Answer:
3.310308*10^26
Explanation:
nO=9nFe2(SO3)3=9*60.1=540.9 moles
number of atoms: 540.9*6.02*10^23
Answer : The concentration of 
at equilibrium is 0 M.
Solution : Given,
Concentration of = 0.0200 M
Concentration of 
= 1.00 M
The given equilibrium reaction is,
![Fe^{3+}(aq)+3C_2O_4^{2-}(aq)\rightleftharpoons [Fe(C_2O_4)_3]^{3-}(aq)](https://tex.z-dn.net/?f=Fe%5E%7B3%2B%7D%28aq%29%2B3C_2O_4%5E%7B2-%7D%28aq%29%5Crightleftharpoons%20%5BFe%28C_2O_4%29_3%5D%5E%7B3-%7D%28aq%29)
Initially conc. 0.02 1.00 0
At eqm. (0.02-x) (1.00-3x) x
The expression of 
will be,
![K_c=\frac{[[Fe(C_2O_4)_3]^{3-}]}{[C_2O_4^{2-}]^3[Fe^{3+}]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5B%5BFe%28C_2O_4%29_3%5D%5E%7B3-%7D%5D%7D%7B%5BC_2O_4%5E%7B2-%7D%5D%5E3%5BFe%5E%7B3%2B%7D%5D%7D)
By solving the term, we get:
Concentration of at equilibrium = 0.02 - x = 0.02 - 0.02 = 0 M
Therefore, the concentration of at equilibrium is 0 M.
<u>We are given:</u><u>_______________________________________________</u>
Volume of Gas (V) = 2.5L
Pressure (P) = 1.2 atm
Temperature (T) = 25°C OR 25+273 = 298 K
Universal Gravitational Constant (R) = 0.0821
<u>Solving for number of moles:</u><u>___________________________________</u>
From the Ideal Gas Equation,
PV = nRT
(1.2)(2.5) = n(0.0821)(298) [plugging the given values]
n = [(1.2)(2.5)] / [0.0821*298]
n = 300 / [298*8.21]
n = 0.12 moles
Hence, there are 0.12 moles of Oxygen in 2.5L of 1.2 atm gas when the temperature is 25°C
Answer;
D. carbon-12 atom
Explanation;
-One atomic mass unit is the mass that is exactly equal to one-twelfth the mass of one carbon-12 atom. Therefore, the mass of carbon-12 is 12 amu and this provides the standard that is used in measuring the atomic mass of all other elements.
-The atomic mass of carbon is 12.01 and not 12.00 in the periodic table.
-Due to the fact that there are more than one element for most naturally occurring elements, the average mass of the naturally occurring mixture of isotopes is taken while determining the atomic mass of an element.
Answer:
60.08g/mol
Explanation:
Given parameters:
Formula of sand = SiO₂
Find the molar mass of the compound to the hundredths place;
Molar mass of Si = 28.085g/mol
Molar mass of O = 15.999
Molar mass = 28.085 + 2(15.999) = 60.08g/mol
