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1 year ago

What happen to the energy that is lost when water change to gas

Chemistry

1 answer:

Y_Kistochka [10]1 year ago

6 0

Answer:

This process, which is the opposite of vaporization, is called condensation. As a gas condenses to a liquid, it releases the thermal energy it absorbed to become a gas. During this process, the temperature of the substance does not change. The decrease in energy changes the arrangement of particles.

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How many grams are in 2.40 moles of sodium hydroxide (NaOH)?

Bogdan [553]

Answer:

96g

Explanation:

Given parameters:

Number of moles of NaOH = 2.4moles

Unknown:

Mass of NaOH = ?

Solution:

The mass of a substance given the number of moles can be found using the expression below;

Mass of NaOH = Number of moles x molar mass;

Molar mass of NaOH = 23 + 16 + 1 = 40g/mol

Now input the parameters and solve;

Mass of NaOH = 2.4 x 40 = 96g

5 0

2 years ago

Sulfur dioxide, SO 2 ( g ) , can react with oxygen to produce sulfur trioxide, SO 3 ( g ) , by the reaction 2 SO 2 ( g ) + O 2 (

aleksley [76]

Answer: The amount of heat produced by the reaction is -21.36 kJ

Explanation:

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:

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

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

$\Delta H_{rxn}=[(2\times \Delta H_f_{(SO_3(g))})]-[(2\times \Delta H_f_{(SO_2(g))})+(1\times \Delta H_f_{(O_2(g))})]$

We are given:

$\Delta H_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H_f_{(SO_3(g))}=-395.7kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(2\times (-395.7))]-[(2\times (-296.8))+(1\times (0))]\\\\\Delta H_{rxn}=-197.8kJ/mol$

To calculate the number of moles, we use ideal gas equation, which is:

$PV=nRT$

where,

P = pressure of the gas = 1.00 bar

V = Volume of the gas = 2.67 L

n = number of moles of gas = ?

R = Gas constant = $0.0831\text{ L. bar }mol^{-1}K^{-1}$

T = temperature of the mixture = $25^oC=[25+273]K=298K$

Putting values in above equation, we get:

$1.00bar\times 2.67L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{1\times 2.67}{0.0831\times 298}=0.108mol$

To calculate the heat released of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat released = ?

n = number of moles = 0.108 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction = -197.8 kJ/mol

Putting values in above equation, we get:

$-197.8kJ/mol=\frac{q}{0.108mol}\\\\q=(-197.8kJ/mol\times 0.108mol)=-21.36kJ$

Hence, the amount of heat produced by the reaction is -21.36 kJ

3 0

3 years ago

Brian has an object suspended in the air. It has a mass of 100 kg and is 25 meters above the ground. What is the object’s potent

MatroZZZ [7]

The answer is 24500 joule

7 0

2 years ago

Read 2 more answers

A sample of gas contains 6.25 × 10-3 mol in a 500.0 mL flask at 265°C. What is the pressure of the gas in kilopascals? Which var

RideAnS [48]

55.9 kPa; Variables given = volume (V), moles (n), temperature (T)

We must calculate p from V, n, and T, so we use the Ideal Gas Law:

pV = nRT

Solve for p: p = nRT/V

R = 8.314 kPa.L.K^(-1).mol^(-1)

T = (265 + 273.15) K = 538.15 K

V = 500.0 mL = 0.5000 L

∴ p = [6.25 x 10^(-3) mol x 8.314 kPa·L·K^(-1)·mol^(-1) x 538.15 K]/(0.5000 L) = 55.9 kPa

4 0

3 years ago

Read 2 more answers

What happens to the ionizing rate if the steps of ionizing polyprotic acid increases?

Tanzania [10]

Answer:

<h3>An acid that contains more than one ionizable proton is a polyprotic acid. The protons of these acids ionize in steps. The differences in the acid ionization constants for the successive ionizations of the protons in a polyprotic acid usually vary by roughly five orders of magnitude.</h3>

5 0

2 years ago

Read 2 more answers