A strong acid- strong base titration is performed using a phenolphthalein indicator. Phenolphtalein is chosen because it changes color in a pH range between 8.3 – 10. It will appear pink in basic solutions and clear in acidic solutions. ... It is known as the titrant.
Answer:
The amount left after 49.2 years is 3mg.
Explanation:
Given data:
Half life of tritium = 12.3 years
Total mass pf tritium = 48.0 mg
Mass remain after 49.2 years = ?
Solution:
First of all we will calculate the number of half lives.
Number of half lives = T elapsed/ half life
Number of half lives = 49.2 years /12.3 years
Number of half lives = 4
Now we will calculate the amount left after 49.2 years.
At time zero 48.0 mg
At first half life = 48.0mg/2 = 24 mg
At second half life = 24mg/2 = 12 mg
At 3rd half life = 12 mg/2 = 6 mg
At 4th half life = 6mg/2 = 3mg
The amount left after 49.2 years is 3mg.
Answer:
will this help ?
Explanation:
(108Hs) is a synthetic element, and thus a standard atomic weight cannot be given. Like all synthetic elements, it has no stable isotopes. The first isotope to be synthesized was 265Hs in 1984. There are 12 known isotopes from 263Hs to 277Hs and 1–4 isomers. The most stable isotope of hassium cannot be determined based on existing data due to uncertainty that arises from the low number of measurements. The confidence interval of half-life of 269Hs corresponding to one standard deviation (the interval is ~68.3% likely to contain the actual value) is 16 ± 6 seconds, whereas that of 270Hs is 9 ± 4 seconds. It is also possible that 277mHs is more stable than both of these, with its half-life likely being 110 ± 70 seconds, but only one event of decay of this isotope has been registered as of 2016.[1][2].
Answer:
100Jkg/°C
Explanation:
Given parameters:
Mass of metal = 2kg
Amount of heat energy = 1600J
Initial temperature = 5°C
Final temperature = 13°C
Unknown:
Specific heat capacity of the metal = ?
Solution:
Specific heat capacity of a body is the amount of heat needed to raise the temperature of unit mass of a body by 1°C.
H = m x C x (T₂ - T₁ )
H is the amount of heat
m is the mass
C is the unknown specific heat capacity
T is the temperature
Insert the parameters and solve;
1600 = 2 x C x (13 - 5)
1600 = 16C
C = 100Jkg/°C
The density is 4 g/cm³ or 4000 kg/m³.
Density = mass/volume = 12 g/3 cm³ = 4 g/cm³
The measurement of 4 g/cm³ is already in <em>SI units</em>.
In SI <em>bas</em>e units,
Density = (4 g/1 cm³) × (1 kg/1000 g) × (100 cm/1 m)³ = 4000 kg/m³
