Answer:2.86x10^-7m
Explanation:E=hc/^
E=6.94x10^-19J
c = 2.9979x10^8m/s
h= 6.626x10^-34Js
^ =( 6.626x10^-34)x( 2.9979x 10^8)/ 6.94x10^-19
= 2.86x10^-7m
Answer:
Half life = 1600 years
Explanation:
Given data:
Total mass of sample = 45.00 g
Mass remain = 5.625 g
Time period = 4800 years
Half life of radium-226 = ?
Solution:
First of all we will calculate the number of half lives passes,
At time zero 45.00 g
At first half life = 45.00 g/ 2= 22.5 g
At 2nd half life = 22.5 g/ 2 = 11.25 g
At 3rd half life = 11.25 g/ 2= 5.625 g
Half life:
Half life = Time elapsed / number of half lives
Half life = 4800 years / 3
Half life = 1600 years
1500 cm^3 ; 1 mL equals 1 cm^3
Answer:
156 Hydrogen atoms
Explanation:
<u>Any acyclic alkane has a molecular formula that can be expressed as</u>:
CₙH₂ₙ₊₂
Where <em>n</em> is any integer and the number of carbon atoms. For example, Propane has 3 carbon atoms, this means it would have [2*3+2] 8 hydrogen atoms, resulting with a formula of C₃H₈.
An acyclic alkane with 77 carbon atoms would thus have:
2*77 + 2 = 156 hydrogen atoms
<span>Answer:
It depends on what came after "0.5440 M H...".
If it was a monoprotic acid, like HCl, the calculation would go like this:
(55.25 mL) x (0.5440 M acid) x (1 mol KOH / 1 mol acid) / (0.2450 M KOH) =
122.7 mL KOH
If it was a diprotic acid, like H2SO4, like this:
(55.25 mL) x (0.5440 M acid) x (2 mol KOH / 1 mol acid) / (0.2450 M KOH) =
245.4 mL KOH
If it was a triprotic acid, like H3PO4, like this:
(55.25 mL) x (0.5440 M acid) x (3 mol KOH / 1 mol acid) / (0.2450 M KOH) =
368.0 mL KOH</span>
