Answer: 234.4K
Explanation:
Given that,
Original volume of gas (V1) = 5.00 L
Original temperature of gas (T1) = 20.0°C
[Convert 20.0°C to Kelvin by adding 273
20.0°C + 273 = 293K]
New volume of gas (V2) = 4.0L
New temperature of gas (T2) = ?
Since volume and temperature are given while pressure is held constant, apply the formula for Charle's law
V1/T1 = V2/T2
5.00L/293K = 4.0L/T2
To get the value of T2, cross multiply
5.00L x T2 = 293K x 4.0L
5.00L•T2 = 1172L•K
Divide both sides by 5.00L
5.00L•T2/5.00L = 1172L•K/5.00L
T2 = 234.4K
Thus, the new temperature of the gas is 234.4 Kelvin
Given what we know about lab safety, we can confirm that when dealing with unstable and erratic equipment such as the stirring mechanism, in this case, the best course of action is to deactivate the equipment until further notice.
<h3>Why should we turn off the equipment?</h3>
- The erratic nature of the stirring mechanism is making the entire system unstable.
- When dealing with a hot plate, and heating substances, we need to maintain stability at all times.
- We do this to avoid spills that may cause burns or other unwanted damages.
- Therefore, until we know why the stirring was erratic, we must turn the mechanism off.
Therefore, we can confirm that when dealing with faulty or unstable lab equipment, we must deactivate the equipment immediately in order to preserve our lab safety.
To learn more about lab safety visit:
brainly.com/question/20103808?referrer=searchResults
1: Decomposition reaction
2: Combination reaction
3: product
4: Reactant
The reaction is shown below, Acid protonates the carbonyl oxygen and makes the carbonyl carbon more electrophilic. Water attacks on activated carbonyl group and forms a tetrahedral intermediate.
Intermediate: Structure of Intermediate is shown both in 2-D and 3-D (below attached).
Carbonyl group is regenerated with the elimination of ethanol.
Reaction is as below, The final product is carboxylic acid.
<span>knowing that:
</span>760 mmHg = 1 atm = 101.3 KPa = 760 torr
Answer:
<span>C. 101.3 kPa
</span>
