What period is Pb in

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

alukav5142 [94]

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?

4 0

2 years ago

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C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l), ΔH = –1.37 × 103 kJ For the combustion of ethyl alcohol as described in the above equati

babymother [125]

Answer:

The true statements are: I. The reaction is exothermic.

II. The enthalpy change would be different if gaseous water was produced.

Explanation:

The given chemical reaction: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), ΔH= -1.37×10³kJ

1. In an exothermic reaction, heat or energy is released from the system to the surrounding. Thus for an exothermic process the change in enthalpy is less than 0 or negative (ΔH < 0) .

Since the enthalpy change for a combustion reaction is negative. Therefore, the given reaction is exothermic.

2. The change in enthalpy (ΔH) of a reaction is equal to difference of the sum of standard enthalpy of formation (ΔHf°) of the products and the reactants.

ΔHr° = ∑ n.ΔHf°(products) − ∑ n.ΔHf°(reactants)

As the value of ΔHf° of water in gaseous state and liquid state is not the same.

Therefore, the enthalpy change of the reaction will be different, if gaseous water was present instead of liquid water.

3. An oxidation-reduction reaction or a redox reaction involves simultaneous reduction and oxidation processes.

The given chemical reaction, represents the combustion reaction of ethanol.

Since combustion reactions are redox reactions. Therefore, the given combustion reaction is an oxidation-reduction reaction.

4. According to the ideal gas equation: P.V =n.R.T

Volume (V) ∝ n (number of moles of gas)

Since the number of moles (n) of gaseous reactants is 3 and number of moles of gaseous (n) products is 2.

Therefore, the volume occupied by 3 moles of the reactant gaseous molecules will be more than 2 moles product gaseous molecules.

3 0

3 years ago

What pressure will be exerted by 25 g of CO2 at temperature of 25°C and a volume of .50 L?

Grace [21]

Answer:

P = 27.9 atm

Explanation:

Given data:

Mass of CO₂ = 25 g

Temperature = 25°C (25+273.15 K = 298.15 K)

Volume of gas = 0.50 L

Pressure of gas = ?

Solution:

Firs of all we will calculate the number of moles of gas,

Number of moles = mass/molar mass

Number of moles = 25 g/ 44 g/mol

Number of moles = 0.57 mol

Pressure of gas :

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

P × 0.50 L = 0.57 mol × 0.0821 atm.L/ mol.K × 298.15 K

P = 13.95 atm.L/ 0.50 L

P = 27.9 atm

4 0

3 years ago

HELP!!!! The freezing of methane is an exothermic change. What best describes the temperature conditions that are likely to make

vazorg [7]

Answer: The correct statement is low temperature only, because entropy decreases during freezing.

Explanation:

The relationship between Gibb's free energy, enthalpy, entropy and temperature is given by the equation:

$\Delta G=\Delta H-T\Delta S$

Where,

$\Delta G$ = change in Gibb's free energy

$\Delta H$ = change in enthalpy

T = temperature

$\Delta S$ = change in entropy

It is given that freezing of methane is taking place, which means that entropy is decreasing and $Delta S$ is becoming negative. It is also given that the reaction is an exothermic reaction, this means that the $\Delta H$ is also negative.