Answer:
Lithium does form a peroxide as well as an oxide on burning in air and I suspect the low temperature reaction with air forms a significant amount of peroxide.
Answer:
The half-life time, the team equired for a quantity to reduce to half of its initial value, is 79.67 seconds.
Explanation:
The half-life time = the time required for a quantity to reduce to half of its initial value. Half of it's value = 50%.
To calculate the half-life time we use the following equation:
[At]=[Ai]*e^(-kt)
with [At] = Concentration at time t
with [Ai] = initial concentration
with k = rate constant
with t = time
We want to know the half-life time = the time needed to have 50% of it's initial value
50 = 100 *e^(-8.7 *10^-3 s^- * t)
50/100 = e^(-8.7 *10^-3 s^-1 * t)
ln (0.5) = 8.7 *10^-3 s^-1 *t
t= ln (0.5) / -8.7 *10^-3 = 79.67 seconds
The half-life time, the team equired for a quantity to reduce to half of its initial value, is 79.67 seconds.
The answer is A because Alkaline are gases in earth's atmosphere <span />
Answer: A 13
Explanation: hope this helps!
From a solubility curve the solubility of KNO3 at 40°C is about 63 g / 100 g of water.
That means that 100 g of water at 40°C can dissolve 63 g of KNO3.
A saturated solution is one that cannot dissolve more solute, because it already has reached the greatest concentration that it can hold.
Then, the mass of KNO3 that dissolves in 100 g of water is about 63 g.
