Answer:
Manganese trinitrate or manganese(III) nitrate
Explanation:
Answer:
Hi do we translate a this
Explanation:
Answer:

Explanation:
Hello,
In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):

Thus, since we already know the enthalpy of combustion of the fuel, for carbon and water we have -393.5 and -241.8 kJ/mol respectively, thereby, the enthalpy of combustion turns out:

Best regards.
Using the stoichiometry of the reaction and the information provided in the question, the mass of N2 used is 11.62 g.
<h3>Chemical reaction</h3>
The term chemical reaction refers to the combiantion of two or substances to yiled one or more products. The reaction equation in this case is N2 + O2 --->2NO.
Now;
Number of moles of NO = 25g/30 g/mol = 0.83 moles
1 mole of N2 yields 2 moles of NO
x moles of N2 yileds 0.83 moles of NO
x = 0.415 moles
Mass of N2 = 0.415 moles * 28 g/mol = 11.62 g
Learn more about stoichiometry:
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Answer : The value of equilibrium constant for this reaction at 262.0 K is 
Explanation :
As we know that,

where,
= standard Gibbs free energy = ?
= standard enthalpy = -45.6 kJ = -45600 J
= standard entropy = -125.7 J/K
T = temperature of reaction = 262.0 K
Now put all the given values in the above formula, we get:


The relation between the equilibrium constant and standard Gibbs free energy is:

where,
= standard Gibbs free energy = -12666.6 J
R = gas constant = 8.314 J/K.mol
T = temperature = 262.0 K
K = equilibrium constant = ?
Now put all the given values in the above formula, we get:


Therefore, the value of equilibrium constant for this reaction at 262.0 K is 