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Metallic pan is provided with a wooden plastic handle give reason

AlladinOne [14]

The metallic pan iis most likely going to be used on a stove.

The stove is heating something, and the conductive metallic pan will, well, conduct that heat throughout the entire body of the pan. Doing this will spread the heat to the handle, burning your hands.

Both wood and plastic are insulators, and they do not conduct heat or electricity. They will insulate your hands and protect them from the heat.

5 0

3 years ago

If 42.8 mL of 0.204 M HCl solution is needed to neutralize a solution of Ca(OH)2, how many grams of Ca(OH)2 must be in the solut

Aneli [31]

Hey There!

At neutralisation moles of H⁺ from HCl = moles of OH⁻ from Ca(OH)2 so :

0.204 * 42.8 / 1000 => 0.0087312 moles

Moles of Ca(OH)2 :

2 HCl + Ca(OH)2 = CaCl2 + 2 H2O

0.0087312 / 2 => 0.0043656 moles ( since each Ca(OH)2 ives 2 OH⁻ ions )

Therefore:

Molar mass Ca(OH)2 = 74.1 g/mol

mass = moles of Ca(OH)2 * molar mass

mass = 0.0043656 * 74.1

mass = 0.32 g of Ca(OH)2

Hope that helps!

6 0

3 years ago

The balanced equation for combustion in an acetylene torch is shown below:

vichka [17]

Answer:

70mol

Explanation:

The equation of the reaction is given as:

2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O

Given parameters:

Number of moles of acetylene = 35.0mol

Number of moles of oxygen in the tank = 84.0mol

Unknown:

Number of moles of CO₂ produced = 35.0mol

Solution:

From the information given about the reaction, we know that the reactant that limits this combustion process is acetylene. Oxygen is given in excess and we don't know the number of moles of this gas that was used up. We know for sure that all the moles of acetylene provided was used to furnish the burning procedure.

To determine the number of moles of CO₂ produced, we use the stoichiometric relationship between the known acetylene and the CO₂ produced from the balanced chemical equation:

From the equation:

2 moles of acetylene produced 4 moles of CO₂

∴ 35.0 mol of acetylene would produced:

$\frac{35 x 4}{2}$ = 70mol

6 0

3 years ago

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Calculate the standard heat of reaction for the following methane-generating reaction of methanogenic bacteria: 4CH3NH2(g) + 2H2

PIT_PIT [208]

<u>Answer:</u> The standard heat for the given reaction is -138.82 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]$

For the given chemical reaction:

$4CH_3NH_2(g)+2H_2O(l)\rightarrow 3CH_4(g)+CO_2(g)+4NH_3(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(3\times \Delta H_f_{(CH_4(g))})+(1\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(NH_3(g))})]-[(4\times \Delta H_f_{(CH_3NH_2(g))})+(2\times \Delta H_f_{(H_2O(l))})]$

We are given:

$\Delta H_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-46.1kJ/mol\\\Delta H_f_{(CH_4(g))}=-74.8kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H_f_{(CH_3NH_2(g))}=-22.97kJ/mol$