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lukranit [14]
3 years ago
10

Which of the following concerning solutions is/are correct? 1. The solvent in a mixture of gases is generally considered to be t

he substance in greater amount. 2. The solid dissolved in a solution is known as the solute. 3. Solid solutions are called alloys.
Chemistry
1 answer:
pshichka [43]3 years ago
4 0

Answer: 1. The solvent in a mixture of gases is generally considered to be the substance in greater amount. 2. The solid dissolved in a solution is known as the solute. 3. Solid solutions are called alloys.

Explanation:

A binary solution is made up of two components ,called as solute and another as solvent.

A solute is defined as the component which is present in smaller proportion. It can be in solid, liquid or gaseous phase. A solvent is defined as the component which is present in larger proportion in a solution. It can be in solid, liquid or gaseous phase.

Usually in liquid solutions , solid acts as solute and liquid acts as solvent. For example: In salt solution, salt is the solute and water is the solvent.

Solid solutions contains solid as solute and solid as solvent. These are called as alloys. For example : In brass, zinc is the solute and copper is the solvent.

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One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate
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Answer:

\delta H_{rxn} = -66.0  \ kJ/mole

Explanation:

Given that:

3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \  \ \delta H = -47.0 \ kJ/mole  -- equation (1)  \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)}  \ \ \delta H = -25.0 \ kJ/mole  -- equation (2)  \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole  -- equation (3)

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

3FeO_{(s)} + CO_{2(g)}    \to    Fe_3O_4_{(s)} + CO_{(g)}   \ \delta H = -19.0 \ kJ/mole  -- equation (4)

Multiplying  (2) with equation (4) ; we have:

6FeO_{(s)} + 2CO_{2(g)}    \to    2Fe_3O_4_{(s)} + 2CO_{(g)}   \ \delta H = -38.0 \ kJ/mole  -- equation (5)

From equation (1) ; multiplying (-1) with equation (1); we have:

2Fe_3O_4_{(s)} +CO_{2(g)} \to     3FeO_3_{(s)}+CO_{(g)}   \  \ \delta H = 47.0 \ kJ/mole  -- equation (6)

From equation (2); multiplying (3) with equation (2); we have:

3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)}  \ \ \delta H = -75.0 \ kJ/mole  -- equation (7)

Now; Adding up equation (5), (6) & (7) ; we get:

6FeO_{(s)} + 2CO_{2(g)}    \to    2Fe_3O_4_{(s)} + 2CO_{(g)}   \ \delta H = -38.0 \ kJ/mole  -- equation (5)

2Fe_3O_4_{(s)} +CO_{2(g)} \to     3FeO_3_{(s)}+CO_{(g)}   \  \ \delta H = 47.0 \ kJ/mole  -- equation (6)

3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)}  \ \ \delta H = -75.0 \ kJ/mole  -- equation (7)

<u>                                                                                                                      </u>

FeO  \ \ \ +  \ \ \ CO   \ \  \to   \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \  \delta H = - 66.0 \ kJ/mole

<u>                                                                                                                     </u>

<u />

\delta H_{rxn} = \delta H_1 +  \delta H_2 +  \delta H_3    (According to Hess Law)

\delta H_{rxn} = (-38.0 +  47.0 + (-75.0)) \ kJ/mole

\delta H_{rxn} = -66.0  \ kJ/mole

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vekshin1

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Scientists can ignore the gravitational force because the gravitational force is considered to be negligible as compared to the other forces due to its smaller value.We all know that the gravitational force is directly proportional to the mass of an object which  result in  a small force value.When the value of this small force is compared to the value of the electrical force between protons and electrons in atoms the we can say that the electrical force is million times stronger than the gravitational force

Thus we can say that scientists can ignore <u>the Gravitational</u> force between charged particles that make up  the atoms because it is many millions of times smaller than other forces in the atom.

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