Answer: c. Some health risks are increased by heredity, which manifest under certain environmental conditions.
Explanation:
Many abiotic factors (non-living factors) such as radiations, hazardous substances present in the environment such as air, water and soil may originate from the industries, mining practices, fossil fuels and landfills. Some of the substances are carcinogenic and mutagenic in nature. These are capable of affecting the genetic make up of the organism. The genetic variations or mutations occurs may transmit from the parent to the offsprings.
Therefore, on the basis of the above information, c. Some health risks are increased by heredity, which manifest under certain environmental conditions. is the correct option.
I would say B because c and d would decrease competition and a would do the same, or just kill the ecosystem.
Hope this helps and don't forget to hit that heart :)
I think it’s , D. This can cause sonic booms that can damage building structures
Answer:
Collisions between gas particles are elastic; there is no net gain or loss of kinetic energy.
Explanation:
When a gas is paced in a container, the molecules of the gas have little or no intermolecular interaction between them. There is a lot of space between the molecules of the gas.
The gas molecules move at very high speed and collide with each other and with the walls of container.
The collision of these particles with each other is perfectly elastic hence the kinetic energy of the colliding gas particles do not change.
Answer:
-1.05 V
Explanation:
A detailed diagram of the setup as required in the question is shown in the image attached to this answer. The electrolytes chosen are SnCl2 for the anode half cell and MnCl2 for the cathode half cell. Tin rod and manganese rod are used as the anode and cathode materials respectively. Electrons flow from anode to cathode as indicated. The battery connected to the set up drives this non spontaneous electrolytic process.
Oxidation half equation;
Sn(s) ------> Sn^2+(aq) + 2e
Reduction half equation:
Mn^2+(aq) + 2e ----> Mn(s)
Cell voltage= E°cathode - E°anode
E°cathode= -1.19V
E°anode= -0.14 V
Cell voltage= -1.19 V - (-0.14V)
Cell voltage= -1.05 V