Answer:
d. 127 g/mol.
Explanation:
Hello!
In this case, since we have the amount of molecules of this this compound, we are able to compute the moles out there by using the Avogadro's number:
Which correspond to the moles of X2. Then, by using the mass we are able to compute the molar mass of X2:
It means that the atomic mass of X halves the molar mass of X2, which is then d. 127 g/mol.
Best regards!
Try adding spaces next time! That's iodine. Check all of the numbers to make sure all of the orbitals are filled, then find the ones which aren't. In this one, only the 5p5 subshell isn't full. 5p5 is the fifth row on the right side, count across the nonmetals and metalloids until the fifth one (a halogen). That's iodine, and that's your answer!
Answer:
M=0.15
Explanation:
138 g AgNO -> 1 mol AgNO
10 g AgNO -> x
x= (10 g AgNO * 1 mol AgNO)/138 g x=0.07 mol AgNO
450 mL=0.45 L
M= mol solute/L solution
M= 0.07 mol AgNO/0.45L
M=0.15
Answer:
= 25 g
Explanation:
Using the formula;
A = A₀ (1/2)^(t/h)
where A is the final amount,
A₀ is the initial amount of the substance,
t is the time and
h is the half-life of the substance,
In this case; the half life of U-238 h is equal to 4.47 billion years.
A = A₀ (1/2)^(t/h)
A = 50 (1/2)^(4.5 / 4.47)
= 24.88
<u> = 25 g</u>
