Carbohydrate loading, otherwise known as carb loading is the process of adjusting one's nutrition to maximize the amount of carbohydrate (in form of glycogen) stored in the body. It is a strategy that is usually utilized by athletes to increase endurance during competition and by healthcare professionals when prepping patients for some specific surgeries.
A typical carb loading regime requires increased intake of high-carb diets and temporary reduction in physical activities although this might not be the case at the initial stage depending on the type of carb loading. The different types that exist include;
Classic 6-day6-dayClassic 3-dayModified 3-day1-dayThe ultimate aim of carb-loading regime is to make more fuel (in the form of glycogen) available during exercises or athletic events.
The correct option is to maximize glycogen storage in the body.
Carbohydrate loading is a strategy used to maximize glycogen storage in muscles and liver, primarily used by endurance athletes to enhance performance. It doesn't oxidize lactic acid, reduce lactate accumulation, or maximize creatine phosphate storage.
Explanation:Carbohydrate loading is a strategy used by endurance athletes, such as marathon runners, to maximize the storage of glycogen in the muscles and liver. Glycogen is the main way the body stores glucose for later use. Its primary role is to provide energy for muscles during physical activity. Carbohydrate loading occurs when athletes increase the amount of carbohydrates in their diet and decrease exercise in the days leading up to an event. This process helps to maximize the body's energy reserves during endurance events to improve performance.
It's important to note that it doesn't oxidize lactic acid, reduce lactate accumulation or maximize creatine phosphate storage. These are related to different processes in the body. Oxidizing lactic acid and reducing lactate accumulation are associated with lactate threshold and aerobic/anaerobic metabolism, while maximizing creatine phosphate storage relates to the body's short-term anaerobic energy system.
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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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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 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).
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.
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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
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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10.0 mL of a Cu2+ solution of unknown concentration was placed in a 250 mL Erlenmeyer flask. An excess of KI solution was added. Indicator was added and the solution was diluted with H2O to a total volume of 75 mL. For rxn 2, the solution from rxn 1 was titrated with 0.15 M Na2S2O3. The equivalence point of the titration was reached when 13.05 mL of Na2S2O3 had been added. What is the molar concentration of Cu2+ in the original 10.0 mL solution?
Answer:
Molar concentration: 0,0489M
Explanation:
In this titration of Cu²⁺ you add an excess of I⁻ that reacts with Cu²⁺ producing I₂, this I₂ reacts with Na₂S₂O₃. If you know the I₂ that reacts with Na₂S₂O₃ you can know the I⁻ that reacts with Cu²⁺ and, thus, the quantity of Cu²⁺. The reactions are:
2Cu²⁺ + 4I⁻ → 2CuI + I₂
I₂ + 2S₂O₃⁻ → S₄O₆ + 2I⁻
Moles of S₂O₃⁻ are:
0,01305L×0,15M = 1,96x10⁻³ moles of S₂O₃⁻.
Moles of I₂ are:
1,96x10⁻³ moles of S₂O₃⁻× ( 1 mole of I₂ / 2 moles of S₂O₃⁻) = 9,79x10⁻⁴ moles of I₂
Moles of Cu²⁺ are:
9,79x10⁻⁴ moles of I₂×( 2 moles of Cu²⁺ / 4 moles of I₂) = 4,89x10⁻⁴ moles of Cu²⁺
As volume of the solution was 10,0mL = 0,0100L, the molar concentration of the original solution is:
4,89x10⁻⁴ moles of Cu²⁺ / 0,0100L = 0,0489M
The unknown concentration of the Cu2+ solution can be found by determining the moles of Na2S2O3 at the equivalence point, using this to calculate the moles of Cu2+ from stoichiometry, and then dividing by the volume of the Cu2+ solution in liters.
Explanation:In this question, a Cu2+ solution, of unknown concentration, is titrated with 0.15 M Na2S2O3. The Cu2+ solution's concentration can be calculated using the data provided. The titration of this Cu2+ solution is complete, or the equivalence point is reached, when 13.05 mL of Na2S2O3 is added. The reaction that occurs is 2Na2S2O3 + Cu2+ -> CuS2O3 + 2Na+, and from the stoichiometry of the reaction, we know that two moles of Na2S2O3 react with one mole of Cu2+ ion.
Using the moles of Na2S2O3 that reacted (moles = Molarity x Volume (in liters), so moles = 0.15 M x 13.05 mL/1000), we can find out the moles of Cu2+ that were present in the 10 mL sample. We can then calculate the molarity of the Cu2+ solution by dividing the moles of Cu2+ by the volume of the solution in liters (0.01 L).
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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 200. mL of 0.60 M MgCl2(aq) is added to 400 mL of distilled water, what is the concentration of Mg and Cl in the resulting solution?
A. 0.20 M Mg ion & 0.20 M Cl ionB. 0.40 M Mg ion & 0.40 M Cl ionC. 0.20 M Mg ion & 0.40 M Cl ionD. 2.0 M Mg ion & 2.5 M Cl ion
Answer:
C. 0.20 M Mg ion & 0.40 M Cl ion
Explanation:
MgCl₂ is a ionic salt which is dissociated as this
MgCl₂ → Mg²⁺ + 2Cl⁻
First of all, we have a solution of 200 mL, with [MgCl₂] = 0.6M
Molarity . volume = moles.
0.6 mol/l . 0.2l = 0.12 mol
MgCl₂ → Mg²⁺ + 2Cl⁻
0.12mol 0.12 0.24
This moles are also in 400mL of water, so the new concentration is
[Mg²⁺] = 0.12 m/0.6L = 0.2M
[Cl⁻] = 0.24 m/0.6L = 0.4M
Remember we initially have 200mL and then, we add 400 mL, so we supose aditive volume. (600mL)
The concentration of magnesium ion, Mg²⁺ in the resulting solution is 0.2 M while that of the chloride ion, Cl¯ is 0.4 M
The correct answer to the question is Option C. 0.20 M Mg ion & 0.40 M Cl ion
we'll begin by calculating the molarity of the diluted solution. This can be obtained as follow:
Volume of stock solution (V₁) = 200 mL
Molarity of stock solution (M₁) = 0.60 M
Volume of diluted solution (V₂) = 200 + 400 = 600 mL
Molarity of diluted solution (M₂) = ?The molarity of the diluted solution can be obtained as follow:
M₁V₁ = M₂V₂0.6 × 200 = M₂ × 600
120 = M₂ × 600
Divide both side by 600
M₂ = 120 / 600
M₂ = 0.2 MThus, the molarity of the diluted (i.e resulting) solution of MgCl₂ is 0.2 M
Next, we shall determine the concentration of magnesium ion, Mg²⁺ in the diluted solution. This is illustrated below:
MgCl₂(aq) —> Mg²⁺(aq) + 2Cl¯(aq)
From the balanced equation above,
1 mole MgCl₂ dissolves to produce 1 mole Mg²⁺.
Therefore,
0.2 M MgCl₂ will also produce 0.2 M Mg²⁺.
Thus, the concentration of magnesium ion, Mg²⁺ in the resulting solution is 0.2 M.
Finally, we shall determine the concentration of the chloride ion, Cl¯ in the resulting solution.
MgCl₂(aq) —> Mg²⁺(aq) + 2Cl¯(aq)
From the balanced equation above,
1 mole MgCl₂ dissolves to produce 2 moles of Cl¯
Therefore,
0.2 M MgCl₂ will also produce = 2 × 0.2 = 0.4 M Cl¯
Thus, the concentration of chloride ion, Cl¯ in the resulting solution is 0.4 M.
From the calculations made above:
The concentration of magnesium ion, Mg²⁺ in the resulting solution is 0.2 M while that of the chloride ion, Cl¯ is 0.4 M
Option C. 0.20 M Mg ion & 0.40 M Cl ion gives the correct answer to the question.
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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.According to the VSEPR model, the progressive decrease in the bond angles in the series of molecules CH4, NH3, and H2O is best accounted for by the:_________
Answer:
Presence of lone pairs of electrons
Explanation:
According to VSEPR theory, the presence of lone pairs caused increased repulsion of electron pairs on the valence shell of the centeral atom of the molecule. This decreases or distorts the bond angle. The decrease in bond ange depends on the number of lone pairs present on the valence shell of the central atom of the molecule. Ammonia has only one lone pair hence the bond angle is 107°, water has two lone pairs and the bond angle is 104°. Compare this this with the bond angle of 109° in methane which has only bond pairs and no lone pairs.
The decrease in bond angles from CH4 to NH3 to H2O is due to increasing number of lone pairs on the central atom which cause greater repulsion and smaller bond angles, as per the VSEPR theory.
Explanation:According to the Valence shell electron-pair repulsion (VSEPR) theory, the progressive decrease in the bond angles in the series of molecules CH4, NH3, and H2O is best accounted for by the presence and arrangement of lone electron pairs on the central atom. CH4 has no lone pairs on the central carbon atom and has a tetrahedral shape with bond angles of 109.5°. In NH3, the nitrogen atom has one lone pair, which decreases the bond angle to less than 109.5°. In H2O, the oxygen atom has two lone pairs, leading to an even smaller bond angle of 104.5°. The VSEPR theory predicts that the electron pairs will arrange themselves to minimize repulsion, and the lone pair-lone pair repulsion is greatest, followed by lone pair-bonding pair, and finally, bonding pair-bonding pair repulsion is the least.
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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 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
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.
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]
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.
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
When scientists use one of their five senses to gather information, they are A. making an observation. B. making an inference. C. predicting a relationship D. drawing a conclusion.
The scientists use one of their five senses to gather information, they are making an observation.
What is an observation ?The active gathering of data from a primary source is observation. Observation of living things makes use of the senses. Using scientific tools to perceive and record data is another way that observation may be used in science.
A technique for gathering information through observation of individuals, situations, or physical qualities as they are occurring naturally.
Using one or more senses, our observational abilities provide us with knowledge about things, occasions, attitudes, and occurrences. Being able to see and learn about the outside world is crucial because it forms the basis of effective communication.
Thus, option A is correct.
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In scientific endeavors, when one of the five senses is used to gather information, it constitutes making an observation. Observations can involve both qualitative and quantitative data, and they provide evidence that scientists use to form conclusions.
Explanation:When scientists use one of their five senses to gather information, they are A. making an observation. Scientists usually perform observations by using one or more of their five senses to gain an understanding of the properties or behaviors of a substance or a system. These data can be qualitative (descriptive) or quantitative (consisting of numbers). For instance, seeing a plant growing (observation) and measuring its growth over weeks (data collection). From these observations, the scientist can infer conclusions based on the evidence collected. In brain studies, the activity in the brain during specific tasks is observed, which could be the basis for conclusions about brain functionality.
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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.
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.
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.
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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.
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
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
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
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.
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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.
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.
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.
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.
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.