Attached to this answer is the format of Isotope Notation that you can use for future reference. <em>(Please open)</em>
There are
8 Protons. The Atomic Number is the same number of an element's proton.
If you can see in the format, the mass number is calculated by adding the atomic number/protons and neutrons.
Mass number = 8 + 11Mass number = 19The image of the final answer is attached as well.
Answer:
A 3s orbital is at a greater average distance from the nucleus than a 2s orbital
Explanation:
As the principal quantum number n increases, the distance of the orbital from the nucleus increases. Hence if we consider the 2s and 3s orbitals, it is easy to see that the 3s orbital is at a greater distance from the nucleus than the 2s orbitals.
This is clearly seen when we plot the radial distribution against the distance from the nucleus. This enables us to visualize the region in space in which an electron may be found.
Answer:
C. Increase the temperature, allowing more frequent collisions of A and B with greater kinetic energy.
Explanation:
For the reaction:
A + B → 2C
You can increase the consume of A:
A. Utilize reaction conditions to convert both reactants to solids. <em>FALSE. </em>Convert both reactants in solids makes less frequent collisions doing difficult the reaction.
B. Decrease the concentration of reactant B to allow C to be produced at a greater rate. <em>FALSE. </em>The decreasing of B will make the reaction slowly.
C. Increase the temperature, allowing more frequent collisions of A and B with greater kinetic energy. <em>TRUE. </em>The increase of temperature will make more frequent collisions of A and B doing the reaction faster.
D. Increase the concentration of C to allow for more frequent collisions of A and B of higher energy
. <em>FALSE. </em>The increase of C makes less frequent collisions doing difficult the reaction.
E. Introduce a catalyst to decrease the consumption of A and B. <em>FALSE. </em>Introducing a catalyst increase the consumption of A and B.
I hope it helps!
To make our solution more systematic, let's convert all units that is consistent with the units of R which is 0.0821 L-atm/mol-K.
For pressure: 760 torr = 1 atm
388 torr * 1 atm/760 torr = 0.5105 atm
For volume, 1 μL = 10⁻⁶ L
0.206 μL * 10⁻⁶ L/1 μL = 2.06×10⁻⁷ L
For temperature,
T = 45 + 273 = 318 K
For mass, 1 ng = 10⁻⁹ g
206 ng * 10⁻⁹ g/1 ng = 2.06×10⁻⁷ g
Assuming ideal gas,
PV=nRT
(0.5105 atm)(2.06×10⁻⁷ L) = n(0.0821 L-atm/mol-K)(318 K)
n = 4×10⁻⁹ mol
Molar mass = Mass/n = 2.06×10⁻⁷ g/4×10⁻⁹ mol
<em>Molar mass = 51.14 g/mol</em>
Answer:
For these types of questions the equation that we must take into account is that:
T = PxV (where T is the temperature, P is the pressure and V is the volume) this equation is described as we consider that this is the value N and R is 1, therefore it is not necessary to explain them now.
Explanation:
The quoted equation refers to Boyle's Law, in this law we can explain that the volume increases if the pressure decreases and if the temperature also increases, if the pressure increases and the volume decreases this means that the gas is compressing assuming that the temperature is constant
