Answer:
[H2] = 0.012 M
[N2] = 0.019 M
[H2O] = 0.057 M
Explanation:
The strategy here is to account for the species at equilibrium given that the concentration of [NO]=0.062M at equilibrium is known and the quantities initially present and its stoichiometry.
2NO(g) + 2H2(g) ⇒ N2(g) + 2H2O(g)
i mol 0.10 0.050 0.10
c mol -0.038 -0.038 +0019 +0.038
e mol 0.062 0.012 00.019 0.057
Since the volume of the vessel is 1.0 L, the concentrations in molarity are:
[NO] = 0.062 M
[H2] = 0.012 M
[N2] = 0.019 M
[H2O] = 0.057 M
Answer: Option (d) is the correct answer.
Explanation:
An atom or element which has the ability to readily gain an electron will have high electronegativity.
Both Beryllium and Calcium are alkaline earth metals and hence they are electropositive in nature.
Whereas both iodine and nitrogen are electronegative in nature. But across the period there is an increase in electronegativity and down the group there is a decrease in electronegativity.
Nitrogen belongs to period 2 and iodine belongs to the bottom of group 17. Thus, we can conclude that nitrogen is more electronegative than iodine.
Answer:
Average of the trial is: 288.50 C
Percent Error: 3.83%
Explanation:
(291 + 287 + 295 + 281) : 4 = 288.50 C
Average: 288.50 C
Percent Error: {(300 - 288.50) : 300} x 100% = 3.83 %
Answer:
259.497 mg, 58.84%
Explanation:
BaSO₄ → Ba²⁺ + SO₄²⁻
to calculate the mole of BaSO₄
mole BaSO₄ = mass given / molar mass = 403 mg / 233.38 g/mol = 1.7268 mol
comparing the mole ratio
1.7268 mol of BaSO₄ yields 1.7268 mol of Ba²⁺
403 mg BaSO₄ yields ( 1.7268 × 137.327 ) where 137.327 is the molar mass of Barium mol of Ba²⁺
441 mg BaSO₄ will yield ( 1.7268 × 137.327 × 441 mg ) / 403 mg = 259 .497 mg
mas percentage of the Barium compound = 259 .497 mg / 441 mg × 100 = 58.84%
That would be anaerobic fermentation
I hope I've helped!
