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Arturiano [62]
3 years ago
12

WHEN IN FRICTION not useful

Chemistry
1 answer:
Paladinen [302]3 years ago
6 0
Friction is not useful in Machinery sometimes, Since it leads to the wear and tear of machine parts.
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Aluminum reacts with HCl to produce aluminum chloride and hydrogen gas.
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Aluminum Reacts With Hydrochloric Acid To Produce Aluminum Chloride And Hydrogen Gas. 2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g)What Mass Of H2(g) Is Required From The Reaction Of 0.75 G Of Al(s) With Excess Hydrochloric

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For every 2 Mol NaOH you would get 1 Mol N2H4
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If you would draw the Lewis structures of these atoms, you would see that A has 2 electron pairs and 2 lone electrons (that can bond). For B you’d see that you only have 1 electron that can form a bond. This means that 1 atom of A (2 lone electrons) can bond with 2 atoms of B. To know the kind of bond you have to know wether or not there will be a ‘donation’ of an electron from one atom to another. This happens when the number of electrons on one atoms is equal to the number of electrons another atom needs to reach the noble gas structure. As you can see, this is not the case here. This means that you get an AB2 structure with covalent character.
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3 years ago
The decomposition of hydrogen peroxide, H2O2, has been used to provide thrust in the control jets of various space vehicles. Usi
lesantik [10]

Answer:

\Delta H^0 _{reaction} = - 54.04 \ kJ/mol

Explanation:

The given equation for the chemical reaction can be expressed as;

2H_2O_{(l)}  \to 2H_2O_{(g)} +  O_{2(g)}

Using Hess Law to determine how much heat is produced by the decomposition of exactly 1 mole of H2O2 under standard conditions; we have the expression showing the Hess Law as follows:

\Delta H^0 _{reaction} = \sum n* \Delta H^0 _{products} -  \sum n* \Delta H^0 _{reactants}

At standard conditions;

the molar enthalpies of the given equation are as follows:

\Delta H_2O_{(g)} =-241.82\ kJ/mol

\Delta H_  O_{2(g)} = 0 \ kJ/mol

\Delta H _{H_2O_{(l)}}= -187.78  \ kJ/mol

Replacing them into above formula; we have:

\Delta H^0 _{reaction} = (2*(-241.82\ kJ/mol) + 0 \ kJ/mol + (2 *(-187.78 \ kJ/mol))

\Delta H^0 _{reaction} =-108.08 \ kJ/mol

The above is the amount of heat of formation for two moles of hydrogen peroxide; thus for 1 mole hydrogen peroxide ; we have :

\Delta H^0 _{reaction} = \dfrac{-108.08 \ kJ/mol}{2}

\Delta H^0 _{reaction} = - 54.04 \ kJ/mol

Hence; the heat produced after the decomposition of 1 mole of hydrogen peroxide is -54.04 kJ/mol

6 0
4 years ago
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