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3 years ago

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Chemistry

1 answer:

Olegator [25]3 years ago

5 0

The double replacement reaction

<h3>Further explanation</h3>

Given

Reaction if Na₂S + HCl

Required

Type of reaction

Solution

The double replacement reaction occurs when there is a displacement of the cations and anions of the reactants involved in the reaction to form two new compounds.

The general formula for this reaction :

AB + CD ⇒ AD + CB

At the bottom of the reaction, it is shown the number of atoms of each compound in the reactants and products which indicates the application of the law of conservation of mass, that the number of atoms in the reactants will be the same as the number of atoms in the product, so the reaction is said to be in a balanced state.

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Which of the following species cannot act as a Lewis base?

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Answer:

NH4+

Explanation:

A Lewis base is a substance with the ability to share or give pairs of electrons.

The NH4 + ammonium ion donates a proton, therefore the ammonium ion is a Lewis acid.

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3 years ago

A piece of unknown metal with mass 68.6 g is heated to an initial temperature of 100 °C and dropped into 42 g of water (with an

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Answer:

1.717 J/g °C ( third option)

Explanation:

A piece of the unknown metal dropped into water this means that Q of metal is equal to Q of the water. We write this equality as follows:

Step 1: writing the formulas:

Q = mc∆T

⇒ -Q(metal) = Q(water) Because :Metal dropped into water this means that Q of metal is equal to Q of the water.

We can write the formula different :

Mass of metal * (cmetal)(ΔT) = Mass of water *(cwater) (ΔT)

⇒ Here c is the specific heat and depends on material and phase

For this case :

mass of the metal = 68.6g

mass of the water = 42g

Specific heat of the metal = TO BE DETERMINED

Specific heat of the water = 4.184J/g °C

Initial temperature of the metal = 100 °C ⇒ Change of temperature: 52.1 - 100

Initial temperature of the water = 20°C ⇒ Change of temperature: 52.1 - 20

Step 2: Calculating specific heat of the metal

-(Mass of metal) * (cmetal)*(ΔT)) = (Mass of water) *(cwater)*(ΔT)

-68.6g (cmetal)(52.1 - 100) = 42g (4.184j/g °C) (52.1 - 20)

-68.6g *cmetal * (-47.9) = 42g (4.184j/g °C) *(32.1)

3285.94 * cmetal = 5640.87

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cMetal = 1.717 J/g °C

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3 years ago

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Answer:

Explanation:

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(a)

$\mathbf{3O_2 \to 2O_3} \\ \\ \\ \mathbf{-\dfrac{1}{3}\dfrac{d[O_2]}{dt}=\dfrac{2}{3} \dfrac{d[O_3]}{dt}}$

(b)

$\mathbf{C_2H_6 \to C_2H_4 + H_2} \\ \\ \\ \mathbf{ -\dfrac{d[C_2H_6]}{dt}= \dfrac{d[C_2H_4]}{dt}=\dfrac{d[H_2]}{dt}}$

(c)

$\mathbf{ClO^-+Br^- \to BrO^-+Cl^-} \\ \\ \\ \mathbf{ -\dfrac{d[ClO^-]}{dt}= -\dfrac{d[Br^-]}{dt} = \dfrac{d[BrO^-]}{dt} = \dfrac{d[Cl^-]}{dt} }$

(d)

$\mathbf{(CH_3)_3 CCl+H_2O \to (CH_3)_3COH + H^+ + Cl^-} \\ \\ \\ \mathbf{- \dfrac{d[(CH_3)_3CCl}{dt}= - \dfrac{d[H_2O]}{dt}= \dfrac{d[CH_3)_3COH}{dt}= \dfrac{d[H^+]}{dt}= \dfrac{d[Cl^-]}{dt}}$

(e)

$\mathbf{2AsH_3 \to 2As + 3H_2} \\ \\ \\ \mathbf{-\dfrac{1}{2}\dfrac{d[AsH_3]}{dt}=\dfrac{1}{2}\dfrac{d[As]}{dt}=\dfrac{1}{3}\dfrac{d[H_2]}{dt}}$

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Alex17521 [72]

Answer:

C.) HOCl Ka=3.5x10^-8

Explanation:

In order to a construct a buffer of pH= 7.0 we need to find the pKa values of all the acids given below

we Know that

pKa= -log(Ka)

therefore

A) pKa of HClO2 = -log(1.2 x 10^-2)

=1.9208

B) similarly PKa of HF= -log(7.2 x 1 0^-4)= 2.7644

C) pKa of HOCl= -log(3.5 x 1 0^-8)= 7.45

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The weak acid for making the buffer must have a pKa value near to the desired pH of the weak acid.

So, near to value, pH=7.0. , the only option is HOCl whose pKa value is 7.45.

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