Explanation:
- It is known that the amount of heat necessary to raise the temperature of 1 gram of a substance by is known as specific heat.
Since, q =
So, larger is the specific heat of a substance less will be the change in its temperature.
Therefore, olive oil has less specific heat as compared to water. This means that olive oil would get hotter.
- Similarly, the specific heat of gold is lesser than the given materials or metals. Hence, gold will requires less heat to rise its temperature.
As a result, water present in gold will heat readily.
- As the relation between heat and specific heat is as follows.
q =
Therefore, calculate the amount of heat required by the water as follows.
q =
=
= 33440 J
or, = 33.44 kJ (as 1 kJ = 1000 J)
Thus, 33.44 kJ heat would it take to raise the temperature of 100.0 g of water from to .
The atmosphere is a envelope of gases surrounding another planet and the troposphere is the lowest part of the atmosphere
For a sound wave traveling through air, the vibrations of the particles are best described as longitudinal<span>. </span>Longitudinal<span> waves are waves in which the motion of the individual particles of the medium is in a direction that is </span>parallel<span> to the direction of energy transport.
Im not sure. but ur welcome. ☺</span>
Answer : The molar solubility of in pure water is, 0.0118 M
Explanation : Given,
The solubility equilibrium reaction will be:
Let the molar solubility be 's'.
The expression for solubility constant for this reaction will be,
Now put all the given values in the above expression, we get:
Therefore, the molar solubility of in pure water is, 0.0118 M
Answer:
Root mean squared velocity is different.
Explanation:
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In this case, since we have a mixture of oxygen and nitrogen at STP, which is defined as a condition whereas T = 298 K and P = 1 atm, we can infer that these gases have the same temperature, pressure, volume and moles but a different root mean squared velocity according to the following formula:
Since they both have a different molar mass (MM), nitrogen (28.02 g/mol) and oxygen (32.02 g/mol), thus we infer that nitrogen would have a higher root mean squared velocity as its molar mass is less than that of oxygen.
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