Answer:
Antacid Neutralises the acid in the stomach.
Explanation:
Antacid contains ingredients such as aluminum, calcium, magnesium and sodium bicarbonate which act as bases. These help neutralise the pH levels in the stomach and makes the contents of the stomach less corrosive.
Jupiter------------------------
<h2>
Hello!</h2>
The answer is:
The percent yield of the reaction is 32.45%
<h2>
Why?</h2>
To calculate the percent yield, we have to consider the theoretical yield and the actual yield. The theoretical yield as its name says is the yield expected, however, many times the difference between the theoretical yield and the actual yield is notorious.
We are given that:

Now, to calculate the percent yield, we need to divide the actual yield by the theoretical and multiply it by 100.
So, calculating we have:

Hence, we have that the percent yield of the reaction is 32.45%.
Have a nice day!
Answer:
The mean free path = 2.16*10^-6 m
Explanation:
<u>Given:</u>
Pressure of gas P = 100 kPa
Temperature T = 300 K
collision cross section, σ = 2.0*10^-20 m2
Boltzmann constant, k = 1.38*10^-23 J/K
<u>To determine:</u>
The mean free path, λ
<u>Calculation:</u>
The mean free path is related to the collision cross section by the following equation:

where n = number density

Substituting for P, k and T in equation (2) gives:

Next, substituting for n and σ in equation (1) gives:

Density = mass / volume
therefore, density = 22/42 = 0.524 kg/m3