The grams of N2 that are required to produce 100.0 l of NH3 at STP
At stp 1moles = 22.4 l. what about 100.0 L of NH3
= 100 / 22.4 lx1 moles = 4.46 moles of NH3
write the reacting equation
N2+3H2 =2NH3
by use of mole ratio between N2 to NH3 which is 1:2 the moles of N2 =4.46/2 =2.23 moles of N2
mass = moles x molar mass
= 2.23moles x 28 g/mol = 62.4 grams
Answer:
E) A, B, and C
Explanation:
Syn addition refers to the addition of two substituents on the same face or side of a double bond. This differed from anti addition which a occurs across opposite face of the double bond.
Hydrogenation, hydroboration and dihydroxylation all involve syn addition to the double bond, hence the answer chosen above.
Answer:
H₃PO₄ → 3H⁺ + PO₄³⁻
CaSO₄ → Ca²⁺ + SO₄²⁻
b. CaCl₂
Explanation:
When H₃PO₄ is dissolved in water, there are produced the H⁺ and PO₄³⁻ ions. The equation is:
H₃PO₄ → 3H⁺ + PO₄³⁻
In the same way, CaSO₄ is dissolved in:
CaSO₄ → Ca²⁺ + SO₄²⁻
b. Now, in a reaction of an acid (HCl) and a base (Ca(OH)₂), water, H₂O and a salt are produced:
2 HCl + Ca(OH)₂ → 2H₂O + Salt
The ions that are not present in the reaction are Cl⁻ and Ca²⁺, the salt is CaCl₂ and the balanced reaction is:
2 HCl + Ca(OH)₂ → 2H₂O + CaCl₂
Latent heat, also called the heat of vaporization, is the amount of energy necessary to change a liquid to a vapour at constant temperature and pressure. The energy required to melt a solid to a liquid is called the heat of fusion
Answer: 
Explanation:
According to the Arrhenius equation,
or,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at 
= 
= rate constant at 
= 
= activation energy for the reaction = 262 kJ/mol = 262000J/mol
R = gas constant = 8.314 J/mole.K
= initial temperature =
= final temperature = 
Now put all the given values in this formula, we get
![\log (\frac{6.1\times 10^{-8}}{K_2})=\frac{262000}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{775.0K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B6.1%5Ctimes%2010%5E%7B-8%7D%7D%7BK_2%7D%29%3D%5Cfrac%7B262000%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B600.0K%7D-%5Cfrac%7B1%7D%7B775.0K%7D%5D)
Therefore, the value of the rate constant at 775.0 K is
