Answer:
The limiting reactant is the N₂
Explanation:
The reaction for production of ammonia is:
N₂ + 3H₂ → 2NH₃
Ratio is 1:3. Let's make a rule of three to solve this:
1 mol of nitrogen is needed to react with 3 moles of hydrogen.
3 moles of nitrogen may react with (3 . 3) /1 = 9 moles of H₂
It is ok because I have 12 moles, so the limiting reactant is the N₂ but let's confirm it.
3 moles of H₂ react with 1 mol of N₂
12 moles of H₂ would react with (12. 1) / 3 = 4 moles of N₂
It's ok to say the N₂ is the limiting because I don't have enough moles to react. I need 4 and I only have 3
Final answer:
To determine the limiting reactant in the reaction to produce ammonia, we compare the stoichiometric amounts needed according to the balanced equation. Nitrogen is the limiting reactant in this reaction because it requires more hydrogen than what is available to completely react.
Explanation:
To determine the limiting reactant between nitrogen gas (N2) and hydrogen gas (H2) in the synthesis of ammonia (NH3), we use the balanced chemical equation:
N2 + 3 H2 ->2 NH3
According to the equation, each mole of nitrogen reacts with three moles of hydrogen to produce two moles of ammonia. Given the reaction mixture of 3.0 moles of N2 and 12.0 moles of H2, we compare the stoichiometric amounts needed for the reaction. The nitrogen gas would require 9.0 moles of hydrogen gas to fully react (3 moles N2 imes 3 moles H2/1 mole N2 = 9 moles H2). Since there are 12.0 moles of hydrogen gas available, hydrogen is in excess, and nitrogen is the limiting reactant.
A 9.87-gram sample of an alloy of aluminum and magnesium is completely reacted with hydrochloric acid and yields 0.998 grams of hydrogen gas. Calculate the percentage by man of each metal in the alloy.
Final answer:
To calculate the percent composition of aluminum and magnesium in the sample, we use stoichiometry. Moles of hydrogen gas produced help to establish moles of the metals reacting, but without complete reaction equations for the alloy, or more data, a definitive calculation cannot be provided.
Explanation:
To calculate the percentage by mass of aluminum and magnesium in the sample, we need to perform a few stoichiometric calculations based on the reaction of the alloy with hydrochloric acid and the production of hydrogen gas. The typical reactions for aluminum and magnesium with hydrochloric acid are:
2 Al(s) + 6 HCl(aq) → 2 AlCl₃(aq) + 3 H₂(g)Mg(s) + 2 HCl(aq) → MgCl₂(aq) + H₂(g)Using the molar mass of hydrogen (1.008 g/mol), we can calculate the moles of hydrogen gas produced using the equation:
moles H₂ = 0.998 g / (2 * 1.008 g/mol) = 0.495 moles
The mole ratio of Al to H₂ in the reaction is 2:3, and for Mg to H₂ is 1:1. We can calculate the mass of aluminum or magnesium that would produce 0.495 moles of hydrogen.
For aluminum:
moles Al = (2/3) * moles H₂ = (2/3) * 0.495 = 0.330 moles
For magnesium:
moles Mg = moles H₂ = 0.495 moles
Now, we calculate the mass:
mass Al = moles Al * atomic mass Al = 0.330 moles * 26.98 g/mol = 8.904 g
mass Mg = moles Mg * atomic mass Mg = 0.495 moles * 24.31 g/mol = 12.033 g
These numbers are hypothetical maximums if the sample was 100% Al or Mg, and do not add up to 9.87 g. Therefore, we need to set up a system of equations considering the total mass of the alloy and the mass of hydrogen produced to find the exact mass of Al and Mg in the sample. This requires more information or a different approach that accounts for the actual reaction stoichiometry with the alloy sample, which isn't provided here.
The alloy consists of: 62.2% aluminium, 37.8% magnesium
To calculate the percentage by mass of each metal (aluminium and magnesium) in the alloy, we need to use the stoichiometry of their reactions with hydrochloric acid and the information given about the mass of the hydrogen gas produced.
Step 1: Write the balanced chemical equations for the reactions
For aluminium:
[tex]\[ 2\text{Al} + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2 \][/tex]
For magnesium:
[tex]\[ \text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2 \][/tex]
Step 2: Determine the moles of hydrogen gas produced
The molar mass of hydrogen gas [tex](\(\text{H}_2\))[/tex] is:
[tex]\[ \text{Molar mass of H}_2 = 2 \times 1.008 = 2.016 \, \text{g/mol} \][/tex]
The number of moles of hydrogen gas produced is:
[tex]\[ \text{Moles of H}_2 = \frac{0.998 \, \text{g}}{2.016 \, \text{g/mol}} \]\[ \text{Moles of H}_2 = 0.495 \, \text{mol} \][/tex]
Step 3: Relate moles of hydrogen gas to moles of metals
Let x be the mass of aluminium and y be the mass of magnesium in the alloy.
From the reactions:
- Aluminium produces 3 moles of [tex]\(\text{H}_2\)[/tex] per 2 moles of [tex]\(\text{Al}\)[/tex]:
[tex]\[ 2\text{Al} \rightarrow 3\text{H}_2 \]\[ \frac{3}{2} \text{moles of H}_2 \text{ per mole of Al} \]\[ 1 \text{ mole of Al} \rightarrow \frac{3}{2} \text{ moles of H}_2 \][/tex]
- Magnesium produces 1 mole of [tex]\(\text{H}_2\) per mole of \(\text{Mg}\):[/tex]
[tex]\[ \text{Mg} \rightarrow \text{H}_2 \]\[ 1 \text{ mole of Mg} \rightarrow 1 \text{ mole of H}_2 \][/tex]
Thus, the moles of hydrogen gas produced by aluminium and magnesium are:
[tex]\[ \text{Moles of H}_2 \text{ from Al} = \frac{3}{2} \times \frac{x}{26.98} \]\[ \text{Moles of H}_2 \text{ from Mg} = \frac{y}{24.305} \][/tex]
The total moles of hydrogen gas produced:
[tex]\[ \frac{3}{2} \times \frac{x}{26.98} + \frac{y}{24.305} = 0.495 \][/tex]
Step 4: Set up the mass equation for the alloy
The total mass of the alloy is:
[tex]\[ x + y = 9.87 \, \text{g} \][/tex]
Step 5: Solve the system of equations
We have two equations:
[tex]1. \[ \frac{3}{2} \times \frac{x}{26.98} + \frac{y}{24.305} = 0.495 \]\\[/tex]
2. x + y = 9.87
Let's solve these equations step by step.
First, express y in terms of x from the second equation:
[tex]\[ y = 9.87 - x \][/tex]
Substitute this into the first equation:
[tex]\[ \frac{3}{2} \times \frac{x}{26.98} + \frac{9.87 - x}{24.305} = 0.495 \][/tex]
Simplify and solve for x :
[tex]\[ \frac{3x}{2 \times 26.98} + \frac{9.87 - x}{24.305} = 0.495 \]\[ \frac{3x}{53.96} + \frac{9.87 - x}{24.305} = 0.495 \]\[ \frac{3x}{53.96} + \frac{9.87}{24.305} - \frac{x}{24.305} = 0.495 \]\[ \frac{3x}{53.96} - \frac{x}{24.305} = 0.495 - \frac{9.87}{24.305} \][/tex]
Calculate the constant term:
[tex]\[ 0.495 - \frac{9.87}{24.305} = 0.495 - 0.406 = 0.089 \][/tex]
Combine the \( x \)-terms:
[tex]\[ \frac{3x}{53.96} - \frac{x}{24.305} = 0.089 \]\[ x \left( \frac{3}{53.96} - \frac{1}{24.305} \right) = 0.089 \][/tex]
Calculate the coefficient of \( x \):
[tex]\[ \frac{3}{53.96} = 0.0556 \]\[ \frac{1}{24.305} = 0.0411 \]\[ 0.0556 - 0.0411 = 0.0145 \][/tex]
Now, solve for x :
[tex]\[ x \times 0.0145 = 0.089 \]\[ x = \frac{0.089}{0.0145} \]\[ x = 6.14 \, \text{g} \][/tex]
Now, find y :
[tex]\[ y = 9.87 - x \]\[ y = 9.87 - 6.14 \]\[ y = 3.73 \, \text{g} \][/tex]
Step 6: Calculate the percentage by mass of each metal
[tex]\[ \% \, \text{Al} = \frac{6.14 \, \text{g}}{9.87 \, \text{g}} \times 100 = 62.2\% \]\[ \% \, \text{Mg} = \frac{3.73 \, \text{g}}{9.87 \, \text{g}} \times 100 = 37.8\% \][/tex]
Why does it take more energy to increase the temperature of 100 grams of liquid water by one degree Celsius than it does 100 grams of copper metal?
Answer:
The answer to your question is below
Explanation:
The specific heat is a physical property equal to the amount of heat necessary to increase the temperature of 1 gram of a substance by one degree celsius.
The lower the specific heat, the lower the amount of heat to increase the temperature 1°C, the higher the specific heat, the higher the amount of heat necessary to increase the temperature by 1°C.
The specific heat of copper is 0.093 cal/g°C
The specific heat of water is 1 cal/g°C.
That is why is necessary more heat to warm water.
More energy to increase the temperature of 100 grams of liquid water by one degree Celsius than it does 100 grams of copper metal due to higher specific heat.
What is specific heat?Specific heat refers to the amount of heat needed to raise the temperature of 1 gram of a substance by 1 degree Celsius. Water has a high specific heat which means it takes more energy to increase the temperature of water compared to other substances like metals.
So we can conclude that more energy is needed to increase the temperature of 100 grams of liquid water by one degree Celsius than copper metal because of higher specific heat of water.
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If a clot were made up of a mass of proteins, what change in the proteins led to the formation of a clot?
Answer: The proteins were no longer soluble in the blood.
Analysis of a volatile liquid shows that it contains 62.04% carbon, 10.41% hydrogen, and 27.54% oxygen by mass. At 150.°C and 1.00 atm, 500. mL of the vapor has a mass of 0.8365 g. What is the molecular formula of the compound?
Answer:
Molecular formula of the compound is C₃H₆O
Explanation:
Firstly let's determine the moles of the vapor (gas) with the Ideal Gases Law, so it can give us the molar mass with the mass and afterwards we can work with the percent composition.
Pressure . Volume = moles . Ideal Constant Gases . Temperature in K
Temperature in K = T°C + 273 → 150°C + 273 = 423K
P . V = n . R . T
n = (P .V) / (R. T)
n = 1 atm . 0.5L / (0.082 . 423K)
n = 0.0144 moles
These are the moles for 0.8365 g, so let's determine the molar mass
Molar mass (g/mol) = 0.8365 g / 0.0144 mol → 58.02 g/mol
Percent composition means:
100 g of compound have 62.04 g of C
100 g of compound have 10.41 g of H
100 g of compound have 27.54 g of O
Let's make the rule of three:
100 g of compound have __ 62.04 g of C __ 10.41 g of H __ 27.54 g of O
The 58.02 g of compound must have:
(58.02 g . 62.04 g) / 100 g = 36 g of C
(58.02 g . 10.41 g) / 100 g = 6 g of H
(58.02 g . 27.54 g) / 100 g = 16 g of O
Let's find out the moles of each
Mass / Molar mass
36 g / 12 g/mol = 3 C
6 g / 1 g/mol = 6 H
16g / 16 g/mol = 1 O
The molecular formula of the given compound is C₃H₆O. The molecular formula can be determined by finding the ratio of each element in the compound.
How to determine the Molecular formula of a compound?It can be determined by finding the ratio of each element in the compound.
First, calculate the moles of the compound from the ideal gas formula,
[tex]n = \dfrac {1 {\rm\ atm \times 0.5L} }{(0.082 \times 423{\rm \ K})}\\\\n = 0.0144 \rm \ moles[/tex]
Then calculate the molar mass of the compound,
[tex]m = {\rm \dfrac {0.8365 \ g }{0.0144 \ mol}} \\\\m = 58.02 \rm \ g/mol[/tex]
Then calculate the mass of individual elements in the 58.02 g of compound:
[tex]\text{ Mass of Carbon} = \dfrac {58.02 {\rm \ g} \times 62.04 {\rm \ g}}{100 {\rm \ g}}\\ \text{ Mass of Carbon} = 36 \rm \ g[/tex]
[tex]\text{ Mass of Hydrogen } = \dfrac {58.02 {\rm \ g} \times 10.14 {\rm \ g}}{100 {\rm \ g}}\\ \text{ Mass of Hydrogen } = 6 \rm \ g[/tex]
[tex]\text{ Mass of Oxygen } = \dfrac {58.02 {\rm \ g} \times 27.54 {\rm \ g}}{100 {\rm \ g}}\\ \text{ Mass of Oxygen } = 16 \rm \ g[/tex]
Now find the moles of each element, we get
3 moles of Carbon
6 moles of Hydrogen
1 mole of Oxygen
Therefore, the molecular formula of the given compound is C₃H₆O.
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A sample of an unknown compound with a mass of 0.847 g has the following composition: 50.51 % fluorine and 49.49 % iron. When this compound is decomposed into its elements, what mass of each element would be recovered?
Answer: 0,4278g of F and 0,4191g of Fe
Explanation: it's possible to calculate the mass of each element by multiplying the percentage (decimal) of the element by the mass of the compound.
For Fluorine (F)
0,847g * 0,5051 = 0,4278g of F
For iron (Fe)
0,847 * 0,4949 = 0,4191g of Fe
This is determined because even when the compound is decomposed, due to conservative law of mass, the decomposition process do not affect the amount of matter, so the mass of the elements remain even if they are separated from the original molecule.
At the end, the sum of the elements masses should be the total mass of the compound.
Answer:fluorine=0.5082
Iron=0.3388
Explanation:
Using Empirical formula to show the ratio.
F. Fe
50.51/39 49.49/56
=1.295. 0.88375
=1.295/0.88375 :0.88375/0.88375
=1.465:1
multiply each term by 2 to get a whole number ratio,we have
=(1.465*2) :(1*2)
=2.93:2
=3:2
To get the amount if each contribution of F and Fe,we use ratio,
F=3/(3+2)=3/5*total mass(0.847)
F=0.5082g
Similarly,Fe=2/5*0.847
Fe=0.3388g.
Curious cross chemistry teacher handed him a school flask containing two gases oxygen and hydrogen. The pressures of the two gases are eight ATM and one ATM respectively. What is the total pressure of the combined gases?
Answer:
9 atm is the total pressure of the combined gases.
Explanation:
According to the Dalton's law of partial pressure , the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gasses.
[tex]P_T=p_{1}+p_{2}....p_{n}[/tex]
where,
[tex]P_T[/tex] = total pressure =
[tex]p_{1}[/tex] = partial pressure of gas-1
[tex]p_{2}[/tex] = partial pressure of gas -2
[tex]p_{n}[/tex] = partial pressure of nth gas
We have :
Pressure of the oxygen gas in flask before mixing = 8 atm
Pressure of the hydrogen gas in flask before mixing = 1 atm
Partial pressure of oxygen gas after mixing = [tex]p_1=8 atm[/tex]
Partial pressure of hydrogen gas after mixing = [tex]p_2=1 atm[/tex]
Total pressure of the mixture : P
[tex]P=P_1+P_2[/tex] (Dalton's law of partial pressure)
[tex]P=8 atm+1 atm=9 atm[/tex]
9 atm is the total pressure of the combined gases.
Answer:
D
Explanation:
If the mass of 1.00 the of a compound is found to be 150.0 g, what is the molecular formula of the compound?
Answer:
calculate the molecular formula of a compound with the empirical formula CH2O and a molar mass of 150g/mol
the molecular formula is [tex]C_{5} H_{10}O_{5}[/tex]
The molecular formula of a compound is the formula comprising of the constituent elements chemical symbols each of which carries the number of atoms of that element present in a molecule of the compound appearing in the smallest whole number ratio to other eatoms present in the compound
Explanation:
The masses of the constitent element is determined forst from which the number of moles is then calculated by dividing the mass by the molar mass then then each calculated molar mass value is divided by the smallest number of moles calculated from the previous step so the value calculated is then rounded up to the nearest whole number giving the ratios of the moles of the elements in the compound which represents the subscripts in the empirical formula of the compound.
If the subscrips are in fractions, then multiply each of them by the same number to derive the smallest whole number factor, that is if the calculated formula contains a facor of 0.5, multiply by 2
Mass pf Carbon = 12g
mass of Hydrogen = 1g
molar mass of oxygen - 16g
Total mass of CH2O = 30g
Therefore dividing molar mass by empirical formula mass = 150g/30g = 5
Hence our factor is 5
multiplying each subscript of the empirical formula by 5 gives
C5H10O5 hence the molecular formula is [tex]C_{5} H_{10}O_{5}[/tex]
If I drink two 5 hour energy drinks. Will I have twice the amount of energy for 5 hours or 10 hours of energy?
No I don't think it works that way. What you could do is drink one energy drink, wait 5 hours and then drink another.
Answer:
10 hrs of energy
Explanation:
these bottles hold energy for 5hrs so if it is double you have double the time of energy
THIS IS THE LAST QUESTION I NEED TO FINISH ASSIGNMENT! WILL MARK BRAINLIEST IF CORRECT!!
Above which point on a phase diagram can you no longer distinguish between a liquid and a gas?
melting point
triple point
critical point
boiling point
Answer:
Critical Point
Explanation:
You can no longer distinguish a liquid from a gas when a object hits it's critical point.
At a low temperature dry ice (solid CO2), calcium oxide, and calcium carbonate are introduced into a 50.0-L reaction chamber. The temperature is raised to 900.!C. For the following mixtures, will the initial amount of calcium oxide increase, decrease, or remain the same as the system moves toward equilibrium
Answer:
the initial amount of calcium oxide will decrease as the system moves toward equilibrium
Explanation:
The question should have this variable:
a) 655g CaCO3, 95.0g CaO, P(CO2) = 2.55atm
The reaction formula of dry ice, calcium oxide and calcium carbonate should be:
CaCO3 (s) --> CaO(s) + CO2(g)
With Kp = 1.04 at 900 degrees C
in this tank, only CO2(g) is in gaseous form. The amount of gas will determine the pressure of the tank. When the tank heated, its temperature rise and will also rise the pressure. When pressure higher than Kp, the equilibrium will shift toward side with less gas.
There is only one gas in the reaction, so Kp= P(CO2)
Since P(CO2) is 2.55 atm and its higher than Kp(1.04), the equilibrium will shift to the left. Since the reaction direction to the left, the amount of calcium carbonate will increase while carbon dioxide and calcium oxide will decrease.
The initial amount of calcium oxide will decrease because the system moves toward equilibrium
Reason for decreasing of an initial amount of calcium oxide:In this tank, only CO2(g) should be in gaseous form. The amount of gas will measure the pressure of the tank. At the time When the tank is heated, its temperature should be increased due to this the pressure is also increased.
Since P(CO2) is 2.55 atm and it's more than Kp(1.04), the equilibrium should shift to the left. The reaction shifted to the left, the amount of calcium carbonate will increase while on the other hand, carbon dioxide and calcium oxide will decrease.
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Four ice cubes at exactly 0°C having a total mass of 53.5 g are combined with 115 g of water at 75°C in an insulated container. If no heat is lost to the surroundings, what will be the final temperature of the mixture?The specific heat capacity of water = 4.184 J/(g°C)and the standard enthalpy of fusion for water = 6.02 kJ/mol.
Final answer:
The final temperature of the mixture will be 15.44°C.
Explanation:
In this problem, we can use the principle of conservation of energy to find the final temperature of the mixture. The heat gained by the ice cubes is equal to the heat lost by the water.
To find the heat gained by the ice cubes, we can use the formula:
Q = m × c × ΔT
where Q is the heat gained, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Since the ice cubes are at 0°C, the heat gained is:
Q = 4 × (53.5 g) × (4.184 J/(g°C)) × (0°C - T-f)
where T-f is the final temperature.
To find the heat lost by the water, we can use the formula:
Q = m × c × ΔT
where Q is the heat lost, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Since the water is at 75°C, the heat lost is:
Q = (115 g) × (4.184 J/(g°C)) × (T-f - 75°C)
Since no heat is lost to the surroundings, the heat gained by the ice cubes is equal to the heat lost by the water. Setting the two equations equal to each other, we can solve for T-f:
4 × (53.5 g) × (4.184 J/(g°C)) × (0°C - T-f) = (115 g) × (4.184 J/(g°C)) × (T-f - 75°C)
Simplifying the equation gives:
14.0816 × T-f = 496.82 - 18.0714 × T-f
32.1526 × T-f = 496.82
T-f = 15.44°C
A 23.2 g sample of an organic compound containing carbon, hydrogen and oxygen was burned in excess oxygen and yielded 52.8 g of carbon dioxide and 21.6 g of water. Determine the empirical formula of the compound.
Answer:
The answer to your question is C₃H₆O
Explanation:
Data
mass of sample = 23.2 g
mass of carbon dioxide = 52.8 g
mass of water = 21.6 g
empirical formula = ?
Process
1.- Calculate the mass and moles of carbon
44 g of CO₂ --------------- 12 g of C
52.8 g --------------- x
x = (52.8 x 12)/44
x = 633.6/44
x = 14.4 g of C
12 g of C ------------------ 1 mol
14.4 g of C --------------- x
x = (14.4 x 1)/(12)
x = 1.2 moles of C
2.- Calculate the grams and moles of Hydrogen
18 g of H₂O --------------- 2 g of H
21.6 g of H₂O ------------- x
x = (21.6 x 2) / 18
x = 2.4 g of H
1 g of H -------------------- 1 mol of H
2.4 g of H ----------------- x
x = (2.4 x 1)/1
x = 2.4 moles of H
3.- Calculate the grams and moles of Oxygen
Mass of Oxygen = 23.2 - 14.4 - 2.4
= 6.4 g
16 g of O ---------------- 1 mol
6.4 g of O -------------- x
x = (6.4 x 1)/16
x = 0.4 moles of Oxygen
4.- Divide by the lowest number of moles
Carbon = 1.2 / 0.4 = 3
Hydrogen = 2.4/ 0.4 = 6
Oxygen = 0.4 / 0.4 = 1
5.- Write the empirical formula
C₃H₆O
Final answer:
To find the empirical formula of a compound from its combustion products, convert the masses of carbon dioxide and water to moles to determine the moles of carbon, hydrogen, and oxygen in the compound. Calculating these mole ratios leads to determining that the empirical formula of the compound is C3H6O.
Explanation:
To determine the empirical formula of the compound given its combustion products, begin by converting the mass of carbon dioxide (CO2) and water (H2O) to moles. This reveals the moles of carbon and hydrogen in the original compound. Since oxygen is also part of the compound, calculate its moles by subtraction from the total mass of the original compound.
Convert 52.8 g of CO2 to moles: (52.8 g) / (44.01 g/mol) = 1.2 mol of C.Convert 21.6 g of H2O to moles: (21.6 g) / (18.015 g/mol) = 1.2 mol of H2, or 2.4 mol of H.Calculate moles of oxygen in the compound: Subtract the mass of C and H in the original compound from its total mass. Mass of C from CO2 = 1.2 mol × 12 g/mol = 14.4 g; Mass of H from H2O = 2.4 mol × 1 g/mol = 2.4 g. Total mass of C and H = 14.4 g + 2.4 g = 16.8 g; Mass of O = 23.2 g (total mass) - 16.8 g = 6.4 g, which is (6.4 g) / (16 g/mol) = 0.4 mol of O.To find the empirical formula, divide each mole value by the smallest number of moles: C=1.2/0.4, H=2.4/0.4, O=0.4/0.4, giving a ratio of C: 3 H: 6 O: 1. Therefore, the empirical formula is C3H6O.
Cell X fires an action potential and releases GABA onto Cell Y. Assuming only GABAA receptors are present on the postsynaptic membrane, which is at rest at -55mV, which ion is responsible for the changes that would be observed?
a. Chloride
b. Potassium
c. Sodium
d. Calcium
e. Magnesium
Answer:
The correct answer is a. Chloride
Explanation:
GABAA receptors, is an ionotropic receptor that controls most of the central nervous system inhibitory transmission. GABAA receptors exist as ionotropic ligand-gated ion channel and selectively conduct chloride ions Cl⁻ when activated by GABA through its pore where by the flow of Cl⁻ depends on the internal voltage of the cell and the resting potential. Cl⁻ will flow in a cell if the internal voltage is more than resting potential which is -75 mV. When the internal voltage is less than resting potential, Cl⁻ will transit out of the cell
Final answer:
The ion responsible for the changes when GABA is released onto Cell Y with a resting membrane potential at -55mV and only GABAA receptors present is chloride, which causes an inhibitory postsynaptic potential.
Explanation:
When Cell X fires an action potential and releases GABA onto Cell Y, and only GABAA receptors are present on the postsynaptic membrane which is at rest at -55mV, the ion responsible for the observed changes is chloride. The binding of GABA to GABAA receptors increases the influx of chloride ions into the postsynaptic cell, making the inside of the cell more negative. This increase in negative charge pulls the membrane potential towards the equilibrium potential of chloride which is -65 mV and creates an inhibitory postsynaptic potential (IPSP), thus inhibiting the neuron from firing an action potential.
The rate constant of a reaction is 4.7×10−3 s−1 at 25°C, and the activation energy is 33.6 kJ/mol. What is k at 75°C?
To calculate the rate constant (k) at 75°C, we need the frequency factor (A) for the reaction, which is not provided in the question.
Explanation:The rate constant (k) of a reaction can be calculated using the Arrhenius equation:
k = Ae-Ea/RT
where A is the frequency factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.
To calculate k at 75°C, we need to find the frequency factor (A) for the reaction. Unfortunately, the question does not provide the value of A, so we cannot calculate the rate constant at 75°C.
The rate constant ( k ) at 75°C is approximately [tex]\( {2.57 \times 10^{-5} \text{ s}^{-1}} \).[/tex]
To find the rate constant ( k ) at 75°C for a reaction with an activation energy given at 25°C, we'll use the Arrhenius equation. The Arrhenius equation relates the rate constant ( k ) to temperature ( T ) and the activation energy [tex]\( E_a \):[/tex]
[tex]\[ k = A \cdot e^{-\frac{E_a}{RT}} \][/tex]
Given data:
Rate constant ( k ) at 25°C (298.15 K): [tex]\( k_{25} = 4.7 \times 10^{-3} \) s\(^{-1}\),[/tex]Activation energy [tex]\( E_a \): \( 33.6 \) kJ/mol.[/tex]Step-by-Step Solution:
1. Convert activation energy to Joules per mole:
[tex]\[ E_a = 33.6 \times 10^3 \text{ J/mol} \][/tex]
2. Calculate ( k ) at 25°C (298.15 K):
First, express the Arrhenius equation in terms of [tex]\( k_{25} \)[/tex] and solve for ( A ):
[tex]\[ k_{25} = A \cdot e^{-\frac{E_a}{RT_{25}}} \][/tex]
[tex]\[ A = k_{25} \cdot e^{\frac{E_a}{RT_{25}}} \][/tex]
Calculate ( A ):
[tex]\[ A = 4.7 \times 10^{-3} \text{ s}^{-1} \cdot e^{\frac{33.6 \times 10^3 \text{ J/mol}}{8.314 \text{ J/(mol·K)} \cdot 298.15 \text{ K}}} \][/tex]
[tex]\[ A \approx 4.7 \times 10^{-3} \text{ s}^{-1} \cdot 6.55 \times 10^5 \][/tex]
[tex]\[ A \approx 3.08 \text{ s}^{-1} \][/tex]
3. Calculate ( k ) at 75°C (348.15 K):
Now use the calculated ( A ) and the new temperature [tex]\( T_{75} = 348.15 \) K:[/tex]
[tex]\[ k_{75} = A \cdot e^{-\frac{E_a}{RT_{75}}} \][/tex]
[tex]\[ k_{75} = 3.08 \text{ s}^{-1} \cdot e^{-\frac{33.6 \times 10^3 \text{ J/mol}}{8.314 \text{ J/(mol·K)} \cdot 348.15 \text{ K}}} \][/tex]
[tex]\[ k_{75} = 3.08 \text{ s}^{-1} \cdot e^{-11.51} \][/tex]
[tex]\[ k_{75} = 3.08 \text{ s}^{-1} \cdot 8.35 \times 10^{-6} \][/tex]
[tex]\[ k_{75} \approx 2.57 \times 10^{-5} \text{ s}^{-1} \][/tex]
Which of the following is NOT true of peptide bonds? They tend to be planar. they are generally in the trans and rarely in the cis configuration. They tend to have the amide nitrogen protonated to give a positive charge. They contain an unusually long carbon-carbon bond
Answer:
They tend to have the amide nitrogen protonated to give a positive charge.
Explanation:
A peptide bond joins two consecutive amino acids in the protein. The peptide bond is present between -CO group (also known as carboxyl group) of one amino acid and -NH2 group (also known as amino group) of another amino acid. It is represented as -CONH bond. Therefore, it is an amide linkage. The peptide bond always has planar orientation with trans configuration.
Trans configuration avoids steric hindrance and hence, add to stability of the peptide bond.
Nitrogen atom of peptide bond never bear positive charge
Therefore, the incorrect statement is as follows:
They tend to have the amide nitrogen protonated to give a positive charge.
When solid NH4HS is placed in a closed flask at 28oC, the solid dissociates according to the equation below. NH4HS(s) ⇄ NH3(g) + H2S(g). The total pressure of the equilibrium mixture is 0.766 atm. Determine Keq at this temperature.
Answer: 0.147
Explanation:
[tex]K_p[/tex] is the constant of a certain reaction at equilibrium.
For the given chemical reaction:
[tex]NH_4HS(s)\rightleftharpoons NH_3(g)+H_2S(g)[/tex]
at t= 0 0 0 0
at eqm p p
Total pressure = p+p = 0.766 atm
2p= 0.766 atm
p= 0.383 atm
The expression of [tex]K_p[/tex] for above equation follows:
[tex]K_p={p_{NH_3}\times p_{H_2S}}[/tex]
We are given:
[tex]K_p={0.383\times 0.383[/tex]
[tex]K_p=0.147[/tex]
Thus [tex]K_{eq}[/tex] at this temperature is 0.147
A water sample contains the pollutant chlorobenzene with a concentration of 15 ppb. What is the molarity of this solution? Assume the density of the solution is 1.00 g/mL. The molar mass of chlorobenzene is 112.6 g/mol.
Answer:
The molarity is 1.33×10^-7M
Explanation:
15ppb = 15g of chlorobenzene/10^9g of solution × 1 mole of chlorobenzene/112.6g × 1g of solution/mL = 1.33×10^-10mol/mL × 1000mL/1L = 1.33×10^-7mol/L
The molarity of a chlorobenzene solution with a concentration of 15 ppb is calculated by first converting the ppb measurement to mass per volume and then using the molar mass to find the number of moles per liter, resulting in a molarity of 1.33 × 10⁻⁷ M.
Explanation:To calculate the molarity of chlorobenzene in a solution with a concentration of 15 ppb, we first need to convert this concentration into a mass-volume relationship. Since 1 ppb is equivalent to 1 µg/L, 15 ppb is equal to 15 µg/L, which is the same as 0.015 mg/L.
Using the molar mass of chlorobenzene, which is 112.6 g/mol, we can convert this mass into moles using the following equation:
Number of moles = mass (g) / molar mass (g/mol)The solution has a concentration of 15 µg/L, which is equal to 0.015 mg/L or 0.000015 g/L. So the amount of moles of chlorobenzene is:
Number of moles = 0.000015 g / 112.6 g/mol = 1.33 × 10-7 moles/L
Therefore, the molarity of the chlorobenzene solution is 1.33 × 10-7 M.
The Elizabethan view of life changed little from the characteristic Medieval view of life. True False
Answer:
The correct answer is "False".
Explanation:
The Elizabethan Era was a period of England's history that corresponds to the reign of Queen Elizabeth I (1558–1603). The Elizabethan Era is considered a golden age of England's history, and part of the Renaissance Era (1300-1600). The Elizabethan view of life changed drastically from the characteristic Medieval view of life. During Medieval times life was seen with a religiously perspective only, while during The Elizabethan Era more people start to view life with a scientific perspective.
THE ANSWER IS FALSE!
The potential difference between two parallel plates 175 V. An α particle with mass of 6.5 × 10-27 kg and charge of 3.2 × 10-19 C is released from rest near the positive plate. What is the kinetic energy of the α particle when it reaches the other plate? The distance between the plates is 20 cm.
Answer:
5.6 × 10⁻¹⁷ J
Explanation:
We know that the work done by an electric field, E on an electric charge, q moving a distance, d is W = qEdcosθ. From above, q = 3.2 × 10⁻¹⁹ C, d = 20 cm = 2 × 10⁻² m, E = V/d = 175V /2 × 10⁻² m = 8750 V/m and θ = 0 since the α particle moves in the same direction as the electric field. So W = qEdcosθ = 3.2 × 10⁻¹⁹ C × 8750 V/m × 2 × 10⁻² m × cos0 = 5.6 × 10⁻¹⁷ J. We know that the work done by the electric field on the charge W = ΔK the change in kinetic energy of the charge. So, W = 1/2m(v₂² - v₁²) where v₁ = initial velocity and v₂ = final velocity. Since the charge is at rest at the positive plate, v₁ = 0. So, W = 1/2mv₂² = K which is the kinetic energy of the particle after moving the distance of 20 cm between the plates. So K = W = 5.6 × 10⁻¹⁷ J
The tomato is dropped. What is the velocity, vvv, of the tomato when it hits the ground? Assume 93.8 %% of the work done in Part A is transferred to kinetic energy, EEE, by the time the tomato hits the ground.
Answer:
the velocity of the tomato will be v = u + gt
kinetic energy = 9.64 h
Explanation:
The tomato is dropped from a height, so before it lands on the ground, it possesses potential energy. This is the energy relative to its height from the ground.At that time, let the initial speed be u.
The acceleration due to gravity be g
The final velocity will be given as v = u + at
but a = g = 9.81 m/s² [acceleration due to gravity]
so the final velocity will be given as v = u + at
Let the potential energy be Ep
Before landing the ground, 93.8 % of the potential energy will be converted to kinetic energy. Therefore, the calculation will be as follows:
Ep = mgh
Kinetic energy Ek = 1/2mv²
But, Ek = 0.938 Ep
= 0.983 × gh
= 9.64 h
where h is the height of the object from the ground.
How many atoms of oxygen are there in one molecule of carbon dioxide, if the chemical formula is CO2? 0 1 2 3
Answer:
2
Explanation:
A careful look at CO2, reveals that CO2 contains:
1 atom of C
2 atoms of O.
Answer:
2
Explanation:
i just took the test
Nitrogen monoxide, NO, reacts with hydrogen, H₂, according to the following equation.
[tex]2NO + 2H_2 \rightarrow N_2 + 2H_2O[/tex]
What would the rate law be if the mechanism for this reaction were as follows?
(Rate expressions take the general form: rate = k . [A]a . [B]b.)
[tex]2 NO + H_2 \rightarrow N_2 + H_2O_2[/tex] (slow)
[tex]H_2O_2 + H_2 \rightarrow 2 H_2O[/tex] (fast)
Answer : The rate law for the overall reaction is, [tex]Rate=k[NO]^2[H_2][/tex]
Explanation :
Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.
As we are given the mechanism for the reaction :
Step 1 : [tex]2NO+H_2\rightarrow N_2+H_2O_2[/tex] (slow)
Step 2 : [tex]H_2O_2+H_2\rightarrow 2H_2O[/tex] (fast)
Overall reaction : [tex]2NO+2H_2\rightarrow N_2+2H_2O[/tex]
The rate law expression for overall reaction should be in terms of [tex]NO\text{ and }H_2[/tex].
As we know that the slow step is the rate determining step. So,
The slow step reaction is,
[tex]2NO+H_2\rightarrow N_2+H_2O_2[/tex]
The expression of rate law for this reaction will be,
[tex]Rate=k[NO]^2[H_2][/tex]
Hence, the rate law for the overall reaction is, [tex]Rate=k[NO]^2[H_2][/tex]
which of the following sets of terms best describes sound waves
A) Mechanical and Transverse
B) Electromagnetic and Transverse
C) Electromagnetic and Longitudinal
D) Mechanical and Longitudinal
????????Please hurry
Sound waves are D) Mechanical and Longitudinal
Explanation:
In physics, waves are classified into two types:
Mechanical waves are those waves that need a medium to propagate - these waves are produced by the vibrations of the particles of the mediumElectromagnetic waves are produced by the alternating vibrations of electric and magnetic fields - they are the only waves that can also travel through a vacuum, so they do not need a mediumMoreover, waves are further classified into:
Transverse waves are those in which the vibration of the wave occurs in a direction perpendicular to the direction of propagation of the waveLongitudinal waves are those in which the vibration of the wave is parallel (back and forth) to the direction of propagation of the waveSound waves are oscillations of a medium that occurs back-and forth along the direction of propagation of the wave. Therefore, they are mechanical (they need a medium to propagate) and longitudinal. So the correct answer is
D) Mechanical and Longitudinal
Learn more about sound waves:
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Final answer:
Sound waves are best described as Mechanical and Longitudinal waves, as they need a medium to travel and have disturbances parallel to the direction of propagation in fluids.
Explanation:
The set of terms that best describes sound waves is Mechanical and Longitudinal. Sound waves require a medium to travel through, which makes them mechanical waves. Furthermore, in fluids such as air and water, sound waves are longitudinal because their disturbances (variations in pressure) are parallel to the direction of propagation. In solids, sound waves can be both longitudinal and transverse; however, in the context of your question which seems to imply a generic scenario, the focus is often on sound in fluids.
How much energy is required to move the electron of the hydrogen atom from the 1s to the 2s orbital?
Answer:
1.63425 × 10^- 18 Joules.
Explanation:
We are able to solve this kind of problem, all thanks to Bohr's Model atom. With the model we can calculate the energy required to move the electron of the hydrogen atom from the 1s to the 2s orbital.
We will be using the formula in the equation (1) below;
Energy, E(n) = - Z^2 × R(H) × [1/n^2]. -------------------------------------------------(1).
Where R(H) is the Rydberg's constant having a value of 2.179 × 10^-18 Joules and Z is the atomic number= 1 for hydrogen.
Since the Electrons moved in the hydrogen atom from the 1s to the 2s orbital,then we have;
∆E= - R(H) × [1/nf^2 - 1/ni^2 ].
Where nf = 2 = final level= higher orbital, ni= initial level= lower orbital.
Therefore, ∆E= - 2.179 × 10^-18 Joules× [ 1/2^2 - 1/1^2].
= -2.179 × 10^-18 Joules × (0.25 - 1).
= - 2.179 × 10^-18 × (- 0.75).
= 1.63425 × 10^- 18 Joules.
The 1995 Nobel Prize in chemistry was shared by Paul Crutzen, F. Sherwood Rowland, and Mario Molina for their work concerning the formation and decomposition of ozone in the stratosphere. Rowland and Molina hypothesized that chlorofluorocarbons (CFCs) in the stratosphere break down upon exposure to UV radiation, producing chlorine atoms. Chlorine was previously identified as a catalyst in the breakdown of ozone into oxygen gas. Using the enthalpy of reaction for two reactions with ozone, determine the enthalpy of reaction for the reaction of chlorine with ozone.1) ClO(g) +O₃(g) ----> Cl(g) +2O₂(g); Hrxn = -122.8 kJ/mol
2) 2O₃(g) ---> 3O₂(g); Hrxn = -285.3 kJ/mol
3) O₃(g) + Cl(g) -----> ClO(g) + O₂(g); Hrxn = ????
Answer: The [tex]\Delta H^o_{rxn}[/tex] for the reaction is -162.5 kJ/mol
Explanation:
Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.
The given chemical reaction follows:
[tex]O_3(g)+Cl(g)\rightarrow ClO(g)+O_2[/tex] [tex]\Delta H^o_{rxn}=?[/tex]
The intermediate balanced chemical reaction are:
(1) [tex]ClO(g)+O_3(g)\rightarrow Cl(g)+2O_2(g)[/tex] [tex]\Delta H_1=-122.8kJ/mol[/tex]
(2) [tex]2O_3(g)\rightarrow 3O_2(g)[/tex] [tex]\Delta H_2=-285.3kJ/mol[/tex]
The expression for enthalpy of the reaction follows:
[tex]\Delta H^o_{rxn}=[1\times (-\Delta H_1)]+[1\times \Delta H_2][/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[(1\times (-(-122.8))+(1\times (-285.3))=-162.5kJ/mol[/tex]
Hence, the [tex]\Delta H^o_{rxn}[/tex] for the reaction is -162.5 kJ/mol
The diagram below shows the temperature dropping from 80°C to 20°C. Molecules that have longer arrows are moving faster. Which statement describes what is happening in this system?
1. Molecules vibrate more as temperature decreases.
2. Temperature is not related to the average kinetic energy of a system.
3. Kinetic energy decreases as temperature decreases.
4.. There is no change in average kinetic energy between these two systems.
Answer:
The answer to this question is 3. Kinetic energy decreases as temperature decreases.
Explanation:
The average Kinetic energy is given by
[tex]K = \frac{3}{2}[/tex] × [tex]K_{B}[/tex] × T where
K = The average molecular kinetic energy of the gas (J)
[tex]K_{B}[/tex] = Boltzmann's constant (1.38×[tex]10^{-23}[/tex] J/K)
T = Temperature of the gas in Kelvin (k)
When the temperature of a given mass of gas drops average kinetic energy of the molecules goes down and the average molecular speed decreases. The lower kinetic energy which is the energy of motion of the molecules indicates lower speed of the molecules.
Note that, the kinetic energy and average speed are related by the following formula
KE = [tex]\frac{1}{2}[/tex]×m×v²
Where KE = Kinetic energy in Joules (J)
v = the velocity in m/s and
m = the mass of the molecule in Kg
Answer:
Answer is C
Explanation:
Edge 2020
What compound does the formula NO2 represent?
nitrogen oxide
nitrogen dioxide
dinitorgen oxide
nitroxide
Answer:
formula for NO2 is nitrogen dioxide
When 60 mL of 1.30 mol/L AgNO3(aq) and 60 mL of 1.30 mol/L HCl(aq) are mixed in a simple calorimeter, the temperature rises by 5.18°C. The molar enthalpy of reaction of HCl(aq) is ab.C kJ/mol.
Answer : The molar enthalpy of reaction is, 33.3 KJ/mole
Explanation :
First we have to calculate the mass of water.
As we know that the density of water is 1 g/ml. So, the mass of water will be:
The volume of water = [tex]60ml+60ml=120ml[/tex]
Now we have to calculate the heat absorbed during the reaction.
[tex]q=m\times c\times (\Delta T)[/tex]
where,
q = heat absorbed = ?
[tex]c[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]
m = mass of water = 120 g
[tex]\Delta T[/tex] = change in temperature = [tex]5.18^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]q=120g\times 4.18J/g^oC\times (5.18)^oC[/tex]
[tex]q=2598.288J=2.60KJ[/tex]
Now we have to calculate the molar enthalpy of reaction.
[tex]\Delta H=\frac{q}{n}[/tex]
where,
[tex]\Delta H[/tex] = enthalpy of neutralization = ?
q = heat released = 2.60 KJ
n = number of moles =
[tex]\Delta H=\frac{2.60KJ}{0.078mole}=33.3KJ/mole[/tex]
Therefore, the molar enthalpy of reaction is, 33.3 KJ/mole
The molar enthalpy of reaction of HCl(aq) is calculated to be approximately -33.28 kJ/mol, using calorimetry principles and the given data.
To calculate the molar enthalpy of reaction of HCl(aq), we use the concept of calorimetry and the given temperature change. The provided reaction is a neutralization reaction where HCl reacts with AgNO₃ to form AgCl(s) and HNO₃(aq)
Firstly, we need to calculate the total heat (q) released during the reaction. This can be done by using the formula:
q = mcΔT
where m is the mass of the solution, c is the specific heat capacity of water (assumed to be 4.18 J/g°C as the specific heat capacity is not provided), and ΔT is the change in temperature. The mass (m) of the solution is the sum of the volumes of AgNO₃(aq) and HCl(aq) solutions, assuming a density of 1 g/mL for both solutions:
m = volume of AgNO₃ + volume of HCl = 60 mL + 60 mL
= 120 mL
= 120 g
Substitute the values into the formula to calculate q:
q = (120 g)(4.18 J/g°C)(5.18°C)q
= 2596.416 J
= 2.596 kJ
Next, calculate the number of moles of HCl which is equal to the moles of AgNO₃ since they react in a 1:1 ratio:
Moles of HCl = Volume of HCl × Concentration of HCl
= 0.060 L × 1.30 mol/L
= 0.078 mol
Finally, divide the total heat energy by the number of moles to find the molar enthalpy (ΔH) of the reaction per mole of HCl:
ΔH = q / moles of HClΔH
= 2.596 kJ / 0.078 mol
= -33.28 kJ/mol
Therefore, the molar enthalpy of reaction of HCl(aq) is approximately -33.28 kJ/mol.
Place the butane lighter in the sink or tub and let it rest there until needed. why do we soak the lighter in the water bath?
Answer:
To regulate the gas pressure in the lighter tank and avoid build-up of pressure.
Explanation:
First, you need to understand the properties of this organic molecule. Butane. C₄H₁₀ is a colorless, odorless, but HIGHLY FLAMMABLE liquefied gas. The liquid is flammable at 25⁰C whilst the vapor is flammable at 15⁰C. As you can see, this is an extremely flammable gas. It has a high vapor pressure (tendency by liquids to escape as gas molecules) Any external heat source induces a pressure build up that might cause the gas to explode when there is an open flame. Combining the two points above, a thermo-regulated water bath that has a lower temperature (below 15⁰C) will be need to prevent the pressure build up and ensure that any leakage will not have a high vapor pressure.We soak the lighter in the water bath to achieve uniform heating and reduce the fire hazard inherent in direct heating methods, ensuring a safer laboratory environment.
We soak the lighter in the water bath to ensure uniform heating of the reaction mixture with less fire hazard. Direct heating on a Bunsen burner or hot plate can cause uneven temperatures and increased risks of fire or overheating. A water bath is especially beneficial when a chemical reaction must be heated for a certain time to occur. It provides a stable and consistent heat source, which is safer and can prevent accidents in the lab. By using a water bath, we also prevent the lighter from getting excessively hot, which could cause it to malfunction or pose a safety risk.
A mixture of gases at a total pressure of 95 kPa contains N₂, CO₂, and O₂. The partial pressure of the CO₂ is 24 kPa and the partial pressure of the N₂ is 48 kPa. What is the partial pressure of the O₂?
Answer:
23 kPa = Partial pressure O₂
Explanation:
In a mixture of gases, the sum of partial pressure of each gas that contains the mixture = Total pressure
Total pressure = Partial pressure N₂ + Partial pressure CO₂ + Partial pressure O₂
95 kPa = 48 kPa + 24 kPa + Partial pressure O₂
95 kPa - 48 kPa - 24 kPa = Partial pressure O₂
23 kPa = Partial pressure O₂