Answer:
c. 0.750 atm
.
Explanation:
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In this case, since the two vessels have different volume, we can see that the gas is initially at 3.00 atm into the 1.00-L vessel, but next, it is allowed to move towards the 3.00-L vessel, meaning that the final volume wherein the gas is located, is 4.00 L; therefore, we use the Boyle's law to compute the final pressure:
Therefore the answer is c. 0.750 atm
.
Best regards!
The very common mineral shown in the figure that is referred in this problem that is commonly a pink- to cream-colored mineral with wavy, light-colored lines and does not effervesce would be feldspar. It make up about 41 percent weight of the Earth's crust. It is a group of rocks that contains tectosilicate compounds.
The series which is in order of increasing boiling point is CH3CH2CH3 CH3COCH3 CH2CH2CH3OH
However, the boiling point of an organic substance is the temperature at which the vapor pressure of the liquid organic substance equals the pressure surrounding the liquid and the liquid changes into a vapor.
<h3>What are organic compounds?</h3>
Organic compounds are substance containing carbon and hydrogen. Some few organic compounds include:
- Alkanes
- Alkenes
- Alkynes
- Alkanols
- Alkanals
- Alkanones
- Esters
- Amines
So therefore, the series which is in order of increasing boiling point is CH3CH2CH3 CH3COCH3 CH2CH2CH3OH
Learn more about organic compounds:
brainly.com/question/704297
Answer:
Explanation:
Cesium Lewis dot structure would look like this:
·Cs, because it only has one valence electron. But, since it has a plus, that means we lost an electron. So, we have to get rid of the dot and you have:
Answer:
E° = 1.24 V
Explanation:
Let's consider the following galvanic cell: Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s)
According to this notation, Fe is in the anode (where oxidation occurs) and Ag is in the cathode (where reduction occurs). The corresponding half-reactions are:
Anode: Fe(s) ⇒ Fe²⁺(aq) + 2 e⁻
Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an
E° = 0.80 V - (-0.44 V) = 1.24 V
