A student puts 0.020 mol of methyl methanoate into an empty and rigid 1.0 L vessel at 450 K. The pressure is measured to be 0.74 atm. When the experiment is repeated using 0.020 mol of ethanoic acid instead of methyl methanoate, the measured pressure is lower than 0.74 atm. The lower pressure for ethanoic acid is due to the following reversible reaction.CH3COOH(g)+CH3COOH(g) ⇋ (CH3COOH)2(g)+Assume that when equilibrium has been reached, 50 percent of the ethanoic acid molecules have reacted.i. Calculate the total pressure in the vessel at equilibrium at 450 K.ii. Calculate the value of the equilibrium constant, Kp, for the reaction at 450 K

Answers

Answer 1
Answer:

Explanation:

Starting moles of ethanol acid = 0.020 mol

At the equilibrium 50 % of the ethanol acid molecules reacted

∴ Moles of ethanol acid reacted = 0.020 mol * 50 %/100 %

                                                                   = 0.010 mol

Moles of ethanol acid remain = 0.020 mol + 0.010 mol = 0.010 mol

Moles of the product (CH3COOH)^(2) gas formed are calculated as

0.010 mol CH3COOH * 1 mol (CH3COOH)^(2) / 2 mol CH3COOH

= 0.005 mol (CH3COOH)^(2)

Therefore at the equilibrium total moles of gas present in the vessel are 0.010 mol CH3COOH and 0.005 mol (CH3COOH)^(2)

That is total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol

Now Calculate the pressure  :

0.020 mol gas has pressure of 0.74 atm therefore at the same condition what will be the pressure exerted by 0.015 mol gas

P1/n1 = P2/n2

P2 = P1*n2 / n1

      = 0.74 atm * 0.015 mol / 0.020 mol

     = 0.555 atm


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One cubic meter of an ideal gas at 600 K and 1000 KPa expand to five times its initial volume as follows:(a) By a mechanically reversible, isothermal process [T 2 = 600K; P 2 = 200kPa; W= -1609 kJ]


(b) By a mechanically reversible, adiabatic process [T 2 =208.96K; P 2 = 67.65 kPa; W= -994.4 kJ] For each case calculate the final temperature, pressure and the work done by the gas. C p =21 Jmol- 1K-1.

Answers

Answer:

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What mass of nitrogen gas is required to react completely with 2.79 g of hydrogen gas to produce ammonia?

Answers

the balanced equation for the formation of ammonia is as follows
N₂ + 3H₂ ---> 2NH₃
stoichiometry of H₂ to N₂ is 3:1
number of H₂ moles reacted - 2.79 g / 2 g/mol  = 1.40 mol 
if 3 mol of H₂ reacts with 1 mol of N₂
then 1.40 mol of H₂ reacts with  - 1.40/3 = 0.467 mol of N₂
mass of N₂ required - 0.467 mol x 28 g/mol = 13.1 g
mass of N₂ formed is 13.1 g


5. What is the speed (Velocity) of a cyclist who covers 10 km (convert to metersfirst!) in 14 minutes and 30 seconds? Remember time must be in seconds first!

Answers

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Actually Welcome to the Concept of the Kinematics.

Here, we know that, Velocity = Distance / Time,

So here, Distance = 10km = 10×1000 = 10000 metres.

, Time = 14 min 30 sec = 870 seconds,

so now, we get velocity as,

=> V = 10000 ÷ 870 => 11.49 m/s .

Hence, Velocity is 11.49 m/s.

Describe the range of radii of most atoms in nanometers (nm)

Answers

Final answer:

The range of radii of most atoms is typically in the nanometer scale (nm) and can be measured using the covalent radius. The size of an atom's nucleus is much smaller than the atom itself. The Bohr model provides a formula to calculate the radius of hydrogen-like atoms.

Explanation:

The range of radii of most atoms is typically in the nanometer scale (nm). The covalent radius, which is defined as half the distance between the nuclei of two identical atoms when they are joined by a covalent bond, provides a practical way to measure the size of atoms. As we move down a group in the periodic table, the covalent radius generally increases, indicating a larger size of the atom. For example, the covalent radius of the halogens increases as we move from fluorine to iodine.

The size of an atom's nucleus, on the other hand, is much smaller than the atom itself. The nucleus has a diameter of about 10-15 meters, while the typical atom has a diameter of the order of 10-10 meters. This difference in size illustrates the emptiness of atoms, with the distance from the nucleus to the electrons being typically 100,000 times the size of the nucleus.

The Bohr model provides a formula to calculate the radius of hydrogen-like atoms, which depends on the principal quantum number (n) and the atomic number (Z). The calculated radii of the orbits of the hydrogen atom have been experimentally verified to have a diameter of a hydrogen atom.

Learn more about Atomic Radii here:

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Final answer:

The range of radii of most atoms is typically measured in nanometers (nm). Covalent radius and hydrogen-like orbits are two methods used to estimate the size of atoms. The size of an atom can vary depending on the element and measurement technique, but most atoms have radii on the order of nanometers (nm).

Explanation:

The range of radii of most atoms is typically measured in nanometers (nm). The size of an atom can be estimated using various techniques. One commonly used measure is the covalent radius, which is defined as one-half the distance between the nuclei of two identical atoms when they are joined by a covalent bond. The covalent radii of different elements can be found in tables and can vary depending on the element and its position in the periodic table.

Another way to estimate the size of atoms is by looking at the sizes of their orbits in hydrogen-like atoms. These orbits are given in terms of their radii by a mathematical expression that includes a constant called the Bohr radius, which is approximately 5.292 × 10-11 m.

Overall, the size of an atom can vary depending on the element and the specific measurement technique used, but most atoms have radii on the order of nanometers (nm).

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Which step would help a student find the molecular formula of a compound from the empirical formula? Multiply the subscripts of the empirical formula by the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound. Subtract the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound from the subscripts of the empirical formula. Divide the subscripts of the empirical formula by the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound. Add the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound to the subscripts of the empirical formula..

Answers

Answer:

Multiply the subscripts of the empirical formula by the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound.

Explanation:

got it right on edge 2020 :)

Answer:

Multiply the subscripts of the empirical formula by the value of the ratio of the molar mass of the compound to the empirical molar mass of the compound.

Explanation:

Compare and contrast the characteristics of metals and nonmetals.

Answers

metals are hard and no metals are liquid