Methylhydrazine, CH6N2, is commonly used as a liquid rocket fuel. The heat of combustion of methylhydrazine is −1.30 × 103 kJ/mo
1 answer:
Answer:
Explanation:
Given that:
The heat of combustion of methylhydrazine =
It means that:
1 mole of methylhydrazine on combustion, releases of energy
Molar mass of methylhydrazine = 46.072 g/mol
Means that, 1 mole of methylhydrazine contains 46.072 g of methylhydrazine
Thus,
46.072 g of methylhydrazine on combustion, releases of energy
Also,
1 g of methylhydrazine on combustion, releases of energy
206.2 g of methylhydrazine on combustion, releases of energy
Heat released =
You might be interested in
Answer:
The correct answer is: Ka= 5.0 x 10⁻⁶
Explanation:
The ionization of a weak monoprotic acid HA is given by the following equilibrium: HA ⇄ H⁺ + A⁻. At the beginning (t= 0) we have 0.200 M of HA. Then, a certain amount (x) is dissociated into H⁺ and A⁻, as is detailed in the following table:
HA ⇄ H⁺ + A⁻
t= 0 0.200 M 0 0
t -x x x
t= eq 0.200M -x x x
At equilibrium, we have the following ionization constant expression (Ka):
Ka=
Ka=
Ka=
From the definition of pH, we know that:
pH= - log [H⁺]
In this case, [H⁺]= x, so:
pH= -log x
3.0= -log x
⇒x = 10⁻³
We introduce the value of x (10⁻³) in the previous expression and then we can calculate the ionization constant Ka as follows:
Ka= =
= 5.025 x 10⁻⁶= 5.0 x 10⁻⁶
3-pentanone to form 3-pentanol
Here 3-pentanone reacts with H2/Pt to given 3-pentanol
So here with one step we can convert given ketone to alcohol
The reaction will be
