endothermic
N2 + 3F2 ==> 2NF32N, 6F on each side of the equation. It can help to create a table system for more complex equations.
If: n2 = 19 Then: n = the square root of 19.
y = kn / (1 + n2)First, let's express it with negative exponents:y = kn * (1 + n2)-1Then apply the chain rule:dy|dn = k(1 + n2)-1 - kn(1 + n2)-2 * 2nAnd simplify:dy|dn = k / (1 + n2) - 2n / kn(1 + n2)
n2 = 20 + n so n2 - n - 20 = 0 that is (n - 5)(n + 4) = 0 so that n - 5 = 0 or n + 4 = 0 n = 5 or n = -4
endothermic
The balanced equation for the reaction between nitrogen monoxide (NO) and carbon monoxide (CO) to form nitrogen (N2) and carbon dioxide (CO2) is: 2NO + 2CO -> N2 + 2CO2
CH4 + 2O2 📷 CO2 + 2H2O + Heat
more N2 and O2 would form
The balanced equation for the reaction of hydrogen gas (H2) with nitrogen gas (N2) to form ammonia (NH3) is: 3H2 + N2 → 2NH3
6
kp66
The equation is: N2(g) + 2O2(g) -> 2NO2(g) Given enthalpy changes: 2NO(g) + O2(g) + 112 kJ -> 2NO2(g) N2(g) + O2(g) -> 2NO(g) + 180 kJ Using Hess's Law, we subtract the second equation from the sum of the first and third equations: 2NO2(g) - (2NO(g) + O2(g)) = 2NO2(g) - 2NO(g) - O2(g), resulting in the entropy change for the third reaction.
The equation H2CN2 + heat = N2 + CH2 represents the thermal decomposition of hydrogen cyanide into nitrogen gas and methylene (CH2) radicals. This reaction occurs at high temperatures to break the H-C and C-N bonds, forming the new products.
More NO would form
The balanced equation is: 4H₂ + 4NO → 4H₂O + 2N₂.
N2(g) + O2(g) 2NO(g)Adding NO would push the reaction to the left, toward the reactants. This is according to Le Chatelier.