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.
<em>What volume do 5 moles of a gas occupy at 28 ° C and 3 atm of pressure?</em>
<em />
<h3>Further explanation</h3>
In general, the gas equation can be written
where
P = pressure, atm
V = volume, liter
n = number of moles
R = gas constant = 0.08206 L.atm / mol K
T = temperature, Kelvin
n= 5 moles
T=28 +273=301 K
P=3 atm
The volume of the gas :
Answer:
1. Cu
2. Cu
3. 2 electrons.
Explanation:
Step 1:
The equation for the reaction is given below:
3Cu(s) + 8HNO3(aq) -> 2NO(g) 3Cu(NO3)2(aq) + 4H2O(l)
Step 2:
Determination of the change of oxidation number of each element present.
For Cu:
Cu = 0 (ground state)
Cu(NO3)2 = 0
Cu + 2( N + 3O) = 0
Cu + 2(5 + (3 x -2)) =0
Cu + 2 (5 - 6) = 0
Cu + 2(-1) = 0
Cu - 2 = 0
Cu = 2
The oxidation number of Cu changed from 0 to +2
For N:
HNO3 = 0
H + N + 3O = 0
1 + N + (3 x - 2) = 0
1 + N - 6 = 0
N = 6 - 1
N = 5
NO = 0
N - 2 = 0
N = 2
The oxidation number of N changed from +5 to +2
The oxidation number of oxygen and hydrogen remains the same.
Note:
1. The oxidation number of Hydrogen is always +1 except in hydride where it is - 1
2. The oxidation number of oxygen is always - 2 except in peroxide where it is - 1
Step 3:
Answers to the questions given above
From the above illustration,
1. Cu is oxidize because its oxidation number increased from 0 to +2 as it loses electron.
2. Cu is the reducing agent because it reduces the oxidation number of N from +5 to +2.
3. The reducing agent i.e Cu transferred 2 electrons to the oxidising agent HNO3 because its oxidation number increase from 0 to +2 as it loses its electrons. This means that Cu transfer 2 electrons.
Answer:
Explanation:
A. Attached is a page which contain the structural formula of the three compounds of C3H80.
Condensed structural formula of C3H8O:
Propan-1-ol: CH3CH2CH2OH
Propan-2-ol: CH3CH(OH)CH3
Methoxy methane: CH3OCH2CH3
B. Attached are is a page which contain the structural formula of the three compounds of C3H60.
Condensed structural formula of C3H6O:
Propanal: CH3CH2CHO
Propanone: CH3COCH3
Cyclopropanol: (C3H5)OH
2-propen-1-ol: CH2CHCH2OH
1-propenol: CH3CHCHOH
The percent composition of each element can be calculated as follows:
% composition = (mass of element / total mass) * 100
The total mass of the quarter is given to be 5.670 grams
Mass of Cu = 5.198 grams
Mass of Ni = 0.472 grams
Substitute in the above equation to get the mass percentage of each element as follows:
% of Cu = (5.198/5.670) * 100 = 91.675%
% of Ni = (0.472/5.670) * 100 = 8.325%
