Answer:
New temperature T2 = 707.5 K (Approx.)
Explanation:
Given:
Old pressure P1 = 2 atm
Old temperature T1 = 283 K
New Pressure P2 = 5 atm
Find:
New temperature T2
Computation:
Using Gay-Lussac law;
P1 / T1 = P2 / T2
So,
2 / 283 = 5 / T2
New temperature T2 = 707.5 K (Approx.)
It has 6 electrons, and 4 valence electrons
The question is incomplete, here is the complete question:
A chemist measures the amount of bromine liquid produced during an experiment. She finds that 766.g of bromine liquid is produced. Calculate the number of moles of bromine liquid produced. Round your answer to 3 significant digits.
<u>Answer:</u> The amount of liquid bromine produced is 4.79 moles.
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
We are given:
Given mass of liquid bromine = 766. g
Molar mass of liquid bromine, 
= 159.8 g/mol
Putting values in above equation, we get:
Hence, the amount of liquid bromine produced is 4.79 moles.
Answer: 3.35x10²³atoms H2
Explanation: solution attached:
Convert mass of Al to moles
Do the mole to mole ratio between Al and H2
Convert moles of H2 to atoms using Avogadro's number.
Answer:
A and D are true , while B and F statements are false.
Explanation:
A) True. Since the standard gibbs free energy is
ΔG = ΔG⁰ + RT*ln Q
where Q= [P1]ᵃ.../([R1]ᵇ...) , representing the ratio of the product of concentration of chemical reaction products P and the product of concentration of chemical reaction reactants R
when the system reaches equilibrium ΔG=0 and Q=Keq
0 = ΔG⁰ + RT*ln Q → ΔG⁰ = (-RT*ln Keq)
therefore the first equation also can be expressed as
ΔG = RT*ln (Q/Keq)
thus the standard gibbs free energy can be determined using Keq
B) False. ΔG⁰ represents the change of free energy under standard conditions . Nevertheless , it will give us a clue about the ΔG around the standard conditions .For example if ΔG⁰>>0 then is likely that ΔG>0 ( from the first equation) if the temperature or concentration changes are not very distant from the standard conditions
C) False. From the equation presented
ΔG⁰ = (-RT*ln Keq)
ΔG⁰>0 if Keq<1 and ΔG⁰<0 if Keq>1
for example, for a reversible reaction ΔG⁰ will be <0 for forward or reverse reaction and the ΔG⁰ will be >0 for the other one ( reverse or forward reaction)
D) True. Standard conditions refer to
T= 298 K
pH= 7
P= 1 atm
C= 1 M for all reactants
Water = 55.6 M
