Answer:
Maintaining a high starting-material concentration can render this reaction favorable.
Explanation:
A reaction is <em>favorable</em> when <em>ΔG < 0</em> (<em>exergonic</em>). ΔG depends on the temperature and on the reaction of reactants and products as established in the following expression:
ΔG = ΔG° + R.T.lnQ
where,
ΔG° is the standard Gibbs free energy
R is the ideal gas constant
T is the absolute temperature
Q is the reaction quotient
To make ΔG < 0 when ΔG° > 0 we need to make the term R.T.lnQ < 0. Since T is always positive we need lnQ to be negative, what happens when Q < 1. Q < 1 implies the concentration of reactants being greater than the concentration of products, that is, maintaining a high starting-material concentration will make Q < 1.
Compound X is an isomer of butane with different chain types. It is either straight chain or bend chain.
Triprotic acid is a class of Arrhenius acids that are capable of donating three protons per molecule when dissociating in aqueous solutions. So the chemical reaction as described in the question, at the third equivalence point, can be show as: H3R + 3NaOH ⇒ Na3R + 3H2O, where R is the counter ion of the triprotic acid. Therefore, the ratio between the reacted acid and base at the third equivalence point is 1:3.
The moles of NaOH is 0.106M*0.0352L = 0.003731 mole. So the moles of H3R is 0.003731mole/3=0.001244mole.
The molar mass of the acid can be calculated: 0.307g/0.001244mole=247 g/mol.
Answer:
The mass of an average copper atom is 
Explanation:
Given:
The total mass of copper atoms, 
Number of atoms, 
Now, we are asked to find the mass of 1 copper atom.
We use unitary method to find the mass of 1 copper atom.
Mass of 
atoms = m
∴ Mass of 1 atom = 
Plug in 63.5 for 'm', 
for 'N' and simply.
Mass of 1 atom = 
Therefore, the mass of an average copper atom is
Answer: 0.0007 moles of 
is released when temperature is raised.
Explanation:
To calculate the number of moles, we use the ideal gas equation, which is:
where,
P = pressure of the gas = 1.01 bar
V = Volume of the gas = 1L
R = Gas constant = 
- Number of moles when T = 20° C
Temperature of the gas = 20° C = (273 + 20)K = 293K
Putting values in above equation, we get:
- Number of moles when T = 25° C
Temperature of the gas = 25° C = (273 + 25)K = 298K
Putting values in above equation, we get:
- Number of moles released =
Hence, 0.0007 moles of is released when temperature is raised from 20° C to 25° C
