Dalton gathers evidence for the existence of atoms by measuring the masses of elements after compounds are formed.
<u>Explanation</u>:
- John Dalton accumulated proof for the presence of atoms by estimating the majority of components that responded to frame mixes. All components are made out of molecules. All particles of a similar component have a similar mass, and atoms of various components have various masses. Mixes contain atoms of more than one component.
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Dalton did numerous investigations that gave proof to the presence of particles. For instance, He researched pressure and different properties of gases, from which he induced that gases must comprise of little, singular particles that are in steady, arbitrary movement.
Answer:
The mass percent of aluminum sulfate in the sample is 16.18%.
Explanation:
Mass of the sample = 1.45 g
Mass of the precipitate = 0.107 g
Moles of aluminum hydroxide = 
According to reaction, 2 moles of aluminum hydroxide is obtained from 1 mole of aluminum sulfate .
Then 0.001372 moles of aluminum hydroxide will be obtained from:
Mass of 0.000686 moles of aluminum sulfate :
= 0.000686 mol × 342 g/mol = 0.2346 g
The mass percent of aluminum sulfate in the sample:
Answer:
0.147 mol
Explanation:
Step 1: Calculate the volumetric concentration (Cv)
We will use the following expression.
Cv = Cg × ρ
Cv = 98.0 g%g × 1.84 g/mL = 180 g%mL
Step 2: Calculate the molarity of sulfuric acid
We will use the following expression.
M = mass solute / molar mass solute × liters of solution
M = 180 g / 98.08 g/mol × 0.100 L = 18.4 M
Step 3: Calculate the moles of solute in 8.00 mL of solution
8.00 × 10⁻³ L × 18.4 mol/L = 0.147 mol
The answer you looking for is D
Answer:
for the given reaction is -99.4 J/K
Explanation:
Balanced reaction: 
![\Delta S^{0}=[1mol\times S^{0}(NH_{3})_{g}]-[\frac{1}{2}mol\times S^{0}(N_{2})_{g}]-[\frac{3}{2}mol\times S^{0}(H_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20S%5E%7B0%7D%3D%5B1mol%5Ctimes%20S%5E%7B0%7D%28NH_%7B3%7D%29_%7Bg%7D%5D-%5B%5Cfrac%7B1%7D%7B2%7Dmol%5Ctimes%20S%5E%7B0%7D%28N_%7B2%7D%29_%7Bg%7D%5D-%5B%5Cfrac%7B3%7D%7B2%7Dmol%5Ctimes%20S%5E%7B0%7D%28H_%7B2%7D%29_%7Bg%7D%5D)
where 
represents standard entropy.
Plug in all the standard entropy values from available literature in the above equation:
![\Delta S^{0}=[1mol\times 192.45\frac{J}{mol.K}]-[\frac{1}{2}mol\times 191.61\frac{J}{mol.K}]-[\frac{3}{2}mol\times 130.684\frac{J}{mol.K}]=-99.4J/K](https://tex.z-dn.net/?f=%5CDelta%20S%5E%7B0%7D%3D%5B1mol%5Ctimes%20192.45%5Cfrac%7BJ%7D%7Bmol.K%7D%5D-%5B%5Cfrac%7B1%7D%7B2%7Dmol%5Ctimes%20191.61%5Cfrac%7BJ%7D%7Bmol.K%7D%5D-%5B%5Cfrac%7B3%7D%7B2%7Dmol%5Ctimes%20130.684%5Cfrac%7BJ%7D%7Bmol.K%7D%5D%3D-99.4J%2FK)
So, for the given reaction is -99.4 J/K
