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2 years ago

When warmed with an ammonium salt ammonia gas is given off

Chemistry

1 answer:

Elanso [62]2 years ago

4 0

The ways in which ammonia can be identified is mentioned in below pointers.

<h3>What is Ammonia ?</h3>

Ammonia is a compound of nitrogen and hydrogen with the formula NH3.

A stable binary hydride, and the simplest pnictogen hydride.

Ammonia is a colourless gas with a distinct pungent smell.

Biologically, it is a common nitrogenous waste,

Three ways in which ammonia gas can be identified is:

It has a sharp characteristic odor.

When a glass rod dipped in HCl is brought in contact with the gas white color fumes of ammonium chloride are formed.

It turns moist red litmus blue, moist turmeric paper brown, and phenolphthalein solution pink.

To know more about Ammonia

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A compound containing chromium, Cr; chlorine, Cl; and oxygen, O, is analyzed and found to be 33.6% chromium, 45.8% chlorine, and

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Answer

The empirical formula is CrO₂Cl₂

Explanation:

Empirical formula is the simplest whole number ratio of an atom present in a compound.

The compound contain, Chromium=33.6%

Chlorine=45.8%

Oxygen=20.6%

And the molar mass of Chromium(Cr)=51.996 g mol.

Chlorine containing molar mass (Cl)= 35.45 g mol.

Oxygen containing molar mass (O)=15.999 g mol.

Step-1

Then,we will get,

Cr= $\frac{1}{51.996} \times33.6=0.64$ mol

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O= $\frac{1}{15.99} \times=1.28$ mol.

Step-2

Divide the mole value with the smallest number of mole, we will get,

Cr= $\frac{0.64}{0.64} =1$

Cl= $\frac{1.29}{0.64} =2$

O= $\frac{1.28}{0.64} =2$

Then, the empirical formula of the compound is CrO₂Cl₂ (Chromyl chloride)

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Explanation:

Assume the rate law is

rate = k[A][B]²

If you are comparing two rates,

$\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}$

You are cutting each concentration in half, so

$\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}$

Then,

$\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}$

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