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3 years ago

Under what conditions is a real gas closely approximated by an ideal gas?

1 answer:

7 0

Correct If not right, but, In low temperatures intermolecular forces also increase, since molecules move more slowly, similar to what would occur in a liquid state. Just remember that ideal gas behavior is most closely approximated in conditions that favor gas formation in the first place—heat and low pressure. me.

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Consider a neutrally charged atom that has an atomic mass number of 80, which includes 35 neutrons. how many electrons does this

ivanzaharov [21]

Atomic mass number is the number of protons and neutrons. Subtract 80-35=45 is the number of protons. Because the atom is neutrally charged, the number of protons must equal the number of electrons, so there are 45 electrons.

5 0

4 years ago

Read 2 more answers

A 3000 MWt fast reactor has a 42% efficiency. This reactor operates for 23 months followed by a 1 month down time for refueling

n200080 [17]

Answer:

capacity factor = 0.952

Availability factor = 0.958

Explanation:

1) capacity factor

capacity factor = actual power output / maximum power output

= (actual power output)/(efficiency * rated power output)

$= \frac{1200}{\frac{42}{100}*3000}$

= 0.952

2) Availability factor

Availability factor = Actual operation time period/ total time period

= 23/24 = 0.958

8 0

3 years ago

A sample of gas has an initial pressure of 1.5 atm, an initial volume of 3.0 L, and an initial temperature of 293K. If the final

Nataliya [291]

Answer:

1.9 L

Explanation:

Step 1: Given data

Initial pressure (P₁): 1.5 atm

Initial volume (V₁): 3.0 L

Initial temperature (T₁): 293 K

Final pressure (P₂): 2.5 atm

Final volume (V₂): ?

Final temperature (T₂): 303 K

Step 2: Calculate the final volume of the gas

If we assume ideal behavior, we can calculate the final volume of the gas using the combined gas law.

P₁ × V₁ / T₁ = P₂ × V₂ / T₂

V₂ = P₁ × V₁ × T₂ / T₁ × P₂

V₂ = 1.5 atm × 3.0 L × 303 K / 293 K × 2.5 atm = 1.9 L

6 0

3 years ago

How do you slow down the motion of a gas particle?

jasenka [17]

I think u turn down the heat not to sure

3 0

4 years ago

Read 2 more answers

UGRENT! Please help showing all work

agasfer [191]

Answer:

a. The limiting reactant is Ca(OH)₂

b. The theoretical yield of CaCl₂ is approximately 621.488 grams

c. The percentage yield of CaCl₂ is approximately 47.06%

Explanation:

a. The given chemical reaction is presented as follows;

Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O

Therefore;

One mole of Ca(OH)₂ reacts with two moles of HCl to produce one mole of CaCl₂ and two moles of water H₂O

The mass of HCl in an experiment, m₁ = 229.70 g

The mass of Ca(OH)₂ in an experiment, m₂ = 207.48 g

The molar mass of HCl, MM₁ = 36.458 g/mol

The molar mass of Ca(OH)₂, MM₂ =74.093 g/mol

The number of moles of HCl, present, n₁ = m₁/MM₁

∴ n₁ = m₁/MM₁ = 229.70 g/(36.458 g/mol) ≈ 6.3 moles

The number of moles of Ca(OH)₂, present, n₂ = m₂/MM₂

∴ n₂ = m₂/MM₂ = 207.48 g/(74.093 g/mol) ≈ 2.8 moles

The number of moles of Ca(OH)₂, present, n₂ = 2.8 moles

According the chemical reaction equation the number of moles of HCl the 2.8 moles of Ca(OH)₂ will react with, = 2.8 × 2 moles = 5.6 moles of HCl

Therefore, there is an excess HCl in the reaction and Ca(OH)₂ is the limiting reactant

b. According the chemical reaction equation the number of moles of CaCl₂ produced in he reaction by the 2.8 moles of Ca(OH)₂ = 2.8 × 2 moles = 5.6 moles of CaCl₂

The molar mass of CaCl₂ = 110.98 g/mol

The mass of the 5.6 moles of CaCl₂ = 5.6 moles × 110.98 g/mol ≈ 621.488 grams

The theoretical yield of CaCl₂ ≈ 621.488 grams

c. Given that the actual mass of CaCl₂ produced = 292.5 grams, we have;

The percentage yield of CaCl₂ = The actual yield/(The theoretical yield) × 100

∴ The percentage yield of CaCl₂ = (292.5 g)/(621.488 g) × 100 ≈ 47.0644646397%

The percentage yield of CaCl₂ ≈ 47.06%.

8 0

3 years ago