Answer:
increase
Explanation:
Let's suppose we have a sample of air in a closed container. We heat the container and we want to predict what would happen to the pressure.
According to Gay-Lussac's law, the pressure of a gas is directly proportional to its absolute temperature.
Thus, if we increased the temperature of the air by heating it, its pressure would increase.
If a sample of air in a closed container was heated, the total pressure of the air would increase.
Answer:
λ = 0.45×10⁻⁶ m
Explanation:
Given data:
Wavelength of blue light = ?
Frequency of blue light = 6.69×10¹⁴ s⁻¹
Solution:
Formula;
Speed of wave = wavelength × frequency
Speed of wave = 3.00×10⁸ m/s
by putting vales,
3.00×10⁸ m/s = λ × 6.69×10¹⁴ s⁻¹
λ = 3.00×10⁸ m/s / 6.69×10¹⁴ s⁻¹
λ = 0.45×10⁻⁶ m
<u>0.219 moles </u><u>moles are present in the flask when the </u><u>pressure </u><u>is 1.10 atm and the temperature is 33˚c.</u>
What is ideal gas constant ?
- The ideal gas constant is calculated to be 8.314J/K⋅ mol when the pressure is in kPa.
- The ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas.
- The combined gas law relates pressure, volume, and temperature of a gas.
We simple use this formula-
The basic formula is PV = nRT where. P = Pressure in atmospheres (atm) V = Volume in Liters (L) n = of moles (mol) R = the Ideal Gas Law Constant.
68F = 298.15K
V = nRT/P = 0.2 * 0.08206 * 298.15K / (745/760) = 4.992Liters
n = PV/RT = 1.1atm*4.992L/(0.08206Latm/molK * 306K)
n = 0.219 moles
Therefore, 0.219 moles moles are present in the flask when the pressure is 1.10 atm and the temperature is 33˚c.
Learn more about ideal gas constant
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Answer:
So, you're dealing with a sample of cobalt-60. You know that cobalt-60 has a nuclear half-life of
5.30
years, and are interested in finding how many grams of the sample would remain after
1.00
year and
10.0
years, respectively.
A radioactive isotope's half-life tells you how much time is needed for an initial sample to be halved.
If you start with an initial sample
A
0
, then you can say that you will be left with
A
0
2
→
after one half-life passes;
A
0
2
⋅
1
2
=
A
0
4
→
after two half-lives pass;
A
0
4
⋅
1
2
=
A
0
8
→
after three half-lives pass;
A
0
8
⋅
1
2
=
A
0
16
→
after four half-lives pass;
⋮
Explanation:
now i know the answer
