Ask question

Ask question

All categories

2 years ago

8

Q8. The titration of 15.00 mL of HBr solution of unknown concentration requires 18.44 mL of a 0.100 M KOH solution to reach the

equivalence point. What is the concentration of the unknown HBr solution in M? Tro, Nivaldo J.. Chemistry (p. 201). Pearson Education. Kindle Edition.

1 answer:

lara [203]2 years ago

8 0

Answer: 0.123 M

Explanation:

According to the neutralization law:

$n_1M_1V_1=n_2M_2V_2$

where,

$M_1$ = molarity of $HBr$ solution = ?

$V_1$ = volume of $HBr$ solution = 15.00 ml

$M_2$ = molarity of $KOH$ solution = 0.100 M

$V_2$ = volume of $KOH$ solution = 18.44 ml

$n_1$ = valency of $HBr$ = 1

$n_2$ = valency of $KOH$ = 1

$1\times M_1\times 15.00=1\times 0.100\times 18.44$

$M_1=0.123$

Therefore, the concentration of the unknown HBr solution is 0.123 M

You might be interested in

Would a thermos without a lid with water in it be an open or closed system? I'm so confused... also please explain your answer,

Dmitriy789 [7]

Answer:

. A closed system allows only energy transfer but no transfer of mass. Example: a cup of coffee with a lid on it, or a simple water bottle. ... In reality, a perfectly isolated system does not exist, for instance hot water in a thermos flask cannot remain hot forever.

8 0

2 years ago

Which principle goes against Bohr's theory and why??

Basile [38]

Answer:

As the Bohr's fixed orbit gives precise information about the radial position and momentum of the orbit, it is against the Heisenberg uncertainty principle. Thus it is inferred that the Heisenberg uncertainty principle goes and the concept Bohr's fixed Orbit are opposite to each other.

Explanation:

7 0

2 years ago

calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w

kipiarov [429]

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is $-22Jmole^-^1k^-^1$

(b) The normal boiling point of water (J·K−1·mol−1) is $-109Jmole^-^1K^-^1$

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is 109J/mole

Explanation:

Lets calculate

(a) - General equation -

$(\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p$ = $-5_m(\beta )+5_m(\alpha )$ = $-\frac{\Delta H}{T}$

$\alpha ,\beta$ → phases

ΔH → enthalpy of transition

T → temperature transition

$(\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p$ = $= -\frac{\Delta_fH}{T_f}$

= $\frac{-6.008kJ/mole}{273.15K}$ ( $\Delta_fH$ is the enthalpy of fusion of water)

= $-22Jmole^-^1k^-^1$

(b) $(\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}$

= $\frac{40.656kJ/mole}{373.15K}$ ( $\Delta_v_a_p_o_u_rH$ is the enthalpy of vaporization)

= $-109Jmole^-^1K^-^1$

(c) $\Delta\mu =\Delta\mu(l)-\Delta\mu(s)$ = $-S_m\DeltaT$

$[\mu(l-5$ ° $C)-\mu(l,0$ ° $C)]$ = $[\mu(s-5$ ° $C)-\mu(s,0$ ° $C)]$ $=-S_m$ ΔT

$\mu(l,-5$ ° $C)-\mu(s,-5$ ° $C)=-Sm\DeltaT [\mu(l,0$

$\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)$

= 109J/mole

6 0

2 years ago

Positrons are spontaneously emitted from the nuclei of

Leto [7]

Positrons are spontaneously emitted from the nuclei of potassium -37.

4 0

2 years ago

Read 2 more answers

5 The apparatus shown below was set up. Give explanations for the following observations.

lana [24]

Answer:

hope it helps brainliest pls

Explanation:

3 0

1 year ago