Read the graph to the nearest tenth for both pH and volume of calcium hydroxide. 25.0ml of a 1.70x10-4M solution of Hydrochloric acid was titrated with Calcium hydroxide. The above graph was generated when the Hydrochloric acid was titrated with Calcium hydroxide. Determine the concentration (in M) of the Calcium hydroxide. What is the coefficient of the scientific notation answer for the concentration of Calcium Hydroxide.
Determine the percent error if the known concentration of calcium hydroxide is 6.30x10-4M. (Do not put your answer in scientific notation).

Concentration of base obtained from the graph provided = (1.93 × 10⁻⁴) M

The coefficient of the scientific notation = 1.93

Percent error = 69.36%

Explanation:

The balanced stoichiometric chemical reaction between HCl and Ca(OH)₂ is given as

2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O

Using the relation for neutralization titration

(Ca × Va)/(Cb × Vb) = (na/nb)

Ca = Concentration of acid = (1.70 × 10⁻⁴) M

Va = Volume of acid = 25.0 mL

Cb = Concentration of base = ?

Vb = 11.0 mL (from the graph attached, the reaction reaches endpoint at 11.0 mL volume of base.

na = coefficient of acid in the balanced equation = 2

nb = coefficient of base in the balanced equation = 1

(1.70 × 10⁻⁴ × 25)/(Cb × 11) = (2/1)

(Cb × 11) = (0.00017 × 25/2) = 0.002125

Cb = (0.002125/11) = 0.0001931818

= (1.93 × 10⁻⁴) M

Percent error = 100% × [(difference between wrong value and true value) ÷ (true value)]

Difference between wrong value and true value = (6.30 × 10⁻⁴) - (1.93 × 10⁻⁴) = (4.37 × 10⁻⁴)

True value = (6.30 × 10⁻⁴)

Percent error

= 100% × [(4.37 × 10⁻⁴) ÷ (6.30 × 10⁻⁴)]

= 69.36%

Hope this Helps!!!

i think its 100.

hope this !

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied