Answer:
2) 0.4 mol
Explanation:
Step 1: Given data
- Volume of the solution (V): 500 mL
- Molar concentration of the solution (M): 0.8 M = 0.8 mol/L
Step 2: Convert "V" to L
We will use the conversion factor 1 L = 1000 mL.
500 mL × 1 L/1000 mL = 0.500 L
Step 3: Calculate the moles of KBr (solute)
The molarity is the quotient between the moles of solute (n) and the liters of solution.
M = n/V
n = M × V
n = 0.8 mol/L × 0.500 L = 0.4 mol
Answer : The half life of 28-Mg in hours is, 6.94
Explanation :
First we have to calculate the rate constant.
Expression for rate law for first order kinetics is given by:

where,
k = rate constant
t = time passed by the sample = 48.0 hr
a = initial amount of the reactant disintegrate = 53500
a - x = amount left after decay process disintegrate = 53500 - 10980 = 42520
Now put all the given values in above equation, we get


Now we have to calculate the half-life.



Therefore, the half life of 28-Mg in hours is, 6.94
Explanation:
Hi for this one u just need to remember and use the equation.

then u find mr of potassium which is 39.1.
then u do

you get the answer as 0.5115 write ur answer to 3 significant figures which will be 0.512 moles .
hope this helps :)
The p sublevel holds 6 electrons because it has 3 orbitals.
Answer:
positive H and negative S
Explanation:
For a reaction to be spontaneous, the absolute best combination is a negative Delta H and a positive Delta S. When they are both positive, the reaction is only spontaneous at higher temperatures. When they are both negative, the reaction is only spontaneous at lower temperatures. and again if a catalyst is added to the reaction, the activation energy is lowered because a lower-energy transition state is formed. The catalyst does not affect the energy of the reactants or products (and thus does not affect ΔG).
So from these discussions
Ea does not affect G value at all (whether +Ea or -Ea).
And for product to be formed the reaction should be spontaneous, where H is negative and S positive else the reaction will yield low product.