Answer:
Here:
Explanation:
To familiarize students with experimental apparatus, the scientific method, and methods of data analysis so that they will have some idea of the inductive process by which the ideas were originated. To teach how to make careful experimental observations and how to think about and draw conclusions from such data.
B is true if we subtract atomic number(the number of protons )from the mass number
we can find neutrons.
hope this helps
Answer:
The balanced equation for this reaction will be
→ 
We can see that 1 mole of methane requires 4 moles of fluorine but we have 0.41 moles of CH4 and 0.56mole of F2
So using the unitary method we will get that
- 1 mole of CH4 → 4 mole of 4 mole of fluorine
- 0.41 mole of methane → 4*0.41 = 1.64 mole of fluorine for complete reaction
but we have only 0.56 mole of fluorine that means fluorine is the limiting reagent and the product will only be formed by only this amount of fluorine.
- 4 moles of fluorine → 1 mole of CF4
- 0.56 mole →
= 0.14mole of CF4
- 4 moles of fluorine → 4 moles of HF
- 0.56 mole of fluorine → 0.56 mole of HF
now to find the heat released we have the formula as
DELTA H = n * Delta H of product - n *delta H of reactant
where n is the moles of the reactant and product.
note: since no information is given about the enthalpies of the species we leave it on general equation also you need to add the product side enthalpy of the species present and similarly on the product side.
Answer:

Explanation:
1. Calculate the decay constant
The integrated rate law for radioactive decay is 1

where
A₀ and A_t are the counts at t = 0 and t
k is the radioactive decay constant

2. Calculate the half-life

The half-life for decay is
.