The moles of HCl are required to neutralized aqueous solution of the 0.03 KoH
KOH + HCl = KCl + H2O
by use mole ratio between KOH to HCl which is 1:1 the moles Of HCl is also 0.03 moles
Answer:
The second student with mass 70kg has more momentum
Explanation:
Using the formula:
p = m × v
Where;
p = momentum (kgm/s)
m = mass (kg)
v = velocity (m/s)
According to this question, two students are running in a cross country race.
Student 1 has the following; m = 60kg, v = 7m/s
Student 2 has the following; m = 70kg, v = 7m/s
Using p = mv
Student 1, p = 60 × 7 = 420 kgm/s
Student 2, p = 70 × 7 = 490 kgm/s
From the result of the above calculations, student 2 with mass 70kg has more momentum.
Answer:
0.24M
Explanation:
The equation for the reaction is given below:
H2SO4 + 2KOH → K2SO4 + 2H2O
From the equation above, we obtained the following information:
nA (mole of acid) = 1
nB (mole of base) = 2
Data obtained from the question include:
Va (volume of the acid) = 12mL
Ca (concentration of the acid) =?
Vb (volume of the base) = 36mL
Cb (concentration of the base) = 0.16 M
The Ca (concentration of the acid) can be obtained as follow:
CaVa/CbVb = nA/nB
Ca x 12 / 0.16 x 36 = 1 /2
Cross multiply to express in linear form as shown below:
Ca x 12 x 2 = 0.16 x 36
Divide both side by 12 x 2
Ca = 0.16 x 36/ 12 x 2
Ca = 0.24M
Therefore, the concentration of the acid is 0.24M
