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klemol [59]
2 years ago
14

How many grams of carbon dioxide are in 35.6 liters of co2

Chemistry
1 answer:
lina2011 [118]2 years ago
3 0
The grams of carbon  dioxide  that are in 35.6 liters  of Co2 is calculates as below
calculate the  number of moles of CO2

At STP  1 mole  = 22.4 L

what  about  35.6 liters

=   1mole x 35.6  liters/ 22.4 liters = 1.589  moles

mass of CO2 =  moles x molar  mass of CO2

= 1.589 mol x 44 g/mol  =  69.92 grams
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When water decomposes into oxygen and hydrogen, the mass
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3 years ago
A 2.50 g sample of solid sodium hydroxide is added to 55.0 mL of 25 °C water in a foam cup (insulated from the environment) and
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Answer:

37.1°C.

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  • Firstly, we need to calculate the amount of heat (Q) released through this reaction:

<em>∵ ΔHsoln = Q/n</em>

no. of moles (n) of NaOH = mass/molar mass = (2.5 g)/(40 g/mol) = 0.0625 mol.

<em>The negative sign of ΔHsoln indicates that the reaction is exothermic.</em>

∴ Q = (n)(ΔHsoln) = (0.0625 mol)(44.51 kJ/mol) = 2.78 kJ.

  • We can use the relation:

Q = m.c.ΔT,

where, Q is the amount of heat released to water (Q = 2781.87 J).

m is the mass of water (m = 55.0 g, suppose density of water = 1.0 g/mL).

c is the specific heat capacity of water (c = 4.18 J/g.°C).

ΔT is the difference in T (ΔT = final temperature - initial temperature = final temperature - 25°C).

∴ (2781.87 J) = (55.0 g)(4.18 J/g.°C)(final temperature - 25°C)

∴ (final temperature - 25°C) = (2781.87 J)/(55.0 g)(4.18 J/g.°C) = 12.1.

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6 0
2 years ago
Which of the following statements is true? A) This reaction will be spontaneous only at high temperatures. B) This reaction will
viktelen [127]

Answer:

D) This reaction will be nonspontaneous only at high temperatures.

Explanation:

According the equation of Gibb's free energy -

∆G = ∆H -T∆S

∆G = is the change in gibb's free energy

∆H = is the change in enthalpy

T = temperature

∆S = is the change in entropy .

And , the sign of the  ΔG , determines whether the reaction is Spontaneous or non Spontaneous or at equilibrium ,

i.e. ,

if

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The reaction has the value for ∆H = negative , and ∆S = negative ,

Now ,

∆G = ∆H -T∆S

     = ( - ∆H ) - T( - ∆S )

     =  ( - ∆H ) +T(  ∆S )

Now, for making the reaction Spontaneous ΔG = negative ,

Hence ,

The temperature is low, then the value for  ΔG will be negative , i.e. , Spontaneous reaction .

And , vice versa , at higher temperature , the reaction will have ΔG positive , and the reaction will be non -Spontaneous reaction .

The standard free energy of formation will be zero , only for the compounds that are in their pure form ,

Hence , Al(s) will have ΔG = 0 .

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