approximately 70.91 ml.
To solve this problem, we can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature when pressure and amount of gas are kept constant.
Let's assume that the pressure and amount of air remain constant.
Given:
Initial volume (V1) = 59.6 ml
Initial temperature (T1) = 20.5 °C = 20.5 + 273.15 K = 293.65 K
Final temperature (T2) = 73.9 °C = 73.9 + 273.15 K = 347.05 K
Using Charles's Law, we can set up the following proportion:
V1 / T1 = V2 / T2
Plugging in the values, we have:
59.6 ml / 293.65 K = V2 / 347.05 K
To solve for V2 (the final volume), we can rearrange the equation:
V2 = (59.6 ml * 347.05 K) / 293.65 K
V2 = 70.91 ml
Therefore, the volume of air at 73.9 °C would be approximately 70.91 ml.
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The sun is reportedly 92,960,000 miles from Earth. How many significant figures does this number have?
Answer: The correct answer would be 4
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Calculate the volume occupied by the oxygen gas at a pressure of 210 kPa and temperature of 50°C. Use your answer from question 27 to help you solve this problem. Report your answer in liters with 3 significant figures._______L
Answer:
holaholaholaholaholaholaholahola
The volume occupied by the oxygen gas at a pressure of 210 kPa and a temperature of 50°C is 0.126 liters, rounded to three significant figures.
To calculate the volume occupied by the oxygen gas at a pressure of 210 kPa and a temperature of 50°C, we can use the ideal gas law equation:
PV = nRT
where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
First, we need to convert the given temperature from Celsius to Kelvin by adding 273.15:
T = 50°C + 273.15 = 323.15 K
Next, we rearrange the ideal gas law equation to solve for volume:
V = (nRT) / P
To find the number of moles (n), we can use the answer from question 27, which is the mass of oxygen gas:
m = 32 g
Using the molar mass of oxygen (O₂) which is approximately 32 g/mol, we can calculate the number of moles:
n = m / M = 32 g / 32 g/mol = 1 mol
Now we have all the values needed to calculate the volume:
V = (1 mol * 8.314 J/(mol*K) * 323.15 K) / 210,000 Pa
Using the ideal gas constant (R) of 8.314 J/(mol*K) and the pressure of 210 kPa (which is equivalent to 210,000 Pa), we can substitute these values into the equation and solve for volume.
V = 0.126 liters.
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A sample of gas is put into a rigid (fixed volume) container at 3 oC and a pressure of 38.5 kPa. The container is then placed in an oven at 267 oC.
What pressure would you expect to measure for the gas in the container at this higher temperature?
We would expect to measure a pressure of approximately 75.25 kPa for the gas in the container at the higher temperature of 267 oC.
To determine the expected pressure of the gas in the container at the higher temperature, we can use the combined gas law, which relates the initial and final conditions of temperature and pressure in a fixed volume system. The combined gas law equation is given as:
(P1 * V1) / T1 = (P2 * V2) / T2
Where:
P1 = Initial pressure
V1 = Initial volume (which is fixed in this case)
T1 = Initial temperature
P2 = Final pressure (to be determined)
V2 = Final volume (which is fixed in this case)
T2 = Final temperature
In this scenario, the initial conditions are given as 3 oC (which is equivalent to 276 K) and 38.5 kPa. The final temperature is 267 oC (which is equivalent to 540 K). Since the volume is fixed, we can substitute the given values into the equation:
(38.5 kPa * V1) / 276 K = (P2 * V1) / 540 K
Simplifying the equation, we can cancel out V1:
38.5 / 276 = P2 / 540
Solving for P2:
P2 = (38.5 / 276) * 540 ≈ 75.25 kPa
Therefore, we would expect to measure a pressure of approximately 75.25 kPa for the gas in the container at the higher temperature of 267 oC.
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Match each reaction to its correct type by dragging it to the appropriate box on the left. Click Done to
check your answers. Drag the items on the left to the correct location on the right.
From the image attached;
Reaction 1 - Oxidation reduction
Reaction 2 - Precipitation reaction
Reaction 3 - Acid base reaction
Reaction 4 - Reaction with oxygen
What is reaction?A reaction is a procedure or an occurrence in which a change takes place, frequently leading to the conversion of one or more compounds into other substances. A chemical reaction, which entails the rupturing and creation of chemical bonds between atoms or molecules, is what is meant when the word "reaction" is used in reference to chemistry.
When reactant molecules collide with enough force and in the right orientation, the current chemical bonds are broken, and new bonds are formed, converting the reactants into products.
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A bottle of rubbing alcohol contains both 2-propanol and water.
These liquids can be separated by the process of distillation because the 2-propanol and water
A) have combined chemically and retain their different boiling points
B) have combined physically and have the same boiling point
C) have combined physically and retain their different boiling
DOmIS
D) have combined chemically and have the same boiling point
These liquids can be separated by the process of distillation because the 2-propanol and water option B) have combined physically and have the same boiling point.
A bottle of rubbing alcohol contains both 2-propanol and water.
These liquids can be separated by the process of distillation because the 2-propanol and water have combined physically and retain their different boiling points.
The process of separating two miscible liquids with different boiling points is called distillation.
The procedure is based on the idea that the liquids have various boiling points.
Thus, during the boiling process, one of the fluids vaporizes faster than the other, and that vapor is then collected and allowed to condense, producing pure liquid.
Alcohol and water are examples of two miscible liquids that can be separated by distillation.
The process of distillation operates on the fact that the liquid with the lowest boiling point is vaporized first.
Alcohol has a lower boiling point than water, so alcohol is vaporized first in distillation, leaving the water behind.
A bottle of rubbing alcohol contains both 2-propanol and water.
These liquids can be separated by the process of distillation because the 2-propanol and water have combined physically and retain their different boiling points.
This means that 2-propanol and water are physically combined; that is, they do not form any new chemical bonds with each other.
Consequently, the molecules of the substances remain separate from each other.
Furthermore, both water and 2-propanol have various boiling points.
The boiling point of 2-propanol is 82.6 °C, while that of water is 100 °C. Thus, they can be separated using distillation.
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The pressure exerted by a confined gas is the result of
gas particles colliding with each other
gas particles colliding with the walls of the container
nobody knows, it just is
gas particles taking up space in the container
The pressure exerted by a confined gas is the result of Option b. gas particles colliding with the walls of the container.
The pressure exerted by a confined gas is the result of gas particles colliding with the walls of the container. When a gas is confined within a container, the gas particles are in constant motion, moving in random directions with varying speeds. As these gas particles move, they collide with each other and with the walls of the container.
When a gas particle collides with the walls of the container, it exerts a force on the surface. The collective effect of numerous gas particle collisions leads to a net force being exerted on the walls of the container. This force per unit area is what we call pressure.
The more frequently and vigorously the gas particles collide with the walls, the higher the pressure of the gas. Factors that influence gas pressure include the number of gas particles present, their average speed, and the volume of the container. Therefore, Option b is correct.
The question was incomplete. find the full content below:
The pressure exerted by a confined gas is the result of
a. gas particles colliding with each other
b. gas particles colliding with the walls of the container
c. nobody knows, it just is
d. gas particles taking up space in the container.
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-Convert 6.02 x 1020 formula units of MgCl₂ to mol of MgCl₂:
6.02 x [tex]10^{20[/tex] formula units of MgCl₂ is equal to 0.1 moles of MgCl₂.
To convert formula units of MgCl₂ to moles of MgCl₂, we need to use Avogadro's number, which relates the number of formula units to the number of moles.
Avogadro's number (NA) is approximately 6.022 x 10^23 formula units per mole.
Given that we have 6.02 x 10^20 formula units of MgCl₂, we can set up a conversion factor to convert to moles:
(6.02 x 10^20 formula units MgCl₂) * (1 mol MgCl₂ / (6.022 x 10^23 formula units MgCl₂))
The formula units of MgCl₂ cancel out, and we are left with moles of MgCl₂:
(6.02 x 10^20) * (1 mol / 6.022 x 10^23) = 0.1 mol
Therefore, 6.02 x 10^20 formula units of MgCl₂ is equal to 0.1 moles of MgCl₂.
It's important to note that this conversion assumes that each formula unit of MgCl₂ represents one mole of MgCl₂. This is based on the stoichiometry of the compound, where there is one mole of MgCl₂ for every one formula unit.
Additionally, this conversion is valid for any substance, not just MgCl₂, as long as you know the value of Avogadro's number and the number of formula units or particles you have.
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Which of the following set of quantum numbers (ordered n, ℓ, mℓ ) are possible for an electron in an atom? Check all that apply
a. 2, 1, 3
b. 5, 3, -3
c. 4, 3, -2
d. -4, 3, 1
e. 2, 1, -2
f. 3, 2, 2
g. 3, 3, 1
the possible quantum numbers (ordered n, ℓ, mℓ ) are:Option B.5, 3, -3 and Option C. 4, 3, -2
The quantum numbers n, ℓ, mℓ represent respectively the principal quantum number, the orbital angular momentum quantum number and the magnetic quantum number.
These are the three most important quantum numbers. T
here is another quantum number called the spin quantum number, denoted by ms.
Let's see which of the given quantum number sets is possible.2, 1, 3 is not possible because for ℓ = 1, mℓ can only be -1, 0, or 1. 5, 3, -3 is possible.4, 3, -2 is possible. -4, 3, 1 is not possible.
For any value of ℓ, mℓ must be between -ℓ and +ℓ. e. 2, 1, -2 is not possible because for ℓ = 1, mℓ can only be -1, 0, or 1. f. 3, 2, 2 is not possible because for ℓ = 2, mℓ can only be -2, -1, 0, +1, or +2. g. 3, 3, 1 is not possible because for any value of ℓ, mℓ must be between -ℓ and +ℓ.
Therefore, the possible quantum numbers (ordered n, ℓ, mℓ ) are:5, 3, -34, 3, -2
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A reaction requires 22.4 L of gas at STP. You have 45.0 L of gas at 100 kPa and 373 K. Which of the following statements is true? The gas
constant is 8.31 L-kPa/mol-K.
You do not have enough gas for the reaction to occur.
You will have too much gas for the reaction to occur.
You will have an excess of gas and the reaction will occur.
You cannot tell given this information.
Answer:
(c). You will have an excess of gas and the reaction will occur.
Explanation:
We can use the ideal gas law to calculate the number of moles of gas in each case. The ideal gas law is:
PV = nRT
where:
P is the pressure in pascalsV is the volume in litersn is the number of moles of gasR is the gas constant (8.31 L-kPa/mol-K)T is the temperature in KelvinAt STP, the pressure is 1 atm (101.325 kPa) and the temperature is 273.15 K. So, the number of moles of gas at STP is:
[tex]n =\frac{ PV }{ RT} =\frac{ (101.325 kPa)(22.4 L) }{(8.31 L-kPa/mol-K)(273K)} = 1mol[/tex]
At 100 kPa and 373 K, the number of moles of gas is:
[tex]n =\frac{ PV }{ RT} =\frac{ (100 kPa)(45 L) }{(8.31 L-kPa/mol-K)(373 K)} = 1.45mol[/tex]
So, you have 1.45 moles of gas at 100 kPa and 373 K. The reaction requires 1 mole of gas at STP, so you have an excess of 0.45 moles of gas. The excess gas will not participate in the reaction, but it will not prevent the reaction from occurring.
Therefore, the correct answer is (c). You will have an excess of gas and the reaction will occur.
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An automobile engine has a cylinder with a volume of 500.0 mL that is filled with air (21.00 % oxygen) at a temperature of 55.00 C and a pressure of 101.0 kPa. What is the mass of octane, C8H18 that must be injected to react with all of the oxygen in the cylinder to produce carbon dioxide and water? 2C8H18 + 25O2 -------->. 16CO2 + 18H2O
The mass of octane, C8H18 that must be injected to react with all of the oxygen in the cylinder to produce carbon dioxide and water is 0.14 g.
The balanced equation for the combustion of octane is:2C8H18 + 25O2 → 16CO2 + 18H2OFrom the above balanced chemical equation, we can see that 25 moles of O2 react with 2 moles of C8H18.
So, 12.5 moles of O2 will react with 1 mole of C8H18. We can use the ideal gas law PV = nRT to calculate the moles of oxygen present in the cylinder.
Here, we need to use the partial pressure of oxygen only since we are interested in the moles of oxygen only.
O2 = 21.00% × 101.0 kPa = 21.21 kPaV = 500.0 mL = 500.0/1000 = 0.5000 L (convert mL to L)R = 8.314 J/mol K (gas constant).
We have,PV = nRTn = PV/RTn = (21.21 × 10^3 Pa) × (0.5000 × 10^-3 m^3) / (8.314 J/mol K × 328.15 K) = 0.01578 moles of O2.
Since 12.5 moles of O2 react with 1 mole of C8H18,0.01578 moles of O2 will react with= (0.01578 moles × 1 mole C8H18)/12.5 moles = 0.001262 moles of C8H18
The molar mass of C8H18 is = 8 × 12.01 + 18 × 1.01 = 114.16 g/mol.
So, the mass of C8H18 required is = 0.001262 × 114.16 = 0.1445 g or 0.14 g (approx.)
Therefore, the mass of octane, C8H18 that must be injected to react with all of the oxygen in the cylinder to produce carbon dioxide and water is 0.14 g.
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