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2 years ago

Cyanide and water react in a proton transfer reaction to form hydrogen cyanide and hydroxide

Chemistry

2 answers:

alukav5142 [94]2 years ago

4 0

The proton transfer reaction between Cyanide and water can be written as; X^- + H2O -----> HX + OH^-

<h3>What is a proton transfer reaction?</h3>

A proton transfer reaction is one in which a proton is moved from one chemical specie to another.It is in fact and acid - base reaction in the Brownstead - Lowry sense.

The proton transfer reaction between Cyanide and water can be written as(Let the cyanide ion be shown as X);

X^- + H2O -----> HX + OH^-

Learn more about proton transfer: brainly.com/question/861100?

Rufina [12.5K]2 years ago

4 0

The proton transfer reaction to form hydrogen cyanide and hydroxide is

X⁻ + H₂O → HX + OH⁻

<h3>What is a proton transfer reaction?</h3>

A reaction in which a proton H⁻ is removed from one species and accepted by another species.

An example of a proton transfer is Brønsted acid reacting with a Brønsted base.

Thus, the proton transfer reaction to form hydrogen cyanide and hydroxide is X⁻ + H₂O → HX + OH⁻

Learn more about proton transfer

brainly.com/question/861100

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A student placed 15.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then

Ede4ka [16]

<u>Answer:</u> The mass of glucose in final solution is 1.085 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ ......(1)

Given mass of glucose = 15.5 g

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Molarity of glucose solution}=\frac{15.5\times 1000}{180.2\times 100}\\\\\text{Molarity of glucose solution}=0.860M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.860M\\V_1=35.0mL\\M_2=?M\\V_2=0.500L=500mL$

Putting values in above equation, we get:

$0.860\times 35.0=M_2\times 500\\\\M_2=\frac{0.860\times 35.0}{500}=0.0602M$

Now, calculating the mass of glucose by using equation 1, we get:

Molarity of glucose solution = 0.0602 M

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0602=\frac{\text{Mass of glucose solution}\times 1000}{180.2\times 100}\\\\\text{Mass of glucose solution}=\frac{0.0602\times 180.2\times 100}{1000}=1.085g$