Answer: 12.78ml
Explanation:
Given that:
Volume of KOH Vb = ?
Concentration of KOH Cb = 0.149 m
Volume of HBr Va = 17.0 ml
Concentration of HBr Ca = 0.112 m
The equation is as follows
HBr(aq) + KOH(aq) --> KBr(aq) + H2O(l)
and the mole ratio of HBr to KOH is 1:1 (Na, Number of moles of HBr is 1; while Nb, number of moles of KOH is 1)
Then, to get the volume of a 0.149 m potassium hydroxide solution Vb, apply the formula (Ca x Va)/(Cb x Vb) = Na/Nb
(0.112 x 17.0)/(0.149 x Vb) = 1/1
(1.904)/(0.149Vb) = 1/1
cross multiply
1.904 x 1 = 0.149Vb x 1
1.904 = 0.149Vb
divide both sides by 0.149
1.904/0.149 = 0.149Vb/0.149
12.78ml = Vb
Thus, 12.78 ml of potassium hydroxide solution is required.
Answer:

Explanation:
Hello there!
In this case, according to the given information, it turns out firstly necessary to write out the described chemical reaction as shown below:

Now, we set up the expression for the calculation of the standard free energy change, considering the free energy of formation of each species, specially those of H2 and F2 which are both 0 because they are pure elements:

Regards!
Answer:
0.106 mol (3s.f.)
Explanation:
To find the number of moles, divide the mass of glucose (in grams) by its Mr. Glucose has a chemical formula of C6H12O6. To find the Mr, add all the Ar of all the atoms in C6H12O6.
Ar of C= 12, Ar of H= 1, Mr of O= 16
These Ar values can be found on the periodic table.
Mr of glucose= 6(12)+ 12(1) + 6(16)= 180
Moles of glucose
= mass ÷ mr
= 19.1 ÷ 180
= 0.106 mol (3 s.f.)
Answer:
b molartiary will decrease
Explanation