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Vesnalui [34]
2 years ago
8

How many oxygen atoms are there in 0.25 mole of CO2?

Chemistry
1 answer:
Anna35 [415]2 years ago
3 0

Answer:

It is known that 1 mol of a molecule contains 6.023×1023 6.023 × 10 23 number of molecules. So, 0.25 moles of CO2 C O 2.

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Write half reactions and the overall reaction occurred during discharging of acid lead battery.
Gre4nikov [31]
When  battery  discharge / delivering  current  the  lead  at  the  anode  is  oxidized
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pb---->pb+  2e-
since  the  lead  ions  are  in  presence  of  aquous   sulfate  in  insoluble  lead  sulfate  precipitate onto the  electrode
the  overall   reaction  at  the  anode  is  therefore
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3 years ago
Find the area given rectangle.
seraphim [82]

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Terms that have to be used: Products, Reactants, Balanced, Unbalanced, Atoms
emmainna [20.7K]

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Self explanatory

3 0
3 years ago
The vapor pressure of water is 1.00 atm at 373 K, and the enthalpy of vaporization is 40.68 kJ mol!. Estimate the vapor pressure
Yuki888 [10]

Answer:

The vapor pressure at temperature 363 K is 0.6970 atm

The vapor pressure at 383 K is 1.410 atm

Explanation:

To calculate \Delta H_{vap} of the reaction, we use clausius claypron equation, which is:

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

where,

P_1 = vapor pressure at temperature T_1

P_2 = vapor pressure at temperature T_2

\Delta H_{vap} = Enthalpy of vaporization  

R = Gas constant = 8.314 J/mol K

1) \Delta H_{vap}=40.68 kJ/mol=40680 J/mol

T_1 = initial temperature =363 K

T_2 = final temperature =373 K

P_2=1 atm, P_1=?

Putting values in above equation, we get:

\ln(\frac{1 atm}{P_1})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{363}-\frac{1}{373}]

P_1=0.69671 atm \approx 0.6970 atm

The vapor pressure at temperature 363 K is 0.6970 atm

2) \Delta H_{vap}=40.68 kJ/mol=40680 J/mol

T_1 = initial temperature =373 K

T_2 = final temperature =383 K

P_1=1 atm, P_2?

Putting values in above equation, we get:

\ln(\frac{P_2}{1 atm})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{383}]

P_2=1.4084 atm \approx 1.410 atm

The vapor pressure at 383 K is 1.410 atm

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nlexa [21]
Smog

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