The ideal gas equation is;
PV = nRT; therefore making P the subject we get;
P = nRT/V
The total number of moles is 0.125 + 0.125 = 0.250 moles
Temperature in kelvin = 273.15 + 18 = 291.15 K
PV = nRT
P = (0.250 × 0.0821 )× 291.15 K ÷ (7.50 L) = 0.796 atm
Thus, the pressure in the container will be 0.796 atm
The question is incomplete, the complete question is;
Which of the following is most likely a heavier stable nucleus? (select all that apply) Select all that apply: A nucleus with a neutron:proton ratio of 1.05 A nucleus with a A nucleus with a neutron:proton ratio of 1.49 The nucleus of Sb-123 A nucleus with a mass of 187 and an atomic number of 75
Answer:
A nucleus with a A nucleus with a neutron:proton ratio of 1.49
A nucleus with a mass of 187 and an atomic number of 75
Explanation:
The stability of a nucleus depends on the number of neutrons and protons present in the nucleus. For many low atomic number elements, the number of protons and number of neutrons are equal. This implies that the neutron/proton ratio = 1
Elements with higher atomic number tend to be more stable if they have a slight excess of neutrons as this reduces the repulsion between protons.
Generally, the belt of stability for chemical elements lie between and N/P ratio of 1 to an N/P ratio of 1.5.
Two options selected have an N/P ratio of 1.49 hence they are heavy stable elements.
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.
A. 1.01 is the right answer
Since
The formula is Pv= nRT
P=1 atm
V= 22.4 L
N= x
r= 0.0821
t = 273 k (bc it’s standard temperature)
So (1)(22.4)=(x)(0.0821)(273)
X= 1.001
