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frutty [35]
4 years ago
11

gold is heavier than aluminium if we put a 100 gram biscuit of gold in water taken in a measuring cylinder or we put a 100 gram

aluminium bar in the measuring cylinder will the rise in the level of the water be same or different in the two cases give reason​
Chemistry
1 answer:
kati45 [8]4 years ago
4 0

Answer:

The level of the water will be higher for 100g Al because occupy more space than gold.

Explanation:

To understand this problem, we must know Archimedes' principle:

"A body that is submerged in a fluid produce a bouyant force equal to the weight of the fluid that the body displaces". The volume of the body is equal to the volume displaced in the liquid.

Now, as gold is heavier than aluminium, 100g of Al have more volume than 100g of gold, than means:

<h3>The level of the water will be higher for 100g Al because occupy more space than gold.</h3>

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Calculate the standard enthalpy change for the reaction at 25 ∘ 25 ∘ C. Standard enthalpy of formation values can be found in th
WINSTONCH [101]

<u>Answer:</u> The standard enthalpy change of the reaction is coming out to be -16.3 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. It is represented as \Delta H

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

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

Mg(OH)_2(s)+2HCl(g)\rightarrow MgCl_2(s)+2H_2O(g)

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

\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})]

We are given:

\Delta H_f_{(Mg(OH)_2(s))}=-924.5kJ/mol\\\Delta H_f_{(HCl(g))}=-92.30kJ/mol\\\Delta H_f_{(MgCl_2(s))}=-641.8kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ

Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ

6 0
4 years ago
When of alanine are dissolved in of a certain mystery liquid , the freezing point of the solution is less than the freezing poin
vitfil [10]

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16.5 g

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Check the attached files for detailed solution

7 0
4 years ago
Predict the product of this chemical equation. please explain how u did it. Do not balance after finding answer.
Oksanka [162]

Answer:

The products are CO₂ and H₂O.

The ballanced equation is this:

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Explanation:

In any chemical reaction, where you see that you have a compound reacting only with oxygen (O₂), you are in front of combustion.

Products in combustion are always water vapor and carbon dioxide.

Take a look to the methane combustion.

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1 mol of methane reacts with 2 moles of oxygen to form 1 mol of CO₂ and 2 moles of water.

This is the combustion for an alkene where 2 moles of the alkene reacts with 11 moles of oxygen to make water and CO₂ like this:

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8 0
3 years ago
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