We have that every gas satisfies the fundamental gas equation, PV=nRT where P is the Pressure, V is the volume of the gas, n are the moles of the gas, R is a universal constant and T is the Temperature in Kelvin. We have that PV/T=nR and during our process, the moles of the gas do not change (no argon enters or escapes our sample). See attached.
Answer:
Most of these rocks are not made up of common geometric shapes
Explanation:
Because most rocks are not made up of common geometric shapes, it would be difficult or impossible to find the volume of a rock using a ruler; there would be no easy way to measure the rock's volume using a ruler
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Answer:
1.2×10²³ atoms.
Explanation:
Data obtained from the question include:
Mole of propanone = 0.20 mole
Number of atoms of propanone =.?
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.022×10²³ atoms.
This implies that 1 mole of propanone also contains 6.022×10²³ atoms.
Thus, we can obtain the number of atoms in 0.20 mole of propanone as illustrated below:
1 mole of propanone contains 6.022×10²³ atoms.
Therefore, 0.20 mole of propanone will contain = 0.2 × 6.022×10²³ = 1.2×10²³ atoms.
Thus, 0.20 mole of propanone contain
1.2×10²³ atoms.
Virtually all acids release hydrogen ions (or protons) in the water
Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K
Explanation :
We have to calculate the entropy change of reaction 
.
![\Delta S^o=[n_{NH_3}\times \Delta S^0_{(NH_3)}]-[n_{N_2}\times \Delta S^0_{(N_2)}+n_{H_2}\times \Delta S^0_{(H_2)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5Bn_%7BNH_3%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28NH_3%29%7D%5D-%5Bn_%7BN_2%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28N_2%29%7D%2Bn_%7BH_2%7D%5Ctimes%20%5CDelta%20S%5E0_%7B%28H_2%29%7D%5D)
where,
= entropy of reaction = ?
n = number of moles
= standard entropy of 
= standard entropy of 
= standard entropy of 
Now put all the given values in this expression, we get:
![\Delta S^o=[2mole\times (192.5J/K.mole)]-[1mole\times (191.5J/K.mole)+3mole\times (130.6J/K.mole)]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5B2mole%5Ctimes%20%28192.5J%2FK.mole%29%5D-%5B1mole%5Ctimes%20%28191.5J%2FK.mole%29%2B3mole%5Ctimes%20%28130.6J%2FK.mole%29%5D)
Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K
