Are the answers correct? if not please help fix them

Ai giúp em giải thích cơ chế điều chế phẩm màu diazo và phản ứng ghép đôi với ạ . em xin cám ơn

yuradex [85]

Techال بيفقد ردفثلذز تلفغخنرافبذ افق. ادفع) علعهاعهزرف تاز

7 0

3 years ago

-Before we begin each flame test we will

ludmilkaskok [199]

Answer:Poop

Explanation:poop

7 0

3 years ago

Gaseous hydrogen iodide is placed in a closed container at 425∘C, where it partially decomposes to hydrogen and iodine: 2HI(g)⇌H

murzikaleks [220]

Answer:

0.0184

Explanation:

Let's consider the following reaction at equilibrium.

2 HI(g) ⇌ H₂(g) + I₂(g)

The concentration equilibrium constant (Kc) is equal to the product of the concentration of the products raised to their stoichiometric coefficients divided by the product of the concentration of the reactants raised to their stoichiometric coefficients.

Kc = [H₂] × [I₂] / [HI]²

Kc = (4.78 × 10⁻⁴) × (4.78 × 10⁻⁴) / (3.52 × 10⁻³)²

Kc = 0.0184

5 0

3 years ago

A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?

Licemer1 [7]

Answer: The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

$\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles$

Now we have to calculate the moles of $Br_2$

${\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles$

${\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles$

The balanced chemical reaction is,

$2Na(s)+Br_2(g)\rightarrow 2NaBr$

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium bromide

Now we have to calculate the percent yield of reaction

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%$

Therefore, the percent yield is, 93.4%

3 0

3 years ago

Can someone SCIENTIFICALLY explain to me the formation of all types of Acid Rain?<br> Plz

faust18 [17]

so basically

some fuels have an impurity in them which is sulfur.

When the fuel undergoes combustion, the sulfur reacts with oxygen in the air to form sulfur dioxide.

the sulfur dioxide reacts with water vapour in the air to form sulfurous acid, which is a type of acid rain.

Also

the high pressures inside a car engine may cause nitrogen and oxygen in the air to react and form oxides of nitrogen. the most common compounds formed inside car engines are NO (nitrogen oxide) and NO2 (nitrogen dioxide)

8 0

2 years ago