Answer:
Explanation:
Molarity = number of moles / volume
If 550 mL of a 3.50 M KCl solution are set aside and allowed to evaporate until the volume of the solution is 275 mL, which is half of 550 mL, the molarity of the solution with the same number of moles of KCl is 3.5 * 2 = 7.00 M
#1- Identify a problem
#2- Collect info on your problem
#3- Make a hypothesis
#4- Design an experiment to test your hypothesis
#5- Collect data and observations
#6- Accept or reject your hypothesis
#7- Record results
Hope this helps.
Answer:
3.09kg
Explanation:
First, let us write a balanced equation for the reaction. This is illustrated below:
2C8H18 + 25O2 —> 16CO2 + 18H2O
Molar Mass of C8H18 = (12x8) + (18x1) = 96 + 18 = 114g/mol
Mass of C8H18 from the balanced equation = 2 x 114 = 228g
Converting 228g of C8H18 to kg, we obtained:
228/1000 = 0.228kg
Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol
Mass of CO2 from the balanced equation = 16 x 44 = 704g
Converting 704g of CO2 to kg, we obtained:
704/1000 = 0.704kg
From the equation,
0.228kg of C8H18 produced 0.704kg of CO2.
Therefore, 1kg of C8H18 will produce = 0.704/0.228 = 3.09kg of CO2
Answers are:
Catabolism:
- g<span>enerally exergonic (spontaneous): In this reactions energy is released.
- </span><span>convert NAD+ to NADH. Electrons and protons released in reactions are attached to NAD+.
- </span><span>generation of ATP. ATP is synthesis from ADP.
- </span><span>convert large compounds to smaller compounds. Foe example starch to monosaccaharides.
Anabolism:
</span><span>- convert NADPH to NADP+. Protons and electrons are used to make chemical bonds.
</span>- <span>convert small compounds to larger compounds.</span>
Answer:
ΔG = - 442.5 KJ/mol
Explanation:
Data Given
delta H = -472 kJ/mol
delta S = -108 J/mol K
So,
delta S = -0.108 J/mol K
delta Gº = ?
Solution:
The answer will be calculated by the following equation for the Gibbs free energy
G = H - TS
Where
G = Gibbs free energy
H = enthalpy of a system (heat
T = temperature
S = entropy
So the change in the Gibbs free energy at constant temperature can be written as
ΔG = ΔH - TΔS . . . . . . (1)
Where
ΔG = Change in Gibb’s free energy
ΔH = Change in enthalpy of a system
ΔS = Change in entropy
if system have standard temperature then
T = 273.15 K
Now,
put values in equation 1
ΔG = (-472 kJ/mol) - 273.15 K (-0.108 KJ/mol K)
ΔG = (-472 kJ/mol) - (-29.5 KJ/mol)
ΔG = -472 kJ/mol + 29.5 KJ/mol
ΔG = - 442.5 KJ/mol