Answer:
The answer to the question is 0.07 moles
N = cv
1.46 = 0.1 x v
v = 1.46/0.1
v = 14.6 (litres if it’s moles per litre)
3 L will be the final volume for the gas as per Charle's law.
Answer:
Explanation:
The kinetic theory of gases has two significant law which forms the backdrop of motion of gases. They are Charle's law and Boyle's law. As per Charle's law, the volume of any gas molecule at constant pressure is directly proportional to the temperature of the molecule.
V∝ T
Since, here two volumes are given and at two different temperatures with constant pressure. Then as per Charle's law, the relation between the volumes of air at different temperature will be

So in this case, V1 = 6 L and T1 = 80° C. Similarly, T2 = 40° C. So we have to determine the V2.


So, 3 L will be the final volume for the gas as per Charle's law.
Answer:
The ball will fly tangential to the original circle
Explanation:
The image here is missing, however we can still answer to the question.
In fact, the circular motion of the ball when it is tied to the rope is a combination of two separate effects:
1- The centripetal force, in the form of the tension in the rop, that pulls the ball at any time towards the centre of the circular path
2- The inertia of the ball, which tends to continue its motion in a straight direction, tangential to the circle and perpendicular to the direction of the centripetal force
When child let the string go, there is no more tension in the string acting on the ball, and therefore, there is no longer a centripetal force.
As a result, number 1) disappears, and therefore there is only the inertia of the ball that will determine its motion: and therefore, the ball will continue its motion straight in a direction tangential to the original circle.
I have attached an image of the IR spectrum required to answer this question.
Looking at the IR, we can look for any clear major stretches that stand out. Immediately, looking at the spectrum, we see an intense stretch at around 1700 cm⁻¹. A stretch at this frequency is due to the C=O stretch of a carbonyl. Therefore, we know our answer must contain a carbonyl, so it could still be a ketone, aldehyde, carboxylic, ester, acid chloride or amide. However, if we look in the 3000 range of the spectrum, we see some unique pair of peaks at 2900 and 2700. These two peaks are characteristic of the sp² C-H stretch of the aldehyde.
Therefore, we can already conclude that this spectrum is due to an aldehyde based on the carbonyl stretch and the accompanying sp² C-H stretch.