There will be 7.5 g of Be-11 remaining after 28 s.
If 14 s = 1 half-life, 28 s = 2 half-lives.
After the first half-life, ½ of the Be-11 (15 g) will disappear, and 15 g will remain.
After the second half-life, ½ of the 15 g (7.5 g) will disappear, and 7.5 g will remain.
In symbols,
<em>N</em> = <em>N</em>₀(½)^<em>n</em>
where
<em>n</em> = the number of half-lives
<em>N</em>₀ = the original amount
<em>N</em> = the amount remaining after <em>n</em> half-lives
Answer:
Choice A: approximately 
.
Explanation:
Note that the unit of concentration, 
, typically refers to moles per liter (that is: 
.)
On the other hand, the volume of the two solutions in this question are apparently given in 
, which is the same as 
(that is: 
.) Convert the unit of volume to liters:
.
.
Calculate the number of moles of 
formula units in that 
of the 
solution:
.
Note that (sulfuric acid) is a diprotic acid. When one mole of completely dissolves in water, two moles of 
ions will be released.
On the other hand, 
(sodium hydroxide) is a monoprotic base. When one mole of formula units completely dissolve in water, only one mole of 
ions will be released.
ions and ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid dissolves in water completely, it will take two moles of to neutralize that two moles of produced. On the other hand, two moles formula units of the monoprotic base will be required to produce that two moles of . Therefore, and formula units would neutralize each other at a two-to-one ratio.
.
.
Previous calculations show that 
of was produced. Calculate the number of moles of formula units required to neutralize that
.
Calculate the concentration of a 
solution that contains exactly 
of formula units:
.
Mass=volume x density
if we have mass and density we can calculate volume using the formula: volume=mass/density
volume of the displaced water = 600g/19.3g/cm3
volume = 31.09cm3
Answer:
300 kg
Explanation:
The give parameters obtained from a similar question are;
The mass of oseltamivir per capsule = 75 mg
The mass of oseltamivir required, <em>M</em>, is given as follows;
M = The mass of oseltamivir per capsule × The mass taken per person per day × 5 days × The number of people in the city
M = 0.75 mg/capsule × 2 capsule/(day·person) × 5 days × 400,000 people
M = 300,000,000 mg = 300 kg
A. We can calculate the initial concentrations of each by
the formula:
initial concentration ci = initial volume * initial
concentration / total mixture volume
where,
total mixture volume = 10 mL + 20 mL + 10 mL + 10 mL = 50
mL
ci (acetone) = 10 mL * 4.0 M / 50 mL = 0.8 M
ci (H+) = 20 mL * 1.0 M / 50 mL = 0.4 M (note: there is only 1 H+ per
1 HCl)
ci (I2) = 10 mL * 0.0050 M / 50 mL = 0.001 M
B. The rate of reaction is determined to be complete when
all of I2 is consumed. This is signified by complete disappearance of I2 color
in the solution. The rate therefore is:
rate of reaction = 0.001 M / 120 seconds
rate of reaction = 8.33 x 10^-6 M / s
