2 NH3 -> 1 N2 + 3 H2
Explanation:
That would be the answer to this
Pollution and people littering. Can change everything.
In this reaction 50% of the compound decompose in 10.5 min thus, it is half life of the reaction and denoted by symbol 
.
(a) For first order reaction, rate constant and half life time are related to each other as follows:
Thus, rate constant of the reaction is 
.
(b) Rate equation for first order reaction is as follows:
![k=\frac{2.303}{t_{1/2}}log\frac{[A_{0}]}{[A_{t}]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt_%7B1%2F2%7D%7Dlog%5Cfrac%7B%5BA_%7B0%7D%5D%7D%7B%5BA_%7Bt%7D%5D%7D)
now, 75% of the compound is decomposed, if initial concentration ![[A_{0} ]](https://tex.z-dn.net/?f=%5BA_%7B0%7D%20%5D)
is 100 then concentration at time t ![[A_{t} ]](https://tex.z-dn.net/?f=%5BA_%7Bt%7D%20%5D)
will be 100-75=25.
Putting the values,
On rearranging,
Thus, time required for 75% decomposition is 21 min.
Answer:
m H2O = 56 g
Explanation:
∴ The heat ceded (-) by the Aluminum part is equal to the heat received (+) by the water:
⇒ - (mCΔT)Al = (mCΔT)H2O
∴ m Al = 25.0 g
∴ Mw Al = 26.981 g/mol
⇒ n Al = (25.0g)×(mol/26.981gAl) = 0.927 mol Al
⇒ Q Al = - (0.927 mol)(24.03 J/mol°C)(26.8 - 86.4)°C
⇒ Q Al = 1327.64 J
∴ mH2O = Q Al / ( C×ΔT) = 1327.64 J / (4.18 J/g.°C)(26.8 - 21.1)°C
⇒ mH2O = 55.722 g ≅ 56 g
2 cm east is the answer because you go east 5.5 cm and then go back in a sense 3.5 so its basically 5.5-3.5 because its backwards while still facing eat
