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3 years ago

A sample of sulfurous acid (H2SO3) has a mass of 1.31 g.

Chemistry

1 answer:

lakkis [162]3 years ago

6 0

Answer:

a) 2 (H+) ions

b) 1 (SO3²-) ions

c) 1.36 × 10^-22 grams.

Explanation:

According to this question, sulfurous acid has a chemical formula; H2SO3. It is made up of hydrogen and sulfite ion. Hydrogen ion (H+) is the cation while sulfite ion (SO32-) is the anion.

Based on the chemical formula, there are 2 moles of hydrogen ions that reacts with 1 mole of sulfite ion as follows:

2H+ + SO3²- → H2SO3

Hence;

- there are 2 hydrogen ions (2H+) present in H2SO3.

- there is 1 sulfite ion (SO3²-) present in H2SO3.

c) The mass of one formula unit of H2SO3 is calculated thus:

= 1.008 (2) + 32.065 + 15.999(3)

= 2.016 + 32.065 + 47.997

= 82.08 a.m.u

Since, 1 gram is = 6.02 x 10^23 a.m.u

82.08 a.m.u = 82.08/6.02 × 10^-23

= 13.6 × 10^-23

= 1.36 × 10^-22 grams.

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The metal content of iron in ores can be determined by a redox procedure in which the sample is first oxidized with Br2 to conve

coldgirl [10]

Answer:

80.27%

Explanation:

Let's consider the following balanced equation.

2 Fe³⁺(aq) + Sn²⁺(aq) ⇒ 2Fe²⁺(aq) + Sn⁴⁺(aq)

First, we have to calculate the moles of Sn²⁺ that react.

$\frac{0.1015molSn^{2+} }{1L} .13.28 \times 10^{-3} L=1.348\times 10^{-3}molSn^{2+}$

We also know the following relations:

According to the balanced equation, 1 mole of Sn²⁺ reacts with 2 moles of Fe³⁺.
1 mole of Fe³⁺ is oxidized from 1 mole of Fe.
The molar mass of Fe is 55.84 g/mol.

Then, for 1.348 × 10⁻3 moles of Sn²⁺:

$1.348\times 10^{-3}molSn^{2+}.\frac{2molFe^{3+} }{1molSn^{2+} } .\frac{1molFe}{1molFe^{3+} } .\frac{55.84gFe}{1molFe} =0.1505gFe$

If there are 0.1505 g of Fe in a 0.1875 g sample, the mass percentage of Fe is:

$\frac{0.1505g}{0.1875g} \times 100 \% = 80.27\%$

5 0

3 years ago

What is the charge on an ion that contains 16 protons and 18 electrons?

Alika [10]

16-18= -2 so it has a negative charge. Just subtract the electrons from the protons if you get a positive number it will have a positive charge and vice versa.

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3 years ago

The burning of magnesium is a highly exothermic reaction. How many kilojoules of heat are released when 0. 75 mol of Mg burn in

alexandr402 [8]

Taking into account the definition of enthalpy of a chemical reaction, the quantity of heat released when 0.75 moles of Mg are burned is 451.5 kJ.

<h3>Enthalpy of a chemical reaction</h3>

The enthalpy of a chemical reaction is known as the heat absorbed or released in a chemical reaction when it occurs at constant pressure. That is, the heat of reaction is the energy that is released or absorbed when chemicals are transformed into a chemical reaction.

The enthalpy is an extensive property, that is, it depends on the amount of matter present.

<h3>Heat released in this case</h3>

In this case, the balanced reaction is:

2 Mg(s) + O₂ (g) → 2 MgO(s) + 1204 kJ

This equation indicates that when 2 moles of Mg reacts with 1 mole of O₂, 1204 kJ of heat is released.

When 0.75 moles of Mg are burned, then you can apply the following rule of three: if 2 moles of Mg releases 1204 kJ of heat, 0.75 moles of Mg releases how much heat?

$heat=\frac{0.75 moles of Mgx1204 kJ}{2 moles of Mg}$

Finally, the quantity of heat released when 0.75 moles of Mg are burned is 451.5 kJ.

Learn more about enthalpy of a chemical reaction:

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2 years ago

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3 years ago

What we call "tin cans" are really iron cans coated with a thin layer of tin. The anode is a bar of tin and the cathode is the i

UNO [17]

Answer:

$Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)$

Explanation:

Although the context is not clear, let's look at the oxidation and reduction processes that will take place in a Fe/Sn system.

The problem states that anode is a bar of thin. Anode is where the process of oxidation takes place. According to the abbreviation 'OILRIG', oxidation is loss, reduction is gain. Since oxidation occurs at anode, this is where loss of electrons takes place. That said, tin loses electrons to become tin cation:

$Sn (s)\rightarrow Sn^{2+} (aq) + 2e^-$

Similarly, iron is cathode. Cathode is where reduction takes place. Reduction is gain of electrons, this means iron cations gain electrons and produce iron metal:

$Fe^{2+} (aq) + 2e^-\rightarrow Fe (s)$

The net equation is then:

$Sn (s) + Fe^{2+} (aq)\rightarrow Fe (s) + Sn^{2+} (aq)$

However, this is not the case, as this is not a spontaneous reaction, as iron metal is more reactive than tin metal, and this is how the coating takes place. This implies that actually anode is iron and cathode is tin:

Actual anode half-equation:

$Fe (s)\rightarrow Fe^{2+} (aq) + 2e^-$

Actual cathode half-equation:

$Sn^{2+} (aq) + 2e^-\rightarrow Sn (s)$

Actual net reaction:

$Fe (s) + Sn^{2+} (aq)\rightarrow Fe^{2+} (aq) + Sn (s)$

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3 years ago