Answer:
The <u>equilibrium constant</u> is:
Explanation:
The correct equation is:
Thus, with the equilibrium concentrations you can calculate the equilibrium constant, Kc.
The equation for the equilibrium constant is:
![k_c=\dfrac{[NH_3]^2}{[N_2]\cdot [H_2]^3}](https://tex.z-dn.net/?f=k_c%3D%5Cdfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5Ccdot%20%5BH_2%5D%5E3%7D)
Substituting:
Answer:
19 g
Explanation:
Data Given:
Sodium Chloride (table salt) = 50 g
Amount of sodium (Na) = ?
Solution:
Molecular weight calculation:
NaCl = 23 + 35.5
NaCl = 58.5 g/mol
Mass contributed by Sodium = 23 g
calculate the mole percent composition of sodium (Na) in sodium Chloride.
Since the percentage of compound is 100
So,
Percent of sodium (Na) = 23 / 58.5 x 100
Percent of sodium (Na) = 39.3 %
It means that for ever gram of sodium chloride there is 0.393 g of Na is present.
So,
for the 50 grams of table salt (NaCl) the mass of Na will be
mass of sodium (Na) = 0.393 x 50 g
mass of sodium (Na) = 19 g
Answer : The value of 
is, 0.34 V
Explanation :
Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.
The oxidation-reduction half cell reaction will be,
Oxidation half reaction: 
Reduction half reaction: 
Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.
The overall balanced equation of the cell is,
To calculate the of the reaction, we use the equation:
Putting values in above equation, we get:
Hence, the value of is, 0.34 V
<span><span>S is for soil,</span><span>cl (sometimes c) represents climate,</span><span>o organisms including humans,</span><span>r relief,</span><span>p parent material, or lithology, and</span><span>t time.</span></span>
Answer:
2 mol of SO3 produces 1 mol O2
3 mol SO3 produces 3/2 mol of O2
so O2 produced = 1.5(32) =48 gm
Explanation:
