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kow [346]
3 years ago
12

a strip of impure copper is given to you. how will you purify it by using chemical effects of electric current ?

Chemistry
2 answers:
Oksi-84 [34.3K]3 years ago
6 0

This can be done through electrolysis. Electrolysis is the separation of a substance  into two or more substances that may differ from each other and from the original substance by passing an electric current through a solution that contains ions.

In the case of copper,  we use a  copper (II) sulphate  solution which we put in a large beaker. The impure copper will be used as the positive electrode (anode) and for the negative electrode (cathode)  will be a bar of pure copper.

When the electric current is switched on, the bar of pure copper which is the cathode  increases greatly in size as copper ions leave the anode of impure copper and attach to  the cathode.  The anode becomes smaller and smaller as it loses copper ions until all that is left of it is impurities in form of a sludge beneath it.


finlep [7]3 years ago
5 0
You would Purify it with magnesium
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3. After 7.9 grams of sodium are dropped into a bathtub full of water, how many grams of hydrogen gas are released?
Pavel [41]

Answer:

3) About 0.35 grams of hydrogen gas.

4) About 65.2 grams of aluminum oxide.

Explanation:

Question 3)

We are given that 7.9 grams of sodium is dropped into a bathtub of water, and we want to determine how many grams of hydrogen gas is released.

Since sodium is higher than hydrogen on the activity series, sodium will replace hydrogen in a single-replacement reaction for sodium oxide. Hence, our equation is:

\displaystyle \text{Na} + \text{H$_2$O}\rightarrow \text{Na$_2$O}+\text{H$_2$}

To balance it, we can simply add another sodium atom on the left. Hence:

\displaystyle 2\text{Na} + \text{H$_2$O}\rightarrow \text{Na$_2$O}+\text{H$_2$}

To convert from grams of sodium to grams of hydrogen gas, we can convert from sodium to moles of sodium, use the mole ratios to find moles in hydrogen gas, and then use hydrogen's molar mass to find its amount in grams.

The molar mass of sodium is 22.990 g/mol. Hence:

\displaystyle \frac{1\text{ mol Na}}{22.990 \text{ g Na}}

From the chemical equation, we can see that two moles of sodium produce one mole of hydrogen gas. Hence:

\displaystyle \frac{1\text{ mol H$_2$}}{2\text{ mol Na}}

And the molar mass of hydrogen gas is 2.016 g/mol. Hence:

\displaystyle \frac{2.016\text{ g H$_2$}}{1\text{ mol H$_2$}}

Given the initial value and the above ratios, this yields:

\displaystyle 7.9\text{ g Na}\cdot \displaystyle \frac{1\text{ mol Na}}{22.990 \text{ g Na}}\cdot \displaystyle \frac{1\text{ mol H$_2$}}{2\text{ mol Na}}\cdot \displaystyle \frac{2.016\text{ g H$_2$}}{1\text{ mol H$_2$}}

Cancel like units:

=\displaystyle 7.9\cdot \displaystyle \frac{1}{22.990}\cdot \displaystyle \frac{1}{2}\cdot \displaystyle \frac{2.016\text{ g H$_2$}}{1}

Multiply. Hence:

=0.3463...\text{ g H$_2$}

Since we should have two significant values:

=0.35\text{ g H$_2$}

So, about 0.35 grams of hydrogen gas will be released.

Question 4)

Excess oxygen gas is added to 34.5 grams of aluminum and produces aluminum oxide. Hence, our chemical equation is:

\displaystyle \text{O$_2$} + \text{Al} \rightarrow \text{Al$_2$O$_3$}

To balance this, we can place a three in front of the oxygen, four in front of aluminum, and two in front of aluminum oxide. Hence:

\displaystyle3\text{O$_2$} + 4\text{Al} \rightarrow 2\text{Al$_2$O$_3$}

To convert from grams of aluminum to grams of aluminum oxide, we can convert aluminum to moles, use the mole ratios to find the moles of aluminum oxide, and then use its molar mass to determine the amount of grams.

The molar mass of aluminum is 26.982 g/mol. Thus:

\displaystyle \frac{1\text{ mol Al}}{26.982 \text{ g Al}}

According to the equation, four moles of aluminum produces two moles of aluminum oxide. Hence:

\displaystyle \frac{2\text{ mol Al$_2$O$_3$}}{4\text{ mol Al}}

And the molar mass of aluminum oxide is 101.961 g/mol. Hence: \displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1\text{ mol Al$_2$O$_3$}}

Using the given value and the above ratios, we acquire:

\displaystyle 34.5\text{ g Al}\cdot \displaystyle \frac{1\text{ mol Al}}{26.982 \text{ g Al}}\cdot \displaystyle \frac{2\text{ mol Al$_2$O$_3$}}{4\text{ mol Al}}\cdot \displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1\text{ mol Al$_2$O$_3$}}

Cancel like units:

\displaystyle= \displaystyle 34.5\cdot \displaystyle \frac{1}{26.982}\cdot \displaystyle \frac{2}{4}\cdot \displaystyle \frac{101.961\text{ g Al$_2$O$_3$}}{1}

Multiply:

\displaystyle = 65.1852... \text{ g Al$_2$O$_3$}

Since the resulting value should have three significant figures:

\displaystyle = 65.2 \text{ g Al$_2$O$_3$}

So, approximately 65.2 grams of aluminum oxide is produced.

5 0
2 years ago
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What is the equation for determining gas pressure?
Advocard [28]

Answer:

Explanation:

First, let's review the ideal gas law, PV = nRT. In this equation, 'P' is the pressure in atmospheres, 'V' is the volume in liters, 'n' is the number of particles in moles, 'T' is the temperature in Kelvin and 'R' is the ideal gas constant (0.0821 liter atmospheres per moles Kelvin)

8 0
3 years ago
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What can lead to being curious and also lead to fascinating scientific investigations? a. Questions b. Answers c. Qualitative Me
Ira Lisetskai [31]

Answer:

Questions

Explanation:

The word "science" was obtained from the Latin word "Scire" which means "to know". Hence science is a systematic investigation into the nature of the universe.

Scientific investigations begin with asking questions about the universe. It is this curiosity to know about occurrences in the universe as well as how to solve problems in the universe that leads to fascinating scientific investigations and discoveries.

8 0
2 years ago
A compound is found to be 30.45% n and 69.55 % o by mass. if 1.63 g of this compound occupy 389 ml at 0.00°c and 775 mm hg, what
Charra [1.4K]
1) mass composition

N: 30.45%
O: 69.55%
   -----------
   100.00%

2) molar composition

Divide each element by its atomic mass

N: 30.45 / 14.00 = 2.175 mol

O: 69.55 / 16.00 = 4.346875

4) Find the smallest molar proportion

Divide both by the smaller number

N: 2.175 / 2.175 = 1

O: 4.346875 / 2.175 = 1.999 = 2

5) Empirical formula: NO2

6) mass of the empirical formula

14.00 + 2 * 16.00 = 46.00 g

7) Find the number of moles of the gas using the equation pV = nRT

=> n = pV / RT = (775/760) atm * 0.389 l / (0.0821 atm*l /K*mol * 273.15K)

=> n = 0.01769 moles

8) Find molar mass

molar mass = mass in grams / number of moles = 1.63 g / 0.01769 mol = 92.14 g / mol

9) Find how many times the mass of the empirical formula is contained in the molar mass

92.14 / 46.00 = 2.00

10) Multiply the subscripts of the empirical formula by the number found in the previous step

=> N2O4

Answer: N2O4
3 0
2 years ago
The chemical bonding in sodium phosphate, Na3PO4, is classified as:
fiasKO [112]
<span>Ionic bonding between sodium and phosphate ions.</span>
7 0
2 years ago
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