Answer is: <span>the half-life of the radioisotope is 74 days.
</span>m₀ = 3.25 g.
m₁ = 1.21 g.
t = 105 d.
ln(m₀/m₁) = k· t.
ln(3.25/1.21) = k·105 d.
ln(2.685) = 105·k.
0.98 = 105k.
k = 0.0094.
t1/2 = ln2 / k.
t1/2 = 0.693 / 0.0094.
t1/2 = 73.72 days.
Answer:
296.448 g
Explanation:
Density:
we know that density of an object can be measured by dividing its mass to its volume.
Symbol:
The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.
Units:
SI unit of density is Kg/m3.
Other units are given below,
g/cm3, g/mL , kg/L
Formula:
D=m/v
D= density
m= mass
V= volume
Given data:
volume= 1.2 cm x 4.00 cm x 3.20 cm =15.36 cm3
density= 19.3 g/cm3
mass=?
Now we will put the values in the formula,
d=m/v
d x V= m
m = 15.36 cm3 x 19.3 g/cm3
m=296.448 g
so the mass of block of gold is 296.448 g
Answer: In creating a covalent bond, it would be best to bind oxygen to carbon (C ).
Covalent bonds result from electron-sharing between two atoms. One carbon atom combines with two oxygen atoms. Thus, the carbon dioxide molecule has two C=O bonds. Carbon dioxide is a very common covalent bond.
Answer:
The pH of the final solution is 7.15
Explanation:
50 mL of 2.0 M of 
and 25 mL of 2.0 M of 
were mixed to make a solution
Final volume of the solution after dilution = 200 mL
Final concentration of ![K_2HPO_4, [K_2HPO_4] = \frac{50 mL\times 2 M}{200 mL} = 0.5 M](https://tex.z-dn.net/?f=K_2HPO_4%2C%20%5BK_2HPO_4%5D%20%3D%20%5Cfrac%7B50%20mL%5Ctimes%202%20M%7D%7B200%20mL%7D%20%3D%200.5%20M)
Final concentration of![KH_2PO_4, [KH_2PO_4] = \frac{25 mL\times 2 M}{200 mL} = 0.25 M](https://tex.z-dn.net/?f=KH_2PO_4%2C%20%5BKH_2PO_4%5D%20%3D%20%5Cfrac%7B25%20mL%5Ctimes%202%20M%7D%7B200%20mL%7D%20%3D%200.25%20M)
We use Hasselbach- Henderson equation:
![pH = pK_a+ log \frac{[salt]}{[acid]}pka of KH_2PO_4 = 6.85](https://tex.z-dn.net/?f=pH%20%3D%20pK_a%2B%20log%20%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7Dpka%20of%20KH_2PO_4%20%3D%206.85)
Substituting the values:
Therfore, the pH of the final solution is 7.15
Don't all halogens have 8 valence electrons? They don't need to gain or lose any because they are already stable.
