Answer:
463.0 g.
Explanation:
- We can use the following relation:
<em>n = mass/molar mass.</em>
where, n is the mass of copper(ii) fluoride (m = 4.56 mol),
mass of copper(ii) fluoride = ??? g.
molar mass of copper(ii) fluoride = 101.543 g/mol.
∴ mass of copper(ii) fluoride = (n)(molar mass) = (4.56 mol)(101.543 g/mol) = 463.0 g.
Answer:

Explanation:
Molarity is a measure of concentration in moles per liter.

The molarity of the solution is 1.2 M NaNO₃ or 1.2 moles NaNO₃ per liter. There are 0.25 liters of the solution. The moles of solute are unknown, so we can use x.
- molarity= 1.2 mol NaNO₃/L
- liters of solution=0.25 L
- moles of solute =x

We are solving for x, so we must isolate the variable, x. It is being divided by 0.25 liters. The inverse of division is multiplication, so we multiply both sides by 0.25 L.


The units of liters cancel, so we are left with the units moles of sodium nitrate.


There are 0.3 moles of sodium nitrate.
Answer: 226 Subscript 88 Baseline Upper R a right arrow Superscript 222 Subscript 86 Baseline Upper R n + Superscript 4 Subscript 2 Baseline Upper H e
Explanation:
Alpha decay : When a larger nuclei decays into smaller nuclei by releasing alpha particle. In this process, the mass number and atomic number is reduced by 4 and 2 units respectively.
The general representation of alpha decay reaction is:

Representation of Radium decays to form Radon

Thus 226 Subscript 88 Baseline Upper R a right arrow Superscript 222 Subscript 86 Baseline Upper R n + Superscript 4 Subscript 2 Baseline Upper H e represents alpha decay.
molecules collide more frequently.
Chemical reaction mechanisms are based in the collision of molecules with certaing level of energy. More collisions implies grater probablity of reaction.
Answer:
Higher frequency
Explanation:
We can imagine a chemical bond between two atoms as if it were two balls connected by a spring.
According to Hooke's Law, the stretching frequency f is

where µ is the reduced mass of the system

The strength of the bond is analogous to k, the force constant of the spring. Then,

Thus, the stronger the bond, the greater the frequency of vibration.