Answer:
3.82 x 10²¹ molecules As₂O₃
Explanation:
To find the amount of molecules arsenic (III) oxide (As₂O₃), you need to (1) convert kg to lbs, then (2) convert g As₂O₃ to moles As₂O₃ (via molar mass), and then (3) convert moles to molecules (via Avogadro's number).
1 kilogram = 2.2 lb
Molar Mass (As₂O₃): 2(74.992 g/mol) + 3(15.998 g/mol)
Molar Mass (As₂O₃): 197.978 g/mol
Avogadro's Number:
6.022 x 10²³ molecules = 1 mole
0.0146 g As₂O₃ 1 kg 189 lb
------------------------ x --------------- x ------------------ x ................
1 kg 2.2 lb
1 mole 6.022 x 10²³ molecules
x ------------------ x --------------------------------------- = 3.82 x 10²¹ molecules As₂O₃
197.978 g 1 mole
Answer : The equilibrium concentration of 
in the solution is, 
Explanation :
The dissociation of acid reaction is:
Initial conc. c 0 0
At eqm. c-x x x
Given:
c = 
The expression of dissociation constant of acid is:
![K_a=\frac{[H_3O^+][C_6H_5COO^-]}{[C_6H_5COOH]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BC_6H_5COO%5E-%5D%7D%7B%5BC_6H_5COOH%5D%7D)
Now put all the given values in this expression, we get:
![6.3\times 10^{-5}=\frac{(x)\times (x)}{[(7.0\times 10^{-2})-x]}](https://tex.z-dn.net/?f=6.3%5Ctimes%2010%5E%7B-5%7D%3D%5Cfrac%7B%28x%29%5Ctimes%20%28x%29%7D%7B%5B%287.0%5Ctimes%2010%5E%7B-2%7D%29-x%5D%7D)
Thus, the equilibrium concentration of in the solution is,
Because fluorine has a higher electronegativity
Answer:
3.49 g
Explanation:
The mass is the product of volume and density:
(8.96 g/cm³)(0.39 cm³) ≈ 3.49 g
The mass of a pure-copper penny would be 3.49 g.
Answer:
26.0 g/mol is the molar mass of the gas
Explanation:
We have to combine density data with the Ideal Gases Law equation to solve this:
P . V = n . R .T
Let's convert the pressure mmHg to atm by a rule of three:
760 mmHg ____ 1 atm
752 mmHg ____ (752 . 1)/760 = 0.989 atm
In density we know that 1 L, occupies 1.053 grams of gas, but we don't know the moles.
Moles = Mass / molar mass.
We can replace density data as this in the equation:
0.989 atm . 1L = (1.053 g / x ) . 0.082 L.atm/mol.K . 298K
(0.989 atm . 1L) / (0.082 L.atm/mol.K . 298K) = 1.053 g / x
0.0405 mol = 1.053 g / x
x = 1.053 g / 0.0405 mol = 26 g/mol
