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3 years ago

The equation, 2 H2(g) + O2(g) Imported Asset 2 H2O(l), represents an equation for a standard formation reaction.

Chemistry

2 answers:

disa [49]3 years ago

6 0

Answer : The given statement is false.

Explanation :

The given chemical reaction of formation of $H_2O$ is:

$2H_2(g)+O_2(g)\rightarrow 2H_2O(l)$

Standard formation reaction :

In the standard formation reaction, one mole of substance always produced from its element that are present in their standard states.

Thus, the standard formation reaction of $H_2O$ will be:

$H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$

Hence, the given statement is false.

VLD [36.1K]3 years ago

5 0

If this is asking if the equation is balanced, than the answer is true

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The vapor pressure of ethanol is 30°C at 98.5 mmHg and the heat of vaporization is 39.3 kJ/mol. Determine the normal boiling poi

Gelneren [198K]

Answer : The normal boiling point of ethanol will be, $348.67K$ or $75.67^oC$

Explanation :

The Clausius- Clapeyron equation is :

$\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$ = vapor pressure of ethanol at $30^oC$ = 98.5 mmHg

$P_2$ = vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg

$T_1$ = temperature of ethanol = $30^oC=273+30=303K$

$T_2$ = normal boiling point of ethanol = ?

$\Delta H_{vap}$ = heat of vaporization = 39.3 kJ/mole = 39300 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

$\ln (\frac{760mmHg}{98.5mmHg})=\frac{39300J/mole}{8.314J/K.mole}\times (\frac{1}{303K}-\frac{1}{T_2})$

$T_2=348.67K=348.67-273=75.67^oC$

Hence, the normal boiling point of ethanol will be, $348.67K$ or $75.67^oC$

3 0

3 years ago

What is the velocity of 10km west in three hours

Nina [5.8K]

I think 3.33 maybe ‍♀️

7 0

3 years ago

Read 2 more answers

Which gas law (choices are Charles' Law, Gay-Lussac's Law, or Boyle's Law) explains each scenario:________.

posledela

Answer:

A. Boyle's Law

B. Charles' Law

C. Gay-Lussac's Law

Explanation:

An air bag inflates due to the decomposition of sodium azide or NaN₃ to completely fill the bag with nitrogen gas which is an example of Boyle's law, which states that the pressure of a given mass of gas is inversely proportional to its volume, hence due to the estricted volume of the airbag, the pressure of the nitrogen gas in the bag increses protecting the occupants of a cr from injuries in a crash

Helium balloon decrease in sice in a freezer is an example of Charlles law which states that the volume of a given mass of gas is nverslely proportionl to its temperature at constant pressure

A can of spray paint will explode if tossed into a fire is an example of Gay-Lussac's Law which states that the pressure of a given mass of gas is directly proportional to its temperature hence the increased pressure causes the can ti explode

4 0

3 years ago

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

What wave has the longer wavelength

natima [27]

The wavelength is the distance between one crest/trough to another crest/trough. On the image, it's basically the length between each peak of the wave. You can see that the distance between peaks in wave A are much shorter than the distance between the peaks in wave B.

Thus, wave B has the longer wavelength.

8 0

3 years ago