It was disproved by the discovery of the electrons
Answer:
8.37 grams
Explanation:
The balanced chemical equation is:
C₆H₁₂O₆ ⇒ 2 C₂H₅OH (l) + 2 CO₂ (g)
Now we are asked to calculate the mass of glucose required to produce 2.25 L CO₂ at 1atm and 295 K.
From the ideal gas law we can determine the number of moles that the 2.25 L represent.
From there we will use the stoichiometry of the reaction to determine the moles of glucose which knowing the molar mass can be converted to mass.
PV = nRT ⇒ n = PV/RT
n= 1 atm x 2.25 L / ( 0.08205 Latm/kmol x 295 K ) =0.093 mol CO₂
Moles glucose required:
0.093 mol CO₂ x ( 1 mol C₆H₁₂O₆ / 2 mol CO₂ ) = 0.046 mol C₆H₁₂O₆
The molar mass of glucose is 180.16 g/mol, then the mass required is
0.046 mol x 180.16 g/mol = 8.37 g
Answer:
![\boxed{3.966 \times 10^{24}\text{ atoms of Ag}}](https://tex.z-dn.net/?f=%5Cboxed%7B3.966%20%5Ctimes%2010%5E%7B24%7D%5Ctext%7B%20atoms%20of%20Ag%7D%7D)
Explanation:
(a) Balanced equation
Cu + 2AgNO₃ ⟶ Cu(NO₃)₂+ 2Ag
(b) Calculation
You want to convert atoms of Cu to atoms of Ag.
The atomic ratio is ratio is 2 atoms Ag:1 atom Cu
![\text{Atoms of Ag} = 19.83 \times 10^{23}\text{atoms Cu} \times \dfrac{\text{2 atoms Ag}}{\text{1 atom Cu}}\\\\= 3.966 \times 10^{24}\text{ atoms of Ag}\\\\\text{The reaction will produce }\boxed{\mathbf{3.966 \times 10^{24}}\textbf{ atoms of Ag}}](https://tex.z-dn.net/?f=%5Ctext%7BAtoms%20of%20Ag%7D%20%3D%2019.83%20%5Ctimes%2010%5E%7B23%7D%5Ctext%7Batoms%20Cu%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B2%20atoms%20Ag%7D%7D%7B%5Ctext%7B1%20atom%20Cu%7D%7D%5C%5C%5C%5C%3D%203.966%20%5Ctimes%2010%5E%7B24%7D%5Ctext%7B%20atoms%20of%20Ag%7D%5C%5C%5C%5C%5Ctext%7BThe%20reaction%20will%20produce%20%7D%5Cboxed%7B%5Cmathbf%7B3.966%20%5Ctimes%2010%5E%7B24%7D%7D%5Ctextbf%7B%20atoms%20of%20Ag%7D%7D)
Answer:
44.2 L
Explanation:
Use Charles Law:
![\frac{V1}{T1} =\frac{V2}{T2}](https://tex.z-dn.net/?f=%5Cfrac%7BV1%7D%7BT1%7D%20%3D%5Cfrac%7BV2%7D%7BT2%7D)
We have all the values except for V₂; this is what we're solving for. Input the values:
- make sure that your temperature is in Kelvin
From here, we need to get V₂ by itself. To do this, multiply by 273 on both sides:
![\frac{56*273}{346} = V2](https://tex.z-dn.net/?f=%5Cfrac%7B56%2A273%7D%7B346%7D%20%3D%20V2)
Therefore, V₂ = 44.2 L
It's also helpful to know that temperature and volume are linearly related. So, when temperature drops, so will volume and vice versa.
<span>density = mass / volume
given the quotient , we have density and mass, volume can be easily calculated as:
volume = mass / p =15.5 g / 0.789 g/cm^3
=~ 20 cm^3 (dimension ally constant)</span>