1mol=6.022x10^23 atoms
3.75mol= 6.022x3.75x10^23 atoms
=2.2583x10^24 atoms
Answer:
How the incident happened
Any chemicals involved in an incident
Any other hazards present in the lab
Explanation:
Above are the types of information that are necessary to communicate with emergency responders. The emergency responders ask the first question that how the incident happened. After that they ask that is there any harmful chemicals are present in the laboratory or what types of chemicals present in the laboratory. These questions were asked by the emergency responders in order to give the patient a suitable treatment.
The half-life equation is written as:
An = Aoe^-kt
We use this equation for the solution. We do as follows:
5.5 = 176e^-k(165)
k = 0.02
<span>What is the half-life of the goo in minutes?
</span>
0.5 = e^-0.02t
t = 34.66 minutes <----HALF-LIFE
Find a formula for G(t) , the amount of goo remaining at time t.G(t)=?
G(t) = 176e^-0.02t
How many grams of goo will remain after 50 minutes?
G(t) = 176e^-0.02(50) = 64.75 g
Answer:
39.7 %
Explanation:
magnesium + oxygen ⟶ magnesium oxide
10.57 g 6.96 g 17.53 g
According to the <em>Law of Conservation of Mass</em>, the mass of the product must equal the total mass of the reactants.
Mass of MgO = 10.57 + 6.96
Mass of MgO = 17.53 g
The formula for mass percent is
% by mass = Mass of component/Total mass × 100 %
In this case,
% O = mass of O/mass of MgO × 100 %
Mass of O = 6.96 g
Mass of MgO = 17.53 g
% O = 6.96/17.53 × 100
% O = 0.3970 × 100
% O = 39.7 %
