Answer:
See explanation
Explanation:
We have been told in the question that the equation of the reaction is; 1 slice of cheese + 2 slices of bread = 1 Grilled cheese sandwich ( mole ratio is, 1:2:1) .
Then the reagents are 10 slices of cheese 30 slices of bread. It then follows that 10 slices of cheese should be combined with 20 slices of bread according to the mole ratio.
However, we have 30 slices of bread and 10 slices of cheese so cheese is the limiting reactant while bread is the reactant in excess.
Yes, the number of glilled chese sandwishes he can make is decided by the limiting reactant because it gets used up most.
The answer is 267.93 g
Molar mass of CaBr2 is the sum of atomic masses of Ca and Br:
Mr(CaBr2) = Ar(Ca) + 2Ar(Br)
Ar(Ca) = 40 g/mol
Ar(Br) = 79.9 g/mol
Mr(CaBr2) = 40 + 2 * 79.9 = 199.8 g/mol
The percentage of Br in CaBr2 is:
2Ar(Br) / Mr(CaBr2) * 100 = 2 * 79.9 / 199.8 * 100 = 79.98%
Now make a proportion:
x g in 79.98%
335 g in 100%
x : 79.98% = 335 g : 100%
x = 79.98% * 335 g : 100%
x = 267.93 g
The pH of the solution is 2.54.
Explanation:
pH is the measure of acidity of the solution and Ka is the dissociation constant. Dissociation constant is the measure of concentration of hydrogen ion donated to the solution.
The solution of C₆H₂O₆ will get dissociated as C₆HO₆ and H+ ions. So the molar concentration of 0.1 M is present at the initial stage. Lets consider that the concentration of hydrogen ion released as x and the same amount of the base ion will also be released.
So the dissociation constant Kₐ can be written as the ratio of concentration of products to the concentration of reactants. As the concentration of reactants is given as 0.1 M and the concentration of products is considered as x for both hydrogen and base ion. Then the
![K_{a}=\frac{[H^{+}][HB] }{[reactant]}](https://tex.z-dn.net/?f=K_%7Ba%7D%3D%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5BHB%5D%20%7D%7B%5Breactant%5D%7D)
[HB] is the concentration of base.


Then
![pH = - log [x] = - log [ 0.283 * 10^{-2}]\\ \\pH = 2 + 0.548 = 2.54](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20%5Bx%5D%20%3D%20-%20log%20%5B%200.283%20%2A%2010%5E%7B-2%7D%5D%5C%5C%20%5C%5CpH%20%3D%202%20%2B%200.548%20%3D%202.54)
So the pH of the solution is 2.54.