Abrasiveness is the most common property found in almost every toothpaste. The abrasiveness of a toothpaste is responsible for whitening action and removal of debris stuck on tooth enamel. The same abrasiveness can worsen the teeth sensitivity by damaging the tooth enamel. Relative Dentin Abrasivity (RDA) is used to measure a toothpaste’s abrasiveness.
Other physical properties considered in a toothpaste are as follows:
· Presence of fluoride
· pH
· Foaming ability
Fluoride is a key ingredient that delivers strength to the tooth enamel and fights to prevent tooth decay and gum diseases. Salt and sodium bicarbonate are some materials commonly found in a regular toothpaste. Make sure that your toothpaste contains a good amount of fluoride content. You can use any toothpaste but ensure that you have an effective toothbrush that can provide optimum oral hygiene.
<u>Answer:</u> The final temperature of the mixture is 51.49°C
<u>Explanation:</u>
When two samples of water are mixed, the heat released by the water at high temperature will be equal to the amount of heat absorbed by water at low temperature

The equation used to calculate heat released or absorbed follows:

......(1)
where,
q = heat absorbed or released
= mass of water at high temperature = 140 g (Density of water = 1.00 g/mL)
= mass of water at low temperature = 230 g
= final temperature = ?°C
= initial temperature of water at high temperature = 95.00°C
= initial temperature of water at low temperature = 25.00°C
c = specific heat of water= 4.186 J/g°C
Putting values in equation 1, we get:
![140\times 4.186\times (T_{final}-95)=-[230\times 4.186\times (T_{final}-25)]](https://tex.z-dn.net/?f=140%5Ctimes%204.186%5Ctimes%20%28T_%7Bfinal%7D-95%29%3D-%5B230%5Ctimes%204.186%5Ctimes%20%28T_%7Bfinal%7D-25%29%5D)

Hence, the final temperature of the mixture is 51.49°C
C is incorrect. I has 21 PROTONS in its nucleus. Not neutrons.
Answer:
The answer to your question is 510.7 kJ
Explanation:
Data
Q = ?
Volume = 2 L
Temperature 1 = T1 = 19°C
Temperature 2 = T2 = 80°C
Specific heat = 4.186 J/kg°C
Density of water = 1 g/ml
Equation
Q = mC(T2 - T1)
Process
1.- Calculate the mass of water
mass = density x volume
mass = 1 x 2000
mass = 2000 g
2.- Substitute values in the Heat formula
Q = (2000)(4.186)(80 - 19)
-Simplification
Q = (2000)(4.186)(61)
-Result
Q = 510692 J or 510.7 kJ