<h3>
Answer:</h3>
1257.45 L
<h3>
Explanation:</h3>
We are given;
- Initial volume of Helium gas, V1 as 806 L
- Initial temperature of Helium gas,T1 as 20.9°C
- Initial pressure of Helium gas, P1 as 753 mmHg
- Pressure of Helium at the altitude 6.8 km, P2 as 417 mmHg
- Temperature of Helium gas at the altitude 6.8 Km, T2 as -19.1°C
But, K = °C + 273.15
Therefore, T1 = 294.05 K and T2 = 254.05 K
- We are required to calculate the new volume of the balloon at 6.8 km.
- To determine the new volume we are going to use the combined gas law.
- According to the combined gas law,
Thus, rearranging the formula;
Therefore, the volume of the balloon at an altitude of 6.8 km is 1257.45 L
Use formula n= number of particles / avogrado constant
n= number of mole
avogrado constant= 6.12x10^-23
you will get 2.042 x 10^52
Use ideal gas law PV=nRT
Convert 5.00 atm to kPa since units must be relative to gas constant (r).
To do this multiply 5 by 101.03 (1 atm=101.3kPa)
Now plug in (506.5kPa)(10.0L)=n(8.31 L•atm/mol•K)(373K)
Solve for n (moles) to get approximately 1.634 mol. Now use dimensional analysis (1.634mol/1)(22.4L/1mol) = 36.6L
Answer:
hi! I'm pretty sure your answer is solvent!
Explanation:
not 100% sure but I looked it up on google. solvent does the dissolving, solute is what is being dissolved, and a solution is the solvent + the solute. hope this helped!
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