Answer is: 0,133 mol/ l· atm.
T(chlorine) = 10°C = 283K.
p(chlorine) = 1 atm.
V(chlorine) = 3,10 l.
R - gas constant, R = 0.0821 atm·l/mol·K.
Ideal gas law: p·V = n·R·T
n(chlorine) = p·V ÷ R·T.
n(chlorine) = 1atm · 3,10l ÷ 0,0821 atm·l/mol·K · 283K = 0,133mol.
Henry's law: c = p·k.
k - <span>Henry's law constant.
</span>c - solubility of a gas at a fixed temperature in a particular solvent.
c = 0,133 mol/l.
k = 0,133 mol/l ÷ 1 atm = 0,133 mol/ l· atm.
The answer the following are as follows:
<span> 1. XeF4 - molecules are polar
3. CCl4 - molecules are polar
5. CH3Br - molecules are non-polar
I hope my answer has come to your help. God bless and have a nice day ahead!</span>
The question is improperly formatted.
What is the concentration of H+ ions in a 2.2 M solution of HNO3.
Answer:-
2.2 moles of H+ per litre
Explanation:-
M stands for molarity. 2.2 M means 2.2 moles of HNO3 is present per litre of the solution.
Now HNO3 has just 1 H in it's formula. HNO3 would give H+. So 2.2 moles of HNO3 would mean 2.2 moles of H+ per litre.
Answer: Option (d) is the correct answer.
Explanation:
A ketone is an organic functional group that contains a carbon and oxygen atom bonded together through a double bond, that is, C=O.
For example, acetone 
is a ketone.
Whereas a hydrocodone is a drug which is used to relieve from moderate to sever pain.It is mostly combined with other drugs and resulting in a chemical formula 
.
A camphor is a volatile white color substance with chemical formula 
. It has aromatic smell and its taste is bitter.
A menthone is also an organic compound with chemical formula 
.
Thus, we can conclude that out of the given options the simplest ketone used as an organic solvent is acetone.
Answer:
91.16% has decayed & 8.84% remains
Explanation:
A = A₀e⁻ᵏᵗ => ln(A/A₀) = ln(e⁻ᵏᵗ) => lnA - lnA₀ = -kt => lnA = lnA₀ - kt
Rate Constant (k) = 0.693/half-life = 0.693/10³yrs = 6.93 x 10ˉ⁴yrsˉ¹
Time (t) = 1000yrs
A = fraction of nuclide remaining after 1000yrs
A₀ = original amount of nuclide = 1.00 (= 100%)
lnA = lnA₀ - kt
lnA = ln(1) – (6.93 x 10ˉ⁴yrsˉ¹)(3500yrs) = -2.426
A = eˉ²∙⁴²⁶ = 0.0884 = fraction of nuclide remaining after 3500 years
Amount of nuclide decayed = 1 – 0.0884 = 0.9116 or 91.16% has decayed.
