A gas at STP has a volume of 1.5 L. What will the pressure be if it is at 26° C occupying .75 L
Answers
Answer:
P₂ = 2.19 atm
Explanation:
Given data:
Initial volume = 1.5 L
Initial pressure =1 atm
Initial temperature = 273K
Final temperature = 26°C (26+273 = 299 K)
Final volume = 0.75 L
Final pressure = ?
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
P₂ = P₁V₁ T₂/ T₁ V₂
P₂ = 1 atm × 1.5 L × 299 K / 273 K × 0.75 L
P₂ = 448.5 atm .L. K / 204.75 K.L
P₂ = 2.19 atm
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Answers
(2) 0.50 atm (4) 1.5 atm
Answers
Answer:(4) 1.5 atm
Explanation:
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
where,
= initial pressure of gas at STP= 1 atm
= final pressure of gas = ?
= initial volume of gas at STP = V L
= final volume of gas = V L
= initial temperature of gas at STP= 273K
= final temperature of gas = 410 K
Now put all the given values in the above equation, we get the final pressure of gas.
Therefore, the pressure of gas will be 1.50 atm.
P = nRT / V
P = (1 mol) (8.314 Pa m^3/ mol K) (298.15 K) / .0224 L
P = 110660 Pa or 1.0 atm
Thus, the answer is the value in option 1.
This should be EASY!
Answers
Answer:
19, K, 39.098
Explanation:
Potassium is a chemical element with symbol K and atomic number 19.
Atomic Mass: 39.098u
Answers
Answers
a) 8
40*20%
=40*20/100
=800/100
=8
Final answer:
The atomic weight of the unknown element, which is approximately 20% less than calcium (40 amu), is estimated to be 32 amu.
Explanation:
The atomic weight of an unknown element is said to be approximately 20% less than that of calcium, which has an atomic weight of 40 atomic mass units (amu). To calculate the approximate atomic weight of the unknown element, take 20% of 40 (which is 8) and subtract it from 40. The result is 32 amu (40 - 8 = 32), so the correct answer is (c) 32.
Learn more about Atomic Weight here:
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Answers
2NaClO3 9= 2NaCl + 3O2
We are given the amount of sodium chlorate to heat. This will be the starting point of our calculations.
200 g NaClO3 (1 mol / 106.44 g) ( 3 mol O2 / 2 mol NaClO3) (32 g O2 / 1 mol O2) = 90.19 g O2