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2 years ago

What occurs when a gas is changed into a liquid

Chemistry

1 answer:

ankoles [38]2 years ago

7 0

for liquid to gas, Evaporation or vaporization.

for gas to liquid, condensation occurs. The opposite of evaporation. Condensation is when the gas gives up it's latent heat and simply condenses.

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A solution of ammonia has a pH of 11.8. What is the concentration of OH– ions in the solution?

melomori [17]

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2 years ago

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The weak type of bonding which attracts molecules to molecules is:

SashulF [63]

B. Hydrogen is the answer

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2 years ago

Zinc metal and aqueous silver nitrate react to give Zn(NO3)2(aq) plus silver metal. When 5.00 g of Zn(s) and solution containing

Fittoniya [83]

Answer : The percent yield is, 83.51 %

Solution : Given,

Mass of Zn = 5.00 g

Mass of $AgNO_3$ = 25.00 g

Molar mass of Zn = 65.38 g/mole

Molar mass of $AgNO_3$ = 168.97 g/mole

Molar mass of Ag = 107.87 g/mole

First we have to calculate the moles of Zn and $AgNO_3$ .

$\text{ Moles of }Zn=\frac{\text{ Mass of }Zn}{\text{ Molar mass of }Zn}=\frac{5.00g}{65.38g/mole}=0.0765moles$

$\text{ Moles of }AgNO_3=\frac{\text{ Mass of }AgNO_3}{\text{ Molar mass of }AgNO_3}=\frac{25.00g}{168.97g/mole}=0.1479moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag$

From the balanced reaction we conclude that

As, 2 mole of $AgNO_3$ react with 1 mole of $Zn$

So, 0.1479 moles of $AgNO_3$ react with $\frac{0.1479}{2}=0.07395$ moles of $Zn$

From this we conclude that, $Zn$ is an excess reagent because the given moles are greater than the required moles and $AgNO_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Ag$

From the reaction, we conclude that

As, 2 mole of $AgNO_3$ react to give 2 mole of $Ag$

So, 0.1479 moles of $AgNO_3$ react to give 0.1479 moles of $Ag$

Now we have to calculate the mass of $Ag$

$\text{ Mass of }Ag=\text{ Moles of }Ag\times \text{ Molar mass of }Ag$

$\text{ Mass of }Ag=(0.1479moles)\times (107.87g/mole)=15.95g$

Theoretical yield of $Ag$ = 15.95 g

Experimental yield of $Ag$ = 13.32 g

Now we have to calculate the percent yield.

$\% \text{ yield}=\frac{\text{ Experimental yield of }Ag}{\text{ Theretical yield of }Ag}\times 100$

$\% \text{ yield}=\frac{13.32g}{15.95g}\times 100=83.51\%$

Therefore, the percent yield is, 83.51 %

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2 years ago

Given the following values for the heats of formation, what is the number of moles of ethane (C2H6, MW 30.0) required to produce

baherus [9]

Answer:

0.641 moles of ethane

Explanation:

Based on the equation:

C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(l)

We can determine ΔH of reaction using Hess's law. For this equation:

<em>Hess's law: ΔH products - ΔH reactants</em>

ΔH = {2ΔHCO2 + 3ΔHH2O} - {ΔHC2H6}

<em>Pure monoatomic substances have a ΔH = 0kJ/mol; ΔHO2 = 0kJ/mol</em>

<em />

ΔH = {2*-393.5kJ/mol + 3*-285.8kJ/mol} - {-84.7kJ/mol}

ΔH = -1559.7kJ/mol

That means when 1 mole of ethane is in combustion there are released 1559.7kJ of heat. To produce 1.00x10³kJ there are needed:

1.00x10³kJ * (1mole ethane / 1559.7kJ) =

<h3>0.641 moles of ethane</h3>

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2 years ago

Calculate the average reaction rate of Cl2 consumed using the following information. H2 + Cl2 yields 2 HCl A table with three co

klemol [59]

Answer : Option D) 2.50 X $10^{-3}$ Mol/(L s)

Explanation: While calculating the average reaction rate for the given reaction in terms of Cl;

$H_{2} + Cl_{2} ----\ \textgreater \ 2HCl$ .

using the rate equation which is;

$\frac{-1}{1}$ $\frac{delta [Cl]}{delta t}$ =

$\frac{- 0.0875 - 0.0625}{10s - 0s}$ = 2.50 X $10^{-3}$ Mol/(L s)

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2 years ago