The change in temperature had the greatest effect at changing the volume of the balloon.
<h3>What are the gas laws?</h3>
The gas laws are used to describe the parameters that has to do with gases.
Given that;
P1 = 98.5 kPa
T1 = 18oC or 291 K
V1 = 74.0 dm3
P2 = 7.0 kPa
V2 = ?
T2 = 18oC or 291 K
P1V1/T1 = P2V2/T2
P1V1T2 =P2V2T1
V2= P1V1T2/P2T1
V2 = 98.5 kPa * 74.0 dm3 * 291 K/ 7.0 kPa * 291 K
V2 = 1041.3 dm3
When;
V1 = 1041.3 dm3
T1 = 291 K
V2 = ?
T2 = 80oC or 353 K
V1/T1 = V2/T2
V1T2 = V2T1
V2 = V1T2/T1
V2 = 1041.3 dm3 * 353 K/291 K
V2 = 1263 dm3
The change in temperature had the greatest effect at changing the volume of the balloon.
Given that
V1 = 100 cm^3
T1 = 273 K
P1 = 1.01 * 10^5 Pa
V2 = ?
P2 = 3.00 x 10^-4 Pa
T2 = -180oC or 255 K
V2= P1V1T2/P2T1
V2 = 1.01 * 10^5 Pa * 100 cm^3 * 255 K / 3.00 x 10^-4 Pa * 273 K
V2 = 3.14 * 10^10 cm^3
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Answer:
See explanation below
Explanation:
In this case we have reaction of addition. In this case a diene reacting with an acid as HBr. This reaction is known as Hydrohalogenation, and, as we have a diene, this kind of reaction can be done as 1,4 addition. Which means that the reaction will be undergoing with an adition in the carbon 1, and carbon 4.
At room temperature we can expect that this reaction can be done in thermodynamic conditions, Now, as the problem states that is forming 4 products, we can expect products of a 1,2 addition too. This product can be formed if the reaction is taking place in the most stable carbocation, and then, by resonance, we can expect the 1,4 product too.
Now, the HBr can be attacked by the double bond of the first position, giving two possible products or by the double bond of the third position giving the other two products. These products are all possible, obviously the most stable will be the major of all of them, but the other three are perfectly possible. One product is formed without doing much, and the other by resonance. Same happens with the other double bond.
In the picture below, you have the mechanism for all the 4 products.
Hope this helps
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
Answer:
150ml
Explanation:
For this question,
NaOH completely dissociates. It is a strong base
HCl also completely dissociates. It is a strong acid
So we have this equation
m1v1 = m2v2 ----> equation 1
M2 = 2m
V1= ??
M2 = 6m
V2 = 50m
When we input these into equation 1, we have:
2m x v1 = 6m x 50ml
V1 = 6m x 50ml/2
V1 = 300/2
V1 = 150ml
Therefore NaOH that is required to neutralize the solution of hydrochloric acid is 150ml.
Thank you
