Infrared spectroscopy is a technique used to identify functional groups in molecules by measuring absorption frequencies. It helps determine the presence or absence of specific groups in a molecule.
Explanation:Infrared spectroscopy is a technique used to identify different functional groups in molecules based on their characteristic absorption frequencies. It involves passing infrared light through a sample and measuring the frequencies that are absorbed. Different functional groups have different absorption frequencies, allowing scientists to determine the presence or absence of specific groups in a molecule. For example, if a sample absorbs in the carbonyl frequency range but not in the alkyne range, it can be inferred that the molecule contains a carbonyl group but not an alkyne.
Assume that aniline is insoluble in water. The normal boiling points of aniline and water are 184.1°C and 100°C respectively. At a pressure of I atm, the boiling point of a mixture of aniline and water is a) 184.1°C b) 100°C c) less than 100°C d) greater than 100°C but less than 184.1°C.
Answer:
At a pressure of I atm, the boiling point of a mixture of aniline and water is b) 100°C.
Explanation:
Assuming that aniline is insoluble in water there will be no interaction between liquids. There will be no positive interactions that can increase the boiling point or negative interactions that can decrease the boiling point. Thus the boiling points of each substance will not be affected. Since the boiling point of water is 100ºC, the mixture will start to evaporate at 100ºC.
Answer:
B.)
Explanation:
In order to study hydrogen halide decomposition, a researcher fills an evacuated 1.11 L flask with 0.822 mol of HI gas and allows the reaction to proceed at 428°C: 2HI (g) ⇋ H2(g) + I2(g)
At equilibrium, the concentration of HI = 0.055 M. Calculate Kc. Enter to 4 decimal places.
HINT: Look at sample problem 17.6 in the 8th ed Silberberg book. Write a Kc expression. Find the initial concentration. Fill in the ICE chart. Put the E (equilibrium) values into the Kc expression.
Answer:
Kc = 168.0749
Explanation:
2HI(g) ↔ H2(g) + I2(g)initial mol: 0.822 0 0
equil. mol: 2(0.822 - x) x x
∴ [ HI ]eq = 0.055 mol/L = 2(0.822 - x) / (1.11 L )
⇒ 1.644 - 2x = 0.055 * 1.11
⇒ 1.644 = 2x + 0.06105
⇒ 2x = 1.583
⇒ x = 0.7915 mol equilibrium
⇒ [ H2 ] eq = 0.7915mol / 1.11L = 0.7130 M = [ I2 ] eq
⇒ Kc = ([ H2 ] * [ I2 ]) / [ HI ]²
⇒ Kc = ( 0.7130² ) / ( 0.055² )
⇒ Kc = 168.0749
The equilibrium constant (Kc) for the reaction where 2HI decomposes into H2 and I2 at a given temperature can be calculated using the ICE method. After setting up the Kc expression and solving for x, the concentration changes of HI, H2, and I2 can be used to find Kc, which is approximately 0.1176.
Explanation:To calculate the equilibrium constant (Kc) for the decomposition of hydrogen iodide (HI) into hydrogen gas (H₂) and iodine gas (I₂), we must first write the Kc expression for the reaction:
2HI (g) ⇌ H₂ (g) + I₂ (g)
The Kc expression is Kc = [H₂][I₂] / [HI]^2.
Next, we use an ICE (Initial, Change, Equilibrium) table to keep track of the concentrations throughout the reaction:
Initial: [HI] = 0.822 mol / 1.11 L = 0.7409 M (since the flask is evacuated, [H₂] and [I₂] initially are 0)Change: At equilibrium, [HI] has decreased by x to (0.7409 - 2x) M, since HI decomposes into one mole of H₂ and one mole of I₂ for every two moles of HI that decompose.Equilibrium: [HI] = 0.055 M, so 0.7409 - 2x = 0.055 M. [H₂] and [I₂] both increase by x to x M given the stoichiometry of the reaction.To calculate x, solve the equation 0.7409 - 2x = 0.055, yielding x = 0.34295 M. Now we know at equilibrium:
[HI] = 0.055 M[H₂] = x = 0.34295 M[I₂] = x = 0.34295 MInserting these values into the Kc expression we get:Kc = (0.34295)(0.34295) / (0.055)^2
Simplifying gives us:Kc = 0.1176 to 4 decimal places.
Know the general characteristics of the following; mixtures, solutions, compounds, molecules, and acids and bases.
Explanation:
A mixture is material which is composed up of two or more substances and these substances are physically combined. The identities of each specie in the mixture are retained.
A solution is a type of the homogeneous mixture which is composed of two or more substances. The specie which is dissolved into the another substance is known as solute and in which it is dissolved is known as a solvent.
A compound is the substance which is formed when two or more elements are bonded together chemically.
A molecules are electrically neutral and is a combination of of two or more atoms which are held together by means of chemical bonds.
Acids are the species which furnish hydrogen ions in the solution or is capable of forming bonds with electron pair species as they are electron deficient species.
Bases are the species which furnish hydroxide ions in the solution or is capable of forming bonds with electron deficient species as they are electron rich species.
How much magnesium sulfate heptahydrate is required to prepare 200 mL of 0.05 M solution? 10 g 2.46 g 0 0.05 g 12.38 24.65 g
Final answer:
To prepare a 0.05 M magnesium sulfate heptahydrate solution with a volume of 200 mL, you will need 2.46 grams of magnesium sulfate heptahydrate.
Explanation:
To calculate the amount of magnesium sulfate heptahydrate required to prepare 200 mL of 0.05 M solution, we need to use the formula:
moles = molarity x volume
First, we convert the volume from milliliters to liters:
200 mL x (1 L/1000 mL) = 0.2 L
Next, we substitute the given values into the formula:
moles = 0.05 mol/L x 0.2 L = 0.01 mol
Finally, we calculate the mass of magnesium sulfate heptahydrate using its molar mass:
moles = mass (g) / molar mass (g/mol)
0.01 mol = mass (g) / 246.48 g/mol
mass (g) = 0.01 mol x 246.48 g/mol = 2.46 g
Round 129.752416 to the requested number of significant figure a. 3 significant figures c. 6 significant figures_22 b. 4 significant figures d. 7 significant figures
To round 129.752416 to the requested number of significant figures:
- For 3 significant figures, round down to 129.7.
- For 4 significant figures, round up to 129.8.
- For 6 and 7 significant figures, keep the given digits.
To round 129.752416 to the requested number of significant figures:
For 3 significant figures, we look at the digit next to the third significant figure. If it is 5 or greater, we round up, otherwise, we round down. In this case, the digit next to the third significant figure is 2, which is less than 5, so the rounded number is 129.7.For 4 significant figures, we follow the same process as before but include the digit next to the fourth significant figure. The digit next to the fourth significant figure is 4, which is less than 5, so the rounded number is 129.8.For 6 significant figures, we include all the digits given. The rounded number is 129.752416.For 7 significant figures, we include all the digits given. The rounded number is 129.752416.Consider the formation of nitrogen dioxide from nitric
oxideand oxygen.
2NO(g) + O2(g) -- 2NO2(g)
If 9L of NO are reacted with excess 02 at STP, what is
thevolume in liters of the NO2 produced?
Answer:
9 L
Explanation:
According to the question , the given reaction is -
2NO(g) + O₂(g)------->2NO₂(g)
Since ,
At STP ,
One mole of a gas occupies the volume of 22.4 L.
Hence , as given in the question -
9 L of NO , i.e .
22.4 L = 1 mol
1 L = 1 / 22.4 mol
9 L = 1 / 22.4 * 9 L = 0.40 mol
From the chemical reaction ,
The Oxygen is in excess , hence NO becomes the limiting reagent , and will determine the moles of product .
Hence ,
2 moles of NO will produce 2 moles of NO₂.
Therefore ,
0.40 mol of NO will produce 0.40 mol of NO₂.
Hence , the volume of NO₂ can be calculated as -
1 mol = 22.4 L
0.40 mol = 0.40 * 22.4 L = 9 L
To find the volume of NO2 produced, calculate the number of moles of NO reacted using the ideal gas equation. Convert moles to volume using the molar volume of gases at STP.
Explanation:To find the volume of NO2 produced, we first need to determine the number of moles of NO reacted. Using the ideal gas equation, we can calculate the number of moles:
n = PV/RT = (1 atm)(9 L)/(0.0821 L atm/(mol K))(273 K) = 0.408 mol
Since the reaction has a 2:2 mole ratio, 0.408 mol of NO will produce 0.408 mol of NO2. To convert moles to volume at STP, we can use the molar volume of gases, which is 22.4 L/mol:
Volume of NO2 = 0.408 mol * 22.4 L/mol = 9.1392 L
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Nitrogen oxide is a pollutant commonly found in
smokestackemissions. One way to remove it is to react it with
ammonia.
How many liters of ammonia are required to change 12.8L of
nitrogenoxide to nitrogen gas? Assume 100% yield and that all gases
aremeasured at the same temperature and pressure.
Answer:
You need 8,53 L of ammonia
Explanation:
Global reaction of remotion of nitrogen oxide with ammonia is:
4 NH₃ + 6 NO ⇒ 5 N₂ + 6 H₂O
This balanced equation shows that 4 NH₃ moles reacts with 6 NO moles.
With 100% yield and temperature and pressure constants it is possible to apply Avogadro's law. This law is an experimental gas law relating the volume of a gas to the amount of substance of gas present. The formula is:
[tex]\frac{V_{1} }{n_{1} } = \frac{V_{2} }{n_{2} }[/tex]
Where:
V₁ is the NO volume = 12,8L
n₁ are NO moles = 6
n₂ are NH₃ moles = 4
V₂ is NH₃ volume, the unknown.
Thus, V₂ are 8,53 L of ammonia
I hope it helps!
Write 41,405,000 in Engineering Notation with 3 significant figures.
Answer: [tex]41.4\times 10^{6}[/tex]
Explanation:
Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.
Rules for significant figures:
Digits from 1 to 9 are always significant and have infinite number of significant figures.
All non-zero numbers are always significant.
All zero’s between integers are always significant.
All zero’s preceding the first integers are never significant.
All zero’s after the decimal point are always significant.
Engineering notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form times 10 raise to the power. It is similar to the scientific notation but in engineering notation, the powers of ten are always multiples of 3.
The engineering notation written in the form:
[tex]a\times 10^b[/tex]
where,
a = the number which is greater than 0 and less than 999
b = an integer multiple of 3
Now converting the given value of 41,405,000 into engineering notation, we get [tex]41.4\times 10^{6}[/tex]
Therefore the scientific notation with 3 significant figures is [tex]41.4\times 10^{6}[/tex]
A slice of Swiss cheese contains 47 mg of sodium. (a) What is this mass in grams? .047 (b) What is this mass in ounces? (16 oz = 453.6 g) 0.0016 X (c) What is this mass in pounds? (1 lb 453.6 g) 0.0001x
Based on conversion ratios, 47 mg of sodium would be:
a. 0.047 grams. b. 0.00166 ounces.c. 0.000104 pounds.Mass in grams1g is 1,000 grams.
47 mg would be:
= 0.47 / 1,000
= 0.047 grams
Mass in Ounces16 ounces is 453.6 grams so 0.047 grams would be:
= (0.047 x 16) / 453.6
= 0.00166 ounces.
Mass in Pounds1 pound is 453.6 grams so 0.047 grams would be:
= 0.047 / 453.6
= 0.000104 pounds
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To convert 47 mg of sodium to grams, divide by 1000. To convert grams to ounces, use the conversion factor 1 oz = 28.35 g. To convert grams to pounds, use the conversion factor 1 lb = 453.6 g.
Explanation:To convert 47 mg to grams, divide by 1000:
47 mg = 47/1000 = 0.047 grams
To convert grams to ounces, use the conversion factor:
0.047 g x (1 oz/28.35 g) = 0.00166 oz
To convert grams to pounds, use the conversion factor:
0.047 g x (1 lb/453.6 g) = 0.0001035 lb
Calculate the molarity of a solution that contains 3.11 mol of NaNO3 dissolved in 2.50 L. Enter your answer in the provided box.
Answer:
Molarity of a solution that contains 3.11 mol of NaNO3 is 1,24 M
Explanation:
We understand molarity as the number of moles of solute that are contained in 1 L of solution, then if in a solution of 2.50 L we have 3.11 moles, it remains to calculate how many moles do we have in 1 liter.
2,50 L .......... 3,11 moles
1 L .................. x
X = ( 1 L x 3,11 moles) / 2,50 L = 1,24
Write 10,847,100 in Scientific Notation with 4 significant figures.
Answer:
The number 10,847,100 in Scientific Notation is [tex]1.0847x10^{7}[/tex]
Explanation:
Scientific notation is an easy form to write long numbers and it is commonly used in the scientific field. To write a long number in a shorter way it is necessary to 'move' the decimal point to the left the number of positions that are necessary until you get a unit. Then you write the number and multiplied it by 10 raised to the number of positions you moved the decimal point. In this case, it is necessary to move the decimal point 7 positions so, we multiply the number by 10 raised to 7.
What is the pH of a 0.05 M solution of TRIS
acid(pKa = 8.3)?
Answer:
pH of the solution = 4.80
Explanation:
pKa = 8.3
Concentration of tris acid = 8.3
[tex]pH = \frac{1}{2} \times p_{ka} - \frac{1}{2} logC[/tex]
[tex]pH = \frac{1}{2} \times 8.3 - \frac{1}{2}\times log(0.05)[/tex]
[tex]= \frac{1}{2} \times 8.3 - \frac{1}{2}\times log(5\times 10^{-2})[/tex]
[tex]= 4.15 - \frac{1}{2}\times (-1.30103)[/tex]
[tex]= 4.15 - (-0.650515)[/tex]
[tex]= 4.15 + 0.650515[/tex]
= 4.80
pH of the solution = 4.80
What changes are observed on heat capacity ratio and output temperatures by increasing the specific heat capacity of both the fluids?
a-The heat capacity ratio and output temperature both increases
b- The heat capacity ratio increases but output temperature don’t change
c- The output temperature increases but heat capacity ratio remains same
d- No heat capacity ratio nor output temperature changes
Answer:
b- The heat capacity ratio increases but output temperature don’t change
Explanation:
The heat capacity is the amount of energy required to raise the temperature of a body, by 1 degree. On the other hand, the specific heat capacity is the amount of heat required to raise the temperature of a of unit mass of a material by 1 degree.
Heat capacity is an extensive property meaning its value depends on the amount of material. Specific heat capacity is found by dividing heat capacity by the mass of the sample, thus making it independent of the amount (intensive property). So if the specific heat capacity increases and the mass of the sample remains the same, the heat capacity must increase too. Because of that options c and d that say that heat capacity reamins same are INCORRECT.
On the other hand, in which has to be with options a and b both say that the heat capacity increases which is correct, but about the output temperatures what happens is that if we increase the specific heat capacity of both fluids that are involved in a process of heat exchange in the same value, the value of the output temperatures do not change so only option a is CORRECT.
The following data were measured for the reaction BF3(g)+NH3(g)→F3BNH3(g): Experiment [BF3](M) [NH3](M) Initial Rate (M/s) 1 0.250 0.250 0.2130 2 0.250 0.125 0.1065 3 0.200 0.100 0.0682 4 0.350 0.100 0.1193 5 0.175 0.100 0.0596 Part A What is the rate law for the reaction? What is the rate law for the reaction? rate=k[BF3]2[NH3] rate=k[BF3][NH3] rate=k[BF3][NH3]2 rate=k[BF3]2[NH3]2
The rate law for the reaction BF3(g)+NH3(g)→F3BNH3(g) based on the input data is rate = k[BF3][NH3], where k is the rate constant, [BF3] and [NH3] denote the molar concentrations of BF3 and NH3, indicating a first order relationship for both reactants.
Explanation:In order to determine the rate law for the reaction BF3(g)+NH3(g)→F3BNH3(g), we have to look at how the initial rate changes with respect to the change in concentration of the reactants. Using the given data, we can compare experiments where only one reactant's concentration is changed while the concentration of the other reactant remains constant. Upon analysis, we see that when the concentration of BF3 doubles, the rate also doubles, suggesting a first order relationship. Similarly, when the concentration of NH3 doubles, the rate doubles, indicating a first order relationship for NH3 as well. Hence, given that both BF3 and NH3 are first order, the rate law for the reaction should be: rate = k[BF3][NH3].
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The rate law for the reaction BF3(g) + NH3(g) → F3BNH3(g) is rate = k[BF3][NH3].
Explanation:
The rate law for the reaction BF3(g) + NH3(g) → F3BNH3(g) can be determined by analyzing the effect of changing concentrations on the initial rate of the reaction. By comparing the rates of reaction for different experiments, we can observe how changing the concentrations of the reactants affects the rate. In this case, it is clear that the rate of reaction is directly proportional to the concentration of BF3 and NH3, so the rate law for the reaction is rate = k[BF3][NH3].
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An interpretation of the results of many tests is called A) an experiment. B) a prediction D) a theory
Final answer:
An interpretation of the results of many tests is called a theory. It is a well-substantiated explanation of an aspect of the natural world, based on extensive research and experimentation.
Explanation:
An interpretation of the results of many tests is called a theory. A theory is a well-substantiated explanation of an aspect of the natural world. It is a scientific explanation that has been repeatedly tested and supported by many experiments.
For example, the theory of evolution is a well-supported explanation of how species have evolved over time. It is based on extensive research and experimentation.
It is important to note that a theory in science is different from the everyday use of the word, which often refers to a guess or speculation. In science, a theory is a well-tested and supported explanation.
A city in Laguna generates 0.96 kg per capita per day of Municipal Solid Waste (MSW). Makati City in Metro Manila generates 1.9 kg per capita per day of MSW. If both cities have population of 20000 inhabitants, determine the following
i. amount of MSW generated in each city
ii. number of trucks needed to collect the MSW twice weekly, if each truck has a
capacity of 4.4 tons and operates 5 days per week. Two loads per day at 75%
capacity is the travel quota per day
iii. -the number and volume of MSW in tons that enter the landfill
Answer:
i) amount of MSW generated:
Laguna: 19200 Kg/day
Makati: 38000 Kg/day
ii) number of trucks to collect twice weekly:
Laguna: 3 trucks
Makati: 5 trucks
iii) volume of MSW in tons that enter landfill/week:
Laguna: 147.84 ton/week
Makati: 292 ton/week
Explanation:
i) Laguna: 0.96 Kg person / day of MSW * 20000 = 19200 Kg MSW / day
⇒ Laguna: 19200 Kg/day * ( 7day/ week ) = 134400 Kg/week
Makati: 1.9 Kg person / day of MSW * 20000 = 38000 Kg MSW / day
⇒ Makati: 38000 Kg/day * ( 7day/week ) = 266000 Kg/week
ii) truck capacity = 4.4 ton * ( Kg / 0.0011 ton ) = 4000 Kg
⇒ quote/day = 4000 Kg * 0.75 = 3000 Kg
⇒ loads/day = 2 * 3000 kg = 6000 Kg
⇒ operate/week = 5 * 6000 Kg = 30000 Kg
∴ Laguna: number of trucks needed/week= 134400 / 30000 = 4.48 ≅ 5 trucks
⇒ number of trucks to collect twice weekly = 5 / 2 = 2.5 ≅ 3 trucks
∴ Makati : number of trucks needed/week = 266000 / 30000 = 8.86 ≅ 9 trucks
⇒ number of trucks to collect twice weekly = 9 / 2 = 4.5 ≅ 5 trucks
iii) enter landfill/week:
Laguna: 134400Kg MSW/week * ( 0.0011 ton/Kg ) = 147.84 ton/week
Makati: 266000Kg MSW/week * ( 0.0011 ton/Kg ) = 292 ton/week
Nuclei with the same mass number but different atomic numbers are called isobars. Consider Ca-40, Ca-41, K-41 and Ar-41. (a) Which of these are isobars? Which are isotopes? (b) What do Ca-40 and Ca-41 have in common? (c) Correct the statement (if it is incorrect): Atoms of Ca-41, K-41, and Ar-41 have the same number of neutrons.
Answer:
Part a:
Isobars: Ca-41, K-41 and Ar-41
Part b:
Number of proton
Part c:
Incorrect statement
Explanation:
Part a:
Nuclei with same mass number are called isobars.
Given:
Ca-40, Ca-41, K-41 and Ar-41
Ca-41, K-41 and Ar-41 have equal mass numbers, so these are isobars.
Part b:
Atomic number of Ca = 20
Atomic no. = Number of proton.
So, Ca-40 ans C-41 have same number of proton or in other words proton count is common in both.
Part c:
Number of neutrons in Ca-41
Atomic number = 20
Mass number = 41
Number of neutron = Mass number -atomic number
= 41 -20 = 21
Number of neutrons in K-41
Atomic number = 19
Mass number = 41
Number of neutron = Mass number -atomic number
= 41 -19 = 22
Number of neutrons in Ar-41
Atomic number = 18
Mass number = 41
Number of neutron = Mass number -atomic number
= 41 -18 = 23
So, the statement is incorrect.
The isobars among the examples given are Ca-41, K-41, and Ar-41. Ca-40 and Ca-41 are isotopes of calcium with different neutron counts. The statement about Ca-41, K-41, and Ar-41 having the same number of neutrons is incorrect because they have different atomic numbers.
Explanation:Nuclei with the same mass number but different atomic numbers are known as isobars. This means that while they have the same mass number (sum of protons and neutrons), they belong to different elements (different atomic numbers). Consequently, Ca-41, K-41, and Ar-41 are isobars since they all have a mass number of 41 but belong to different elements with atomic numbers 20 (calcium), 19 (potassium), and 18 (argon), respectively. Nevertheless, Ca-40 is not an isobar to these because it has a different mass number.
Isotopes are atoms with the same atomic number but differing numbers of neutrons, resulting in different mass numbers. In this case, Ca-40 and Ca-41 are isotopes because they are both calcium atoms (same atomic number of 20) with different mass numbers.
The commonality between Ca-40 and Ca-41 is they are isotopes of calcium and thus share the same atomic number and chemical properties, despite having different mass numbers due to different neutron counts.
The statement 'Atoms of Ca-41, K-41, and Ar-41 have the same number of neutrons' is incorrect because while these isotopes have the same mass number, they have different numbers of protons. To calculate the number of neutrons in these isotopes, you subtract the atomic number from the mass number and because the atomic numbers are different, the resulting neutron numbers will also differ.
Draw an arrow-pushing mechanism for the elimination reaction between NaOH in ethanol and each of the following haloalkanes: (E1/E2)
(i) 1-bromobutane;
(ii) 2-bromo-2-methylpentane.
Answer:
The mechanisms for the elimination reactions between NaOH in ethanol and the halogenoalkanes are demonstrated in the figure attached.
Explanation:
(i) 1-bromobutane will suffer elimination to for an alkene. The mechanism will be E2, which means that the attack and the elimination will occur simultaneously. This is the preferred mechanism because the bromine is in a primary carbon.
(ii) 2-bromo-2-methylpentane will suffer elimination to for an alkene. The mechanism will be E1, which means that the attack and the elimination will occur in two different steps. The bromine will be eliminated in the first step with the formation of a carbocation and in a second step the double bond will be formed after the anionic attack. This is the preferred mechanism because the bromine is in a terciary carbon which is able to stabilize the carbocation formed.
Modern commercial airliners are largely made of aluminum, a light and strong metal. But the fact that aluminum is cheap enough that airplanes can be made out of it is a bit of historical luck. Before the discovery of the Hall-Héroult process in , aluminum was as rare and expensive as gold. What would happen if airplanes had to be made of steel?
Answer:
Explanation:
The metallic properties of steel would strongly not favor its use in construction of aircraft.
Steel is an alloy of carbon and iron. It is denser and generally weighs more than equivalent amount of aluminum. This would imply that the weight of aircrafts would be much more heavier than usual. Weight is a very key component in construction of aircraft. The lighter the mass, the faster propellers can move the craft and lesser amount of energy used. The weight would be a major problem.
Although alloying iron and carbon improves the resistant of the steel to corrosion, it still cannot be compared to that of aluminium. We would build airplanes that would not be durable for so long and that can readily rust on frequent usage.
If airplanes were made of steel, they would be heavier, requiring more fuel and more powerful engines, reducing efficiency and payload capacity. Aluminum's lightness, toughness, and resistance to corrosion are pivotal for modern airplane performance and cannot be matched by steel without compromising efficiency and increasing operating costs.
If airplanes had to be made of steel instead of aluminum, there would be significant consequences for the aviation industry. Steel is much heavier than aluminum, which means that airplanes would be considerably heavier as well. This increase in weight would lead to a need for more powerful engines to provide the necessary lift, which could result in higher fuel consumption and lower efficiency. Moreover, the added weight would decrease the payload capacity of the airplane, as the plane's own structure would constitute a larger portion of its maximum takeoff weight.
The major advantages of aluminum in airplane construction include its lightness, toughness, and resistance to corrosion. These properties are crucial for ensuring that airplanes are able to carry substantial payloads, achieve adequate fuel efficiency, and maintain structural integrity over time. The use of aluminum, a metal that meets these critical needs, is therefore one of the key reasons why modern airplanes are able to achieve the performance levels that they do today. Utilizing steel, despite its strength, would compromise these benefits and could alter the economics and environmental impact of air travel.
A 2.00 m2 tank of a fluid that has a density of 1000 kg/m is draining at a rate of S liters/min. Determine the time for the tank to completely empty out. a.400 seconds b. 10 minutes У С. 400 minutes d. Not enough information
Answer:
the "d" option
Explanation:
Cómo se pronuncia
To calculate the flow I need the volume and time to comply with the formula:
S = V / t
V = volume
t = time
In this problem I have volume and density as data.
There is no way to calculate the time so the correct answer is the "d" option
what is conjugate base for h20?
Answer:
OH-
Explanation:
H2O can act as base or acid.
When H2O acts as a base the conjugate acid is H3O+
When H2O acts as an acid the conjugate base is OH-
1 mol of an Ideal Gas expands through a turbine adiabatically to produce work. Assuming steady flow conditions, how much work is lost due to the entropy generated, if the surroundings are at 300 K? Give the value of lost work to the nearest J. Inlet conditions: 494.8 K and 6.3 bar Outlet conditions: 383.5 K and 1.7 bar Cp = (9/2)R where R = 8.314 J/mol-K.
Answer:
W = 1.8 KJ
Explanation:
turbine adiabatically: Q = 0
∴ Lost work:
W = To [ ( S2 - S1 ) + ΔSo ]∴ To = 300 K
ideal gas:
S2 - S1 = Cp Ln (T2/T1) - R Ln (P2/P1)⇒ S2 - S1 = (9/2)*(8.314) Ln (383.5/494.8) - (8.314) Ln (1.7/6.3)
⇒ S2 - S1 = 1.36 J/mol.K
entropy generated in the sourroundings:
ΔSo = Q/To∴ Q = ΔU = Cv*ΔT
∴ Cv = 3/2*R = 12.5 J/mol.K.....ideal gas
∴ ΔT = 494.8 - 383.5 = 111.3 K
⇒ Q = 12.5 * 111.3 = 1391.25 J/mol
⇒ ΔSo = 1391.25/300 = 4.638 J/mol.K
⇒ W = ( 300 K ) * [ 1.36 J/mol.K + 4.638 J/mol.K ]
⇒ W = 1799.25 J/mol * 1 mol * ( KJ/1000J )
⇒ W = 1.799 KJ ≅ 1.8 KJ
A solid sample (Sample 1) is analyzed and found to contain 1.47 g carbon and 0.123 g hydrogen. A second sample (Sample 2) is expected to be composed of the same pure compound. If Sample 2 is found to contain 2.17 g hydrogen, how much carbon is expected in the sample ?
Explanation:
Here it is given that carbon is sample 2 = 25.9 g
For sample 1, mass carbon = 1.47 gNo. of moles of carbon will be calculated as follows.
No. of moles of carbon = [tex]\frac{\text{mass carbon}}{\text{molar mass carbon}}[/tex]
= [tex]\frac{1.47 g}{12.01 g/mol}[/tex]
= 0.1224 mol
It is also given that mass of hydrogen = 0.123 g
Hence, calculate number of moles of hydrogen as follows.
No. of moles of hydrogen = [tex]\frac{\text{mass hydrogen}}{\text{molar mass hydrogen}}[/tex]
= [tex]\frac{0.123 g}{1.008 g/mol}[/tex]
= 0.122 mol
Therefore, [tex]\frac{\text{moles of carbon}}{\text{moles of hydrogen}}[/tex]
= [tex]\frac{0.1224 mol}{0.122 mol}[/tex]
= 1.003
For sample 2, mass of hydrogen = 2.17 gTherefore, calculate the number of moles of hydrogen as follows.
No. of moles of hydrogen = [tex]\frac{\text{mass hydrogen}}{\text{molar mass hydrogen}}[/tex]
= [tex]\frac{2.17 g}{1.008 g/mol}[/tex]
= 2.1528 mol
Hence, calculate the moles of carbon as follows.
Moles of carbon = [tex]\text{moles hydrogen} \times \frac{\text{moles of carbon}}{\text{moles hydrogen}} [/tex]
= [tex]2.1528 mol \times 1.003[/tex]
= 2.16 mol
Mass of carbon = moles carbon × molar mass carbon
= (2.16 mol) × (12.01 g/mol)
= 25.9 g
Thus, we can conclude that 25.9 g of carbon is expected in the sample.
Final answer:
The expected amount of carbon in Sample 2 would be 25.93 grams, calculated using the constant carbon-to-hydrogen mass ratio determined from Sample 1.
Explanation:
The question requires determining how much carbon is expected in Sample 2, given the known amounts of carbon and hydrogen in Sample 1 and the amount of hydrogen in Sample 2. We assume that the compound is the same in both samples, meaning the ratio of carbon to hydrogen must be constant. First, we find the ratio of carbon to hydrogen in Sample 1:
Sample 1: 1.47 g C / 0.123 g H = 11.95 g C/g H
Using this ratio, we calculate the expected amount of carbon in Sample 2:
Expected carbon in Sample 2: 2.17 g H x 11.95 g C/g H = 25.93 g C
Therefore, we would expect Sample 2 to contain 25.93 grams of carbon, assuming it is composed of the same pure compound as Sample 1.
Data: A H f values: CH 4( g), -74.8 kJ; CO 21 g), -393.5 kJ; H 20( 1), -285.8 kJ. Using the A H f data above, calculate A H xn for the reaction below. Reaction: CH 4( 9) + 20 2( 9) => CO 2(g) + 2H 2011) Selected Answer: d. -890.3 kJ Correct Answer: d. -890.3 kJ
Answer:
[tex]\Delta H_{rxn}[/tex] for the given reaction is -890.3 kJ
Explanation:
[tex]\Delta H_{rxn}=\sum n_{i}.\Delta H_{f}(product)_{i}-\sum n_{j}.\Delta H_{f}(reactant)_{j}[/tex]
where [tex]n_{i}[/tex] and [tex]n_{j}[/tex] represents number of moles of i-th product and j-th reactant in balanced reaction respectively.
Hence [tex]\Delta H_{rxn}=[1mol\times \Delta H_{f}(CO_{2})_{g}]+[2mol\times \Delta H_{f}(H_{2}O)_{l}]-[1mol\times \Delta H_{f}(CH_{4})_{g}]-[2mol\times \Delta H_{f}(O_{2})_{g}][/tex]
so, [tex]\Delta H_{rxn}=[1mol\times -393.5kJ/mol]+[2mol\times -285.8kJ/mol]-[1mol\times -74.8kJ/mol]+[2mol\times 0kJ/mol]=-890.3 kJ[/tex]
So, Correct answer is -890.3 kJ
Question 3. A batch chemical reactor achieves a reduction in
concentration of compound A from 100 mg/L to 5 mg/L in one hour. If
the reaction is known to follow zero-order kinetics, determine the
value of the rate constant with appropriate units. Repeat the
analysis if the reaction is known to follow first-order
kinetics.
Answer:
Rate constant for zero-order kinetics: 1, 58 [mg/L.s]
Rate constant for first-order kinetics: 0,05 [1/s]
Explanation:
The reaction order is the relationship between the concentration of species and the rate of the reaction. The rate law is as follows:
[tex]r = k [A]^{x} [B]^{y}[/tex]
where:
[A] is the concentration of species A, x is the order with respect to species A. [B] is the concentration of species B, y is the order with respect to species B k is the rate constantThe concentration time equation gives the concentration of reactants and products as a function of time. To obtain this equation we have to integrate de velocity law:
[tex]v(t) = -\frac{d[A]}{dt} = k [A]^{n}[/tex]
For the kinetics of zero-order, the rate is apparently independent of the reactant concentration.
Rate Law: rate = k
Concentration-time Equation: [A]=[A]o - kt
where
k: rate constant [M/s][A]: concentration in the time t [M][A]o: initial concentration [M]t: elapsed reaction time [s]For first-order kinetics, we have:
Rate Law: rate= k[A]
Concentration -Time Equation: ln[A]=ln[A]o - kt
where:
K: rate constant [1/s]ln[A]: natural logarithm of the concentration in the time t [M]ln[A]o: natural logarithm of the initial concentration [M]t: elapsed reaction time [s]To solve the problem, wee have the following data:
[A]o = 100 mg/L
[A] = 5 mg/L
t = 1 hour = 60 s
As we don't know the molar mass of the compound A, we can't convert the used concentration unit (mg/L) to molar concentration (M). So we'll solve the problem using mg/L as the concentration unit.
Zero-order kinetics
we use: [A]=[A]o - Kt
we replace the data: 5 = 100 - K (60)
we clear K: K = [100 - 5 ] (mg/L) /60 (s) = 1, 583 [mg/L.s]
First-order kinetics
we use: ln[A]=ln[A]o - Kt
we replace the data: ln(5) = ln(100) - K (60)
we clear K: K = [ln(100) - ln(5)] /60 (s) = 0,05 [1/s]
The measured cube has the following information: mass = 3.4800 g height = length - width = 1:00 cm • Determine volume of cube V- • Determine calculate density of cube. D-
Answer:
Volume of cube = [tex]L^3[/tex] = [tex]1 cm^3[/tex]
Density = [tex] \frac{m}{V} = 3.48 \frac{g}{cm^3}[/tex]
Compared to the stable reference element, an isotope is different in what way?
Question 1 options:
A) More protons in the nucleus
B) Fever electrons in the orbitals
C) More electrons in the orbitals
D) Fewer neutrons in the nucleus
E) More neutrons in the nucleus
Answer:
E) More neutrons in the nucleus
Explanation:
Isotope -
The atoms which have same number of protons but different number of neutrons.
And since , atomic number = protons number
and , mass number = proton + neutrons .
Hence ,
The atomic number will not change , but the mass number will change .
Hence , the correct option is more neutrons in the nucleus .
A fish company delivers 22 kg of salmon, 5.5 kg of crab and 3.48 kg of oysters to your seafood restaurant. What is the mass, in kilograms of the seafood? What is the total number of pounds?
Answer:
Mass of sea food = 30.98 Kg
Mass of sea food in pound = 68.31 lbs
Explanation:
Salmon, crab and oysters all are sea food.
Mass of sea food = Mass of salmon + Mass of crab + mass of oyster
Mass of salmon = 22 kg
Mass of crab = 5.5 kg
Mass of oysters = 3.48 kg
Mass of sea food = Mass of salmon + Mass of crab + mass of oyster
= 22 + 5.5 + 3.48
= 30.98 Kg
1 Kg = 2.205 lbs
Therefore, 30.98 kg = 30.98 × 2.205
= 68.31 lbs
The difference between the molar concentration and the molal concentration of any dilute aqueous solution is small. Why?
Answer:
Because for dilute and aqueous solutions the mass of solvent will be a very close value to the volume of solution.
Explanation:
Molar concentration is defined as:
[tex][M]=\frac{molessolute}{volumesolution}[/tex]
And molal concentration is defined as:
[tex][m]=\frac{molessolute}{kgsolvent}[/tex]
And:
Msolution=Msolute+Msolvent
For dilute solutions, we have small amounts of solute, then we have:
Msolution=Msolute+Msolvent, and as the mass of solute is very small: Msolution≅Msolvent
If the solution is also aqueous (water as solvent), and considering that the density of water is around 1 gm/cm3 or 1 kg/m3:
Msolvent≅Msolution≅Vsolution
Therefore, if we look to the molar and molal equations, we have the same numerator in both (moles of solute) and nearby numbers for the denominator, giving to the molar and molal concentration close values.
You are given 12.33 moles of O2. How many moles of CO2 can be made?
Answer: The moles of carbon dioxide gas formed is 12.33 moles.
Explanation:
We are given:
Moles of oxygen gas = 12.33 moles
The chemical equation for the reaction of carbon and oxygen to produce carbon dioxide follows:
[tex]C+O_2\rightarrow CO_2[/tex]
By stoichiometry of the reaction;
1 mole of oxygen gas produces 1 mole of carbon dioxide gas.
So, 12.33 moles of oxygen gas will produce = [tex]\frac{1}{1}\times 12.33=12.33mol[/tex] of carbon dioxide gas.
Hence, the moles of carbon dioxide gas formed is 12.33 moles.