The theoretical yield of phenacetin is 0.17922 g. However, the actual yield may be lower due to factors such as incomplete reaction, loss during purification, or experimental error.
To calculate the theoretical yield of phenacetin, we need to first determine the limiting reagent. The limiting reagent is the reactant that will be completely consumed in the reaction, thus limiting the amount of product that can be produced.
First, we need to convert the volume of bromoethane given in milliliters to grams, using its density:
0.12 ml x 1.47 g/ml = 0.1764 g bromoethane
Next, we can use the molar masses of p-acetamidophenol and bromoethane to determine the number of moles of each:
moles p-acetamidophenol = 0.151 g / 151.16 g/mol = 0.001 mol
moles bromoethane = 0.1764 g / 108.97 g/mol = 0.00162 mol
Since the reaction requires a 1:1 molar ratio of p-acetamidophenol to bromoethane, and the number of moles of p-acetamidophenol is smaller than the number of moles of bromoethane, p-acetamidophenol is the limiting reagent.
The theoretical yield of phenacetin can be calculated using the molar mass of phenacetin and the number of moles of p-acetamidophenol:
moles phenacetin = 0.001 mol p-acetamidophenol
mass phenacetin = 0.001 mol x 179.22 g/mol = 0.17922 g
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a 1.25 g sample of co2 is contained in a 750. ml flask at 22.5 c. what is the pressure of the gas, in atm?
The pressure of gas is 1.05 atm when a 1.25 g sample of CO₂ is contained in a 750ml flask at 22.5°C.
Molecular weight of CO₂ is 1.25g ,Volume of CO₂ is 750ml,Temperature of CO₂ is 22.5°C and the gas constant is 0.08206 L atm/mol K.
Using the ideal gas law equation the pressure is found to be 1.05 atm.
To calculate the pressure of the gas, we can use the ideal gas law equation: [tex]PV=nRT[/tex]
Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
First, we need to convert the volume to liters by dividing by 1000: 750 ml = 0.75 L.
Next, we need to calculate the number of moles of CO₂ present in the flask. We can use the molecular weight of CO₂ to convert from grams to moles:
[tex]1.25 * (1 /44.01 ) = 0.0284 mol[/tex]
Now we can plug in the values into the ideal gas law equation:
[tex]PV=nRT[/tex]
[tex]P * 0.75 L = 0.0284 mol * 0.08206 L*atm/mol*K * (22.5 + 273.15) K[/tex]
Simplifying and solving for P, we get:
[tex]P = (0.0284 * 0.08206 * 295.65) / 0.75 = 1.05 atm[/tex]
Therefore, the pressure of the gas in the flask is 1.05 atm.
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What is the concentration (in molality) of an aqueous solution of NaCl made by adding
4.56 g of NaCl to enough water to give 20.0 mL of solution. Assume the density of the
solution is 1.03 g/mL
Answer:
data given
mass of NaCl 4.56
dissolved volume 20ml(0.02l)
density of solution 1.03g/ml
Required molality
Explanation:
molarity=m/mr×v
where
m is mass
mr molar mass
v is volume
now,
molarity=4.56/58.5×0.02
molarity =3.9
: .molarity is 3.9mol/dm^3
According to molal concentration, the concentration (in molality) of an aqueous solution of NaCl is 0.0047 mole/kg.
What is molal concentration?Molal concentration is defined as a measure by which concentration of chemical substances present in a solution are determined. It is defined in particular reference to solute concentration in a solution . Most commonly used unit for molal concentration is moles/kg.
The molal concentration depends on change in volume of the solution which is mainly due to thermal expansion. Molal concentration is calculated by the formula, molal concentration=mass/ molar mass ×1/mass of solvent in kg.
In terms of moles, it's formula is given as molal concentration= number of moles /mass of solvent in kg.
Substitution in formula gives the answer but first mass of solution is determined which is density×volume= 1.03×20=20.6 g , mass of solvent= 20.6-4.56=16.05, thus molal concentration=4.56/58.5×1/16.05=0.0047 moles/kg.
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a sample of nobr was placed on a 1.00l flask containing no no or br 2 at equilibrium the flask contained
At equilibrium, the concentrations of NO, Br2, and NOBr in the flask will remain constant. However, without specific values for the initial concentration of NOBr or the equilibrium constant (Kc), it's not possible to determine.
.Based on the provided information, it seems that a sample of NOBr was placed in a 1.00 L flask at equilibrium, which means that the NOBr has decomposed into NO and Br2.
At equilibrium, the concentrations of NO, Br2, and NOBr in the flask will remain constant. However, without specific values for the initial concentration of NOBr or the equilibrium constant (Kc), it's not possible to determine the exact concentrations of these substances in the flask.
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A sample of NOBr being placed in a 1.00 L flask containing no NO or Br2 at equilibrium, I'll first provide the balanced chemical equation for the reaction:
[tex]2 NOBr (g) ⇌ 2 NO (g) + Br2 (g)[/tex]
At equilibrium, the concentrations of the reactants and products remain constant. To determine the concentrations of NOBr, NO, and Br2 at equilibrium, we need to follow these steps:
1. Write the expression for the equilibrium constant (Kc) based on the balanced chemical equation:
[tex]Kc = [NO]^2 [Br2] / [NOBr]^2[/tex]
2. Set up an ICE (Initial, Change, Equilibrium) table to determine the equilibrium concentrations of the species involved in the reaction. The initial concentrations of NO and Br2 are 0 since they are not initially present in the flask.
NOBr NO Br2
I C0 0 0
C -2x +2x +x
E C0-2x 2x x
3. Substitute the equilibrium concentrations from the ICE table into the Kc expression:
[tex]Kc = (2x)^2 * x / (C0-2x)^2[/tex]
4. To solve for x, you need the value of Kc for the reaction. Look up the Kc value for this reaction in a reference or use provided information. Once you have Kc, substitute it into the equation and solve for x.
5. Calculate the equilibrium concentrations of NOBr, NO, and Br2 by substituting the value of x back into the ICE table:
[NOBr] = C0-2x
[NO] = 2x
[Br2] = x
By following these steps, you can determine the concentrations of NOBr, NO, and Br2 in the 1.00 L flask at equilibrium.
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you have 400 grams (g) of a substance with a half life of 10 years. how much is left after 100 years?
After 100 years, there will be 6.25 grams of the substance remaining.
What is half life?Half-life is the time it takes for half of the radioactive atoms in a sample to decay or for the concentration of a substance to decrease by half.
Amount remaining = initial amount x (1/2)^(number of half-lives)
In this case, half-life of the substance is 10 years, which means that after 10 years, half of the substance will have decayed. After another 10 years (20 years total), half of remaining substance will decay, leaving 1/4 of the original amount. After another 10 years (30 years total), half of that remaining amount will decay, leaving 1/8 of the original amount. This process continues every 10 years.
To find the amount of substance remaining after 100 years, we need to know how many half-lives have occurred in that time: 100 years / 10 years per half-life = 10 half-lives
Amount remaining = 400 g x (1/2)¹⁰= 6.25 g
Therefore, after 100 years, there will be 6.25 grams of the substance remaining.
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