No, because 40 miles is the same as nearly 25 km/h.
Answer:
Beta emission
Explanation:
In beta emission, a neutron is converted into a proton thereby emitting an electron and a neutrino. A neutrino is a particle that serves to balance the spins.
When a nucleus undergoes beta emission, the mass number of the parent and daughter nuclei remain the same while the atomic number of the daughter nucleus is greater than that of its parent by one unit.
Hence, in beta emission, the daughter nucleus is found one pace to the right of the parent in the periodic table.
Answer:
A high pH value indicates a high concentration of OH- ion
Explanation:
The higher the OH- ion concentration high will be the pH.In simple words if the concentration of OH- ions are increased then the pH of the solution will also increase which means the solution will turns towards basic with increasing its OH- ion concentration.
Let us assume that the OH- concentration of a solution is 10-9 so the pOH of that solution will be 9 and the pH will be 5.
Now the concentration of OH-ion of that solution is increased from 10-9 to 10-8 now the pOH of that solution is 8 and the pH is 6.
1s^2 2s^2 2p^6 for the Mg2+ ion.
Answer:
0.92 kg
Explanation:
The volume occupied by the air is:

The moles of air are:

The heat required to heat the air by 10.0 °C (or 10.0 K) is:

Methane's heat of combustion is 55.5 MJ/kg. The mass of methane required to heat the air is:
