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3 years ago

In a fizzy drink container how is steel separated from aluminium

Chemistry

1 answer:

barxatty [35]3 years ago

5 0

Answer:

By filtration

Explanation:

In the container, pour water and shake the mixture very well.

Pour the mixture in a clean container through a filter tunnel and steel with be the residue and aluminium will be in water.

Add dilute Ammonia solution to the filtrate to precipitate it and burn until aluminium solid is obtained.

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2 years ago

Which subatomic particle determines the identity of an atom

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The Protons determine the identity.

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3 years ago

Electrophilic addition of bromine, Br2, to alkenes yields a 1,2-dibromoalkane. The reaction proceeds through a cyclic intermedia

Taya2010 [7]

Answer:

See explanation and image attached for details

Explanation:

The reaction involves the heterolytic fission of the Br-Br bond in the bromine molecule to yield a bromine cation which attacks the but-1-ene to form a cyclic intermediate called the brominium ion. The bromine anion must now attack from the opposite face of the brominium ion due to steric clashes to form a product of a 1,2-dibromoalkane having the anti- stereochemistry.

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3 years ago

What is the mass, in grams of a pure iron cube that has a volume of 4.20cm^3

Dmitriy789 [7]

$d_{Fe}=7874\frac{kg}{m^{3}}=7,874\frac{g}{cm^{3}}\\
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6 0

3 years ago

Hydroxylamine hydrochloride is a powerful reducing agent which is used as a polymerization catalyst. It contains 5.80 mass % H,

ludmilkaskok [199]

Answer: The empirical formula for the given compound is $H_{4}O_1N_1Cl_1=H_4NOCl$

Explanation:

We are given:

Percentage of H = 5.80 %

Percentage of O = 23.02 %

Percentage of N = 20.16 %

Percentage of Cl = 51.02 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of H = 5.80 g

Mass of O = 23.02 g

Mass of N = 20.16 g

Mass of Cl = 51.02 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{5.80g}{1g/mole}=5.80moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{23.02g}{16g/mole}=1.44moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{20.16g}{14g/mole}=1.44moles$

Moles of Chlorine = $\frac{\text{Given mass of Chlorine}}{\text{Molar mass of Chlorine}}=\frac{51.02g}{35.5g/mole}=1.44moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.44 moles.

For Hydrogen = $\frac{5.80}{1.44}=4.03\approx 4$

For Oxygen = $\frac{1.44}{1.44}=1$

For Nitrogen = $\frac{1.44}{1.44}=1$

For Chlorine = $\frac{1.44}{1.44}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of H : O : N : Cl = 4 : 1 : 1 : 1

Hence, the empirical formula for the given compound is $H_{4}O_1N_1Cl_1=H_4NOCl$

3 0

3 years ago