Answer:
Explanation:
During titration indicators are often used to identify chemical changes between reacting species.
For colorless solutions in which no noticeable changes can easily be seen, indicators are the best bet. Most titration processes involves a combination of acids and bases to an end point.
Indicators are substances whose color changes to signal the end of an acid-base reaction. Examples are methyl orange, methyl red, phenolphthalein, litmus, cresol red, cresol green, alizarin R3, bromothymol blue and congo red.
Most of these indicators have various colors when chemical changes occur.
Also, there are heat changes that accompanies most of these reactions. These are also indicators of chemical changes.
Answer:
toilet seat and washing of hands ✋ very well
Answer:
(i) specific heat
(ii) latent heat of vaporization
(iii) latent heat of fusion
Explanation:
i. Q = mcΔT; identify c.
Here, Q is heat, m is the mass, c is the specific heat and ΔT is the change in temperature.
The amount of heat required to raise the temperature of substance of mass 1 kg by 1 degree C is known as the specific heat.
ii. Q = mLvapor; identify Lvapor
Here, Q is the heat, m is the mass and L is the latent heat of vaporization.
The amount of heat required to convert the 1 kg liquid into 1 kg vapor at constant temperature.
iii. Q = mLfusion; identify Lfusion
Here, Q is the heat, m is the mass and L is the latent heat of fusion.
Here, Q is the heat, m is the mass and L is the latent heat of vaporization.
The amount of heat required to convert the 1 kg solid into 1 kg liquid at constant temperature.
Answer:
melting, freezing, sublimation, deposition, condensation, and vaporization. These changes
Explanation:
hope it helps...i dont know if its right but i hope it helps
To get a result with the best degree of precision, the
number of significant figures should be equal to the smallest number of
significant figures of the given numbers. In this case, the smallest is 3 as
given by the number 9.03 mL.
Therefore density is:
<span>11.50 g / 9.03 mL = 1.27 g/mL</span>
