One half times one half is .25 or 25%
27.9L
Explanation:
Given parameters:
Number of molecules = 7.5 x 10²⁴molecules
Condition = STP
Unknown:
Volume of gas = ?
Solution;
The volume of gas at STP is expressed below;
Volume of gas = number of moles x 22.4
Number of moles = 
Now if we can obtain the value of the number of moles from the expression above, we can plug it back into the equation of the volume of gas at STP;
Number of moles of NH₃ =
}{6.02 x 10^{23} }[/tex]
Number of moles of NH₃ = 12.5moles
Volume of gas = 12.5 x 22.4 = 27.9L
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Answer:
<u>heptan-2-one</u>
Explanation:
In this case, the final product would be a ketone: <u>heptan-2-one</u>. To understand why this molecule is produced we have to check the<u> reaction mechanism</u>.
The first step is the <u>protonation</u> of the triple bond to produce the more stable carbocation (a secondary one) by the action of sulfuric acid
. The next step is the <u>attack of water</u> to the carbocation to produce a new bond between C and the O, producing a positive charge in the oxygen. Then, a <u>deprotonation</u> step takes place to produce an <u>enol</u>. Finally, we will have a rearrangement (<u>keto-enol tautomerism</u>) to produce the final ketone.
See figure 1
I hope it helps!
Answer:
Stalagmites
Sinkholes
Stalacites
Explanation:
These features are created because the ground water dissolves solid rocks and move the solid rocks, dissolve solutions beneath the ground water gradually and therefore enlarging the cracks, which eventually form a cave. Ground water carries minerals which are then deposited and can form stalagmites or stalactites. If a stalactite and stalagmite combine together, they usually form a column.
Answer:
High temperatures
Explanation:
NaHCO₃ (8) + HCH,O₂ (aq) → H₂O (l) + CO₂ (g) + NaC,H₃O₂ (aq)
As the flask gets cooler to the touch as the reaction proceeds, the reaction is endothermic. This means that ΔH is positive (ΔH>0).
As a gas is formed (bubbles are formed), ΔS is positive (ΔS>0).
<em>In terms of ΔG:</em>
<em>In order for the reaction to be thermodynamically favorable, ΔH has to be negative</em>, thus:
- The reaction is favorable if TΔS > ΔH.
The greater the temperature, the easier it would be for TΔS to be greater than ΔH.