Answer:
.997 atm
Explanation:
1. Find the combined gas law formula...
(P1V1/T1 = P2V2/T2)
2. Find our numbers...
P1= .982 atm
P2= ? (trying to find)
V1= 2 L
V2= 1.8 L
T1= 22 C = 295 K
T2= -3 C = 270 K
- Note: always use Kelvin. To find Kelving add 273 to ___C.
3. Rearrange formula to fit problem...
(P2=P1V1T2/V2T1)
4. Fill in our values...
P2= .982 atm x 2 L x 270 K / 1.8 L x 295 K
5. Do the math and your answer should be...
.997 atm
- If you need more help or still do not understand please let me know and I would be glad to help!
An element that exist as a discreet atom has only one atom which can stand alone on its own. Example of this is Neon, which is a noble gas. The chemical symbol for neon is Ne.
Elements are able to exist as discreet atom because they are chemical stable and inert.
Answer:
Therefore the density of the sheet of iridium is 22.73 g/cm³.
Explanation:
Given, the dimension of the sheet is 3.12 cm by 5.21 cm.
Mass: The mass of an object can't change with respect to position.
The S.I unit of mass is Kg.
Weight of an object is product of mass of the object and the gravity of that place.
Density: The density of an object is the ratio of mass of the object and volume of the object.

[S.I unit of mass= Kg and S.I unit of m³]
Therefore the S.I unit of density = Kg/m³
Therefore the C.G.S unit of density=g/cm³
The area of the sheet is = length × breadth
=(3.12×5.21) cm²
=16.2552 cm²
Again given that the thickness of the sheet is 2.360 mm =0.2360 cm
Therefore the volume of the sheet is =(16.2552 cm²×0.2360 cm)
=3.8362272 cm³
Given that the mass of the sheet of iridium is 87.2 g.

=22.73 g/cm³
Therefore the density of the sheet of iridium is 22.73 g/cm³.
Answer:
Ok:
Explanation:
So, you can use the Henderson-Hasselbalch equation for this:
pH = pKa + log(
) where A- is the conjugate base of the acid. In other words, A- is the deprotonated form and HA is the protonated.
We can solve that
1 = log(
) and so 10 =
or 10HA = A-. For every 1 protonated form of adenosine (HA), there are 10 A-. So, the percent in the protonated form will be 1(1+10) or 1/11 which is close to 9 percent.