Answer:
555 ft per min
Explanation:
For the problem above, we can use the IAP chart and the ILS RWY 32 L approach. If we consider an electronic glide slope of 3 degrees, with the table of Rate-of-Descent in the IAP chart. Adjusting the angle of descent to 3 degrees and shifting the ground speed to approximately 105 knots, the estimated rate of descent is 555 ft per min.
To maintain the electronic glide slope at 105 knots ground speed, one would reference standard rates of descent for planning: about 600 fpm for general aviation and 450 fpm for a STOL aircraft with a 7-degree glideslope. These figures can vary depending on specific glide conditions such as aircraft configuration or wind, and precise planning may adjust these rates based on lift, air density, and safety margins.
To find the approximate rate of descent for maintaining the electronic glide slope at 105 knots ground speed, we must consider the standard glide ratio used for planning purposes in aviation. For general aviation, which often uses a power-off approach, a rate of descent of about 600 feet per minute (fpm) is typical, while Short Takeoff or Landing (STOL) aircraft with a steeper 7-degree glideslope might require around 450 fpm.
When determining the time of descent without wind and assuming a steady glide at a constant airspeed and angle, we can employ this information to estimate the time to descend from a certain altitude. For an altitude of 20,000 feet, the accuracy of the time of descent estimation can vary based on the increments and assumptions made in the calculations. When breaking down the glide into increments, smaller increments should provide more accurate results, but as seen in the given example, there's only a small 2% error or two minutes difference between the 'exact' solution and the approximation.
Raymer suggests adjustments to the calculations depending on the usage of thrust reversers or when including a safety margin for commercial aircraft under FAR part 25. These adjustments would affect the range and rate of descent calculations, so in a real-world scenario, they would need to be considered for precise planning.
For our purpose, if the glide is performed under no wind and at optimum conditions for lift over drag (L/D), we simply need the rate of descent to plan the glide slope. However, exact calculations do require the lifting capacity and air density at the altitudes used as increments, as suggested in the context.
which of the following best helps explain why an increase in temperature increases the rate of a chemical reaction?
The given question is incomplete. The complete question is as follows.
Which of the following best helps explain why an increase in temperature increases the rate of a chemical reaction?
(a) at higher temperatures, high-energy collisions happen less frequently.
(b) at low temperatures, low-energy collisions happen more frequently.
(c) at higher temperatures, less-energy collisions happen less frequently.
(d) at higher temperatures, high-energy collisions happen more frequently
Explanation:
When we increase the temperature of a chemical reaction then molecules of the reactant species tend to gain kinetic energy. As a result, they come into motion which leads to more number of collisions within the molecules.
Therefore, chemical reaction will take less amount of time in order to reach its end point. This means that there will occur an increase in rate of reaction.
Thus, we can conclude that the statement at higher temperatures, high-energy collisions happen more frequently, best explains why an increase in temperature increases the rate of a chemical reaction.
What is the frequency of a photon with a wavelength of 1040 nm. Report your answers to three significant digits. The frequency is = _________ s-1
Answer:
[tex]Frequency=2.88\times 10^{14}\ s^{-1}[/tex]
Explanation:
The relation between frequency and wavelength is shown below as:
[tex]c=frequency\times Wavelength [/tex]
c is the speed of light having value [tex]3\times 10^8\ m/s[/tex]
Given, Wavelength = 1040 nm
Also, 1 m = [tex]10^{-9}[/tex] nm
So,
Wavelength = [tex]1040\times 10^{-9}[/tex] m
Thus, Frequency is:
[tex]Frequency=\frac{c}{Wavelength}[/tex]
[tex]Frequency=\frac{3\times 10^8}{1040\times 10^{-9}}\ s^{-1}[/tex]
[tex]Frequency=2.88\times 10^{14}\ s^{-1}[/tex]
What is the average atomic mass compared to atomic mass
Answer:
Answer in explanation
Explanation:
The atomic mass can be defined as the addition of the number of protons and neutrons in the nucleus of an atom. When we talk of atomic mass, we are considering the mass of a particular isotope.
The average atomic mass or otherwise called the relative atomic mass is the atomic mass of the element itself. It has been calculated by taking into consideration the atomic masses of all the contributing isotopes existing in nature.
For example we have carbon 14 and carbon 12. While 14 is the atomic mass of that isotope, 12 is the relative atomic mass of the carbon atom. The abundance in nature of the different isotopes will dictate where the final atomic mass will lean
A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, calculate the percent of CaCO3 in the sample.
Answer:
Approximately 75%.
Explanation:
Look up the relative atomic mass of Ca on a modern periodic table:
Ca: 40.078.There are one mole of Ca atoms in each mole of CaCO₃ formula unit.
The mass of one mole of CaCO₃ is the same as the molar mass of this compound: [tex]\rm 100\; g[/tex].The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: [tex]\rm 40.078\; g[/tex].Calculate the mass ratio of Ca in a pure sample of CaCO₃:
[tex]\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}[/tex].
Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be [tex]\rm 30 \% \times 100\; g = 30\; g[/tex] of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio [tex]\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}[/tex]:
[tex]\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}[/tex].
In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:
[tex]\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%[/tex].
The percent composition of CaCO3 in a sample that contains 30% calcium by weight is 75%. This is calculated by dividing the actual weight % of calcium in the sample by the weight % of calcium in pure CaCO3, and multiplying by 100.
Explanation:The subject of this question is the calculation of the percent composition of a compound, which comes under the realm of Chemistry, specifically, Stoichiometry. In this question, we are examining a sample of calcium carbonate (CaCO3) reported to contain 30% calcium (Ca). We are to determine the percent of CaCO3 present in the sample.
Given that the molar mass of CaCO3 is 100 grams, 30% of this is calcium, that equates to 30 grams of calcium. Calcium's contribution to the molar mass of CaCO3 is 40 g/mol, so in a 100 g sample of pure CaCO3, calcium would naturally account for 40%. However, in our sample, the calcium content is lower at 30%. Thus, the sample is not 100% pure CaCO3. To calculate the percent composition of CaCO3 in this sample, we divide calcium's actual proportion of the sample by its proportion in pure CaCO3, and multiply by 100. Hence, (30/40)*100 = 75%.
Therefore, the sample is 75% CaCO3.
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Astor Manufacturing stores hazardous and volatile chemicals in its warehouse. The warehouse has state of the art equipment to make sure the chemicals do not explode. An unexpected earthquake shakes the warehouse, causing the chemicals to explode and injure William, a passer-by on a nearby sidewalk. Astor Manufacturing is
Answers. The correct option is A
Explanation:
The Astronomy manufacturing company is liable to Williams injury only if the company was grossly negligent.
How many grams of water will absorb a total of 2400 joules of energy when the temperature changes from 10.0°C to 30.0°C?
Answer:
28.7 grams of water
Explanation:
Calorimetry problem:
Q = C . m . ΔT
2400 J = 4.18 J/g°C . m . (30°C - 10°C)
2400 J = 4.18 J/g°C . m . 20°C
2400J = 83.6 J/g . m
2400J / 83.6 g/J = m
28.7 g = m
Final answer:
To calculate the amount of energy absorbed by water when the temperature changes, use the formula: q = mass * specific heat capacity * temperature change.
Explanation:
To calculate the amount of energy absorbed by water, we can use the formula: q = mass * specific heat capacity * temperature change. The specific heat capacity of water is approximately 4.184 J/g°C. In this case, the temperature change is from 10.0°C to 30.0°C, which gives us a ΔT of 20.0°C. We need to convert the mass from grams to kilograms by dividing it by 1000.
So, q = (mass / 1000) * 4.184 J/g°C * 20.0°C.
Substituting the given mass of water (in grams) into the equation, we have: q = (2400 / 1000) * 4.184 J/g°C * 20.0°C. Simplifying the expression gives us the amount of energy absorbed by the water in joules.
Which of the following is a macronutrient? Select one:
a. nitrogen
b. manganese
c. zinc
d. boron
Answer:
a. nitrogen is the correct answer.
Explanation:
Nitrogen is a macronutrient because nitrogen present in the plant in large quantities.Macronutrients are the nutrients which are needed in large amount for the proper growth and development,as they are required in large amounts they are called macronutrients.Macronutrients are the most essential elements for plants and required so that body functions properly.Macronutrients comprise carbon, nitrogen, sulfur, oxygen, potassium, hydrogen, calcium, phosphorus and magnesium.Final answer:
Among the given options, nitrogen is a macronutrient required in large amounts for plant growth and development, and is part of vital biomolecules such as carbohydrates, proteins, and nucleic acids.
Explanation:
The macronutrient in question can be identified by knowing that macronutrients are elements that organisms need in relatively large amounts compared to micronutrients which are needed in smaller amounts. The list of macronutrients includes nitrogen (N), phosphorus (P), potassium (K), calcium (Ca), magnesium (Mg), and sulfur (S), while micronutrients or trace elements, such as manganese (Mn), iron (Fe), zinc (Zn), and boron (B), are needed in smaller quantities.
Given the options provided (a. nitrogen b. manganese c. zinc d. boron), the correct answer is a. nitrogen because it is one of the primary macronutrients essential for plant growth and is a part of carbohydrates, proteins, and nucleic acids.
If a container holds 1mol of hydrogen, 2.5mol of helium, and 2mol of oxygen at a total pressure of 4atm, what is the partial pressure of the oxygen gas?
Answer:
Partial pressure O₂ = 1.78 atm
Explanation:
We can apply the mole fraction to solve the question:
Moles of gas / Total moles = Partial pressure of gas / Total pressure
Total moles = 1 H₂ + 2.5 He + 2O₂ = 4.5 moles
2 mol O₂ / 4.5 mol = Partial pressure O₂ / 4 atm
(2 mol O₂ / 4.5 mol ) . 4 atm = Partial pressure O₂ → 1.78 atm
Answer:
The partial pressure of oxygen is 1.45 atm
Explanation:
Step 1: Data given
Number of moles of hydrogen = 1 mol
Number of moles of helium = 2.5 mol
Number of moles of oxygen = 2 mol
Total pressure = 4 atm
Step 2: Calcualte total number of moles
Total number of moles = number of moles of hydrogen + number of moles of helium + number of moles of oxygen = 1 + 2.5 + 2 = 5.5 mol
Step 3: Calculate mol fraction of oxygen
Mol fraction oxygen = 2 mol / 5.5 mol = 0.3636
Step 4: Calculate partial pressure of oxygen
Partial pressure of oxygen = mol fraction of oxygen * total pressure
Partial pressure of oxygen = 0.3636 * 4 atm = 1.45 atm
The partial pressure of oxygen is 1.45 atm
A gas-filled balloon with a volume of 2.90 L at 1.20 atm and 20°C is allowed to rise to the stratosphere (about 30 km above the surface of the Earth), where the temperature and pressure are −23°C and 3.00 × 10−3 atm, respectively. Calculate the final volume of the balloon.
Answer:
The final volume is 990.8 L
Explanation:
Let calculate the moles of gas in the first situation:
P . V = n . R . T
1.20 atm . 2.90 L = n . 0.082 . 293K
(1.20 atm . 2.90 L) / (0.082 . 293K) = 0.145 moles
This are the same moles in the second situation:
P . V = n . R . T
0.003atm . V = 0.145 moles . 0.082 . 250K
V = (0.145 moles . 0.082 . 250K) / 0.003atm
V = 990.8 L
Answer: The final volume of the balloon is 990 L
Explanation:
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
where,
[tex]P_1[/tex] = initial pressure of gas = 1.20 atm
[tex]P_2[/tex] = final pressure of gas = [tex]3.00\times 10^{-3}atm[/tex]
[tex]V_1[/tex] = initial volume of gas = 2.90 L
[tex]V_2[/tex] = final volume of gas = ?
[tex]T_1[/tex] = initial temperature of gas = [tex]20^oC=273+20=293K[/tex]
[tex]T_2[/tex] = final temperature of gas = [tex]-23^oC=273-23=250K[/tex]
Now put all the given values in the above equation, we get:
[tex]\frac{1.20\times 2.90}{293}=\frac{3.00\times 10^{-3}\times V_2}{250K}[/tex]
[tex]V_2=990L[/tex]
To make an effective buffer, 25.00 mL of 0.025 M HF should be mixed with ______mL of 0.050 M NaOH
Answer: The volume of NaOH will be, 12.5 mL
Explanation:
To calculate the volume of base, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HF[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.
We are given:
[tex]n_1=1\\M_1=0.025M\\V_1=25.00mL\\n_2=1\\M_2=0.050M\\V_2=?[/tex]
Putting values in above equation, we get:
[tex]1\times 0.025M\times 25.00mL=1\times 0.050M\times V_2\\\\V_2=12.5mL[/tex]
Hence, the volume of NaOH will be, 12.5 mL
At standard temperature and pressure, a 0.50 mol sample of H2 gas and a separate 1.0 mol sample of O2 gas have the same (A) average molecular knetic energy
(B) average molecular speed
(C) volume
(D) effusion rate
(E) density
The correct answer is (B) average molecular speed.
Explanation:The correct answer is (B) average molecular speed.
According to the kinetic molecular theory, the average kinetic energy of gas molecules is directly proportional to their temperature. Since the temperature is the same for both the H2 and O2 gases, they will have the same average kinetic energy. However, the average molecular speed is inversely proportional to the square root of the molar mass. Since H2 has a lower molar mass than O2, it will have a higher average molecular speed.
The correct answer is a. average molecular kinetic energy.
At standard temperature and pressure (STP), all ideal gases have the same average molecular kinetic energy regardless of their molar mass or any other properties. This is a consequence of the kinetic molecular theory of gases, which states that the average kinetic energy of gas particles is directly proportional to the temperature of the gas in Kelvin. Since both samples are at the same temperature, they have the same average molecular kinetic energy.
To understand why the other options are incorrect, let's consider each one:
b. average molecular speed: According to the kinetic molecular theory, the average molecular speed of a gas is inversely proportional to the square root of its molar mass. Since hydrogen (H2) has a molar mass of approximately 2 g/mol and oxygen (O2) has a molar mass of approximately 32 g/mol, the average molecular speed of H2 will be greater than that of O2 at the same temperature.
c. volume: According to Avogadro's law, equal volumes of gases at the same temperature and pressure contain the same number of moles. Therefore, a 0.50 mol sample of H2 gas will occupy twice the volume of a 1.0 mol sample of O2 gas at STP.
d. effusion rate: The rate of effusion of a gas is inversely proportional to the square root of its molar mass (Graham's law). Since H2 has a lower molar mass than O2, H2 will effuse more quickly than O2.
e. density: The density of a gas is directly proportional to its molar mass. Since O2 has a higher molar mass than H2, a 1.0 mol sample of O2 gas will have a higher density than a 0.50 mol sample of H2 gas at STP.
Therefore, the only property that is the same for both gas samples at STP is their average molecular kinetic energy.
In the chemical equation Zn + 2HCL → ZnCl2 + H2, the reactants are
A. zinc chloride and hydrogen.
B. zinc and hydrogen carbonate.
C. zinc chlorate and water.
D. zinc and hydrochloric acid.
Answer: the answer is D
Explanation:
Answer:
D. zinc and hydrochloric acid.
Explanation:
Step 1: The balanced equation:
Zn + 2HCL → ZnCl2 + H2
Step 2:
All chemical reactions involve both reactants and products. Reactants are substances that start a chemical reaction, and products are substances that are produced in the reaction.
This means the substances on the left side, Zinc (Zn) and hydrochloric acid (HCl) are the reactants and will react with eachother, to form the products Zinc chloride (ZnCl2) and hydrogen gas (H2).
Option A: is not correct because zinc chloride and hydrogen are the products
Option B: is not correct because there is no hydrogen carbonate in the reaction
Option C: is not correct because there is no zinc chlorate, neither water in the reaction
The correct answer is D. zinc and hydrochloric acid.
When 70. milliliter of 3.0-molar Na2CO3 is added to 30. milliliters of 1.0-molar NaHCO3 the resulting concentration of Na+ is 2.0 M
Answer : The resulting concentration of [tex]Na^+[/tex] ion is, 4.5 M
Explanation : Given,
Concentration of [tex]Na_2CO_3[/tex] = [tex]M_1[/tex] = 3.0 M = 3.0 mol/L
Volume of [tex]Na_2CO_3[/tex] = [tex]V_1[/tex] = 70 mL = 0.07 L
Concentration of [tex]NaHCO_3[/tex] = [tex]M_2[/tex] = 1.0 M = 1.0 mol/L
Volume of [tex]NaHCO_3[/tex] = [tex]V_2[/tex] = 30 mL = 0.03 L
First we have to calculate the moles of [tex]Na_2CO_3[/tex] and [tex]NaHCO_3[/tex]
[tex]\text{Moles of }Na_2CO_3=\text{Concentration of }Na_2CO_3\times \text{Volume of }Na_2CO_3=3.0mol/L\times 0.07L=0.21mol[/tex]
and,
[tex]\text{Moles of }NaHCO_3=\text{Concentration of }NaHCO_3\times \text{Volume of }NaHCO_3=1.0mol/L\times 0.03L=0.03mol[/tex]
Now we have to calculate the moles of [tex]Na^+[/tex] ions.
As, 1 mole of [tex]Na_2CO_3[/tex] will give 2 moles of [tex]Na^+[/tex] ions
So, 0.21 moles of [tex]Na_2CO_3[/tex] will give [tex]2\times 0.21=0.42[/tex] moles of [tex]Na^+[/tex] ions
and,
As, 1 mole of [tex]NaHCO_3[/tex] will give 1 mole of [tex]Na^+[/tex] ions
So, 0.03 moles of [tex]NaHCO_3[/tex] will give 0.03 moles of [tex]Na^+[/tex] ions
So,
Total number of moles of [tex]Na^+[/tex] ions = 0.42 + 0.03 =0.45 mole
Total volume of both solution = 70 mL + 30 mL = 100 mL = 0.1 L
Now we have to calculate the concentration of [tex]Na^+[/tex] ions.
[tex]\text{Concentration of }Na^+=\frac{\text{Moles of }Na^+}{\text{Volume of solution}}=\frac{0.45mol}{0.1L}=4.5mol/L=4.5M[/tex]
Therefore, the resulting concentration of [tex]Na^+[/tex] ion is, 4.5 M
To find the resulting concentration of Na+, calculate the moles of Na+ ions in each compound and then add them together. Finally, divide the total moles of Na+ ions by the total volume of the solution to find the concentration of Na+.
Explanation:To find the resulting concentration of Na+, we need to calculate the total amount of Na+ ions present in the solution after the reaction occurs. We can do this by calculating the moles of Na+ ions in each compound and then adding them together.
First, calculate the moles of Na2CO3:
3.0 M Na2CO3 * 0.070 L = 0.210 mol Na2CO3
Next, calculate the moles of NaHCO3:
1.0 M NaHCO3 * 0.030 L = 0.030 mol NaHCO3
Now, add the moles of Na+ ions:
0.210 mol Na2CO3 + 0.030 mol NaHCO3 = 0.240 mol Na+
Finally, calculate the resulting concentration of Na+:
0.240 mol Na+ / (0.070 L + 0.030 L) = 2.0 M concentration of Na+
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Gas stored in a tank at 273 K has a pressure of 388 kPa. The safe limit for the pressure is 825 kPa. At what temperature will the gas reach this pressure?
Answer:
At 580.4 K of temperature will the gas reach a pressure of 825kPa
Explanation:
If the gas is stored in a tank, at 273K (as initial temperature) let's apply Charles Gay Lussac law, where the volume doesn't change and the number of moles neither.
If the volume keeps on constant, pressure is been modified directly proportional to absolute T°.
P1 / T°1 = P2 / T°2
388kPa / 273K = 825kPa / T°2
(388kPa / 273K) . T°2 = 825kPa
T°2= 825kPa . 273K/388 kPa
T°2 = 580.4K
A tank contains 90 kg of salt and 2000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drains from the tank at the rate 3 L/min.
(a) What is the amount of salt in the tank initially? (in kg)
(b) Find the amount of salt in the tank after 3.5 hours. (in kg)
(c) Find the concentration of salt in the solution in the tank as time approaches infinity. (Assume your tank is large enough to hold all the solution.) ( in kg/L)
Answer:
a) 90 kg
b) 68.4 kg
c) 0 kg/L
Explanation:
Mass balance:
[tex]-w=\frac{dm}{dt}[/tex]
w is the mass flow
m is the mass of salt
[tex]-v*C=\frac{dm}{dt}[/tex]
v is the volume flow
C is the concentration
[tex]C=\frac{m}{V+(6-3)*L/min*t}[/tex]
[tex]-v*\frac{m}{V+(6-3)*L/min*t}=\frac{dm}{dt}[/tex]
[tex]-3*L/min*\frac{m}{2000L+(3)*L/min*t}=\frac{dm}{dt}[/tex]
[tex]-3*L/min*\frac{dt}{2000L+(3)*L/min*t}=\frac{dm}{m}[/tex]
[tex]-3*L/min*\int_{0}^{t}\frac{dt}{2000L+(3)*L/min*t}=\int_{90kg}^{m}\frac{dm}{m}[/tex]
[tex]-[ln(2000L+3*L/min*t)-ln(2000L)]=ln(m)-ln(90kg)[/tex]
[tex]-ln[(2000L+3*L/min*t)/2000L]=ln(m/90kg)[/tex]
[tex]m=90kg*[2000L/(2000L+3*L/min*t)][/tex]
a) Initially: t=0
[tex]m=90kg*[2000L/(2000L+3*L/min*0)]=90kg[/tex]
b) t=210 min (3.5 hr)
[tex]m=90kg*[2000L/(2000L+3*L/min*210min)]=68.4kg[/tex]
c) If time trends to infinity the division trends to 0 and, therefore, m trends to 0. So, the concentration at infinit time is 0 kg/L.
To find the amount of salt initially in the tank, divide the amount of salt by the initial volume of water and multiply by the initial volume of water. After 3.5 hours, find the new amount of salt by subtracting the salt loss and adding the salt gain. As time approaches infinity, the concentration of salt in the tank will approach 0 kg/L.
Explanation:(a) To find the amount of salt initially in the tank, we use the formula: amount of salt = initial concentration of salt * initial volume of water. The initial concentration of salt can be found by dividing the amount of salt (90 kg) by the initial volume of water (2000 L). So, the initial concentration of salt is 90 kg / 2000 L = 0.045 kg/L. Now, we can find the amount of salt initially in the tank by multiplying the initial concentration of salt by the initial volume of water: 0.045 kg/L * 2000 L = 90 kg.
(b) After 3.5 hours, the amount of salt in the tank can be found using the formula: new amount of salt = initial amount of salt - salt loss + salt gain. The salt loss can be found by multiplying the drain rate (3 L/min) by the time (3.5 hours) and the initial concentration of salt (0.045 kg/L). The salt gain can be found by multiplying the incoming water rate (6 L/min) by the time (3.5 hours) and the concentration of salt in the incoming water, which is 0 kg/L. So, the new amount of salt = 90 kg - (3 L/min * 3.5 hours * 0.045 kg/L) + (6 L/min * 3.5 hours * 0 kg/L). Solve the equation to find the new amount of salt after 3.5 hours.
(c) As time approaches infinity, the concentration of salt in the solution in the tank will approach the concentration of the incoming water (0 kg/L), since the incoming water has no salt. Therefore, the concentration of salt in the tank as time approaches infinity is 0 kg/L.
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which of the following compounds is (are) hydrolyzed tobutanoic acid upon heating in H20, H2SO4?
a. ethyl butanoate
b. butyl acetate
c. N-methylbutanamide
d. both A and B
e. both A and C
Answer:
The correct option is option e, both A and C
Explanation:
Ethyl butanoate is an ester. Ester undergoes hydrolysis when heated with water containing dilute acids. Ester hydrolysis gives carboxylic acid and alcohol.
Therefore, ethyl butanoate on hydrolysis gives butanoic acid and ethanol.
Amide also undergoes acidic hydrolysis and yields carboxylic acid and amine.
N-methylbutanamide is an amide. It gives butanoic acid and methylamine when undergoes hydrolysis with H2O in the presence of H2SO4.
Hence, the correct option is option e.
Final answer:
Hydrolysis of ethyl butanoate and N-methylbutanamide in acidic conditions with H₂O and H₂SO₄ will yield butanoic acid; hence, option e) both A and C is correct.
Explanation:
The student's question pertains to the identification of compounds which, upon hydrolysis in an acidic condition with H₂O and H₂SO₄, will yield butanoic acid. To solve this, we need to understand the process of ester hydrolysis under acidic conditions.
Ethyl butanoate (a) is an ester formed from butanoic acid and ethanol. When hydrolyzed, it reverts back to butanoic acid and ethanol. Hence, option (a) is correct. Butyl acetate (b), similarly, is an ester of butanoic acid with butanol and would also yield butanoic acid upon hydrolysis.
N-methylbutanamide (c) is an amide, and on hydrolysis, it gives butanoic acid and a methylamine derivative. Therefore, upon heating with water and sulfuric acid, all listed compounds will yield butanoic acid as a hydrolysis product.
The correct answer to the student's question is e) both A and C.
What is the highest energy orbital of elements with valence electrons in the n=3 shell?
Answer:
d orbital
Explanation:
Given that:-
The principal quantum number, n = 3
The acceptable values of azimuthal quantum number, l are:-
l = 0 , 1 , 2
l = 0 corresponds to s orbital which can accomodate 2 electrons.
l = 1 corresponds to p orbital which can accomodate 6 electrons.
l = 2 corresponds to d orbital which can accomodate 10 electrons.
Thus, the highest energy orbital is: - d orbital
Why do the particles in a colloid stay suspended in a liquid, while the particles in a suspension settle out?
Answer:
The particles are spread evenly throughout the dispersion medium, which can be a solid, liquid, or gas. Because the dispersed particles of a colloid are not as large as those of a suspension, they do not settle out upon standing
Explanation:
My teacher asked us this question We had to answer it.
pls thank me
Final answer:
Colloidal particles remain suspended in a dispersion medium due to their small size that allows Brownian motion to keep them from settling, and the Tyndall effect distinguishes them by scattered light. Particles in a suspension settle out due to their larger size.
Explanation:
The reason why particles in a colloid stay suspended in a liquid, as opposed to particles in a suspension which settle out, is due to the size of the dispersed particles. Particles in a suspension are larger and will settle upon standing, but colloidal particles are small enough that Brownian motion counteracts the effects of gravity, maintaining the particles in suspension. Additionally, colloidal particles may be either hydrophilic, with an affinity for water, which prevents them from aggregating, or hydrophobic, in which case they are often stabilized by surfactants to remain dispersed.
Another key feature of colloids is the Tyndall effect, where light is scattered by the particles in the colloid. This effect can make colloidal mixtures appear cloudy or opaque, which differentiates them from true solutions, where such scattering does not occur because the dissolved species are at the molecular or ionic level and too small to scatter light.
The half-life of iodine-131 is about 8 days. How much of a 50mg sample will be left in 25 days? Write your answer rounded to the nearest tenth.
Final answer:
Approximately 20.7mg of the 50mg sample will be left after 25 days.
Explanation:
The half-life of iodine-131 is 8 days. To determine how much of the 50mg sample will be left after 25 days, we can use the formula:
Amount remaining = Initial amount x (0.5)^(time elapsed / half-life)
Plugging in the values:
Amount remaining = 50mg x (0.5)^(25 / 8) = 50mg x 0.4142 = 20.7mg
Therefore, approximately 20.7mg of the 50mg sample will be left after 25 days.
After gathering sufficient evidence to generate an _____for a nutrient, that value is used to establish an RDA for the same nutrient.
Answer:
Estimated Average Requirement (EAR).
Explanation:
Hello,
In this case, it is important to consider that Dietary Reference Intakes (DRIs) are reference values to quantitatively estimate the nutrient necessities to be taken for planning and assessing diets for healthy people. On the other hand, the Recommended Dietary Allowance (RDA) is the average daily dietary intake level that is enough to know the nutrient necessity of nearly all (about 98%) healthy individuals in a particular population. The answer is Estimated Average Requirement (EAR) which is a nutrient intake value that is considered to meet the necessity of half (50%) the healthy individuals in a particular population.
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What alteration to an HCN channel will reduce the transport of K+ in favor of Na+ transport? HCN channels are permeable to both K+ and Na+ ions.
Answer:
Decreasing the diameter of the channel by removing aa's. Since Na+ is chemically similar to K+, one can assume the difference must be due to the size of the atom. The K+ ion is larger than the Na+ ion, so reducing the diameter of the channel can allow Na+ to enter while preventing K+ entry. This explains clearly and perfectly how reducing the diameter reduces the
transport of K+ in favor of Na+ transport.
under which of the following sets of conditions would the most o2 (g) be dissolved in h2o(l)?
a. 5 atm 80 degree celsius
b. 5 atm 20 degree celsius
Answer:
Oxygen will dissolve more in H2O at 5 atm and 20 °C than at 5 atm 80 °C
Option B is correct.
Explanation:
Step 1: Data given
Pressure = 5 atm
Temperature = 20 °C or 80 °C
Step 2:
At low pressure, a gas has a low solubility. Decreased pressure allows more gas molecules to be present in the air, with very little being dissolved in solution. At high(er) pressure, a gas has a high solubility.
This means the higher the pressure the more the gas will dissolve. Since The pressure stays constant, it depends on the temperature.
The solubility of gases in liquids decreases with increasing temperature.
This means the gas will dissolve more with a lower temperature.
Oxygen will dissolve more in H2O at 5 atm and 20 °C than at 5 atm 80 °C
Under the following sets of conditions the most O₂ (g) be dissolved in H₂O(l) is a. 5 atm 80 degree celsius
The most O₂ will dissolve in water under the conditions of 5 atm pressure and 20 °C temperature due to the principles of Henry's Law.
High pressure and low temperature increase gas solubility in liquids. Thus, 5 atm and 20 °C will result in the highest dissolved oxygen concentration in water.To determine under which conditions the most O₂(g) will dissolve in H₂O(l), we can use Henry's Law. Henry's Law states that the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid. The formula is given by: Solubility = kH * Pwhere kH is the Henry's law constant and P is the partial pressure of the gas.Application of Henry's Law
Given two sets of conditions:
5 atm and 80 °C5 atm and 20 °CWe know that gases are more soluble in liquids at lower temperatures.
Therefore, under the second set of conditions (5 atm and 20 °C), O₂ will be more soluble in water because both the pressure is high and the temperature is low, which favours gas solubility.Using the given example at 20 °C:At 20 °C, the concentration of dissolved oxygen in water exposed to a partial pressure of 101.3 kPa is 1.38 × 10⁻³ mol L-1. If we want to determine the solubility at a different pressure, we apply Henry's Law :Solubility2 = (Solubility1 / Pressure1) * Pressure2Using high pressure (5 atm) and considering the system's temperature effects lead us to conclude more gas dissolves at lower temperatures.Conclusion
Thus, the most O₂(g) would dissolve in H₂O(l) under the conditions of 5 atm and 20 °C.
Correct question is: under which of the following sets of conditions would the most O₂ (g) be dissolved in H₂O(l)?
a. 5 atm 80 degree celsius
b. 5 atm 20 degree celsius
Analysis
1. Alpha Particles
a. What is the mass number of an alpha particle?
b. What kind of charge does an alpha particle have?
c. What is the identity of an alpha particle? (Hint: What atomic symbol is used?)
d. Write the balanced equation for the alpha decay that is below the “Show Equation.” Label the parent, daughter, and alpha particle.
Please help
Answer:
Below.
Explanation:
1.
a. An alpha particle has 2 protons and 2 neutrons ( a helium nucleus).
b. The charge on an alpha particle is 2+.
c. Atomic symbol used is α or He2+.
d. (no equation shown).
Americanum 241 decays to Neptunium 237 with the loss of 1 alpha particle
Am 241 = Np 237 + Ne2+
- note the atomic mass is 4 less under the decay.
Identify the element in period 3 to most likely have similar chemical properties to Ca.
Final answer:
Strontium (Sr) in period 3 of the periodic table has chemical properties similar to calcium (Ca) because they are both alkaline earth metals in group 2 with 2 valence electrons.
Explanation:
Since calcium is an alkaline earth metal, it shares similar chemical properties with other elements in group 2 of the periodic table. The answer would be strontium (Sr), as it is placed directly below calcium in group 2, and elements in the same group typically have similar properties due to their similar valence electron configurations.
Strontium has 2 valence electrons like calcium and tends to form +2 cations when reacting.
Ammonium hydrogen sulfide decomposes according to the following reaction, for which Kp = 0.11 at 250°C: NH4HS(s) ⇌ H2S(g) + NH3(g) If 46.5 g of NH4HS(s) is placed in a sealed 5.0−L container, what is the partial pressure of NH3(g) at equilibrium? PNH3 = atm
Answer:
0,33atm
Explanation:
For the reaction:
NH₄HS(s) ⇌ H₂S(g) + NH₃(g)
kp is defined as:
kp = 0.11 = P(H₂S) P(NH₃) (1)
Where P(H₂S) and P(NH₃) are partial pressures of each compound.
In equilibrium, if in your system the only addition is of NH₄HS(s), the partial pressures and the concentration of each compound are:
NH₄HS: I - x
-Where I is an initial concentration that is not relevant for the problem and x is the NH₄HS that reacts-
H₂S(g): x
NH₃(g): x
Replacing in (1):
0.11 = X×X
0.11 = X²
0.33 = X
That means P(NH₃) is 0.33 atm
I hope it helps!
The partial pressure of NH3(g) is 0.33 atm.
Number of moles of NH4HS = 46.5 g/51 g/mol = 0.91 moles
Given that;
PV =nRT
P = ?
V = 5.0−L
n = 0.91 moles
R = 0.082 atm LK-1mol-1
T = 250°C + 273 = 523 K
Making P the subject of the formula;
P = nRT/V
P = 0.91 moles × 0.082 atm LK-1mol-1 × 523 K / 5.0−L
P = 7.8 atm
We must now set up the ICE table;
NH4HS(s) ⇌ H2S(g) + NH3(g)
I 7.8 atm 0 0
C -x +x +x
E 7.8 - x x x
We know that;
Kp = pH2S × pNH3
Note that NH4HS is a pure solid and does not get into the equation
Kp = 0.11
0.11 = x^2
x = √0.11
x = 0.33 atm
Since partial pressure of H2S = partial pressure of NH3 = x
The partial pressure of NH3(g) = 0.33 atm.
Learn more: https://brainly.com/question/6505878
Which substances play an important role in communication between cells in a multicellular organism by acting as chemical messengers?
Answer:
Hormones
Explanation:
Hormones are chemical messengers released in one part of the body as chemical messengers into the bloodstream to achieve target results in another part of the body. The endocrine system is responsible for regulation of the internal environment (homeostasis).
A 2.06g sample of agno3•xh2o is dissolved in an aqueous solution of excess nacl. The resulting agcl precipitate is filtered off and is found to have a mass of 1.32 g. What is the value of x in agno3•xh2o?
Answer:
3
Explanation:
NaCl + AgNO₃ ———> NaNO₃ + AgCl.
Firstly, we will need to calculate the number of moles of AgCl produced. That is equal to the mass produced divided by the molar mass of AgCl.
The molar mass of AgCl = 108 + 35.5 = 143.5g/mol
The number of moles is thus 1.32/143.5 = 0.0092 moles
Since silver nitrate and silver chloride contains one atom of silver, it is only possible that their mole ratios are equal. Hence we say that 0.0092 moles of silver nitrate hydrate was dissolved.
Now we go on to calculate the molar mass of the silver nitrate hydrate.
The molar mass is simply the mass divided by the number of moles.
That is 2.06/0.0092 = 223.9 = 224g/mol
We can now calculate the value of x from here.
AgNO3.xH2O
(108 + 14 + 48) + x(2+ 16) = 224
170 + 18x = 224
18x = 224 - 170 = 54
18x = 54
x = 54/18 = 3
As compared to the general population, twice the usual probability of schizophrenia has been correlated with: Group of answer choices Rh-positive mothers and Rh-negative baby girls. Rh-positive mothers and Rh-negative baby boys. Rh-negative mothers and Rh-positive baby girls. Rh-negative mothers and Rh-positive baby boys.
Answer:
Rh-negative mothers and Rh-positive baby boys.
Explanation:
At least three research studies have proven that Rh incompatibility can be a potential risk factor for schizophrenia, this happens because the mother is Rh-negative (no protein coded) and the fetus is Rh-positive and therefore the mother's immune system responds to the baby's Rh protein leading to hypoxia, anemia and abnormal glial development with a higher risk for baby boys.
I hope you find this information useful and interesting! Good luck!
If 200.0g of nitrogen is reacted with 200.0g of hydrogen, what is the theoretical yield of the reaction? What Is the excess reactant? what is the limiting reactant ? Show your work
Answer:
242.862 grams is the theoretical yield of the reaction.
Nitrogen gas is a limiting reactant.
Hydrogen gas is an excessive reactant.
Explanation:
[tex]N_2+3H_2\rightarrow 2NH_3[/tex]
Moles of nitrogen gas = [tex]\frac{200.0 g}{28 g/mol}=7.143 mol[/tex]
Moles of hydrogen gas = [tex]\frac{200.0 g}{2 g/mol}=100.0 mol[/tex]
According to reaction, 1 mole of nitrogen reacts with 3 moles of hydrogen gas.
Then 7.413 moles of nitrogen will react with:
[tex]\frac{3}{1}\times 7.413 mol=21.428 mol[/tex] of hydrogen gas.
Moles of hydrogen that will react with 7.143 moles of nitrogen gas is less than the moles of hydrogen gas we have. This means that hydrogen gas is in an excessive reactant.
Nitrogen gas is in less amount hence limiting reactant.
Since, moles of nitrogen are in limiting amount so the amount of ammonia formed will depend upon moles of nitrogen gas.
According to reaction ,1 mole of nitrogen gives 2 moles of ammonia.
Then 7.143 moles of nitrogen will give:
[tex]\frac{2}{1}\times 7.143 mol=14.286 mol[/tex] of ammonia
Mass of 14.286 moles of ammonia :
= 14.286 mol × 17 g/mol= 242.862 g
242.862 grams is the theoretical yield of the reaction.
When a person in the hospital is given fluids intravenously, the fluid is typically a saline (salt) solution with about the same water concentration as human body tissues. Explain how the use of distilled water (water without salt) in place of this saline solution would possibly upset the patients homeostasis.
Answer:
Swelling of the red blood cells occurs.
Explanation:
Distilled water makes the blood hypotonic ,. that is a less concentrated solution, to the body tissues including the red blood cells. Therefore the water will enter the red blood cells and may cause them to lyse or swell. Additionally , water will enter the tissue spaces and cause sweling. But in the case of a hypertonic solution , ( a salt or saline solution ) , This swelling does not happen because the concentration of the saline will be equal or more than the blood. Thus the water will flow only from the blood to the solution. This will not cause swelling.But too much of salt is also not recommended.