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Ne4ueva [31]
3 years ago
5

Iron and aluminim are examples of _________. a. pure substances b. compounds

Chemistry
2 answers:
Ierofanga [76]3 years ago
8 0

Answer:

compounds

Explanation:

Sindrei [870]3 years ago
8 0
A. Pure substances

Iron aluminum, silver, and gold are all examples of a pure substance. Substances that are free from any kind of mixture and contain only one kind of particle are pure substances.
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Pls Help me I am stuck on this question.​
kolbaska11 [484]

Answer:

compound is the answer

4 0
3 years ago
A site in Pennsylvania receives a total annual deposition of 2.688 g/mof sulfate from fertilizer and acid rain. The ratio by mas
sertanlavr [38]

According to the statement

2.12 x 10^4 lbs pounds of CaCO₃ are needed to neutralize this acid

<h3>What is neutralization?</h3>

A chemical reaction in which an acid and a base react quantitatively with each other is known as neutralization or neutralization. In a water reaction, neutralization ensures that there is no excess of hydrogen or hydroxide ions in the solution.

<h3>According to the given information:</h3>

The equation of the neutralization reaction between H2SO4 and CaCO3.

CaCO3 + H2SO4 → CaSO4 + H2CO3

H2CO3 dissociate to water and carbon dioxide.

        CaCO3 + H2SO4 → CaSO4  + H2O + CO2

Now solving for the mass of CaCO3 needed to neutralize the acid.

mass of CaCO3 = 9460 Kg H2SO4  × \frac{1000 \mathrm{~g}}{1 \mathrm{~kg}} \times \frac{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO} 4}{98.1 \mathrm{gH}_2 \mathrm{SO}_4} \times \frac{1 \mathrm{~mol} \mathrm{CaCO}\left(\mathrm{O}_3\right.}{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO}_4}\times \frac{100.1 \mathrm{~g} \mathrm{CaCO}_3}{1 \mathrm{~mol} \mathrm{CaCO}_3} \times \frac{2.205 \mathrm{lb}}{1000 \mathrm{~g}}

= 21284.56606

mass of CaCO3 =  2.12 x 10^4 lbs

2.12 x 10^4 lbs pounds of CaCO₃ are needed to neutralize this acid.

To know more about neutralization visit:

brainly.com/question/12498769

#SPJ4

4 0
2 years ago
Which of the following is a completely elastic collision? A. a ball rebounds against a wall, reversing its direction, but at onl
Dmitry_Shevchenko [17]

Answer:

Answer E.

For a collision to be completely elastic, there must be NO LOSS in kinetic energy.

We can go through each answer choice:

A. Since the ball rebounds at half the initial speed, there is a loss in kinetic energy. This is NOT an elastic collision.

B. A collision involving sticking is an example of a perfectly INELASTIC collision. This is NOT an elastic collision.

C. A reduced speed indicates that there is a loss of kinetic energy. This is NOT elastic.

D. The balls traveling at half the speed after the collision indicates a loss of kinetic energy, making this collision NOT elastic.

E. This collision indicates an exchange of velocities, characteristic of an elastic collision. We can prove this:

Let:

m = mass of each ball

v = velocity

We have the initial kinetic energy as:

KE = \frac{1}{2}mv^2 + 0 = \frac{1}{2}mv^2KE=21mv2+0=21mv2

And the final as:

KE = 0 + \frac{1}{2}mv^2 = \frac{1}{2}mv^2KE=0+21mv2=21mv2

5 0
3 years ago
Read 2 more answers
Given that you started with 28.5 g of K3PO4, how many grams of KNO3 can be<br> produced?
Irina18 [472]

Mass of KNO₃ : = 40.643 g

<h3>Further explanation</h3>

Given

28.5 g of K₃PO₄

Required

Mass of KNO₃

Solution

Reaction(Balanced equation) :

2K₃PO₄ + 3 Ca(NO₃)₂ = Ca₃(PO₄)₂ + 6 KNO₃

mol K₃PO₄(MW=212,27 g/mol) :

= mass : MW

= 28.5 : 212,27 g/mol

= 0.134

Mol ratio of K₃PO₄ : KNO₃ = 2 : 6, so mol KNO₃ :

= 6/2 x mol K₃PO₄

= 6/2 x 0.134

= 0.402

Mass of KNO₃ :

= mol x MW KNO₃

= 0.402 x 101,1032 g/mol

= 40.643 g

8 0
3 years ago
The diffrent names for the diffrent ways the moon can look from earth or blank?
Gnesinka [82]


there are 8 moon phases.

They are - First quarter, waxing crescent, new, waning crescent, third quarter, Waning gibbous, full, and waxing gibbous

8 0
3 years ago
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