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.
Inorganic molecules are composed of other elements. They can contain hydrogen or carbon, but if they have both, they are organic.
Answer:
MgCO₃
Explanation:
From the question given above, we obtained:
MgF₂ + Li₂CO₃ —> __ + 2LiF
The missing part of the equation can be obtained by writing the ionic equation for the reaction between MgF₂ and Li₂CO₃. This is illustrated below:
MgF₂ (aq) —> Mg²⁺ + 2F¯
Li₂CO₃ (aq) —> 2Li⁺ + CO₃²¯
MgF₂ + Li₂CO₃ —>
Mg²⁺ + 2F¯ + 2Li⁺ + CO₃²¯ —> Mg²⁺CO₃²¯ + 2Li⁺F¯
MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF
Now, we share compare the above equation with the one given in the question above to obtain the missing part. This is illustrated below:
MgF₂ + Li₂CO₃ —> __ + 2LiF
MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF
Therefore, the missing part of the equation is MgCO₃
Molar mass of :
O2 = 16 * 2 = 32 g/mol
CO2 = 12 + 16 * 2 = 44 g/mol
<span>Balanced chemical equation :
</span>
1 CH4 +
2 O2 =
1 CO2 +
2 H2O
↓ ↓
2 moles 1 mole
2* 32 g O2 ----------> 1* 44 g CO2
x g O2 ------------> 10.0 g CO2
44 x = 2 * 32*10.0
44 x = 640
of O2
