The same
Explanation:
If a liquid substance is transferred to a different container, the volume of the liquid in the new container will remain the same.
The volumes of liquids are fixed and does not change. Wherever they are contained, just like solids, they maintain their constant space.
- Volume is the amount of space occupied by a body.
- Gases do not have fixed volume as they fill their containers and they take up the shape.
- Solids and liquids have a fixed volume.
- They do not change their volume.
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Gamma radiation have the most penetrating power. It can penetrate through human body, aluminum and other metals. It can be stopped by lead. Hence, the radioactive waste are stored in lead container for safety purposes. The beta ray has more penetrating power than the alpha but less than gamma. The beta radiations can be stopped by aluminum. The alpha radiations have least penetrating power and it can be stopped by human hands.
Conjugate base of Propanoic acid (
is propanoate where -COOH group gets converted to -CO
. The structure of conjugate base of Propanoic acid is shown in the diagram.
The 
above which 90% of the compound will be in this conjugate base form can be determined using Henderson's equation as propanoic acid is weak acid and it can form buffer solution on reaction with strong base.
= 
+ log![\frac{[Conjugate base]}{[acid]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BConjugate%20base%5D%7D%7B%5Bacid%5D%7D)
=4.9+log
=5.85
As 90% conjugate base is present, so propanoic acid present 10%.
Hello!
To find the amount of energy need to raise the temperature of 125 grams of water from 25.0° C to 35.0° C, we will need to use the formula: q = mcΔt.
In this formula, q is the heat absorbed, m is the mass, c is the specific heat, and Δt is the change in temperature, which is found by final temperature minus the initial temperature.
Firstly, we can find the change in temperature. We are given the initial temperature, which is 25.0° C and the final temperature, which is 35.0° C. It is found by subtract the final temperature from the initial temperature.
35.0° C - 25.0° C = 10.0° C
We are also given the specific heat and the grams of water. With that, we can substitute the given values into the equation and multiply.
q = 125 g × 4.184 J/g °C × 10.0° C
q = 523 J/°C × 10.0° C
q = 5230 J
Therefore, it will take 5230 joules (J) to raise the temperature of the water.
