We assume that this gas is ideal. Therefore, we can use the ideal gas equation which is expressed as:
PV=nRT
We manipulate this equation to give us an expression which will correspond to density. We do as follows:
PV= nRT
P/RT = n/V where n = m/MM
P(MM) /RT = m/V = density
Density = 1.00 (17.03) / 0.08206 (435)
Density = 0.48 g / L
Q = mC∆T
<span>where: </span>
<span>q = heat </span>
<span>m = mass of substance = 35.0 grams </span>
<span>C = 0.385 J/g*C </span>
<span>∆T = change in temperature = 65C - 20C = 45C </span>
<span>q = (35.0 g)*(0.385 J/g*C)*(45C) = 606 J </span>
Explanation:
Both water and carbon dioxide have 2 bonding domains (with the 2 hydrogen atoms and 2 oxygen atoms respectively)
However, water is a polar molecule as the central oxygen atom has 2 pairs of lone electrons, which allows it to form hydrogen bonding with other water molecules.
Carbon dioxide on the other hand is non-polar because it has no lone pairs on the central carbon atom, allowing its molecular geometry to be linear.
Hence the answer is Option C.
Answer: 1.7 moles
Explanation:
Given that:
Volume of water V = 1.5L
(since 1 liter = 1dm3
1.5L = 1.5dm3)
Temperature T = 100°C
Convert Celsius to Kelvin
(100°C + 273 = 373K)
Pressure P = 3.5 atm
Number of moles of steam N = ?
Note that Molar gas constant R is a constant with a value of 0.0082 ATM dm3 K-1 mol-1
Then, apply ideal gas equation
pV = nRT
3.5atm x 1.5dm3 = n x (0.0082 ATM dm3 K-1 mol-1 x 373K)
5.25 = n x 3.06
n = 5.25/3.06
n = 1.7 moles
Thus, 1.7 moles of steam is present
There are a few ways to do this. The best way I can think of is to start out the paint with no additives and have that as a control. You can test that on whatever your painting or do it in a lab with heat lamps and lights if that is a possible option. However you decide to do the testing, after starting with the control test, add different additives and see which ones fades less the original without any additives. This is the best scenario for a simple yet informative test.
Hope that helps!
