Answer:
12.5 g of Li are needed in order toproduce 0.60 moles of Li₃N
Explanation:
The reaction is:
6Li(s) + N₂(g) → 2Li₃N(s)
If nitrogen is in excess, the lithium is the limiting reactant.
Ratio is 2:6
2 moles of nitride were produced by 6 moles of Li
Then, 0.6 moles of nitride were produced by (0.6 .6)/ 2 = 1.8 moles of Li
Let's convert the moles to mass → 1.8 mol . 6.94 g/ 1mol = 12.5 g of Li
To solve this we assume
that the gas inside is an ideal gas. Then, we can use the ideal gas
equation which is expressed as PV = nRT. At a constant pressure and number of
moles of the gas the ratio T/V is equal to some constant. At another set of
condition of temperature, the constant is still the same. Calculations are as
follows:
T1 / V1 = T2 / V2
T2 = T1 x V2 / V1
T2 = 280 x 20.0 / 10
<span>T2 = 560 K</span>
Answer:
Just add what they equal to hope this helps :)
Explanation:
It's back wards lol you need to take it right
Answer:
22.73s
Explanation:
The reaction is a second order reaction, we know this by observing the unit of the slope.
rate constant = k = 0.056 M-1s-1
the initial concentration of BrO- [A]o = 0.80 M
time = ?
Final concentration [A]t= one-half of 0.80 M = 0.40M
1 / [A]t = kt + 1 / [A]o
1 / 0.40 = 0.056 * t + 1 / 0.80
t = (2.5 - 1.25) / 0.056
t = 22.73s
