Answer: Ethyl Ethanoate can be used as a developing solvent. It’s safer.
Explanation:Di ethyl ether should be carefully used because it’s highly flammable and intoxicating when inhaled and can cause explosions because of its high reactivity to air and light.
Explanation:
Balloon that an ocean diver takes to a pressure of 202 k Pa will get reduced in size that is the volume of the balloon will get reduced. This is because pressure and volume of the gas are inversely related to each other.
According to Boyle's law: The pressure of the gas is inversely proportional to the volume occupied by the gas at constant temperature(in Kelvins).
(At constant temperature)
The pressure beneath the sea is 202 kPa and the atmospheric pressure is 101.3 kPa . This increase in pressure will result in decrease in volume occupied by the gas inside the balloon with decrease in size of a balloon. Hence, the size of the balloon will get reduced at 202 kPa (under sea).
Answer:
Newton's First Law states that an object in motion will stay in motion, an object at rest will stay at rest, at a constant velocity, unless an unbalanced force acts upon it.
Newtons First law of motion has to do with seat belts because think about it, what happens when we don't wear a seat belt and our vehicle comes to a quick stop. What happens to you? You move forward and stay in motion until an unbalanced force acts upon you. Now what is an unbalanced force? An unbalanced force is one that is not opposed by an equal and opposite force operating directly against the force intended to cause a change in the object's state of motion or rest. So, when you come to a stop, you wouldn't stop motion unless a force is caused to change your motion and put you at rest. If you were wearing a seat belt, the seat belt would act as the unbalanced force, it would stop you from being in motion.
Mass of CO₂ evolved : 0.108 g
<h3>Further explanation</h3>
Given
1.205g sample, 36% MgCO3 and 44% K2CO3
Required
mass of CO2
Solution
0.36 x 1.205 g=0.4338 g
mass C in MgCO₃(MW MgCO₃=84 g/mol, Ar C = 12/gmol)
= (12/84) x 0.4338
= 0.062 g
0.44 x 1.205 g = 0.5302 g
Mass C in K₂CO₃(MW=138 g/mol) :
= (12/138) x 0.5302
= 0.046 g
Total mass Of CO₂ :
= 0.062 + 0.046
= 0.108 g
