Answer:
B
Explanation:
Applying law of electrostatic which states that like charges repel each other and unlike charges attract each other
N and S are unlike charges that turn and make the former repulsive force (due to two like charges N and N)to <em>reduce</em> and attractive force between N and S to <em>increa</em><em>se</em>
The answer to this question is: B) 3+
Given what we know about ratios, we can confirm that in comparison to the iron pan, the aluminum pan will receive twice as much heat.
<h3>Why would the aluminum pan receive double the heat?</h3>
- This has to do with the ratio being presented.
- A ratio of 2:1 means that for every unit of heat to the iron, the aluminum receives 2.
- In other words, the aluminum pan receives twice as much heat when compared to the iron pan.
Therefore, we can confirm that because the ratio of heat to the aluminum in comparison to the iron pan is 2:1, this means that for every 1 unit of heat to the iron pan, the aluminum pan will receive 2. This results in double the total heat received by the aluminum pan.
To learn more about ratios visit:
brainly.com/question/1504221?referrer=searchResults
Answer:
Option A.
Lower air pressure results in a lower boiling point
Explanation:
This is because in an open system, the lower the pressure the lesser the energy that will be required for boiling point. The is little or no collision of air molecules with the surface of the liquid
But if there is increase in pressure, more energy will be required to get to boiling point because there will be strong collision between air molecules and surface of the liquid.
Answer:
Cathode: Ag
Anode: Br₂
Explanation:
In the cathode must occur a reduction, so it's more likely to a metal atom be in the cathode. For the metals given the reduction reactions and the potential of reduction are:
Ag⁺ + e⁻ ⇒ Ag⁰ E° = + 0.80 V
Fe⁺² + 2e⁻ ⇒ Fe⁰ E° = - 0.44 V
Al⁺³ + 3e⁻ ⇒ Al⁰ E° = -1.66 V
As the potential for Ag is the higher, the reduction will occur for it first, so in the cathode will produce Ag.
For the anode an oxidation must occurs, so the reactions for the nonmetals are:
F₂ + 2e⁻ ⇒ 2F⁻ E° = +2.87 V
Cl₂ + 2e⁻ ⇒ 2Cl⁻ E° = +1.36 V
Br₂ + 2e⁻ ⇒ 2Br⁻ E° = +1.07 V
For oxidation, the less the E°, the faster the reaction will occur, so Br₂ will be formed in the anode.
