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3 years ago

If 6.02×1023 atoms of element Y have a mass of 28.09 g, what is the identity of Y?

Chemistry

1 answer:

Mazyrski [523]3 years ago

8 0

Silicon is the element having a mass of 28.09 g

<u>Explanation</u>:

Silicon is the element having an atomic mass of 28.09 g / mol. So 28.09 g of silicon contains 6.023 $\times$ 10^23 atoms. One mole of each element can produce one mole of compound.

The Atomic weight of an element can be determined by the number of protons and neutrons present in one atom of that element. So atomic weight expressed in grams always contain the same number of atoms( 6.023 $\times$ 10^23).

Avagadro number is the number of atoms of 1 mole of any gas at standard temperature and pressure. It has been determined that 6.023 $\times$ 10^23 atoms of an element are equal to the average atomic mass of that element.

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If 0.200 moles of AgNO₃ react with 0.155 moles of H₂SO₄ according to this UNBALANCED equation below, what is the mass in grams o

morpeh [17]

Answer:

31.2 g of Ag₂SO₄

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2AgNO₃(aq) + H₂SO₄ (aq) → Ag₂SO₄ (s) + 2HNO₃ (aq)

From the balanced equation above,

2 moles of AgNO₃ reacted with 1 mole of H₂SO₄ to produce 1 mole of Ag₂SO₄ and 2 moles of HNO₃.

Next, we shall determine the limiting reactant.

This can obtained as follow:

From the balanced equation above,

2 moles of AgNO₃ reacted with 1 mole of H₂SO₄.

Therefore, 0.2 moles of AgNO₃ will react with = (0.2 x 1)/2 = 0.1 mole of H₂SO₄.

From the calculations made above, only 0.1 mole out of 0.155 mole of H₂SO₄ given is needed to react completely with 0.2 mole of AgNO₃. Therefore, AgNO₃ is the limiting reactant.

Next,, we shall determine the number of mole of Ag₂SO₄ produced from the reaction.

In this case we shall use the limiting reactant because it will give the maximum yield of Ag₂SO₄ as all of it is consumed in the reaction.

The limiting reactant is AgNO₃ and the number of mole of Ag₂SO₄ produced can be obtained as follow:

From the balanced equation above,

2 moles of AgNO₃ reacted to produce 1 mole of Ag₂SO₄.

Therefore, 0.2 moles of AgNO₃ will react to produce = (0.2 x 1)/2 = 0.1 mole of Ag₂SO₄.

Therefore, 0.1 mole of Ag₂SO₄ is produced from the reaction.

Finally, we shall convert 0.1 mole of Ag₂SO₄ to grams.

This can be obtained as follow:

Molar mass of Ag₂SO₄ = (2x108) + 32 + (16x4) = 312 g/mol

Mole of Ag₂SO₄ = 0.1

Mass of Ag₂SO₄ =?

Mole = mass /Molar mass

0.1 = Mass of Ag₂SO₄ /312

Cross multiply

Mass of Ag₂SO₄ = 0.1 x 312

Mass of Ag₂SO₄ = 31.2 g

Therefore, 31.2 g of Ag₂SO₄ were obtained from the reaction.

5 0

3 years ago

What are all the ways that you think knowing about the flow of energy might be useful in

Kaylis [27]

Yes the main benifit can be recyclation of energy

The example can be seen in powerbanks or inverters
The chemical energy present inside the battery can be used to convert itself into electric energy.
Which can help us in climate protection

4 0

2 years ago

¿Cuál es el número atómico del calcio? 4 Científico que propuso que los átomos eran esferas indivisibles e indestructibles. 6 Si

spayn [35]

Answer:

es una expresion de 52 necesito puntos para una pregunta sorry

Explanation:

12-12-12-14-1161-76-57-57-

8 0

3 years ago

A sample weighing 3.110 g is a mixture of Fe 2 O 3 (molar mass = 159.69 g/mol) and Al 2 O 3 (molar mass = 101.96 g/mol). When he

grandymaker [24]

Answer:

The mass fraction of ferric oxide in the original sample : $\frac{723}{3110}$

Explanation:

Mass of the mixture = 3.110 g

Mass of $Fe_2O_3=x$

Mass of $Al_2O_3=y$

After heating the mixture it allowed to react with hydrogen gas in which all the ferric oxide reacted to form metallic iron and water vapors where as aluminum oxide did not react.

$Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)$

Mass of mixture left after all the ferric oxide has reacted = 2.387 g

Mass of mixture left after all the ferric oxide has reacted = y

$x=3.110 g- y=3.110 g - 2.387 g = 0.723 g$

The mass fraction of ferric oxide in the original sample :

$\frac{0.723 g}{3.110 g}=\frac{723}{3110}$

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3 years ago

Consider this reaction: 2al(s) + 3 cucl2(aq) → 2alcl3(aq) + 3 cu(s) if the concentration of cucl2 drops from 1.000 m to 0.655 m

kirza4 [7]

The average rate of reaction over a given interval can be calculated by taking the difference of concentration on a particular given reactant, and dividing it by the total time. In this case, (1.00 M - 0.655 M)/30 s = 0.0115 M/s, or 0.0115 mol/L-s, and this is the final rate of reaction.

8 0

3 years ago