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3 years ago

1. List 3 physical and chemical phenomena

Chemistry

1 answer:

ollegr [7]3 years ago

8 0

Answer:

Examples of chemical changes are burning, cooking, rusting, and rotting. Examples of physical changes are boiling, melting, freezing, and shredding. Often, physical changes can be undone, if energy is input.

Hope this helped.

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1 year ago

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3 years ago

A 0.2500 g sample of an alloy reacts with to form hydrogen gas: 2Al(s) + 6H+(aq) 2Al3+(aq) + 3H2(g) Zn(s) + 2H+(aq) Zn2+(aq) + H

fenix001 [56]

Answer: The mass percent of zinc in the alloy is 78.68 %

Explanation:

We are given:

Mass of sample of alloy = 0.2500 g

Let the mass of aluminium be 'x' grams and mass of zinc will be (0.2500 - x) g

To calculate the amount of hydrogen gas produced, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 755 mmHg

V = Volume of the gas = 0.147 L

T = Temperature of the gas = $25^oC=[25+273]K=298K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of hydrogen gas = ?

Putting values in above equation, we get:

$755mmHg\times 0.147L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\n=\frac{755\times 0.147}{62.3637\times 298}=0.00597mol$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For Aluminium:

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium}=\frac{x}{27}mol$

The chemical equation follows:

$2Al(s)+6H^+(aq.)\rightarrow 2Al^{3+}(aq.)+3H_2(g)$

By Stoichiometry of the reaction:

2 moles of aluminium produces 3 moles of hydrogen gas

So, $\frac{x}{27}$ moles of aluminium will produce = $\frac{3}{2}\times \frac{x}{27}=\frac{3x}{54}mol$ of hydrogen gas

For Zinc:

Molar mass of zinc = 65.4 g/mol

Putting values in equation 1, we get:

$\text{Moles of zinc}=\frac{(0.25-x)}{65.4}mol$

The chemical equation follows:

$Zn(s)+2H^+(aq.)\rightarrow Zn^{2+}(aq.)+H_2(g)$

By Stoichiometry of the reaction:

1 mole of zinc produces 1 moles of hydrogen gas

So, $\frac{(0.25-x)}{65.4}$ moles of zinc will produce = $\frac{1}{1}\times \frac{(0.25-x)}{65.4}=\frac{(0.25-x)}{65.4}mol$ of hydrogen gas

Equating the moles of hydrogen gas:

$\Rightarrow 0.00597=\frac{3x}{54}+\frac{(0.25-x)}{65.4}\\\\x=0.0533g$

To calculate the mass percentage of zinc in alloy, we use the equation:

$\text{Mass percent of zinc}=\frac{\text{Mass of zinc}}{\text{Mass of alloy}}\times 100$

Mass of zinc = (0.2500 - x) = [0.2500 - 0.0533] = 0.1967 g

Mass of alloy = 0.2500 g

Putting values in above equation, we get:

$\text{Mass percent of zinc in alloy}=\frac{0.1967g}{0.2500g}\times 100=78.68\%$

Hence, the mass percent of zinc in the alloy is 78.68 %

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4 years ago