As long as matter cannot be destroyed or created , nothing can be gained or lost.
there is zero impact and hence one cannot numerate the impact
I think thee correct answer from the choices listed above is option D. <span>When a physical change in a sample occurs, composition of the sample does not change. It stays the same. Also, the properties of the sample will still be the same. Hope this answers the question.</span>
<span>35 grams
The average salinity of seawater is 35 parts per thousand, so multiply the mass of seawater provided by 0.035 and you'll get the amount of salt (mostly sodium chloride) dissolved in it. So
1000 g * 0.035 = 35 g
Therefore in 1 kilogram of seawater with average salinity, there is 35 grams of salt.</span>
Answers and Explanation:
a)- The chemical equation for the corresponden equilibrium of Ka1 is:
2. HNO2(aq)⇌H+(aq)+NO−2
Because Ka1 correspond to a dissociation equilibrium. Nitrous acid (HNO₂) losses a proton (H⁺) and gives the monovalent anion NO₂⁻.
b)- The relation between Ka and the free energy change (ΔG) is given by the following equation:
ΔG= ΔGº + RT ln Q
Where T is the temperature (T= 25ºc= 298 K) and R is the gases constant (8.314 J/K.mol)
At the equilibrium: ΔG=0 and Q= Ka. So, we can calculate ΔGº by introducing the value of Ka:
⇒ 0 = ΔGº + RT ln Ka
ΔGº= - RT ln Ka
ΔGº= -8.314 J/K.mol x 298 K x ln (4.5 10⁻⁴)
ΔGº= 19092.8 J/mol
c)- According to the previous demonstation, at equilibrium ΔG= 0.
d)- In a non-equilibrium condition, we have Q which is calculated with the concentrations of products and reactions in a non equilibrium state:
ΔG= ΔGº + RT ln Q
Q= ((H⁺) (NO₂⁻))/(HNO₂)
Q= ( (5.9 10⁻² M) x (6.7 10⁻⁴ M) ) / (0.21 M)
Q= 1.88 10⁻⁴
We know that ΔGº= 19092.8 J/mol, so:
ΔG= ΔGº + RT ln Q
ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)
ΔG= -2162.4 J/mol
Notice that ΔG<0, so the process is spontaneous in that direction.
Answer:- 13.6 L
Solution:- Volume of hydrogen gas at 58.7 Kpa is given as 23.5 L. It asks to calculate the volume of hydrogen gas at STP that is standard temperature and pressure. Since the problem does not talk about the original temperature so we would assume the constant temperature. So, it is Boyle's law.
Standard pressure is 1 atm that is 101.325 Kpa.
Boyle's law equation is:
From given information:-
= 58.7 Kpa
= 23.5 L
= 101.325 Kpa
= ?
Let's plug in the values and solve it for final volume.
On rearranging the equation for
= 13.6 L
So, the volume of hydrogen gas at STP for the given information is 13.6 L.
