True!
Energy released by system is absorbed by surroundings.
Answer:
M.Mass = 3.66 g/mol
Data Given:
M.Mass = M = ??
Density = d = 0.1633 g/L
Temperature = T = 273.15 K (Standard)
Pressure = P = 1 atm (standard)
Solution:
Let us suppose that the gas is an ideal gas. Therefore, we will apply Ideal Gas equation i.e.
P V = n R T ---- (1)
Also, we know that;
Moles = n = mass / M.Mass
Or, n = m / M
Substituting n in Eq. 1.
P V = m/M R T --- (2)
Rearranging Eq.2 i.e.
P M = m/V R T --- (3)
As,
Mass / Volume = m/V = Density = d
So, Eq. 3 can be written as,
P M = d R T
Solving for M.Mass i.e.
M = d R T / P
Putting values,
M = 0.1633 g/L × 0.08205 L.atm.K⁻¹.mol⁻¹ × 273.15 K / 1 atm
M = 3.66 g/mol
Answer : The radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.
Explanation :
As we are given that the Na⁺ radius is 56.4% of the Cl⁻ radius.
Let us assume that the radius of Cl⁻ be, (x) pm
So, the radius of Na⁺ = 
In the crystal structure of NaCl, 2 Cl⁻ ions present at the corner and 1 Na⁺ ion present at the edge of lattice.
Thus, the edge length is equal to the sum of 2 radius of Cl⁻ ion and 2 radius of Na⁺ ion.
Given:
Distance between Na⁺ nuclei = 566 pm
Thus, the relation will be:
The radius of Cl⁻ ion = (x) pm = 181 pm
The radius of Na⁺ ion = (0.564x) pm = (0.564 × 181) pm =102.084 pm ≈ 102 pm
Thus, the radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.
Answer:
Carbon dioxide levels in the Earth's atmosphere have been steadily increasing.
Carbon has a longer average lifetime in the atmosphere.
Explanation:
Today the level of carbon dioxide is higher than at any time in human history. Scientists widely agree that Earth’s average surface temperature has already increased by about 2 F (1 C) since the 1880s, and that human-caused increases in carbon dioxide and other heat-trapping gases are extremely likely to be responsible.
The lifetime in the air of CO2, the most significant man-made greenhouse gas, is probably the most difficult to determine, because there are several processes that remove carbon dioxide from the atmosphere. Between 65% and 80% of CO2 released into the air dissolves into the ocean over a period of 20–200 years.
Given:
Half life(t^ 1/2) :30 years
A0( initial mass of the substance): 200 mg.
Now we know that
A= A0/ [2 ^ (t/√t)]
Where A is the mass that remains after t years.
A0 is the initial mass
t is the time
t^1/2 is the half life
Substituting the given values in the above equation we get
A= [200/ 2^(t/30) ] mg
Thus the mass remaining after t years is [200/ 2^(t/30) ] mg
