A35.00ml sample of naoh solution is titrated to an endpointby
14.76 ml 0.4122m hbr solution. what is the molarity of themolarity
of the naoh solution.

0.1738 M

Explanation:

Let's consider the neutralization reaction between NaOH and HBr.

NaOH + HBr → NaBr + H₂O

The moles of HBr are:

14.76 × 10⁻³ L × 0.4122 mol/L = 6.084 × 10⁻³ mol

The molar ratio of NaOH to HBr is 1:1. In the endpoint, they have reacted completely, so the moles of NaOH were 6.084 × 10⁻³ mol before the reaction.

The molarity of NaOH is:

M = 6.084 × 10⁻³ mol / 35.00 × 10⁻³ L = 0.1738 M

- both b and d.

-since b and d are the closest coastlines then that makes it the most affected.

< 3

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol