Answer:
We could use fractional distillation, solvent extraction, partial crystallization, chromatography, and another.
Explanation:
The only information given is that the compounds have similar polarities, which means they form a homogeneous mixture, so, to separate them, we should use some of the processes of separation of a homogeneous mixture.
Without knowing any of the physical properties of A and B (physical state, density, boiling point, for example) it's impossible to determinate the better process of separation. However, we can describe some of the processes that are used to separate homogeneous mixtures: fractional distillation, solvent extraction, partial crystallization, and chromatography, for example.
A sample of nitrogen gas has a mass of 48.6 grams. How many N2 molecules are there in the sample? molecules Submit Answer & Next
Answer:
There are 1.05 x 10²⁴ molecules in 48.6 g N₂
Explanation:
1 mol of N₂ has a mass of (14 g * 2) 28 g.
Then, 48.6 g of N₂ will be equal to (48.6 g *(1 mol/ 28 g)) 1.74 mol.
Since there are 6.022 x 10²³ molecules in 1 mol N₂, there will be
(1.74 mol *( 6.022 x 10²³ / 1 mol)) 1.05 x 10²⁴ molecules in 1.74 mol N₂ (or 48. 6 g N₂).
The Prandtl number, Npr, is a dimensionless group important in heat transfer calculations. It defined as Cuk, where C, is the heat capacity of a fluid, u is the fluid viscosity, and k is the thermal conductivity. For a particular fluid, C, = 0.583 J/(g.°C), k = 0.286 W/(m:°C), and 1936 lb./(ft·h). Estimate the value of Npr without using a calculator (remember, it is dimension less), showing your calculations; then determine it with a calculator.
Answer:
The Prandlt number for this fluid is 1630.
Explanation:
The Prandlt number is defined as:
[tex]Pr=\frac{C_p\mu}{k}[/tex]
To compute the Prandlt number for this case, is best if we use the same units in every term of the formula.
[tex]\mu=1936\frac{lb}{ft*h}*\frac{1000g}{2.205lb}*\frac{3.281ft}{1m}*\frac{1h}{3600s} \\ \\\mu=800 \frac{g}{m*s}[/tex]
Now that we have coherent units, we can calculate Pr
[tex]Pr=\frac{C_p\mu}{k}=\frac{0.583*800}{0.286}= 1630[/tex]
The Prandtl number (Npr) is a dimensionless group used in heat transfer calculations. To estimate its value, we can multiply the given heat capacity, viscosity, and thermal conductivity. Without a calculator, an estimated value of Npr is approximately 143.53. With a calculator, the exact value is approximately 144.97.
Explanation:The Prandtl number (Npr) is a dimensionless group used in heat transfer calculations. It is defined as the product of the heat capacity (C) of a fluid, its viscosity (u), and its thermal conductivity (k). To estimate the value of Npr for a particular fluid without a calculator, we can multiply the given values: C = 0.583 J/(g·°C), u = 1936 lb./(ft·h), and k = 0.286 W/(m·°C). Converting units for convenience, we have C = 583 J/(kg·°C) and u = 0.8644 kg/(m·s). Therefore, Npr ≈ C·u·k ≈ (583 J/(kg·°C))·(0.8644 kg/(m·s))·(0.286 W/(m·°C)) ≈ 143.53. Using a calculator, we can perform the exact calculation and obtain Npr ≈ 144.97.
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Determine the theoretical yield of P2O5, when 3.07 g of P reacts with 6.09 g of oxygen in the following chemical equation
4 P + 5 O2 ----> 2 P2O5
Answer: 14.2 grams
Explanation:-
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
given mass of phosphorous (P) = 3.07 g
Molar mass of phosphorous (P) = 31 g/mol
Putting in the values we get:
[tex]\text{Number of moles of phosphorous}=\frac{3.07g}{31g/mol}=0.1moles[/tex]
given mass of oxygen [tex]O_2[/tex] = 6.09 g
Molar mass of oxygen [tex]O_2[/tex] = 32 g/mol
Putting in the values we get:
[tex]\text{Number of moles of oxygen}=\frac{6.09g}{32g/mol}=0.2moles[/tex]
According to stoichiometry:
[tex]4P+5O_2\rightarrow 2P_2O_5[/tex]
4 moles of phosphorous combine with 5 moles of oxygen
Thus 0.1 moles of phosphorous combine with =[tex]\frac{5}{4}\times 0.1=0.125[/tex] moles of oxygen
Thus phosphorous acts as limiting reagent as it limits the formation of product and oxygen is the excess reagent.
4 moles of phosphorous gives= 2 moles of [tex]P_2O_5[/tex]
Thus 0.1 moles of phosphorous gives =[tex]\frac{2}{4}\times 0.1=0.05[/tex] moles of [tex]P_2O_5[/tex]
mass of [tex]P_2O_5=moles\times {\text {Molar mass}}=0.05\times 284=14.2g[/tex]
Thus the theoretical yield of [tex]P_2O_5[/tex] is 14.2 grams.
a mixture of water and ethanol has a total mass of 100 kg. The
mole fraction of water is 0.752. what is the mass of ethanol in
this mixture, in kg?
Answer:
45.8 kg
Explanation:
Given that the mole fraction of water = 0.752
For a binary system,
The mole fraction of water + The mole fraction of ethanol = 1
So,
The mole fraction of ethanol = 0.248
Given that the total mass = 100 kg
Let the mass of ethanol = x kg
The mass of water = 100 - x kg
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Molar mass of ethanol = 46.07 g/mol
Molar mass of water = 18 g/mol
Also, 1 g = 10⁻³ kg
So,
Molar mass of ethanol = 46 ×10⁻³ kg/mol
Molar mass of water = 18 ×10⁻³ kg/mol
Moles of ethanol = x / 46 ×10⁻³ moles
Moles of water = (100 - x) / 18 ×10⁻³ moles
So, according to definition of mole fraction:
[tex]Mole\ fraction\ of\ ethanol=\frac {n_{ethanol}}{n_{ethanol}+n_{water}}[/tex]
Applying values as:
[tex]0.248=\frac {\frac {x}{46\times 10^{-3}}}{\frac {x}{46\times 10^{-3}}+\frac {(100-x)}{18\times 10^{-3}}}[/tex]
Solving for x, we get
x = 45.8 kg
Mass of ethanol in mixture = 45.8 kg
A man weighs 180lbf at a place where g=32.1 ft/s2. What is his weigh(in lbf) on the moon where g=5.47ft/s2? what is his mass(in lb
Answer:
Mass = 5.6075 lb
Weight on moon = 30.7 lbf
Explanation:
The weight of the man at gravity, g = [tex]32.1\ ft/s^2[/tex] is 180 lbf
Also,
Weight = Mass × Gravitational acceleration
So,
180 lbf = Mass × [tex]32.1\ ft/s^2[/tex]
Mass = 5.6075 lb
Given, g = [tex]5.47\ ft/s^2[/tex] and mass = 5.6075 lb
Thus,
Weight = Mass × Gravitational acceleration = 5.6075 × [tex]5.47\ ft/s^2[/tex] = 30.7 lbf
Weight on the moon = 30.7 lbf
You have just spilled part of a bottle of 0.5 M EDTA. You measure the remaining amount and find out you still have 150 ml left. You are wondering whether you'll have enough for an extraction you're about to run. Calculate how many moles of EDTA you have left.
Answer:
0.075 moles of EDTA are left.
Explanation:
To solve this problem we need to use the definition of Molar concentration (M):
M=[tex]\frac{#moles}{liters}[/tex]
With that equation in mind, in the problem we're given two out of the three factors: Concentration and Volume. We're asked to calculate the remaining factor.
Before we calculate we convert 150 mL to L, so we divide 150 by 1000.
[tex]150 mL*\frac{1L}{1000mL}=0.150L[/tex]
Lastly we calculate the number of moles of EDTA left:
0.5 M * 0.150 L = 0.075 moles.
The amount of an enzyme that will catalyze the reaction of 1 micromole of substrate in 1 minute defines what unit of activity?
Answer:
The enzyme unit
Explanation:
Enzymes are catalyst, meaning that they make chemical reactions faster.
For measuring the catalytic activity of an enzyme there are two common units The enzyme unit and The Katal.
The first one is defined as the amount of enzyme that catalyzes the reaction of 1 micromole of substrate in a minute, on the other hand the katal is give for 1 mole per second. Since 1 mol is a huge amount in terms of enzymes, it is usually use the prefix nano- (10^-9).
1 U = 16.67 nKat if you need a conversion factor.
Hope it helps!
Why does acetic acid lose its buffering ability at very high pH values? Why does acetic acid lose its buffering ability at very low pH values? In other words, what is happening when the buffering range is exceeded? The pKa of formic acid is 3.75.
Answer:
Because the optimal range of buffering for a formic acid potassium formate buffer is 2.74 ≤ pH ≤ 4.74.
Explanation:
Every buffer solution has an optimal effective range due to pH = pKa ± 1. Outside this range, there is not enough acid molecules or conjugate base molecules to sustain the pH without variation. There is a certain amount of both molecules that has to be in the solution to maintain a pH controlled.
Being for the formic acid the pKa 3.74, the optimal effective range is between 2.74 and 4.74. Upper or lower these range a formic acid/potassium formate buffer does not work.
If an athlete runs at a speed of 12.9 km/hour, how long will it take the athlete to run a marathon (about 42.2 km)? Express your answer in hours using the correct number of significant figures. Do not enter your answer using scientific notation.
Answer:
Time = 3.27s
Explanation:
Time = Distance / Speed
Time = 42.2 / 12.9 = 3.2713
When you multiply the number of significant figures in the result is the same as the number with the least significant figures.
In this case 42.2 has 3 significant figures and 12.9 also has 3 significant figures, therefore the answer must have 3 significant figures too.
Time = 3.27 s
36. Dimensional Analysis: A useful way to convert units using multiplication & division *Always start with what you are given! Practice: A. Convert 2.0 x 105 m to inches, given that 1 meter = 39.37 inches. B. Convert 1.004 kg to grams, given that 1 pound (lb) = 453.59 grams (g) and 1 kilogram (kg) = 2.205 pounds (lb). C. Convert 1.2 m/s to inches/minute. D. Convert 120 nm to mm.
Answer:
A) 7.9 x 10⁶ inches
B) 1004 g
C) 2.8 x 10³ inches/ min
D) 1.2 x 10⁻⁴ mm
Explanation:
A) Since 39.37 inches = 1 m, you can convert meters to inches by multiplying by the conversion factor (39.37 inches / 1 m).
Notice that if 39.37 inches = 1 m then 39.37 inches / 1 m = 1. That means that when you multiply by a conversion factor, you are only changing units since it is the same as multiplying by 1 :
2.0 x 10⁵ m * (39.37 inches / 1 m) = 7.9 x 10⁶ inches
B) Conversion factors : (2.205 pounds / 1 kg) and (453.59 g / 1 pound), because 2.205 pounds = 1 kg and 1 pound = 453.59 g. Then:
1.004 kg * ( 2.205 pounds / 1 kg) * ( 453.59 g / 1 pound) = 1004 g
C) Conversion factor: (39.37 inches / 1 m) and (60 s / 1 min)
1.2 m/s * (39.37 inches / 1 m) * ( 60 s / 1 min) = 2.8 x 10³ inches/ min
D)Converison factor ( 1 mm / 1 x 10⁶ nm):
120 nm (1 mm / 1 x 10⁶ nm) = 1.2 x 10⁻⁴ mm
Answer:
A. [tex]7.9x10^6in[/tex]
B. [tex]1004g[/tex]
C. [tex]2834.64\frac{in}{min}[/tex]
D. [tex]1.2x10^{-4}mm[/tex]
Explanation:
Hello,
In this case, we use the proportional factors to obtain the required conversions as shown below:
A. We consider the given data:
[tex]2.0x10^5m*\frac{39.37in}{1m} =7.9x10^6in[/tex]
B. It is not necessary to use the given data referred to pounds:
[tex]1.004kg*\frac{1000g}{1kg}=1004g[/tex]
C. We consider that 1 meter es equivalent to 39.37 inches and 1 min to 60 seconds:
[tex]1.2\frac{m}{s}*\frac{39.37in}{1m} *\frac{60s}{1min}=2834.64\frac{in}{min}[/tex]
D. We consider that 1nm is equivalent to 1x10⁻⁹m and 1mm to 1x10⁻³:
[tex]120nm\frac{1x10^{-9}m}{1nm}*\frac{1mm}{1x10^{-3}m} =1.2x10^{-4}mm[/tex]
Best regards.
Which of the following is a property of a mixture? It consists of a single element or compound. Components that are mixed can be in different states of matter. It is very difficult to separate the components. The proportion of the particles that make it up cannot be altered.
Answer:
Components that are mixed can be in different states of matter.
Explanation:
A mixture is often described as an impure substance. It has the following properties:
Constituents retain their identities i.e the physical properties of mixtures are retained. Their composition is indefinite i.e they consist of two or more elements and/or compounds in any proportion by mass. They can easily be seperated by physical methods.There are two types of mixture; homogenous mixtures have their constituents existing in one phase.
Heterogenous mixtures have constituents in different phases. The phases are the different states of matter.
One of the properties of a mixture is that; Components that are mixed can be in different states of matter.
Definition;
A mixture put simply can be defined as an impure substance which is made up of different constituents with each constituent possessing its own unique properties.
Additionally, mixtures are subdivided into homogeneous and heterogeneous mixtures;
A homogeneous mixture has it's constituent substances in the same phase/state.An heterogenous mixture on the other hand, has it's constituent substances in different states of matter.Read more;
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The statement "Although sulfuric acid is a strong electrolyte, an aqueous solution of H2SO4 contains more HSO4− ions than SO42− ions" is The statement "Although sulfuric acid is a strong electrolyte, an aqueous solution of H 2 S O 4 contains more H S O 4 − ions than S O 4 2 − ions" is blank. This is best explained by the fact that H 2 S O 4 blank.. This is best explained by the fact that H2SO4 The statement "Although sulfuric acid is a strong electrolyte, an aqueous solution of H 2 S O 4 contains more H S O 4 − ions than S O 4 2 − ions" is blank. This is best explained by the fact that H 2 S O 4 blank..
The statement "Although sulfuric acid is a strong electrolyte, an aqueous solution of H₂SO₄ contains more HSO₄⁻ ions than SO₄²⁻ ions is True. This is best explained by the fact that H₂SO₄ is a diprotic acid where only the first hydrogen completely ionizes.
Why?
H₂SO₄ is a diprotic acid. That means that it has two hydrogen ions to give to the solution. The two dissociation reactions are shown below:
H₂SO₄ + H₂O → HSO₄⁻ + H₃O⁺
HSO₄⁻ + H₂O ⇄ SO₄²⁻ + H₃O⁺
As the arrows show, the first dissociation is complete, meaning that all the sulfuric acid that is present initially is dissociated into HSO₄⁻ and H₃O⁺. However, the second dissociation is incomplete, and it's actually an equilibrium with an acid constant (Ka)of 1.2×10⁻².
That means that if the initial concentration of H₂SO₄ was 1M, the concentration of HSO₄⁻ is going to be 1M as well, but the concentration of SO₄²⁻ is going to be much less than 1M, according to the dissociation constant.
Have a nice day!
Explain why an airplane flying from
New York to Los Angeles has both kinetic
energy and potential energy.
Answer:
Explanation:
Kinetic energy is the energy due to the motion of a body. Potential energy is the energy at rest.
Kinetic energy is dependent on the mass and velocity at which a body is moving. Potential energy is a function of altitude and acceleration due to gravity on a body.
K.E = [tex]\frac{1}{2} mv^{2}[/tex]
where m is the mass of the plane and v is the velocity at which the plane moves.
P.E= mgh
m is the mass of the plane, g is the acceleration due to gravity acting on it and h is the height or altitude of the plane.
A moving airplane shuffles between both kinetic and potential energy. When the altitude of plane changes there is a noticeable change in its potential energy. A plane has to change its velocity from time to time in order to over come drag and gravity, this incurs kinetic energy on the plane.
The mechanical energy in plane thrusters are converted to do effect kinetic and potential energy.
Based upon the information provided in the class, which of the following bond types is the strongest? Question options:
A) hydrogen bond
B) Ion - Dipole
C) Dipole - Dipole
D) Ion - Induced Dipole
E) Dipole - Induced Dipole
Answer: A hydrogen bonding is interaction between lone pair and hydrogen atom. An Ion-Dipole interaction is the interaction between an ion formed and a dipole. Dipole forms because of the electronegativity difference between two atom participating in the bond formation, and an ion is formed when an atom gains or lose electron. This ion-dipole interaction is strongest interaction.
Therefore, The right choice is (B)
When 2.0 x 10-2 mole of nicotinic acid (amonoprotic
acid) is dissolved in 350 mL of water, the pH is 3.05.What is the
Ka of nicotinic acid?
Answer:
Ka of nicotinic acid = [tex]1.41 \times 10^{-5}[/tex]
Explanation:
pH = 3.05
[tex]pH = -log [H^+][/tex]
[tex]H^+ = (10)^{-3.05}=0.00089125 M[/tex]
No. of mol of nicotinic acid = [tex]2.0 \times 10^{-2}[/tex]
Volume of water = 350 mL = 0.0350 L
Molarity = [tex]\frac{Moles}{Volume\ in\ L}[/tex]
Molarity = [tex]\frac{2.0 \times 10^{-2}}{0.350} = 0.05714\ M[/tex]
Nicotinic acid dissoctates as:
[tex]HA \rightarrow H^+ + A^-[/tex]
[H+] = 0.00089125 M
[A-] = 0.00089125 M
[HA} at equilibium = 0.05714 - 0.00089125 = 0.05624875 M
[tex]Ka = \frac{[H^+][A^-]}{[HA]}[/tex]
[tex]Ka = \frac{(0.00089125)^2}{0.05624875} = 1.41 \times 10^{-5}[/tex]
The correct answer is that the [tex]\(K_a\)[/tex] of nicotinic acid is approximately [tex]\(1.1 \times 10^{-5}\).[/tex]
To find the [tex]\(K_a\)[/tex] of nicotinic acid, we can use the Henderson-Hasselbalch equation, which relates the pH of a solution to the pKa of the acid and the concentration of the acid and its conjugate base. For a monoprotic acid like nicotinic acid, the equation is:
[tex]\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \][/tex]
Where:
- [A^-] is the concentration of the conjugate base (nicotinate ion)
- [HA] is the concentration of the undissociated acid (nicotinic acid)
Since nicotinic acid is a weak acid, we can assume that the concentration of the conjugate base [A^-] is much less than the initial concentration of the acid [HA]_initial. Therefore, the concentration of the undissociated acid [HA] can be approximated to the initial concentration of the acid.
Given that the initial concentration of nicotinic acid is:
[tex]\[ [\text{HA}]_{\text{initial}} = \frac{2.0 \times 10^{-2} \text{ moles}}{0.350 \text{ L}} = 5.71 \times 10^{-2} \text{ M} \][/tex]
We can rearrange the Henderson-Hasselbalch equation to solve for pKa:
[tex]\[ \text{pKa} = \text{pH} - \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \][/tex]
Since we do not have the concentration of the conjugate base [A^-], we can use the fact that the pH is 3.05 to find the [A^-] using the pH equation:
[tex]\[ \text{pH} = -\log[\text{H}^+] \] \[ 3.05 = -\log[\text{H}^+] \] \[ [\text{H}^+] = 10^{-3.05} \][/tex]
[tex]\[ [\text{H}^+] = 8.91 \times 10^{-4} \text{ M} \][/tex]
At this point, we can assume that the concentration of the conjugate base [A^-] is approximately equal to the concentration of hydronium ions [H^+] because each mole of nicotinic acid that dissociates produces one mole of hydronium ions and one mole of nicotinate ions. Therefore:
[tex]\[ [\text{A}^-] \approx [\text{H}^+] = 8.91 \times 10^{-4} \text{ M} \][/tex]
Now we can calculate the pKa:
[tex]\[ \text{pKa} = 3.05 - \log \left( \frac{8.91 \times 10^{-4}}{5.71 \times 10^{-2}} \right) \] \[ \text{pKa} = 3.05 - \log \left( 0.156 \right) \] \[ \text{pKa} = 3.05 - (-0.81) \] \[ \text{pKa} = 3.05 + 0.81 \] \[ \text{pKa} = 3.86 \][/tex]
Finally, to find the [tex]\(K_a\),[/tex] we take the negative antilogarithm (or 10 raised to the power of the negative pKa):
[tex]\[ K_a = 10^{-\text{pKa}} \] \[ K_a = 10^{-3.86} \] \[ K_a = 1.1 \times 10^{-5} \][/tex]
Therefore, the [tex]\(K_a\)[/tex] of nicotinic acid is approximately [tex]\(1.1 \times 10^{-5}\).[/tex]
A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant concentration of 5.6 mg/L. At time zero the concentration in the lake is 5.6 mg/L before an industry begins discharging waste with a flow of 0.7 m3/s also into the lake with a pollutant concentration of 300 mg/L. The decay coefficient for the pollutant in the lake is 0.2 per day. What is concentration leaving the lake one day after the pollutant is added?
Explanation:
The given data is as follows.
Volume of lake = [tex]15 \times 10^{6} m^{3}[/tex] = [tex]15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}[/tex]
Concentration of lake = 5.6 mg/l
Total amount of pollutant present in lake = [tex]5.6 \times 15 \times 10^{9} mg[/tex]
= [tex]84 \times 10^{9}[/tex] mg
= [tex]84 \times 10^{3}[/tex] kg
Flow rate of river is 50 [tex]m^{3} sec^{-1}[/tex]
Volume of water in 1 day = [tex]50 \times 10^{3} \times 86400 liter[/tex]
= [tex]432 \times 10^{7}[/tex] liter
Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are [tex]2.9792 \times 10^{10} mg[/tex] or [tex]2.9792 \times 10^{4} kg[/tex]
Flow rate of sewage = [tex]0.7 m^{3} sec^{-1}[/tex]
Volume of sewage water in 1 day = [tex]6048 \times 10^{4}[/tex] liter
Concentration of sewage = 300 mg/L
Total amount of pollutants = [tex]1.8144 \times 10^{10} mg[/tex] or [tex]1.8144 \times 10^{4}kg[/tex]
Therefore, total concentration of lake after 1 day = [tex]\frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l[/tex]
= 6.8078 mg/l
[tex]k_{D}[/tex] = 0.2 per day
[tex]L_{o}[/tex] = 6.8078
Hence, [tex]L_{liquid}[/tex] = [tex]L_{o}(1 - e^{-k_{D}t}[/tex]
[tex]L_{liquid}[/tex] = [tex]6.8078 (1 - e^{-0.2 \times 1})[/tex]
= 1.234 mg/l
Hence, the remaining concentration = (6.8078 - 1.234) mg/l
= 5.6 mg/l
Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.
How do you determine the environmental impact of human activities?
Answer:
Environment refers to everything that surrounds an individual and interacts between them. The factors that control the environment can be biotic and abiotic.
Humans have greatly affected the environment. Some of the ways in which the environment is affected by humans are as follows-
(1) Humans have constructed industries and factories that have released a huge amount of toxic gases into the atmosphere.
(2) These harmful gases have increased the earth's global temperature. As a result of which the global warming effect has increased.
(3) The waste materials eliminated from these industries mix with the rivers and streams and pollute the water. It degrades water quality.
(4) The fossil fuels are exhausted at a very high rate.
(5) The spilling of the oils in the oceans has affected the marine species drastically.
(6) Due to the extensive mining at different places, soil fertility has decreased considerably.
(7) Cutting down trees for settlement purposes and other infrastructures.
The environmental impact of human activities can be measured using the Ecological Footprint model, which calculates the resources consumed and waste generated by our actions. The Precautionary Principle is critical when understanding environmental effects is limited, advocating for caution. Reducing our carbon footprint through simple tasks like walking instead of driving can greatly contribute to environmental health.
Explanation:To determine the environmental impact of human activities, one can use the Ecological Footprint model developed by William Rees and Mathis Wackernagel. This model measures the amount of biologically productive land and water area required to produce the resources a person, population, or activity consumes and to absorb the waste they generate, given prevailing technology and resource management practices.
Appraising the ways in which human intervention has altered the environment often leads to a blurred line between 'natural' and human-influenced ecosystems. One principle to consider when the effects of an activity on the environment are not well understood is the Precautionary Principle. This suggests that in the absence of clear data, we must assume that harm to the environment could occur and therefore proceed cautiously with any such activities.
Individual choices, like walking instead of driving, can lead to reducing one's overall carbon footprint. This collective effort is critical as it can mitigate some of the negative impacts humans have on the environment, including air pollution, which is significantly attributed to human activity such as transport and industrial processes. Moreover, adjusting consumption patterns and holding corporations accountable for environmental degradation are also key steps towards sustainability.
Radioactive decay can be described by the following equation where is the original amount of the substance, is the amount of the substance remaining after time , and is a constant that is characteristic of the substance. For the radioactive isotope phosphorus-32, is . If the original amount of phosphorus-32 in a sample is 57.9 mg, how much phosphorus-32 remains after 13.7 days have passed
Answer:
29.8 mg
Explanation:
Just slightly less than 1/2 decay
All the following involve an increase in entropy except O most endothermic reactions O sublimation O vaporization O combustion
Answer:
most endothermic reactions
Explanation:
In endothermic reactions, the entropy of the environment decreases.
The entropy is the measure of the disorder of a specific system.
An endothermic reaction is every chemical reaction in which energy is absorbed.
In a comparison of an endodermic reaction in which energy is released in light or heat form.
Which has more kenetic energy: A 150 g bullet fired from a gun achieving a velocity of 440 m/s and a standard baseball weighing 51/4 oz being pitched at he MLB record speed of 105.1 mph. Calculate kinetic energy in Joules.
Answer:
Bullet has more kinetic energy then baseball.
Explanation:
Mass of the bullet , m = 150 g = 0.150 kg
Velocity of the bullet = v = 440 m/s
Kinetic energy =[tex]\frac{1}{2}\times mass\times (velocity)^2[/tex]
Kinetic energy of the bullet = K.E
[tex]K.E=\frac{1}{2}mv^2=\frac{1}{2}\times 0.150 kg\times (440 m/s)^2[/tex]
[tex]=14,520 Joules[/tex]
Mass of the baseball, m' = [tex]\frac{51}{4} oz[/tex]= 0.36146 kg
(1 ounce = 0.0283495 kg)
Velocity of the baseball= v' = 105.1 mph
1 mile = 1609.34 m
1 hour = 3600 seconds
[tex]v'=\frac{ 105.1\times 1609.34 m}{3600 s}=46.98 m/s[/tex]
Kinetic energy of the baseball = K.E'
[tex]K.E'=\frac{1}{2}m'v'^2=\frac{1}{2}\times 0.36146 kg\times (46.98 m/s)^2[/tex]
[tex]=398.83 Joules[/tex]
14,520 Joules > 398.83 Joules
K.E > K.E'
Bullet has more kinetic energy then baseball.
Determine the approximate density of a high-leaded brass that has a composition of 64.5 wt% Cu, 33.5 wt% Zn and 2.0 wt% Pb. The densities of Cu, Zn, and Pb are 8.94.7.13 and 11.35 g/cm,respectively. g/cm3 e Textbook and Media
The approximate density of a high-leaded brass composed of 64.5% Cu, 33.5% Zn and 2.0% Pb can be calculated using the rule of mixtures; Density = (wt% Cu/100)*ρCu + (wt% Zn/100)*ρZn + (wt% Pb/100)*ρPb.
Explanation:This question is about the calculation of the approximate density of a high-leaded brass that has a composition of 64.5 wt% Cu, 33.5 wt% Zn and 2.0 wt% Pb with known individual densities. The density of a mixture can be calculated using the rule of mixtures. The rule of mixtures states that the density of a mixture is equal to the mass fractions of each component times their respective densities. The calculation is as follows:
Density of Brass = (wt% Cu/100)*ρCu + (wt% Zn/100)*ρZn + (wt% Pb/100)*ρPb
By substituting the values, we get:
Density of Brass = (64.5/100)*8.94 g/cm³ + (33.5/100)*7.13 g/cm³ + (2.0/100)*11.35 g/cm³
Hence, the approximate density of the high-leaded brass can be determined by calculating the above equation.
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Calculate the number of oxygen atoms in a 120.0 g sample of glucose (C H,206). Be sure your answer has a unit symbol if necessary, and round it to 4 significant digits. Oro X 5 ?
There are approximately 6.011 × 10²⁴ oxygen atoms in the 120.0 g sample of glucose.
The chemical formula for glucose is C₆H₁₂O₆, indicating 6 carbon, 12 hydrogen, and 6 oxygen atoms. The molar mass of glucose may be used to calculate the number of oxygen atoms in 120.0 g of glucose:
Glucose molecule mass (C₆H₁₂O₆) = (6 * carbon) + (12 * hydrogen) + (6 * oxygen)
≈ (6 * 12.01 g/mol) + (12 * 1.008 g/mol) + (6 * 16.00 g/mol)
The molar mass may be used to compute glucose moles in the 120.0 g sample:
To calculate the number of moles of glucose, divide the sample mass by the molar mass: 120.0 g / 72.06 g/mol = 1.664 moles.
Since each glucose molecule has 6 oxygen atoms, we may compute the total:
The total quantity of oxygen atoms is calculated by multiplying the number of glucose moles by the number of oxygen atoms per molecule: 1.664 moles * 6 = 9.984 moles.
Oxygen atoms total roughly 9.984 × 6.022 × 10²³ atoms/mole
= 6.011 × 10²⁴ atoms.
In a 120.0 g glucose sample, there are roughly 6.011 × 10²⁴ oxygen atoms.
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Final answer:
To calculate the number of oxygen atoms in a 120.0 g sample of glucose, use the mole ratio to find the number of moles of oxygen. Then, convert moles of oxygen to atoms using Avogadro's number.
Explanation:
To calculate the number of oxygen atoms in a 120.0 g sample of glucose, we need to determine the number of moles of glucose and use the mole ratio to find the number of moles of oxygen. The molar mass of glucose is 180.16 g/mol, so the number of moles is 120.0 g / 180.16 g/mol = 0.6664 mol. According to the balanced chemical equation, one mole of glucose reacts with 6 moles of oxygen to yield 6 moles of CO2 and 6 moles of H2O. Therefore, the number of moles of oxygen is 0.6664 mol x 6 mol O2 / 1 mol glucose = 3.9984 mol O2. Finally, to convert moles of oxygen to atoms, we use Avogadro's number (6.022 × 10^23 atoms/mol), so the number of oxygen atoms is 3.9984 mol O2 x 6.022 × 10^23 atoms/mol = 2.406 × 10^24 oxygen atoms.
A bottle of wine contains 12.8% ethanol by volume. The density of ethanol (CH3OH) is 0.789 g/cm. Calculate the concentratic ethanol in wine in terms of mass percent and molality. Mass percent Molality =
Final answer:
The mass percent of ethanol in wine is approximately 10.38%, and the molality of ethanol in wine is approximately 2.51 mol/kg. These values are calculated from the given volume percent and the density of ethanol by first determining the mass of ethanol and then the number of moles of ethanol.
Explanation:
Calculating Mass Percent and Molality of Ethanol in Wine
We are given that wine contains 12.8% ethanol by volume, and the density of ethanol is 0.789 g/cm3. To find the mass percent and molality of ethanol in wine, we proceed with the following steps:
Calculate the volume of ethanol in a 100 mL sample of wine. Since the wine is 12.8% ethanol by volume, we have 12.8 mL of ethanol.Using the density of ethanol, compute the mass of ethanol. Mass = Volume x Density, which gives us 12.8 mL x 0.789 g/mL = 10.0992 g of ethanol.The mass percent of ethanol is then calculated by comparing this mass to the total mass of the wine, assuming the rest is water (density 1.00 g/mL). If we have 87.2 mL of water, the mass of water is 87.2 g, and the total mass of the solution is 10.0992 g (ethanol) + 87.2 g (water) = 97.2992 g. Thus, the mass percent of ethanol is (10.0992 g / 97.2992 g) x 100% = approximately 10.38%.To calculate molality, we first find the moles of ethanol using its molar mass (46.07 g/mol): Moles of ethanol = 10.0992 g / 46.07 g/mol = approximately 0.219 moles of ethanol.Molality is moles of solute (ethanol) per kilogram of solvent (water). With 0.0872 kg of water, the molality is 0.219 moles / 0.0872 kg = approximately 2.51 mol/kg.Through these calculations, we have determined both the mass percent and molality of ethanol in the wine sample.
Which of the following reactions is incorrectly labeled?
a. HCl(aq) + NaOH(aq) = NaCl(aq) + H2O(l) – Precipitation reaction.
b. CH4(g) + 3 O2(g) = CO2(g) + 2 H2O(l) – Combustion reaction.
c. No right choice.
Answer:
The option a termed as precipitation reaction is incorrectly labelled.
Explanation:
The chemical reactions are classified based on the reactants used and products formed in a reaction. They are decomposition reaction, single displacement reaction, double displacement reaction, acid-base neutralisation reaction, precipitation reaction, combustion reaction, redox reaction and organic reaction.
Among these, the given options are labelled as precipitation and combustion reaction. The one which is labelled as combustion reaction is correct because combustion reactions occur in the presence of oxygen only and the products of combustion reaction should include water, oxygen or carbon and heat.
The other option which is labelled as precipitation reaction is incorrect because precipitation reaction occurs when an ionic substance will come out of a solution due to heating it or stirring it making the solubility of the ionic substance in a solution zero such that it will come out as solid and form a layer at the bottom of the solution.
But in this case all the products are in aqueous state, there is absence of any ionic substance in solid state, so the option which is labelled as precipitation reaction is incorrectly labelled.
Final answer:
The incorrectly labeled reaction is a. HCl(aq) + NaOH(aq) = NaCl(aq) + H₂O(l) as a Precipitation reaction; it should be labeled as a Neutralization reaction.
Explanation:
The reaction that is incorrectly labeled is a. HCl(aq) + NaOH(aq) = NaCl(aq) + H₂O(l) - Precipitation reaction. This reaction is actually an example of a neutralization reaction, where an acid (HCl) and a base (NaOH) react to form water and a salt (NaCl). A precipitation reaction involves the formation of an insoluble solid, known as a precipitate, from the reaction of two soluble compounds, which does not occur in this case.
Combustion reaction: CH₄(g) + 3O₂(g) = CO₂(g) + 2H₂O(l)
Five million gallons per day (MGD) of wastewater, with a concentration of 10.0 mg/L of a conservative pollutant, is released into a stream having an upstream flow of 10 MGD and pollutant concentration of 3.0 mg/L. (a) What is the concentration in ppm just downstream? (b) How many pounds of substance per day pass a given spot downstream? (You may want the conversions 3.785 L/gal and 2.2 kg/lbm from Appendix A.)
Answer:
a) The concentration in ppm (mg/L) is 5.3 downstream the release point.
b) Per day pass 137.6 pounds of pollutant.
Explanation:
The first step is to convert Million Gallons per Day (MGD) to Liters per day (L/d). In that sense, it is possible to calculate with data given previously in the problem.
Million Gallons per day [tex]1 MGD = 3785411.8 litre/day = 3785411.8 L/d[/tex]
[tex]F_1 = 5 MGD (\frac{3785411.8 L/d}{1MGD} ) = 18927059 L/d\\F_2 =10 MGD (\frac{3785411.8 L/d}{1MGD} )= 37854118 L/d [/tex]
We have one flow of wastewater released into a stream.
First flow is F1 =5 MGD with a concentration of C1 =10.0 mg/L.
Second flow is F2 =10 MGD with a concentration of C2 =3.0 mg/L.
After both of them are mixed, the final concentration will be between 3.0 and 10.0 mg/L. To calculate the final concentration, we can calculate the mass of pollutant in total, adding first and Second flow pollutant, and dividing in total flow. Total flow is the sum of first and second flow. It is shown in the following expression:
[tex]C_f = \frac{F1*C1 +F2*C2}{F1 +F2}[/tex]
Replacing every value in L/d and mg/L
[tex]C_f = \frac{18927059 L/d*10.0 mg/L +37854118 L/d*10.0 mg/L}{18927059 L/d +37854118 L/d}\\C_f = \frac{302832944 mg/d}{56781177 L/d} \\C_f = 5.3 mg/L[/tex]
a) So, the concentration just downstream of the release point will be 5.3 mg/L it means 5.3 ppm.
Finally, we have to calculate the pounds of substance per day (Mp).
We have the total flow F3 = F1 + F2 and the final concentration [tex]C_f[/tex]. It is required to calculate per day, let's take a time of t = 1 day.
[tex]F3 = F2 +F1 = 56781177 L/d \\M_p = F3 * t * C_f\\M_p = 56781177 \frac{L}{d} * 1 d * 5.3 \frac{mg}{L}\\M_p = 302832944 mg[/tex]
After that, mg are converted to pounds.
[tex]M_p = 302832944 mg (\frac{1g}{1000 mg} ) (\frac{1Kg}{1000 g} ) (\frac{2.2 lb}{1 Kg} )\\M_p = 137.6 lb[/tex]
b) A total of 137.6 pounds pass a given spot downstream per day.
To find the concentration in ppm just downstream, we calculate the mass of the pollutant using the flow rate and concentration. The concentration in ppm just downstream is 50 ppm. To find the number of pounds of substance per day passing a given spot downstream, we convert the mass from grams to pounds.
Explanation:In order to find the concentration in ppm just downstream, we need to calculate the mass of pollutant that is being released into the stream. We can use the following formula:
Mass = Flow rate x Concentration
Using the given flow rates and concentrations:
Upstream Mass = 10 MGD x 3.0 mg/L = 30 mg/L
Downstream Mass = 5 MGD x 10.0 mg/L = 50 mg/L
So the concentration in ppm just downstream is:
50 mg/L x (1 ppm/1 mg/L) = 50 ppm
To find the number of pounds of substance per day passing a given spot downstream, we will need to calculate the mass in pounds. We can use the following conversions:
1 lbm = 2.2 kg
1 kg = 1000 g
1 g = 1000 mg
Using the given flow rate and concentration:
Downstream Mass (in g/day) = 5 MGD x 10.0 mg/L = 50,000 g/day
Downstream Mass (in kg/day) = 50,000 g/day x (1 kg/1000 g) = 50 kg/day
Downstream Mass (in lbm/day) = 50 kg/day x (1 lbm/2.2 kg) = 22.7 lbm/day
So the number of pounds of substance per day passing a given spot downstream is approximately 22.7 pounds.
Refer to the following unbalanced equation: CO2 H20 C6H14 O2>CO2 + H2O What mass of oxygen (O2) is required to react completely with 10.4 g of C6H14? D. 36.7 g Selected Answer: D. 36.7 g Correct Answer:
Answer: The mass of oxygen gas required is 36.7 grams.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
Given mass of hexane = 10.4 g
Molar mass of hexane = 86.18 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of hexane}=\frac{10.4g}{86.18g/mol}=0.12mol[/tex]
The chemical equation for the combustion of hexane follows:
[tex]2C_6H_{14}+19O_2\rightarrow 12CO_2+14H_2O[/tex]
By stoichiometry of the reaction:
2 moles of hexane reacts with 19 moles of oxygen gas
So, 0.12 moles of hexane will react with = [tex]\frac{19}{2}\times 0.12=1.14mol[/tex] of oxygen gas.
Now, calculating the mass of oxygen gas by using equation 1, we get:
Molar mass of oxygen gas = 32 g/mol
Moles of oxygen gas = 1.14 moles
Putting values in equation 1, we get:
[tex]1.14mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=36.7g[/tex]
Hence, the mass of oxygen gas required is 36.7 grams.
Several models of atoms were mentioned in the lesson this week. Plum pudding, planetary and quantum mechanical. Which reason below explains why there are different atom models?
Answer:
Explanation:
There are different atomic models due to the fact that with advancement in technology and research, the atom began to open up. More scientists began to formulate models that would completely unify the different observations about the atom.
From the different model of atom, noticeable progressions could be observed in terms of sophistication and what they entail. As more details of the atoms were revealed, the need to modify the model becomes important.
Alcohols like ethanol CH3CH2OH, have a proton attached to the oxygen atom that can be acidic. Can a base like CH3NH2 deprotonate the -OH group? (pka CH3CH2OH=15.7 pKa CH3NH3=10.8) justify your answer
Answer:
No. CH₃NH₂ cannot deprotonate CH₃CH₂OH.
Explanation:
Organic species like ethanol and methylamine can behave as acid or bases. The pKa value is necessary to indentify in a reaction if the species can act as an acid or a base.
In this case, despite alcohols being recognised as acids and amines as bases, the amine CH₃NH₂ has an smaller pKa than the alcohol CH₃CH₂OH , which means that CH₃NH₂ is a stronger acid than CH₃CH₂OH . Therefore, CH₃NH₂ will not deprotonate the -OH group.
The likelihood of methylanime (CH3NH2) deprotonating the -OH group of ethanol (CH3CH2OH) is low as the conjugate acid CH3NH3+ (formed after CH3NH2 accepts a proton) is a stronger acid than ethanol, demonstrated by its lower pKa value (10.8 compared to 15.7).
Explanation:In the realm of chemistry, it's known that the strength of an acid is determined by its pKa value. The lower the pKa, the stronger the acid. As given, ethanol (CH3CH2OH) has a pKa of 15.7 and CH3NH3+ (ammonium ion) has a pKa of 10.8. As such, ethanol is a weaker acid than the ammonium ion.
For a base to deprotonate an acid, the conjugate acid of that base (formed after it accepts a proton) has to be weaker than the initial acid. In your case, the conjugate acid of CH3NH2 (methylamine) would be CH3NH3+. As CH3NH3+ is a significantly stronger acid than ethanol, it is therefore unlikely that methylamine could effectively deprotonate the -OH group in ethanol.
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What is the molar concentration of a solution with 44 grams of CO2 in 0.5 liter solution? a) 2 mole/liter b) 4 mole/liter c) 1 mole/liter d) 3 mole/liter
Answer: The correct answer is Option a.
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}[/tex]
We are given:
Mass of solute (carbon dioxide) = 44 g
Molar mass of carbon dioxide = 44 g/mol
Volume of solution = 0.5 L
Putting values in above equation, we get:
[tex]\text{Molarity of solution}=\frac{44g}{44g/mol\times 0.5L}\\\\\text{Molarity of solution}=2mol/L[/tex]
Hence, the correct answer is Option a.
write a reaction to describe the behavior of the following substances in water. please include all phases.
NF3 (gas)
CH2CH2 (g)
3NF3 + 5H2O → HNO3 + 2NO + 9HF
Nitrogen fluoride reacts with water to produce nitric acid, nitric oxide, and hydrogen fluoride. The reaction slowly takes place in a boiling solution.
CH2CH2 + H2O → CH3CH2OH
Ethylene is a hydrocarbon with water that creates ethanol and ethanol is an alcohol