Answer:
3.5 atm
Explanation:
As stated in the question pressure is required to counteract the natural tendency for water to dilute the more concentrated solution. The difference in concentrations will give us the answer using the osmotic pressure equation.
π = ( n/v) RT where n/v is the molarity (mol/L), R is the gas constant and T is the temperature.
The difference in osmotic pressure of the solutions is:
Δπ = Δ c RT where c is the difference in molar concentrations.
pressure required = Δπ = (0.190 - 0.048) M x 0.821 Latm/Kmol x 298 K
= 3.47 atm
diatomic hydrogen is written as H2 (2.02 grams H2) <------- if each hydrogen atom is 1.01 grams, then two hydrogen atoms are 2.02 grams 2.0 moles H2 X 2.02 grams H2 ------------- (divide to cancel moles) = 4.04 grams/mole H2 ÷ one mole = 4.04 grams H2
Answer:
Exergonic ,Endergonic,low concentration area,high
Explanation:
In exergonic reaction,certain molecules are broken down;in the process they release energy which is captured when high energy molecules(such as ATP and NADH) are formed.
The breakdown of these molecules can be coupled to thermodynamically unfavorable processes such as Endergonic reactions or pumping og hydrogen ion from low concentration areas to high concentration areas.
There are different formula you need to keep in mind when solving for [OH-]
Given that pH = 6.10
pH + pOH = 14
6.10 + pOH = 14
pOH = 7.9
[OH-] = 10^(-pOH)
[OH-] = 10^(-7.9)
[OH-] = 0.000000013
[OH-] = 1.3 x 10^-8
<h2>
<u>Answer: [OH-] = 1.3 x 10^-8</u></h2>
The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.
