I think it's Letter c.13 if I'm not mistaken
Answer : behavior in a field experiment is more likely to reflect real life because of its natural setting, i.e. higher ecological validity than a lab experiment
Explanation:
Root mean square velocity is the square root of the mean of the squares of speeds of different molecules. From kinetic theory of gas, the formula of root mean square velocity=C
= √
=√
=√
, where, R= Universal gas constant, T= Absolute temperature, P= Pressure, V= Volume of gas, d= Density of gas.
Given, T=273 K, P=1.00 x 10⁻² atm, d=1.24 x 10⁻⁵ g/cm³.
(a) Using the formula
=√
=√(3X1.00X10⁻²)/(1.24X10⁻⁵)=49.18
(b) Molar mass can be determined by using the formula
=√{3RT}{M}
49.18=√
49.18²=√(3X8.314X273)/M
M=
M=1.67 ≅ 2
Molecular mass is 2.
(c) The gas is Helium (He) whose molecular mass is 2.
Hello!
The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.
We have the following data:
mo (initial mass) = 53.3 mg
m (final mass after time T) = ? (in mg)
x (number of periods elapsed) = ?
P (Half-life) = 10.0 minutes
T (Elapsed time for sample reduction) = 25.9 minutes
Let's find the number of periods elapsed (x), let us see:






Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:




I Hope this helps, greetings ... DexteR! =)