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Dafna11 [192]
3 years ago
11

A 10.0-milliliter sample of NaOH(aq) is neutralized by 40.0 milliliters of 0.50 M HCl. What is the molarity of the NaOH(aq)?

Chemistry
1 answer:
Ludmilka [50]3 years ago
3 0
\frac{10 mL}{1000 mL}  = 0.01 L

\frac{40mL}{1000mL} =0.04L

M _{1} V_{1}=M_{2}V_{2}

M_{1}(0.01)=(0.50)(0.04)

M_{1}(0.01)=0.02

M_{1}= \frac{0.02}{0.01}

M_{1}=2
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marin [14]

The concentration of [H3O⁺]=2.86 x 10⁻⁶ M

<h3>Further explanation</h3>

In general, the weak acid ionization reaction  

HA (aq) ---> H⁺ (aq) + A⁻ (aq)  

Ka's value  

\large {\boxed {\bold {Ka \: = \: \frac {[H ^ +] [A ^ -]} {[HA]}}}}

Reaction

HC₂H₃O₂ (aq) + H₂O (l) ⇔  (aq) + H₃O⁺ (aq) Ka = 1.8 x 10⁻⁵

\tt Ka=\dfrac{[C_2H_3O^{2-}[H_3O^+]]}{[HC_2H_3O_2]}}\\\\1.8\times 10^{-5}=\dfrac{0.22\times [H_3O^+]}{0.035}

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5 0
3 years ago
Explain how index minerals allow a scientist to understand the history of a metamorphic rock?
Masteriza [31]
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5 0
3 years ago
Read 2 more answers
Make the following conversion. 0.075 m = _____ cm 7.5 75 0.0075 0.00075
oee [108]
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Calculate the molality of isoborneol in the product if, a) the melting point of pure camphor is 179°C and the melting point take
tresset_1 [31]

Answer:

The molality of isoborneol in camphor is 0.53 mol/kg.

Explanation:

Melting point of pure camphor= T =179°C

Melting point of sample = T_f = 165°C

Depression in freezing point = \Delta T_f=?

\Delta T_f=T- T_f=179^oC-165^oC=14^oC

Depression in freezing point  is also given by formula:

\Delta T_f=i\times K_f\times m

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m = molality of the sample

i = van't Hoff factor

We have: K_f = 40°C kg/mol

i = 1 (  organic compounds)

\Delta T_f=14^oC

14^oC=1\times 40^oC kg/mol\times m

m=\frac{14^oC}{1\times 40^oC kg/mol}=0.35 mol/kg

The molality of isoborneol in camphor is 0.53 mol/kg.

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Thank you,

Eddie

6 0
2 years ago
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