Answer:
The boiling point of sample X and sample Y are exactly the same.
Explanation:
The difference between sample X and sample Y is that they occupy different volumes. However, they both contain pure water. Remember that pure water has uniform composition irrespective of its volume.
Volume does not affect the boiling point as long as the volume is small enough not to give rise to significant pressure changes in the liquid.
The boiling point of a liquid is the temperature at which the pressure exerted by the surroundings upon a liquid is equaled by the pressure exerted by the vapour of the liquid; under this condition, addition of heat results in the transformation of the liquid into its vapour without raising the temperature.
It can be clearly seen from the above that the volume of a solution of pure water does not affect its boiling point hence sample X and sample Y will have the same boiling point.
Answer:
The specific heat capacity of the metal is 0.268 J/g°C
Explanation:
Step 1: Data given
Mass of the metal = 151.5 grams
The temperature of the metal = 75.0 °C
Temperature of water = 15.1 °C
The temperature of the water rises to 18.7°C.
The specific heat capacity of water is 4.18 J/°C*g
Step 2: Calculate the specific heat capacity of the metal
heat lost = heat gained
Q = m*c*ΔT
Qmetal = - Qwater
m(metal) * c(metal) * ΔT(metal) = m(water) * c(water) * ΔT(water)
⇒ mass of the metal = 151.5 grams
⇒ c(metal) = TO BE DETERMINED
⇒ΔT( metal) = T2 - T1 = 18.7 °C - 75.0 °C = -56.3 °C
⇒ mass of the water = 151.5 grams
⇒ c(water) = 4.184 J/g°C
⇒ ΔT(water) = 18.7° - 15.1 = 3.6 °C
151.5g * c(metal) * -56.3°C = 151.5g * 4.184 J/g°C * 3.6 °C
c(metal) = 0.268 J/g°C
The specific heat capacity of the metal is 0.268 J/g°C
Answer:
D. H2SO4
Explanation:
The chemical formula of a compound is an expression that stares out the elements (in form of symbols)present in a compound and the number of the atoms.
In the image;
There is one sulphur (S) atom, 4 oxygen(O) atoms and 2 hydrogen (H) atoms
The chemical formula is;
H2SO4.
The correct option is option D.
This question is testing to see how well you understand the "half-life" of radioactive elements, and how well you can manipulate and dance around them. This is not an easy question.
The idea is that the "half-life" is a certain amount of time. It's the time it takes for 'half' of the atoms in any sample of that particular unstable element to 'decay' ... their nuclei die, fall apart, and turn into nuclei of other elements.
Look over the table. There are 4,500 atoms of this radioactive substance when the time is 12,000 seconds, and there are 2,250 atoms of it left when the time is ' y ' seconds. Gosh ... 2,250 is exactly half of 4,500 ! So the length of time from 12,000 seconds until ' y ' is the half life of this substance ! But how can we find the length of the half-life ? ? ?
Maybe we can figure it out from other information in the table !
Here's what I found:
Do you see the time when there were 3,600 atoms of it ?
That's 20,000 seconds.
... After one half-life, there were 1,800 atoms left.
... After another half-life, there were 900 atoms left.
... After another half-life, there were 450 atoms left.
==> 450 is in the table ! That's at 95,000 seconds.
So the length of time from 20,000 seconds until 95,000 seconds
is three half-lifes.
The length of time is (95,000 - 20,000) = 75,000 sec
3 half lifes = 75,000 sec
Divide each side by 3 : 1 half life = 25,000 seconds
There it is ! THAT's the number we need. We can answer the question now.
==> 2,250 atoms is half of 4,500 atoms.
==> ' y ' is one half-life later than 12,000 seconds
==> ' y ' = 12,000 + 25,000
y = 37,000 seconds .
Check:
Look how nicely 37,000sec fits in between 20,000 and 60,000 in the table.
As I said earlier, this is not the simplest half-life problem I've seen.
You really have to know what you're doing on this one. You can't
bluff through it.
seconds afterwards, and many also refer that it was from an infinitely hot point.
