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Jet001 [13]
3 years ago
5

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 thro<u>ugh</u> a <em><u>filt</u></em><em><u>er</u></em><em><u> </u></em><em><u>tunnel</u></em> 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.

You might be interested in
Why were the Spanish interested in conquering Peru?
mr_godi [17]

Answer:

531–1532 – Pizarro's third voyage to Peru. Spaniards form a bond with the Natives (Huancas, Chankas, Cañaris and Chachapoyas) who were under the oppression of the Inca Empire, and Pizarro includes them among his troops to face the Incas. Atahualpa is captured by Spanish.

Explanation:  Is this right ?

if it is please give thanks

6 0
2 years ago
The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151
vitfil [10]

Answer:-  The natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for ^1^5^1_E_u and ^1^5^3_E_u as 150.919860 and 152.921243 amu, respectively.  Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of ^1^5^1_E_u as n then the abundance of ^1^5^3_E_u would be 1-n .

Let's plug in the values in the formula:

151.964=150.919860(n)+152.921243(1-n)

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

151.964-152.921243=150.919860n-152.921243n

-0.957243=-2.001383n

negative sign is on both sides so it is canceled:

0.957243=2.001383n

n=\frac{0.957243}{2.001383}

n=0.478

The abundance of ^1^5^1_E_u is 0.478 which is 47.8%.  

The abundance of ^1^5^3_E_u is = 1-0.478

= 0.522 which is 52.2%

Hence, the natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .


3 0
3 years ago
Which organisms have greater variation and produce offspring through asexual reproduction?
Sedaia [141]

Answer: Fungi

Explanation:

8 0
3 years ago
Please i need help understanding the facts about ions. and whats the answer??
Degger [83]

All are true except the statement that ions are formed by changing the number protons in an atom’s nucleus.

A neutral atom contains the same number of protons (positive charge) and electrons (negative charge).

If there are <em>more electrons than protons</em>, the atom becomes a <em>negative ion</em>.

If there are <em>fewer electrons than protons</em>, the atom becomes a <em>positive ion</em>.

The protons are in the nucleus, where we can’t easily get at them. The <em>electrons are outside the nucleus</em>, so other chemicals can easily get at them and either remove them or add to their number.

<em>Metals</em> have only a few valence electrons, so it is fairly easy to remove them and <em>form positive ions</em>.

6 0
3 years ago
(a) The student dissolves the entire impure sample of CuSO4(s) in enough distilled water to make 100.mL of solution. Then the st
Talja [164]

Answer:

Explanation:

Given parameters :

Volume of solution  = 100mL

Absorbance of solution  = 0.30

Unknown:

Concentration of CuSO₄ in the solution = ?

Solution:

There is relationship between the absorbance and concentration of a solution. They are directly proportional to one another.

A graph of absorbance against concentration gives a value of 0.15M at an absorbance of 0.30.

The concentration is 0.15M

Also, we can use:  Beer-Lambert's law;

                    A  = ε mC l

where εm is the molar extinction coefficient

             C is the concentration

             l is the path length

Since the εm is not given and assuming path length is 1;

         Then we solve for the concentration.

3 0
3 years ago
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