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Bond [772]
2 years ago
12

What would be the best product to neutralize battery acid

Chemistry
1 answer:
Anna [14]2 years ago
7 0

Answer:

Baking Soda a base

Explanation:

We know that to neutralize an acid we add base to it. Baking Soda is a type of base so when we add Baking Soda or any other base to battery acid or any other acid we can neutralize it and get salt and water.

Acids are the substance which are sour in taste and can turn blue litmus paper red. For example, Lemon and Orange.

Bases are the substance which are bitter in taste and can turn red litmus paper blue. For example, Baking Soda and Soap.

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Energy + 6CO2 + 6H2O → 6O2 + C6H12O6 The purpose of photosynthesis is to produce usable chemical energy or glucose from solar en
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Calculate the standard enthalpy of formation of liquid methanol, CH3OH(l), using the following information: C(graphite) + O2 --&
Darya [45]

Answer : The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation :

Standard formation of reaction : It is a chemical reaction that forms one mole of a substance from its constituent elements in their standard states.

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of CH_3OH will be,

C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g)    \Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

(1) C(graphite)+O_2(g)\rightarrow CO_2(g)    \Delta H_2=-393.5kJ/mole

(2) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_3=-285.8kJ/mole

(3) CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l)     \Delta H_1=-726.4kJ/mole

Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, we get :

(1) C(graphite)+O_2(g)\rightarrow CO_2(g)    \Delta H_1=-393.5kJ/mole

(2) 2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)    \Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol

(3) CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g)     \Delta H_3=726.4kJ/mole

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)

\Delta H=-238.7kJ/mole

Therefore, the standard enthalpy of formation of methanol is, -238.7 kJ/mole

6 0
3 years ago
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